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General Physics rio s_ in fo er si ta rio s
General Physics Particle Kinematics Dynamics of Particle Circular Motion Energy and Momentum
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(Volume 2)
Chandigarh • Delhi • Chennai https://t.me/universitarios_infohttps://www.jamarana.com
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Production Editor: Nitkiran Bedi Composition: Tantla Composition Services, Chandigarh Printer: Sanat Printers
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The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertant omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book.
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Copyright © 2010 Dorling Kindersley (India) Pvt. Ltd
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This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated, without the publisher’s prior written consent, in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise) without the prior written permission of both the copyright owner and the abovementioned publisher of this book. ISBN 978-81-317-3444-5 10 9 8 7 6 5 4 3 2 1 Published by Dorling Kindersley (India) Pvt. Ltd, licensees of Pearson Education in South Asia.
Head Office: 7th Floor, Knowledge Boulevard, A-8-(A) Noida-201309, India. Registered Office: 11 Community Centre, Panchsheel Park, New Delhi 110 017, India.
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Chapter 2
Gravitations
Chapter 3
Elasticity
Chapter 4
Hydrostatics and Hydrodynamics
Chapter 5
Simple Harmonic Motion
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v 1.1–1.115 2.1–2.54 3.1–3.33 4.1–4.64 5.1–5.76
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Rotational Dynamics
Chapter 1
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Preface
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Contents
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For a science student, physics is the most important subject as it requires logical reasoning and high imagination. Without improving the level of physics, it is difficult to achieve a goal with the kind of competition that exists today. This five part volume covers all parts of general physics—Mechanics, Heat, Wave, Light, Electromagnetism and Modern Physics—which is written in accordance with the latest syllabus of the IIT-JEE and AIEEE. There is no single book that is available in the market that contains a large amount of solved examples.
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■ Entire syllabus is covered in five volumes. ■ Content of each chapter is well defined and builds new concepts from the scratch. ■ Each chapter describes the theory in a simple and lucid style. ■ Covers a wide spectrum of questions to enable the student to develop enough expertise to tackle any problem. ■ Helps students in building analytical and quantitative skills, which, in turn, develop confidence in problem solving. ■ Practice exercises are given at the end of each chapter. ■ Numerous diagrams in every chapter. After studying the entire chapter, students will be able to learn different tricks and techniques of problem solving with suitable level of analytical ability. Suggestions for improving the book are always welcome.
All the best!
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S. C. PANDEY
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1.1 KINEMATICS OF A RIGID BODY
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Rotational Dynamics
→
r12 = Constant
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A body of finite shape is said to be rigid, if it does not suffer any deformation under the increase of the external force or torque. In the given figure, if 1 and 2 are points on the body, then for the body to be rigid.
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Example A sheet of plywood is in motion on a horizontal floor such that the velocity of two points 1 and 2 are v1 and v2 respectively, as shown in the figure, then (a) v1 cos θ1 > v2 cos θ2 (b) v1 cos θ1 = v2 cos θ2 (c) v2 cos θ2 > v1 cos θ1 Ans. v1 cos θ1 = v2 cos θ2 Translational motion of a rigid body Consider a rigid body in motion. At a certain instant of time, the position vectors of any → → two points say 1 and 2 are r1 and r2 respectively with respect to ground from (x, y, z). →
→
→
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Rotational Dynamics
Differentiating with respect to time, we get →
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v1 = v2 + ω × r12
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→ → → → → = a2 + α × r12 + ω × ω × r12
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Again differentiating with respect to time, we get → → dω → → → a1 = a2 + × r12 + ω × v12 dt
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If point 2 becomes the centre of rotation, then →
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→ → → → ω× ω × r12 = ω2 − r12 Hence, from equation (ii), we have
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ω is in the z direction. → → → → →→ ω k × ω k × j r12 = ω k × (− i ) ω r12
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1.2 TRANSLATIONAL MOTION →
In this motion, r12 = Constant
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vCM = v (velocity of centre of mass) aCM = a Note: (i) All the kinematic terms of the centre of mass of rigid body represent its translational motion only. (ii) From the point of view of kinematics, ∑ Fi = mac can be written for the translational of the rigid body. https://t.me/universitarios_infohttps://www.jamarana.com
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Rotational Dynamics
So, for the linear motion of the body
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And similarly for curvilinear motion
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(iii) In (ii), the first term of the right hand side arrange due to pure translational with point 2 and second term arises due to pure rotation of 1 about point 2. (iv) If point 2 becomes the centre of mass (C.M.) of the rigid body, we can understand the general motion or plane motion of rigid body as translational with centre of mass and rotation about centre of mass.
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Example A plywood moves on a horizontal plane as shown in the figure, find the velocity of C.
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Rotation of a rigid body about a point If r2 = constant and zero.
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Rotational Dynamics →
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In the Fig. (a), the direction of ω changes with time or the line joining 1 and 2 forms a cone during its rotation. Such type of rotation is fixed point rotation (but not the fixed axis rotation). In Fig. (b), sense of ω vector is directed along a fixed line. Hence, the fixed point rotation about the point 2 is also the fixed axis rotation, which passes from the point → 2 and perpendicular the plane of disc. In the sense of ω in Fig. (b), the point 2 is also the centre of circle which the position 1 inscribed. The Fig. (c) also represents the fixed point rotation as well as the fixed axis rotation but the point 2 is not the centre of the circle which the particle inscribed.
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Note: (i) The sense of ω is also the direction of axis of rotation.
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Rotational Dynamics
(ii) In all the three figures, centre of the circle inscribed by all the particles lie on a line know as axis of rotation. In Fig. (b) all the circle are concentric. (iii) Perpendicular from the point on the axis of rotation is its radius. (iv) ω is same for all the points of rigid body in pure rotation. (v) ω is independent of axis of rotation. (vi) As ω (or dθ) and α is same for all the points of a rigid body in pure rotation, then it could define the angular kinematical term at any point of a rigid body. Therefore, all the formula of angular kinematics is equally hold for the rigid body. →
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It is similar to the angular motion in circular motion chapter.
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Example A flywheel of radius 30 cm starts from rest and accelerates with constant acceleration of 0.5 rad/sec2. Compute the tangential radial and resultant acceleration of a point on its circumference,
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(i) Initially, θ = 0, so ωi = 0
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Tangential acceleration = at = R α = (0.3)(0.5) = 0.15 m/sec2 Net acceleration, anet. = (at ) 2 + (an) 2 = 02 + (.15) 2 = 0.15 m/sec2
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Rotational Dynamics
1.7
2π (ii) After, θ = 120º 3 2π 2π ωf2 = ωi2 + 2α θ = 0 + 2 (0.5) = 3 3 2π rad/sec. 3
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It is a kind of general motion in which the motion of rigid body is confined in a plane motion say X-Y plane and Z co-ordinate of all the points is constant. The plane motion can be regarded as sum of the translational and rotational motion for simplicity.
v = vc + w × r12
→ → a1 = ac + ω2 − r12
Example, Motion of slipping stick along of the corner of plane motion. Y$
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vA l sin θ
vB = vA tan θ Example The end A of the rod AB is being pulled on the floor with a constant velocity v0 as shown. Taking the length of the rod as l, calculate: (a) The velocity of end B; (b) The angular velocity of the rod; and (c) The velocity of centre of mass of the rod at the instant when θ = 37º.
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Rotational Dynamics
1.9
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1.10
Rotational Dynamics
1.4 INSTANTANEOUS CENTRE OF ROTATION OF ZERO VELOCITY The velocity of a point of a rigid body in a plane motion can be visualized or imagine in pure rotation. The point about which the plane motion can be imagined pure rotation is known as instantaneous centre of rotation. The centre of rotation may lie outside the rigid body. Let us denote the centre of rotation by 2. Velocity of any point of the rigid body in plane motion can be written as → → → vi = ω × ri 2 i ≠ 2
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Equation (i) indicates that the point 2 is perpendicular to any point of rigid body at the instant. It is also clear from equation (i) that when the direction of atleast → two points of rigid body is same as ω then the instantaneous centre of rotation is perpendicular to the velocity vector.
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Rotational Dynamics
1.11
→ → vc = ω − k × vc2 j →
→
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Example A rotating disc moves in positive direction of x-axis. Find the equation y (x) describing the position of the instantaneous axis of rotation, if at the initial moment, the axis C of the disc was located at the point O after which is moved.
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(i) With a constant velocity v, while the disc started rotating counterclockwise with a constant angular acceleration α (the initial angular velocity is equal to zero). (ii) With a constant acceleration a (and the zero initial velocity) while the disc rotates counter-clockwise with a constant angular velocity ω.
) Solution (i) ω = ω0 + α t
2
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(ii) Centre moves with constant acceleration a, initial velocity is zero. 1 x = at 2 2 ∴
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2x = 2 ax a
vc 2ax = ω ω
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1.13
Rotational Dynamics
1.5 ROLLING MOTION Rolling motion can be considered as combination of rational and translational motion. For the analysis of rolling motion, we deal translation separately and rotation separately and then we combine the result to analyse the overall motion. $ 2 Z
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Consider a uniform disc rolling on a horizontal surface. Velocity of its center of mass is v and its angular speed is ω as shown: A, B and C are three points on the disc. Due to the translational motion each point A, B and C will move with center of mass have tangential velocity w R, R is readius of disc. When the two motions are combined, resultant velocities of different points are given by
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Rotational Dynamics
Example A ball of radius R = 10 cm rolls without slipping down an inclined plane, so that its centre moves with constant acceleration a = 2.50 cm/s2, t = 2.00 sec. After the beginning of motion its position corresponding to that is shown in the figure. Find: (i) The velocity of the points A, B and O. (ii) The acceleration of these points. $ &
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Let the fixed co-ordinate axis x, y and z as shown in the figure. As the ball is rolling without slipping along the rigid surface, therefore
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At the position corresponding to that of figure in accordance with the problem, ac = a, So vc = at vc at = , R k a and, α = R ω=
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Rotational Dynamics →
→
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Now, vA = vc + ω × rAC → → → → = vc i + ω − k × R j = (vc + ωR) i →
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vc2 (− j ) R
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1.15
1.16
Rotational Dynamics
a 2t 2 = 2a i + (− j ) R ∴
aA = 4 a 2 +
at 2 a 4t 4 = 2 a 1 + R2 2R
2
→ → → → → Similarly, aB = aC + ω2 − rBC + α × rBC → → → → = a i + ω2R − i + α − k × R i
na .c om
v2 → = a − C i + αR − j R 2
ar a
m
1.6 SUMMATION OF ANGULAR VELOCITY
er s
w
ita
.ja
2
rio s_ in fo
at 2 a 2t 2 2 aB = a − + a = a 1 + 1 − R R
So,
er si ta rio s
iv Z
ht tp
s: //
t.m e/
un
Z
%
ht tp s:
//t
ht
.m
tp
e/
s:
un iv
//w
w
Relative angular velocity in rigid body Let us consider a rigid body rotating about an axis OA which itself rotates about fixed axis OB as shown in the figure.
Z
$
2
Let the angular velocity of OA with respect to the OB is ω′ and the angular velocity at axis OB is ω0. Then, the angular velocity of rigid body is ω = ω′ = ω0 Again consider a special case in which →
→
angular velocity = ω0 and ω′ = Constant. →
→ d ω′ In this case, angular acceleration α = d ω = = ω′ ω0 dt dt
→
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Rotational Dynamics
1.17
Example A round cone with half-angle α = 30º and the radius of the base R = 5.0 cm rolls uniformly and without slipping over a horizontal plane as shown in the figure. The cone apex is hinged at the point O which is on the same level with the point C, the cone base centre. The velocity of point C is v = 10.0 cm/s. Find the modulii of (i) The vector of the angular velocity of the cone and the angle it forms with the vertical. (ii) The vector of the angular acceleration of the cone.
ar a m
ita
.ja
er si ta rio s
w
er s
w
) Solution
&
D
rio s_ in fo
na .c om
2
iv
→
e/
→
.m
→
ω = ω′ + ω0
Z
un
2
t.m e/
Z
Z
s: //
ht tp s:
//t
ht
tp
s:
un iv
//w
(i) Let the axis of the cone (OC) rotates in anticlockwise sense with constant → angular velocity ω′ and the cone→itself about it’s own axis (OC) in clockwise sense with angular velocity ω0 . Then, the angular velocity of the cone,
ht tp
T T
2
D 5
X
$ →
→
As the rolling is pure, magnitude of the vector ω′ and ω0 can be easily found from the figure. ω′ =
v v , ω0 = R cot α R
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...(ii) https://t.me/universitarios
1.18
Rotational Dynamics →
→
As ω′ ⊥ ω0 , from equations (i) and (ii), we have 2
v v ω = ω′2 + ω02 = + R cot α R =
2
v = 2.3 rad/sec. R cos α
(ii) Vector of angular acceleration →
→
d ω d (ω′ + ω0 ) α= = dt dt →
→
rio s_ in fo
na .c om
→ as ω′ = constant
→
ar a
The vector ω0 which rotates about the OO′ axis with the angular velocity ω′ retains its magnitude. This increment in the time interval dt is equal to
ita
.ja
m
|dω0| = ω0 . ω′ dt
→
er si ta rio s
→
//w
w
er s
→
Thus, α = ω′ × ω0
un iv
w
or in vector form dω0 = (ω′ × ω0) dt
e/
.m
iv
v v v2 = tan α = 2.3 rad/sec 2 R cot α R R2
s: //
v v v2 = tan α = 2.3 rad/sec 2 R cot α R R2
t.m e/
un
//t
∴α=
Example Find the velocity of centre of the body which rolls without slipping on a horizontal floor in terms of v, r and R.
ht tp
α=
α = ω′ ω0 (as ω′ ⊥ ω0)
ht tp s:
ht
tp
s:
The magnitude of the vector α is equal to
5 U
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Rotational Dynamics
1.19
) Solution →
→
→
v0 = ω × r02 → → → v = ω − k × (R − r ) j
3 5GT
Z
v ω= R −r
or
5
T
3
→ v Ri R −r
na .c om
vC =
Note: PP′ = R dθ = ds
Y
P GV
er s
w
ita
.ja
Rω = v Rα = a
ar a
d θ ds = dt dt
m
R
rio s_ in fo
→ → → → vC = ω × rC2 = ω − k × R j
er si ta rio s
iv
0
5
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
un iv
//w
w
Example In the figure shown, a block of mass m and given the velocity v. Calculate the velocity of centre of mass of inclined sphere.
U
1
ht tp
s: //
5 '
U
5 7
U
T
3
P Y
) Solution Let, the angular velocity of sphere 3 is ω. ∴
v = (R3 – r3) ω3
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Rotational Dynamics
v R 3 − r3
or
ω3 =
∴
vT = ω3 2R 3 =
2vR 3 (R 3 − r3 )
vN = ω2 r2 ∴
ω2 =
vN 2vR 3 = r2 (R 3 − r3) r2
vM = ω2 × R 2 =
2vR 2 R 3 = vD (R 3 − r3 )r2
2vR 2 R 3 (R 3 − r3) r2 (R1 + r1)
2v R1 R2 R3 m/sec. (R3 − r3) r2 (R 1 + r1 )
er s
w
Torque About a Point
ita
.ja
m
ar a
vC1 = ω1 R1 =
rio s_ in fo
ω1 =
na .c om
vD = ω1 × (R1 + r1)
er si ta rio s
iv
)
T
T
s: //
t.m e/
ht tp s:
un
//t
2
U
.m
tp
e/
s:
un iv
//w
w
Let a point influence by a force F and the position of some point of application of the force (of the particle). Then the moment of force F about the point O,
ht
0
→
→
ht tp
1.20
τ0 = r × F |τ0| = F (r sin θ) = F × OM directed towards points and line of action of force. Let there are a number of forces acting on a system and we want the net torque about a certain point O. Then, from the principle of super-position (or vector addition) →
→
→
τ = τ0 = ∑ ri × Fi = ∑ τ0 where, τ0 = Torque about point O.
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Rotational Dynamics
1.21
Torque About an Axis Let there is û axis (x, y, z) passing through the point O. Then, the moment of force or torque about the axis (û axis). → → → → ^ ^ τu = u . τ 0 = u . r × F
ux u y uz τu = x y z δx δy δz →
^
→
→
na .c om
if u is collinear to r or F, then δu = 0
er si ta rio s
→
.m
un iv
→
e/
→
τ z = k . τ0
rio s_ in fo
er s
→
= mgr j
→
iv
= k . mgr j
un
//t
ht
tp
s:
//w
w
w
ita
.ja
m
ar a
Example A circular disc rotates in horizontal plane about a fixed axis passing through its centre and perpendicular to the axis. A particle of mass m which is at rest with respect to disc and which is at a distance r from the centre of disc. Find the torque of the weight of the particle about the point O, and also about the axis of rotation of the disc. → → → → ) Solution τ0 = r × F = rc × F − k
t.m e/
ht tp s:
=0
ht tp
s: //
Example In the position shown in the figure, find the net torque about the axis passing through the point O in the sense of angular velocity (or axis of rotation) and also about the axis OA. 2
T O
$
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PJ U
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Rotational Dynamics →
) Solution τ0 = → → = l sin θ i × mg − j →
= −lmg sin θ k →
→
τ z = k .(−lmg sin θ k ) = – lmg sin θ.
5
ar a
Z
un iv e/
P
iv
.m t.m e/
un
//t ht tp s:
7
er si ta rio s
er s
w w //w s: tp
→
5
ita
.ja
m
7
rio s_ in fo
na .c om
Example Find the net torque of the tension forces about the axis of rotation of the disc.
ht
→
s: //
) Solution τT = r × F
0
→
→
= R1 (− i ) × T(− j )
ht tp
1.22
→
= T R1 k →
→
τT 0 = R 2 i × T0 (− j ) →
= − R 2 T0 k ∴
τ0 = (τT + τT0) →
= (TR1 − T0 R 2 ) k →
→
τ z = k . (TR1 − T0 R 2) k = (TR1 – R2 T0)
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Rotational Dynamics
1.23
Angular momentum of a particle about a point Let a particle of mass m follows a curve path. At a certain instant of time, its position →
→
with respect to a reference point O is r and its velocity with respect to point O is v . →
→
→
→
LO = r × p
→
→
LO = r × (m v )
Pĺ Y
rio s_ in fo
na .c om
U
ar a
The angular momentum for a system of particle with respect to the point O, →
m
LO = Σ LiO
→
→
un iv
w
//w
→
→
ht
e/
tp
Lz = O
.m
s:
If k is collinear with r and p . Then
er si ta rio s
→
ita
w
→
Lz = k . L O = k . ( r × p )
er s
.ja
Angular momentum about an axis passing through to the point O,
ht tp s:
un
//t
iv
From the principle of the super-position, the total angular momentum of a system of particle about the z-axis,
t.m e/
Lz = ∑ Li
→
→
→
L0 = r × p
ht tp
s: //
Relationship between torque and angular momentum about the same point. (corresponding to a particle)
→
→
→ d p d L0 d r = × p + r× dt dt dt
→ → d L0 → = v× m v + r × F dt
d L0 = τ0 dt If,
τ0 = 0, then L0 = constant Στ x =
dl dlx dl , Στ y = y , Στ z = z dt dt dt
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Rotational Dynamics
Example The angular momentum of a particle relative to a certain point O varies with time as L0 = a + bt2, where a and b are constant vectors, with a ⊥ b. Find the force moment τ0 relative to the point O acting on the particle when the angle between the vector τ0 and L0 equals 45º.
) Solution τ0 =
d L0 = 2 bt dt
...(i)
Given that the angle between τ0 and L0 is 45º
na .c om
1 2b 2t 3 = 2 2bt . a 2 + b 2 t 4
1 b 2t 4 = 2 2 4 2 a +b t
a2 = b2 t4
w
e/
un
//t
iv
.m
a b
un iv
w
//w
s: tp
τ0 = 2b .
er s
a b Hence form equation (i), t=
or
ita
.ja
m
or
ar a
Squaring we get,
er si ta rio s
or
rio s_ in fo
τ0 .L0 2bt.[a + bt 2 ] = |τ0||L0| 2bt . a 2 + b 2 t4
cos 45° =
∴
ht
Angular momentum about a point and about an axis in polar co-ordinate system →
t.m e/
ht tp s: →
u^ u^ θ k = rρ rθ rz
s: //
L0 = r × p
=
ht tp
1.24
U
XT XU
Pρ Pθ Pz
Lz = rρ Pθ = ρ mvθ = ρm ρ ωz = mρ2 ωz Similarly, for the torque
N
2
τz = ρFθ. where, ρ = perpendicular distance from the point mass to the axis. Note: Projection of Lz and τz are indeed independent choice of the point O on the z-axis relative to which are τ0 and L0.
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Rotational Dynamics
1.25
Angular momentum of a rigid body rotating about a stationary axis Lz = ΣLiz
=
= Σ mi ρi2 ωz = ωz Σmi ρi2
Y ± U P ±
= Iωz
na .c om
where, Σmi ρi2 = I - is known as M.I. of solid body or rigid body about z-axis. In general, continuous mass of a rigid body about an axis can be defined
rio s_ in fo
I = ∫ r 2 dm
ar a
Moment of inertia of rigid body depends on the distribution of mass relative to the axis as well as on the axis of rotation.
iv
er si ta rio s
&
.m
e/
;
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Example Three particles, each of mass m are situated at the vertices of an equilateral triangle ABC of side L as shown in the figure. Find the moment of inertia of the system about the line AX perpendicular to AB in the plane of ABC.
%
/
ht tp
s: //
$
) Solution Perpendicular distance of A from AX = 0
∴
Perpendicular distance of B from AX = L L Perpendicular distance of C from AX = 2 Moment of inertia of the particle at A = 0
Moment of inertia of the particle at B = mL2, and
L Moment of inertia of the particle at C = m 2
2
The moment of inertia of the three-particle system about AX is 2
2
L 5mL 0 + mL2 + m = 4 2
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1.26
Rotational Dynamics
Radius of gyration If M is the mass and I is the moment of inertia of a rigid body, then the radius of gyration (K) of a body is given by I M ⇒ I = MK2 K=
un iv e/
iv
.m
Annular cylinder (or ring) about cylinder axis, I = 1 M(R12 + R 2 2 ) 2
t.m e/
ht tp s:
un
//t
ht
er si ta rio s
er s
w w //w s: tp
Hoop about cylinder axis. I = MR2
$[LV
s: //
$[LV
ht tp
5
5 5
(b)
ita
.ja
5
$[LV
rio s_ in fo
(a)
m
ar a
na .c om
1.7 MOMENT OF INERTIA OF SOME IMPORTANT BODIES
(c)
(d) /
/ 5
Solid cylinder (or disc) about cylinder axis. 1 I = MR 2 2
Solid cylinder (or disc) about centre diameter 1 1 I = MR 2 + ML2 4 12
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Rotational Dynamics
1.27
$[LV
(e)
/
Thin rod about axis through centre perpendicular to length, I =
1 ML2 12
$[LV /
rio s_ in fo
na .c om
(f)
1 3
(h)
5
ht tp
s: //
t.m e/
ht tp s:
un
iv
5
//t
ht
.m
(g)
e/
tp
s:
un iv
//w
w
er s
w
$[LV
er si ta rio s
ita
.ja
m
ar a
Thin rod about axis through one end perpendicular to the length, I = ML2
Solid sphere about any diameter 2 I = MR 2 5
(i)
Thin spherical shell about any diameter 2 I = MR2 3
Hoop about any diameter. I = 1 MR 2 2
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Rotational Dynamics $[LV
(j)
E
D
1 M(a 2 + b 2 ) 12
rio s_ in fo
I=
ar a
na .c om
Rectangular plate about perpendicular axis through centre.
ita
er s
w
) Solution
.ja
m
Example Find the moment of inertia of a uniform hollow sphere about a diameter.
M (2πR sin θ) (R dθ) 4πR2
t.m e/
M sin θ d θ 2
ht tp
s: //
=
un
//t
ht tp s:
Mass of the ring =
M 4πR2
iv
.m
tp
Mass per unit area of the sphere =
er si ta rio s
un iv
= (2 πR sin θ) (R dθ)
e/
s:
//w
w
Let, mass be m and radius be R, Then area of the ring
ht
1.28
5 GT T
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Rotational Dynamics
1.29
Moment of inertia of this elements ring about OX, M M dI = sin θ d θ (R sin θ)2 = R2 sin3 θ d θ 2 2
0
π
=∫ 0
MR2 sin 3 θ . d θ 2 MR2 (1 − cos2 θ) sin θ d θ 2 π
MR2 2
COS 3θ −COS θ − 3 0
=
MR 2 2
1 2 2 1 + 3 = 3 MR .
na .c om
=
rio s_ in fo
π
I=∫
m
ar a
Two Important Theorems on Moment of Inertia
er si ta rio s
un iv
.m
e/
I = ICM + md2
iv
(ii) Perpendicular axis theorem Consider a lamina (cross-section of rigid body). From the figure
s: //
ht tp
ri2 = xi2 + yi2
t.m e/
ht tp s:
un
//t
ht
tp
s:
//w
w
er s
w
ita
.ja
(i) Theorem of parallel axis The rotational inertia of any body about an arbitrary axis equals the rotational inertia about a parallel axis through the centre of mass plus the total mass times the squared distance between the two axis. Mathematically, the parallel axis theorem has the following form:
]
\
U
[\ [
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Rotational Dynamics
Σmi ri2 = Σmi xi2 + Σmi yi2
Multiplying both sides by Σmi, we get
Iz = Ix + Iy
where, Iz = Moment of inertia of the rigid body about an axis passing through centre of mass.
a
s
a
r
Fundamental Equation of Rotational Dynamics of Rigid Body ( otation bout a tationary xis) In this case, if axis of rotation is denoted by z-axis, then we have Lz = Iω dω = Iα dt
m
rio
ar
s_ i
where, α = angular acceleration of the rigid body.
nf o
τz = I
an
a.
co m
Differentiating with respect to time, we get
w
si
ta
.ja
Kinematic Energy of Rigid Body Rotating About a Fixed Axis
s io ta r
iv
si
un
er
e/
//t .m
1 = Σ mi ρi2 ω2 2
ht
tp s
:// w
w
er
1 T = Σ mi vi2 2
.m
tp
ht tp
s:
1 dW = dT = d Iω2 2
//t
ht
s:
e/ un
iv
1 = Iω2 , 2 where, I = Σ mi ρi2 The work performed by external forces from work-energy theorem,
= I ωz . dωz dW = I ωz . αz dt = I . αz ωz dt = τz dθ where, τz = Iαz and dθ = ωz dt ∆t
W = ∫ τz . d θ
0
Kinetic energy of a rigid body in plane motion,
1 T = T + mvc2 2
1.30
1 1 = IC ω2 + mvC2 2 2
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Rotational Dynamics
1.31
Impulse momentum theorem →
∆ L = ∫ τ dt This equation holds equal in the all the frame (inertial or non-inertial frame) In the non-inertial frame, →
→
τ = Σ τext + Σ τ (system)
Total Momentum of External Force →
→
t.m e/
→
→
τ0 = (r0 + r1 ′) × Fi
→
→
ht tp
→
s: //
τ0 = Σ ri × Fi
→
2
un
//t ht tp s: →
er si ta rio s
un iv
e/ .m
tp
U
U
iv
er s
w //w s:
2
ht →
ita
m w
.ja
U
5LJLG ERG\
rio s_ in fo
)U
ar a
na .c om
Let τ0 be the total moment of force relative to the point O and τ01 relative to the → point O′, whose radius vector is equal to r 0 .
τ0 = (r1 × Fi ) + r0 × Fi →
→
→
τ0 = τ01 + (r0 × F) If F = 0, τ0 = τ01 In other words, when F is equal to zero, the total momentum of external forces do not depend on its choice of the point relative to which it is to be determines. In particular, such is the case when a couple acts on the system with respect to the C frame, which possesses an important characteristic.
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Rotational Dynamics
This reference frame is rigidly fixed to centre of inertia of the system of particles. Since in the general case, the C frame is non-inertial, resultant of the external force must include not only the external forces of introduction but also the inertial forces, i.e., →
...(i)
F = F net + Fin
→
→
Σ τic = 0 →
→
→
ar a
Σ ri × (− mi ) ac = Σmi ri × (− ac )
m
Because in the C frame itself,
ita
.ja
→
rc = 0
rio s_ in fo
na .c om
On the other hand, the system of particles as a whole rest in the C frame. Therefore, → in this frame F = 0 Taking into account the equation (i) we conclude that in the C frame, the total moment of forces (including inertial forces) does not depends on the choice of the point O, and another important conclusion that in C frame the total moment of inertial forces relative to the centre of mass is always equal to zero.
→
t.m e/
Σ τic ( z ) = Iα z
er si ta rio s
un
ht tp s:
//t
iv
→
Σ Fic = m ac ,
.m
ht
tp
e/
s:
un iv
//w
w
er s
w
∴ ∑τic = 0 Note that the angular acceleration αz as well as ωz are equal in both reference frames. Since, the C frame translate relative to the inertial frame (in the frame of ground earth). Basic equation of dynamics of the solid or a rigid body in plane motion is
s: //
How to solve a problem:
ht tp
(i) Draw the free body diagram of each body. (ii) On the basis of force diagram, analyse direction of kinematical term and then write the chemical dependence of the body. (iii) For the translation (linear motion) Σ Fix = m ac x , ΣFiy = m acy
If necessary put τic = 0 For the curve linear motion Σ Fit = m atc , Σ Fin = manc
And if necessary put τic = 0 (iv) For the pure rotation
1.32
Σ τic = I αzc
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Rotational Dynamics
1.33
If necessary put ΣFix = 0, ΣFiy = 0 or ΣFit = 0, ΣFin = 0 (v) In the plane motion use →
Σ Fi = m ac Example A sphere of mass m rolls without slipping on an inclined plane of inclination θ. Find the linear acceleration of the sphere and the force of friction acting on it. What should be the minimum coefficient of friction to support pure rolling?
) Solution Applying the law of motion equation
I
ita
.ja
m
ar a
1
un iv
//w s:
e/
iv
.m s: //
2 a fr = mr 2 5 r
a=
f =
ht tp
2 ma 5 From equations (i) and (iii), we get or
...(iii)
5 2 g sin θ and f = mg sin θ 7 7
For pure roller, µN > f or
µ mg cos θ >
or
µ>
...(ii)
t.m e/
ht tp s:
un
//t
tp ht
a r fr = I α
er si ta rio s
w
er s
w
PJ
T
As the sphere rolls without slipping, thus α=
...(i)
rio s_ in fo
na .c om
mg sin θ – f = ma
2 mg sin θ 7
2 tan θ 7
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1.34
Rotational Dynamics
3. Two identical solid spheres of mass M and radius R are joined together and the combination is rotated about an axis tangent to one sphere and perpendicular to the line connecting them. What is the rotational inertia of the combination?
1. A wheel having moment of inertia 2 kg-m2 about its axis, rotates at 50 rpm about this axis. Find the torque that can stop the wheel in one minute.
)
) Solution Initial angular velocity = 50 rpm 5π rad/sec. 3 Using, ω = ω0 + αt =
ω − ω0 −5π /3 −π = rad/sec 2 = rad/sec 2 t 60 36 The torque that can produces this deceleration is −π τ = I (α) = (2 kg − m 2) rad / sec 2 36 π = N−m 18
co m
α=
5
a.
an
rio
ar
m
) Solution Like masses, rotational inertias of )
si ta
.ja
rio s
ta
si
er
iv
m
2 MR2 + MR2 = 1.4 MR2 5 And for the second sphere, I2 = ICM + Md2
un
//t .
=
tp s
NJ P
:// t
.m
ht tp
e/
s:
iv
un
e/
tp s
ht
NJ P
solid objects add like scalars, therefore total for the two sphere, I = I1 + I2. For the first sphere (the one closer to the axis of rotation) we have, from the parallel axis theorem. I1 = ICM + Md2
er
w
w
:// w
=
)
ht
) Solution Taking moment about pivoted
I = I1 + I 2
1 (m2 g − m1 g ) × = I . α 2
= 1.4 MR2 + 9.4 MR2 = 10.8 MR2
2
α=
7/4
1 2 = 8.4 rad/sec 2.
∴
(5 − 2) × 9.8 ×
1 1 7 = m1 + m2 = kg.m 2 4 2 2
4. A uniform sphere of weight W and radius r is being held by a rope attached to a frictionless wall a distance L above the centre of the sphere as shown in the figure. Find: (i) Tension in the rope. (ii) The force exerted on the sphere by the wall.
2
Moment of inertia about pivoted
2 MR2 + M(3R)2 = 9.4 MR2 5
2. A light rod of length 1 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends as shown in the figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.
5
s_ in fo
WORKED OUT EXAMPLES
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Rotational Dynamics
1.35
with L1 = 20.0 cm and L2 = 80.0 cm. The rod is held in the horizontal position shown in the figure and then released. Calculate the linear acceleration of the two blocks as they start to move.
/
/
/
U P
) Solution )
P
) Solution Taking torque about point of pivot,
nf o
we have (mgL2 – mgL1) = I . α ...(i) Moment of inertia about pivot mL12 + mL22 = m(L12 + L22) Putting the value I in equation (i), we get
s
rio
9.8 × .60 = 7.13 rad/sec 2 (.20) 2 + (.80) 2
er s
iv
e/ un
6. In the given figure, calculate the linear acceleration of the blocks. Mass of the block A = 10 kg Mass of the block B = 8 kg Mass of disc shaped pulley = 2 kg
.m //t ht
1/2
r2 T = W(1 + tan 2 θ)1/2 = W 1 + 2 L
tp
s:
r L
α=
iv un e/ .m
ht tp :
From
∴
Linear accelerations are a1 = α L1 = 0.20 × 7.13 = 1.426 m/sec2 a2 = α L2 = 0.80 × 7.13 = 5.70 m/sec2
//t
1
But from the figure, tan θ =
mg(L2 – L1) = m (L12 + L22) . α
ita
w w 7
s:
ht
tp s:
//w
T
er si
.ja
m
ta rio
...(i)
s_ i
W cos θ
a.
T=
ar an
T = W sec θ
or
or
T cos θ = W
)
co
m
Analysis of the forces From ΣFiy = 0
ΣFix = 0 T sin θ = N
r2 r N = W 1 + 2 . 2 L ( r + L2 )1/2 r L
$
) Solution According to the law of motion,
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m2g – T2 = m2a T1 – m1g = m1a
5. Figure given below shows two blocks each of mass m suspended from the ends of a rigid weightless rod of length L1 + L2,
=W
%
)
1/2
∴
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...(i) ...(ii)
1.36
Rotational Dynamics
Now,
Σ τic = I α
1 (T2 R − T1R) = MR2 . α 2 7
7
PJ
∴
a = αR
in f
an a
rio
ar //t
ht
s:
ht
tp
s
First Method: Let the particle after time t reached the point P. Thus, the velocity and position of point P relative to origin O are. →
→
→
v = [v0 cos α i + (v0 sin α − gt ) j ]
1 r = v0 cos αti + v0 sin α t − g t 2 2
→
)
.m
tp
) Solution
(i) What is the angular acceleration of the rod when it is at an angle θ to the vertical? (ii) What is the tangential linear acceleration of the free end when the rod is horizontal? The moment of inertia 1 of a rod about one end is ML2. 3
ita
rs
ve
un i
e/
s:
(10 − 8) × 10 20 = m/sec 2. 10 + 8 + 1 19
7. A uniform rod of length L and mass M is pivoted freely at one end.
rio
iv un e/
.m //t
ht
=
(m2 − m1) g M m2 + m1 + 2
si
er
w //w
a=
tp
s:
M m2 g − m1 g = m2 + m1 + a 2 or
8. A ball of mass m is thrown at an angle α to the horizontal with the initial velocity v0. Find the time dependance of the magnitude of the ball’s angular momentum vector relative to the point from which the ball is thrown. Find the angular momentum L at the highest point of the trajectory if m = 130 g, α = 45º, and v0 = 25 m/s. The air drag is to be neglected.
ta
m .ja
7
w
7
α=
o
.c
3 g sin θ rad/sec 2 2L When the rod is horizontal, π 3g θ = and α = 2 2L 3g at = α L = m/sec 2. ∴ 2
or
s_
1 Ma ...(iii) 2 From equation (i), (ii) and (iii), we get T2 – T1 =
MgL 1 sin θ = ML2 α 2 3
om
or
Iα =
PJ
PJ
<
) Solution Take torque about pivoted end of rod
3
)
MgL sin θ Torque due to weight = 2 and, torque due to rotational motion = Iα.
D 2
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;
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j
Rotational Dynamics
Angular momentum, →
→
L(t )= Thus, L(t) =
→
L0 = r × p
mv0 gt 2 cos α 2
Thus, angular momentum at maximum height
→
1.37
→
= r×mv
t v0 sin α = → g 2 1 2 = m v0 cos αt i + v0 sin αt − gt j × v0 cos α i + (v0 sin α − gt ) j 2 2 t mv0 g v0 sin α L = cos α → → → 2 g 1 2 → s αt i + v0 sin αt − gt 2 j × v0 cos α i + (v0 sin α − gt ) j 2 mv03 2 = → sin α .cos α 1 2g = m v0 .v0 sin α ⋅ cos αt − v0 gt 2 cos − v02 sin α cos αt + gt 2 v0 cos α k 0 2 = 37 kg-m2/sec. i.e., at m t =
→
→
in fo
rio
s
er si
ta
un t.m ) Solution
→ → → → → 1→ → L(t ) = r × P = v 0 t + g t 2 × m( v 0 + g t ) 2
Second Method:
(i) mg – T = ma ...(i) Torque about the centre T.R=I.α Since tangential acceleration, a = αR, where α is angular acceleration
;
P )
D
ht
PJ
tp s: //
\
5
0
e/
tp s ht
:// t
iv
.m
ht tp
mv 3 = 0 sin 2 α cos α 2g = 37 kg-m2/sec Y
ta rio s_
e/ un
.ja
w w
w
s: //
2
v sin α 1 mg . v0 . 0 cos α 2 g
9. A light thread with a body of mass m tied at its end and is wound on a uniform solid cylinder of mass M and radius R as shown in the figure. At a moment t = 0, the system is set in motion. Assuming the friction in the axle of the cylinder to be negligible, find the time dependence of (i) the angular velocity of the cylinder. (ii) the kinetic energy of the whole system.
iv er si
v0 sin α g
At the highest point, t = L0 =
ar an m
1 = mg t 2 v0 cos α 2
a.
1 → in α ⋅ cos αt − v0 gt 2 cos − v02 sin α cos αt + gt 2 v0 cos α k 0 2
co m
→
π → 1 π → = mv0 gt 2 sin + α − k + mv0 gt 2 sin + α −∴k T.R= 1 MR 2 . a 2 2 2 2 R
1 → = mv0 gt 2 cos α − k 2 https://t.me/universitarios_infohttps://www.jamarana.com
1 I = MR 2. 2 1 T = Ma 2
π → 1 π → mv0 gt 2 sin + α − k + mv0 gt 2 sin + α − k 2 2 2
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...(i)
1.38
Rotational Dynamics
P
0
)
0
7
)
) Solution
...(iii)
1 mR2 . α 2
I
I
rio s_ in fo
)
For the cylinder, linear acceleration + angular acceleration = block acceleration. i.e., ac + α . R = a ...(iv) From equations (ii) and (iii), we get
er si ta rio s
ita er s
1 1 K = Iω2 + mv 2 2 2
w
(ii) Total kinetic energy
w
2mgt ω= αt = R(M + 2m)
ar a
a 2mg = R R(M + 2m)
m
∴ α=
na .c om
D&
2mg M + 2m
f.R=I.α =
Ma = ma 2
or a =
mg −
.ja
From equations (i) and (ii), we get
...(i) ...(ii)
F – f = Ma f = mac
PJ
un
iv
t.m e/
ht tp
un iv
e/
.m
s: //
ht tp s:
//t
ht
tp
s:
//w
f 2f 3f + = m m m 2 2 1 1 2mgt 1 2mgt 2 2 3 f = . MR . or R a= + m. 2 2 R(M + 2m) 2 R(M + 2m) m ma 2 2 ∴ f = ...(v) 2mgt 1 2mgt 2 1 1 3 = . MR2 . + m . R 2 2 From equations (i) and (v), we get R(M + 2m) 2 R(M + 2m) ma F− = Ma mg 2 t 2 3 = . M F 21 + ∴ a= 2m m M+ 3 10. A block of mass M rests on a horizontal 11. A rod of length l is constrained to move frictionless surface. A uniform solid in vertical plane containing its length cylinder with a mass m rests on the upper between two smooth perpendicular surface of the block as shown in the figure. planes. When it is released from the verA horizontal force F is now applied to the tical position by pulling its bottom gently, block. As a consequence, the cylinder find its angular acceleration and angurolls without slipping toward the rear of lar velocity as a function of the angle θ the block as the block accelerates. Derive made by it with horizontal as shown in an expression for the acceleration of the the figure. block in terms of F, M and m. ac + α.R =
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Rotational Dynamics
1.39
From equations (ii) and (v)
l N B − mg = m [α cos θ − ω2 sin θ] 2
N B = mg +
or
ml [α cos θ − ω2 sin θ] 2 ...(vii)
Putting this value in equation (iii), we get
T
ml ml [(−α sin θ − ω2 cos θ)sin θ] − mg + 2 2
)
) Solution
ΣFx = max NA = max ...(i) ΣFy = may or NB – mg = may ...(ii) Taking torque about centre of the gravity, we have τG = (NA sin θ – NB cos θ)
[α cos θ − ω2 sin θ]cos θ =
m
fo
co
a.
s
rio
ta
αl l α − = g cos θ 2 6
rs i
−
ve
or
un i
e/
−4αl = g cos θ 6
or
or
3 3g g cos θ = − cos θ 2l 2l
α = −ω
ω=
dω dθ
d ω 3g = cos θ d θ 2l
l l Now, x = cos θ and y = sin θ 2 2
α=−
Here, α is negative, so our reference is reverse.
tp s
ht
or
:// t.m
ht tp
1 2 ml ⋅ α 12
ml α (NA sin θ – NB cos θ) = 6
αl 2 l lα l ω2 sin θ − ω2 cos 2 θ.sin θ − g cos θ − cos 2 θ + s 2 2 2 2
−
er si
iv
e/ un
or
.m
%
s: //t
ht
T
PJ
or
ta
.ja w
:// tp s
D[
1%
I ⋅ α=
rio
m
s_
ar an
in
l lα l ω2 lα − α cos 2 θ − cos 2 θ + sin θ ⋅ cos θ = 2 2 2 6
αl 2 l lα l ω2 lα sin θ − ω2 cos 2 θ.sin θ − g cos θ − cos 2 θ + sin θ ⋅ cos θ = 2 2 2 2 6
w
−
l [−α sin 2 θ − ω2 cos θ − sin θ] − g cos θ 2
or
w
$ 1$
mlα 6
...(v)
ml [−α sin θ − ω2 cos θ] 2
...(vi)
∴
3g 2l
θ
∫ cos θ d θ
90°
ω2 3g θ = − [sin θ]90 ° 2 2l 3g 3g ° = [sin θ]90 [1 − sin θ] θ = 2l 2l
or
From equations, (i) and (iv) NA =
0
d2y l = [α cos θ − ω2 sin θ] dt 2 2
...(iv)
d 2x l = [− α sin θ − ω2 cos θ] dt 2 2
∴
and
ω
Integrating, ∫ ω d ω = −
ω=
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1.40
at rest. The rod is then given an angular g velocity 2 . Assuming that the rings L are free to slide on the rod, find the distance between the points where the rings strike the floor in the subsequent motion.
12. A uniform thin bar of 4 kg and length 2 metres is bent to make a square. Calculate its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the square.
Rotational Dynamics
)
) Solution Total mass of the bar is m kg and length l metres.
)
) Solution
'
&
Y
Z
Y
na .c om
ita
...(i)
er si ta rio s
2
iv
ML2 3ML2 = 2 2
un
I = ML2 + 2ML2 = 3ML2
g 3 ML2 .2 I0 ω0 2 L = g ω= = 2 I 3ML L
g . L = gL L From energy conservation,
v2 = ω L =
1 1 1 I0 ω02 = Iω2 + .2M(v12 +v 22 ) 2 2 2
∴
3 g g ML2 . 4 . = ML2 . + 2 MgL + 2Mv12 2 L L
or
or 6gL – 3gL = 2v12 ⇒
13. A uniform rod of length 2 L and mass 3 M is free to rotate in a horizontal plane about a vertical axis passing through its mid point O which is fixed at a height L above the horizontal floor. Two small rings each of the mass M are located on the rod on opposite sides of O and at disL tance from O. The system is initially 2
t.m e/ s: //
∴
ht tp
ml 2 1 I = 4 l1 = × kg.m 2 16 3 Putting the value of m and l, we get 4× 4 1 I= × kg.m 2 = 0.333 kg.m2 16 3
e/
.m
ml 2 1 = × kg.m 2 64 3 Similarly, total moment of inertia
1 L .3M.(2L) 2 + 2M. 12 2
= ML2 +
and,
//t
1 1 12 + 4
s:
tp
ml 2 64
I0 ω0 = I ω
where, I0 =
un iv
2
m 1 l m l . + . [Parallel axis theo4 12 4 4 8 rem]
ht
=
//w
2
I1 =
ht tp s:
w
Moment of inertia of bar of side AB
er s
w
.ja
∴
Y
/
Let the angular velocity just after the ring slip-off is ω. From conservation of angular momentum
m
m Then, mass of one side of square is 4 l and length is metres. 4
/
rio s_ in fo
%
ar a
$
Y
v1 =
3 gL 2
Time of fall of each ring is given by 1 2 gt = L 2
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1.41
Rotational Dynamics
2L g
∴
=
Required distance between the points
2µg (1 + µ) R (1 + µ 2 )
t=
⇒
I
= 2 (L + v1t ) 2 + (v2t ) 2
1 1 5 1
= 2L (1 + 3) 2 + 2 I
= 2L 6 + 2 3
PJ
14. A uniform cylinder of radius R is spinned about its axis to the angular velocity ω0 and then placed into a corner shown in the figure. The coefficient of friction between the corner walls and the cylinder is equal to µ. How many turns will the cylinder accomplish before it stops?
Kinematics equation of the rigid body is ω02 – ω2 = 2 α θ From the problem ω = 0
m
in f
s_
rio s
er si ta
iv
e/ un iv
) Solution The given problem is shown in the )
According to the rigid body, torque about the centre
µmg µ 2 mg and, f 2 = 2 1+ µ 1 + µ2
f1 =
15. A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end ‘A’ of the rod with a velocity v0 in a direction perpendicular to AB. The collision is completely elastic. After the collision the particle comes to rest m (i) Find the ratio . M (ii) A point P on the rod is at rest immediately after the collision. Find the distance AP. (iii) Find the linear speed of the point P πL after the collision. at a time (3v0 )
∴
ω02 R(1 + µ 2 ) 8πµg (1 + µ)
.m
ht tp s: //t
f1 + µf1 µ
un e/
t.m
tp s
ht
://
...(i) ...(ii)
mg =
n=
er
w
s: //w ht tp
)
∑Fy = 0 mg = N1 + f2 f1 = N 2 f1 = µN1 And, f2 = µN2 From equation (i)
si
.ja w
∴
) Solution From the Law of motion
ω02 ω02 ω2 R(1 + µ 2 ) = = 0 2α 2. 2µg (1 + µ) 4µg (1 + µ) R 1 + µ2
But θ = 2 π n
5
θ=
ta rio
∴
o
ar an a. co m
= 6.15 L
1 ( f1 + f 2 )R = Iα = mR2α 2
or
µmg µ 2 mg 2 + 1 + µ 2 1 + µ 2 2( f1 + f 2 ) α= = mR mR
figure. %
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Y&
3 [ Z[
$ Y2 P
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1.42
Let the given point P on the rod be situated as shown in the figure. It is given that the point P remains at rest immediately after the collision of the object with the rod. This is possible only when the effect of collision is to cause a linear motion of the rod with velocity vc as shown in the figure and also a counterclockwise rotation of the rod about the axis passing through its centre of mass. Let x be the distance of the point from its centre of mass, then the above condition will be satisfied when at the point P. ∴ vc = ωx ...(i) where ω the angular velocity of the rod. From the conservation of momentum, mv0 = Mvc ...(ii) From the conservation of angular momentum about the centre of mass,
Using equations (ii) and (vi) in equation (i), we get
Distance of P from A, PA = x +
in time
co m
s
si ta iv er un e/
//t .m
Y2 Y2
3
Y2
The position of the rod is as shown in the figure.
πL , the point P is under the in3 v0 fluence of two velocities. Velocities due to the linear motion of the rod (acting in the upward direction) is At time
1 m vc = v0 = v0 4 M
2
m 3m + M M
m 1 = M 4
&
s:
ht tp
1 1 mv 1 1 6mv0 mv0 2 = M 0 + ML2 2 2 M 2 12 ML
or
rio
un
e/ t.m
://
...(vi) Using equations (ii) and (vi) in equation (v), we get
1=
fo
ta r
si
iv er
w w
tp s
ht
mv0 (L/2) mv0 (L /2) 6 mv0 = = Ic (1/12)ML2 ML
m 1 = , we get M 4
6v πL π θ= 0 = 4L 3v0 2
...(v)
From equations (iii) and (iv), we get
2
io s_ in
ar
m
.ja
//w
s:
ht tp
1 1 1 mv02 ) = Mvc2 + Iω2 ( 2 2 2
or
6mv0 πL = ML 3v0
From equation (vii), since
...(iv)
Also, it is given that the collision of the object with the rod is elastic. Therefore, from the principle of conservation of energy, we have
ω=
πL is 3v0
θ = ωt
a.
an
1 ML2 12
Ic =
L 2
L L 2 + = L ...(viii) 6 2 3 The angle through which the rod has rotated =
L mv0 = Ic ω ...(iii) 2 where, Ic is the moment of inertia of the rod about the axis passing through its centre of mass. It is given by the expression.
vc mv0 /M L = = ω 6mv0 /ML 6
x=
Rotational Dynamics
...(vii)
Velocity due to the rotation motion of the rod (acting in the horizontal direction) is
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1.43
Rotational Dynamics
From equations (i) and (ii), we have
6mv0 L vr = ωx = mL 6
mg sin α −
1 m = v0 = v0 4 M Resultant of these two velocities is 2
1 9.8 × g sin α 2 = = 1.6 m/s 2 a= 0.45 I 1 + 2 1 + 0.2 × (.03) 2 mr
2
v v v v = vc2 + vr2 = 0 + 0 = 0 2 2 4 4
17. A man pushes a cylinder of mass m1 with the help of plank of mass m2 as shown in the figure. There is no slipping at any contact. The horizontal component of the force applied by the man is F. Find:
o
rio s si ta
er
un
)
e/ .m //t
1 )
...(i)
ht tp s:
cylinder is shown in the figure.
D
I PJ
P
iv
.m
//t s: tp ht
ΣFi = ma or mg sin α – T = ma Torque about the centre is given by a T.r=I.α=I. r a r2
I
...(ii)
D I
7
Applying the kinematic law f1 + f2 = m1 a1
...(i)
T=I
PJ
or
1 a1 = αR and I = m1R2 2
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Also (f1 – f2) R = I α D
or
P
) Solution Free body diagram of plank and
D
D
)
ita
e/
s: ht tp
P
) Solution
)
un i
//w
ve
rs
w w
.ja
m
ar
rio s_
(i) The acceleration of the plank and the centre of the mass of the cylinder. (ii) The magnitudes and directions of frictional forces at contact point.
an
in f
a.
co
m
16. A spool with a thread wound on it is placed on an inclined smooth plane set at an angle α = 30º to the horizontal. The free end of the thread is attached to the wall as shown in the figure. The mass of the spool is m = 200 g, its moment of inertia relative to its own axis I = 0.45 g m2, the radius of the wound thread layer r = 3.0 cm. Find the acceleration of the spool axis.
The velocity vector will be directed at an angle of 45º with respect to the horizontal.
I mg sin α = 2 + m a r
or ∴
Ia = ma r2
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...(ii)
1.44
Rotational Dynamics
From equations (i) and (ii), we get
x=
R sin θ
1 f1 − f 2 = m1a1 ...(iii) 2 Then, from the equations (i) and (iii), we get
1 f 2 = m1a1 4 Equation of the plank F – f2 = m2a+2 and a2 = 2 α R = 2 a1
...(v)
When x = 2.25
and
om .c
si
Y$ [
rs
ve
ni
= (0.438) 2 + (0.216) 2 = 0.52 m/sec.
19. A spool with thread wound on it of mass m, rests on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to I = KmR2, where K is the numerical factor, and R is the outside radius of the spool. The radius of the wound thread layer is equal to r. The pool is pulled without sliding by the thread with a constant force F directed at an angle α to the horizontal. Find: (i) The projection of the acceleration vector of the spool axis on the x-axis. (ii) The work performed by the force F during the first t seconds after the beginning of motion.
T
ita rio s
)
dx → = vA dt
From the figure
v = (vBx ) 2 + (vBy ) 2
e/ u
$
→
) Solution
vBx = [0.6 + 1.35 × 0.6 (–0.2)] i vBy = 1.35 × 0.8 × 0.2
.m
//t
s:
tp ht Y$
[
dyB R cos θ . d θ = dt dt
er ∴
//t
s:
ht tp P
P
dxB dx R sin θ . d θ = + and dt dt dt
iv
un
ht
18. The end A of the slender rod AB against a fixed semi-cylindrical surface as shown is pulled with vA = 0.6 m/sec (→). If x = 2.25 m, determine ωAB of AB and vB of the end B.
∴ and
.m e/
tp s:
m1F . 3m1 + 8m2
%
fo
ta rio s_ in
ar
//w
w
w
.ja
m
Putting the value of a1 in equations (iv) and (v), we get 3m1F 3 4F f1 = m1. = 4 3m1 + 8m2 3m1 + 8m2
f2 =
dθ 0.6 × 0.6 × 0.6 =ω=− dt 1.35 × 0.8
= – 0.2 rad/sec (anti-clockwise) For point B xB = x – R cos θ and yB = R sin θ
an a
4F a1 = 3m1 + 8m2
⇒
∴
3 F = m1a1 + 2m2 a1 4
F – f1 = 2m2 a1 ⇒
R 1.35 = = 0.6 x 2.25 cos θ = 0.8 sin θ =
and,
sin 2 θvA dθ =− dt R cos θ
⇒
...(iv)
3 f1 = m1a1 4
dx R cos θ d θ =− dt sin 2 θ dt
∴
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Rotational Dynamics
1 1 v = m(ax t ) 2 + I. C 2 2 R
1 1 = max 2t 2 + KmR 2 ax2t 2 2 2
[
=
)
2
(i) ΣFx = max
r F2t 2 cos α − R . Work done = 2m (1 + K)
rio s_
in
fo
rio s
P
5
s: //
e/ un i
ve
t.m
tp
%
//t
.m
ht tp
[
s:
0
) Solution When the cylinder reached at
1 1 = mvC 2 + Iω2 2 2
or or ∴
bottom point, the block M traveled through a distance x. Since, the cylinder is slip of the track, thus, does not possess angular velocity. mxm + MxM = 0
(ii) The friction is no required any work because friction is static. ∴ Work done = Change in kinetic energy
)
r F cos α − R ax = m(1 + K)
∴
ht
tp
From equations (i) and (ii), we get r Fcos α − F. = K max + max R
rs ita
e/ u
$
) D
ht U
si ta
.ja m
s:
5
//w w
r − K max R
ax R
w
or F ⋅ r − f ⋅ R = K mR 2 ⋅ or f = F ⋅
om
.c
a R
20. A block of mass M with a semi-circular track of radius R, rests on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips on the semi-circular frictionless track. How far has the block moved when the cylinder reaches the bottom (point B) of the track? How fast is the block moving when the cylinder reaches the bottom of the track?
ni ve r
Fr − f ⋅ R = I ⋅α = I ⋅
ar an a
where, f = Frictional force, because no sliding occurs. Torque about the centre of spool
...(i)
or F cos α – f = max
m 2 2 t ax [1 + K] 2
Putting the value of ax, we get
) Solution
2
) D
U
5
1.45
m[xmM + xM] + MxM = 0 m[(R – r) – x] – Mx = 0
x=
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m(R − r ) m+M
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1.46
Rotational Dynamics
92
5
D
)
) Solution The cylinder does not jump, it means that normal reaction is always present.
s_ ita rio ∴
ita rio
rs
ni ve
.m e/ u s:
//t
where ω0 =
ht tp
2 g (R − r )m 2 M(M + m)
s
//t
ht tp s:
v2 2 =
PJ
v0 mR 2 and I = R 2
mv0 2 +
mv02 mv 2 = mv12 + 1 − 2mgR (1 − cos α) 2 2
v2 =
2 g (R − r ) .m M(M + m) 2 mv02 mv 2 mv0 + = mv12 + 1 − 2mgR (1 − cos α) 2 2 21. A uniform solid cylinder of radius R = 3 2 3 2 15 cm rolls over a horizontal plane passv0 = v1 − 2 gR (1 − cos α) or 2 2 ing into an inclined plane forming an angle α = 30º with the horizontal shown 4 v0 2 = v12 − gR (1 − cos α) ...(ii) or in the figure. Find the maximum value 3 of the velocity v0 which still permits the From equations (i) and (ii), we get cylinder to roll on to the inclined plane NR 4 section without a jump. The sliding is v0 2 = gR cos α − − gR (1 − cos α) m 3 assumed to be absent.
⇒
in fo
.c
e/ u
1Mv 1 + Mv2 2 2 m 2
Now 2 g (R − r )m 2 = (M 2 + Mm) v2 2 ∴
D
1 1 mv0 2 + I ω0 2 2 2 1 1 2 = mv1 + Iω2 − mg R(1 − cos α) 2 2
.m
ht
=
2
Here, O is the instantaneous centre of rotation and its velocity is v1. From the energy conservation
er s
2
1 M 1 mg (R − r ) = m v2 + Mv2 2 2 m 2
ni v
tp s: //w
w
w
Substituting the value v1 in equation (i), we get
2 2 2
...(i) 1
an a
.ja m
M v1 = v2 m
ar
1 1 mg (R − r ) = mv12 + Mv2 2 2 2 From the conservation of momentum mv1 – Mv2 = 0
mv12 R
om
Hence, the block moved in negative x-direction. From the energy conservation
mg cos α − N =
∴
[
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Rotational Dynamics $
1 NR = gR (7 cos α − 4) − 3 m
1 = gR (7 cos α − 4) 3
→
fo
in
s_ ita
ve rs
un i
e/
.m
//t
s:
ht tp
ita rio s
(ii) Since, the wedge does not undergo any rotation, the magnitude of torque due to F should be completely balanced by the torque due to N. Hence,
where, h is the perpendicular distance between the centre of mass of the wedge and the line of action of the force F.
2mv 3 ∆t
2mv → In the vector form N = mg + k 3 ∆t
23. In the arrangement shown in the figure, a block A of mass m is suspended by a string. The other end of the string is wound over a pulley B having a coaxial axel C. The axel-pulley system is suspended symmetrically by two light strings, one end of each is fixed to the ceiling and the other end would over the axel. The mass of the pulley-axel system
)
During the collision, the total downward force acting on the wedge,
4mv τ = Fh = h 3 ∆t
) Solution The collision is elastic and the sphere
is fixed so velocity of wedge is same and return to the original path. Hence, change in momentum causes the impulse force. Now, change in momentum in x-direction = 2 mv ∴ F cos 30º . t = 2 mv 2mv 4mv = F= or 3 3t ×t 2
4mv 3 → 4mv 1 → . i− × k 3t 2 3t 2
rio
ve rs e/ un i
tp
ht
s:
//t
(i) Find the force F and also the normal force N exerted by the table on the wedge during the time ∆t. (ii) Let h denotes the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge, during time interval ∆t.
→
N = mg + Fsin 30° = mg +
&
.m
%
→
→ 2mv → k = 2mv i − 3 → 1 → k = 2mv i − 3
a. c
ar an
m
w
//w ht tp s:
[
&
)
om
=
.ja w Y
)[
F = (Fcos α i − Fsin α k )
$
5
)
In the vector form
%
22. A wedge of mass m and triangular crosssection (AB=BC=CA=2R) is moving with a → constant velocity – v i towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time ∆t, during which the sphere exerts a constant force F on the wedge.
\
5
Hence, maximum value at v0
]
1.47
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Rotational Dynamics
% 7
&
f = ma1 f = µN = µmg ∴ a1 = µg ∴ v1 = µgt Torque about sphere surface I . α = f . R = µmg . R
) Solution )
ar a
m
un iv
er s
ita
.ja
w
w
//w
e/
t.m e/
5
I
I
s: //
5
P
un
ht tp s:
5
iv
.m
//t
ht
tp
s:
%
0
ht tp
&
P
$
∴
2 I = mR 2 5 α=
µmg .R 5µg = 2 mR 2 2R 5
Angular velocity, ω = αt =
Iac = 3ma + Mac R
Velocity of board, v2 = v − µgt .
m M
For the pure rolling, v2 = v1 + ωR
Mg + 3mg −
...(i)
5µg 2R
Since, there is no slipping, therefore, a = ac + 2 αR = ac + 2ac = 3ac From equations (i) and (iii), multiplying equation (iii) by 3, we get Mg + 3mg – (2TA + 2T) = 3ma + Mac ...(v) From the equations (ii) and (v), we have
where,
→
where, ac = acceleration of centre of mass of the cylinder Torque about common centre a TA 2R + 2TR = I. α = I c …(ii) R Motion of block mg – TA = ma ...(iii) where, a = Acceleration of block
...(i)
Mg + TA – 2T = Mac
)
) Solution
$
P
na .c om
7$
PJ
er si ta rio s
5
3(Mg + 3mg ) I M + 9m + 2 R
24. A board of mass M, whose upper surface is rough and under surface is smooth, rests on a smooth horizontal plane. A sphere of mass m is placed on board and the board is suddenly given a velocity v in the direction of its length. Find the time after which the sphere begins pure rolling, if the coefficient of friction between the board and the sphere is µ.
7 5
a=
rio s_ in fo
or
or
a a Mg + 3mg = 3ma + M + I 2 3 3R
is M and radius of the axel and the pulley are R and 2R respectively. The strings do not step over the pulley and axel. Find the acceleration of block A when the system is left free. The moment of inertia of the pulley-axel system about the common axis is I.
1.48
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Rotational Dynamics
(ii) Work done = change in kinetic energy Since, initial kinetic energy = 0 Therefore, final kinetic energy = work done =Fx
m 5µgt = µgt + .R M 2R
Thus, v − µgt .
µmg 5µgt t + µgt + M 2 v t= 7 m + µg 2 M
1 = F at 2 2 1 F(3m1 + 2m2 ) 2 =F. t 2 m1 (m1 + m2 )
25. A uniform solid cylinder A of mass m1 can freely rotate a horizontal axis fixed to a mount B of mass m2 shown in the figure. A constant horizontal force F is applied to the end K of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent. Find: (i) The acceleration of the point K. (ii) The kinetic energy of this system t seconds after the beginning of motion.
F 2 t 2 (3m1 + 2m2 ) m1 (m1 + m2 )
26. A small body A is fixed to the inside of a thin rigid hoop of radius R and mass equal to that of the body A. The hoop rolls without slipping over a horizontal plane. At the moments when the body A gets into the lower position the centre of the hoop moves with velocity v0 as shown in the figure. At what value of v0, will the hoop move without bouncing?
m
=
fo
in
s iv
un
ht
tp
s: //
(i) The acceleration of cylinder and the mount is same. F a= ...(i) m1 + m2
$
Torque about at the centre of cylinder F. R = I . α F.R 2F ∴ α= = 1 m1R 2 m1R 2 Acceleration of point K = a + αR
) Solution Clearly, the tendency of bouncing of )
)
5
Y2
e/ t.m
tp
s: //
)
ht
) Solution
si ta er
e/ t.m
tp ht %
si ta
un
s: //
iv
w
$
rio
w
er
w
.ja
m
rio
s_
ar an
the loop will be maximum when the small body A, will be at the highest point of the hoop during its rolling motion. Let the velocity of centre of mass of the hoop equals v at this position. The static friction does not work on the hoop, so from conservation of energy. 1
F 2F = + m1 + m2 m1
1
PJ PJ 1
Y2
F(3m1 + 2m2 ) m1 (m1 + m2 )
or
N = mg – N′
=
a.
co
∴
v=
or
1.49
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1.50
= mg + mg − ⇒
2
$ 5
2mgR − NR v = m 2
= 2 gR −
mv mv 2 = 2 mg − R R
NR m
& 3
...(iii)
Rotational Dynamics
4 &
From equations (i) and (iii), we get NR 3 2 gR − = v0 2 − 2 g R m
$
om
o
in f
rio s
ta
rs i
ve
ni
e/ u
Moment of inertia of disc about PQ 2
R I 2 = IC + m (using parallel 4 axis theorem)
//t .m
Initially, the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position
2
25mR 2 5R I1 = m = 16 4
ht
4
Total potential energy decreases 5mgR mgR = + 2 2 = 3mgR Moment of inertial of m about PQ,
tp s:
&
R mgR = mg 2 × = 4 2
ve
un i
.m e/
s: //t ht tp 5
3
rs ita rio s_
ar
.ja m
w
w
w
://
tp s
ht
$
Decrease in potential energy of disc from centre C to C′
3N m
27. A uniform circular disc has radius R and mass m. A particle also of mass m is fixed at a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed horizontal chord PQ R from the centre that is at a distance 4 C of the disc. The line AC is perpendicular to PQ.
v0 = 8gR −
5 5mgR = mg 2 × R = 4 2
)
) Solution The final position of rotation when particle reaches the lowest position is shown in the figure.
=
mR 2 mR 2 5mR 2 + = 4 16 16
Hence, total moment of inertia about PQ = I1 + I 2 25 5mR 2 15 mR 2 + = mR 2 16 16 8 1 Thus, kinetic energy = Iω 2 2 1 15 = . mR 2 ω 2 2 8 =
∴
Decrease in potential energy of the particle
3NR m
an a. c
v0 2 = 8 gR −
or
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1.51
Rotational Dynamics
Now, Decrease in potential energy = Increases in kinetic energy
∴
Ml ω = m(v − v1 ) 6
The linear speed of the particle, at which it reaches the lowest position is,
5R = 5 gR 4
in
s_
rio ta
m(v − v1 ) 3m(v − v1 ) + M M
s
(4m − M) .v (4m + M)
er si
v1 =
ta
or
iv
t.m
e/
un
mv = mv1 + Mv2
Substituting the value v1, we get
t.m
or
s: // tp ht
(4m − M) 2mv .M ∴ Mv2 = mv 1 − = (4m − M) (4m + M) 2m v2 = v 4m + M
From equation (iii) Ml ω 4m − M = m(v − v1 ) = m v − v 6 4m + M
(i) Let after impact, the velocity of puck be v1 and the centre of mass be v2. From the conservation of linear momentum mv = mv1 + Mv2 ...(i) From conservation of angular momentum, we have
(4m − M) . v + Mv2 (4m − M)
ht
Y
)
2M = m v 4m + M
∴ ω=
...(iv)
or Mv = –Mv1 + mv – mv1 + 3mv – 3mv1
mv = m
) Solution
m (v − v1 ) M
(ii) From equations (i),
s: // tp P
v = −v1 +
er si
iv
un e/
s: tp ht FHQWUH
...(ii)
From equations (i) and (iv), we get
ar
.ja
//w
w
w
0
ωl − v1 2 v
fo
co m
a.
an
v = −v1 + v2 + 3
m
(i) What is the velocity of the puck after impact? (ii) What is the velocity of the centre of mass and the angular velocity of the stick after impact?
v2 +
Since, e = 1, therefore, ωl ...(iii) v = −v1 + v2 + 2 From equations (ii) and (iii), we get
28. A metre stick lies on a frictionless horizontal table. It has a mass M and is free to move in any way on the table. A hockey puck m, moving as shown with speed v collide elastically with the stick.
v =ω
vn − vn e= 2 1 = v−0
rio
16 g 5R
ω=
or
or
15 mR 2 ω 2 16
3mgR =
or
2 mv1l I = Ml mvl = Iω + 12 2 2 Ml ω mv = + mv1 6
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1.52
29. A billiard ball initially at rest is given a sharp impulse by a cue. The cue is held horizontally at a distance h above the centre line as shown in the figure. The ball leaves the cue with a speed v0 and because of its forward “English” (rolling and sliding) eventually acquires a final 9 v0 4R speed of . Prove that h = , where 7 5 R is the radius of the ball.
µmgR =α Ic
or
µ mgR ω = ω0 − α t = ω 0 − ...(v) t Ic Since, after time t, the ball is in pure rolling, therefore, ∴
5
a. c
&
rio
ita
rio s
ita
er s
iv
un or
e/ ∴
5
mv 9 2R v0 = 0 h − R 7 I0 7
mv0 9 2R v0 = h− R 2 7 7 2 mR 5
9 5 2R = h− 7 2R 7
or 1
P1
2R 18R = 35 h − 7
f . R = Icα
or
35(7 h − 2R) 7 18 R = 35 h – 10R 18R =
∴
or
In final position, only friction torque is required to decrease the angular velocity. ∴ f = ma or µmg = ma or a = µg ∴ vt = v0 + µgt ...(iv)
or
PJ
v = ωR
2 But, Ic = mR 2 , because ball in solid sphere 5
.m //t
s: tp
ht K
.m
//t tp s:
ht
...(iii)
mv0 h Ic
mv0 h 2mv0 R − Ic Ic
ve
e/ un i
...(ii)
2mv0 R 7Ic
mv0 h mv0 2R ω = = h− Ic Ic 7
Now,
From equations (i) and (ii), we get
ω0 =
=
rs
w
//w w
⇒
ht
tp
s:
...(i)
µ mgR 2v0 . Ic 7µg
s_ in
an ar .ja m
)
or mv0 – 0 = F ∆t From angular momentum theorem Ic ω0 = τz . ∆t Ic ω0 = Fh . ∆t Fh . ∆t ω0 = Ic
2v0 7µg
fo
ω = ω0 − = ω0 −
) Solution From impulse momentum theorem Px = F ∆t
t=
From equation (v)
om
K
⇒
9v0 = v0 + µgt 7
v = ωR From equation (iv)
Rotational Dynamics
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4 R 5 https://t.me/universitarios
30. The axis of the cylinder given in the figure is fixed. The cylinder is initially at rest. The block of mass M is initially moving to the right without friction and with speed v1. If passes over the cylinder to the dashed position. When it first makes contact with the cylinder, it slips on the cylinder, but the friction is large enough, so that slipping ceases before M loses contact with the cylinder. The cylinder has a radius R and a rotational inertia I. Find the final speed v2 in term of v1, M, I and R. This can be done easily by using the relation between impulse and change in momentum.
)
the rod initially rest on its centre of mass, so there is no horizontal displacement. The velocity of centre of mass
v2 =
rio
ta
si
rio
s
er
2
l θ 1 4v 1 . 2sin 2 = ml 2 . 2 c2 + mvc 2 2 2 24 l sin θ 2
er
si ta
iv un e/
un
iv
or gl sin 2
//t .m
θ (1 + 3sin 2 θ)vc2 = 2 6sin 2 θ
e/
s:
t.m :// tp s
mv1 I m + 2 R
θ (sin 2 .sin θ) ∴ vc = 6 gl 2 1 + 3sin θ
32. Two cylinders having radii R1 and R2 and rotational inertias I1 and I2 respectively, are supported by axis perpendicular to the plane shown in the figure. The large cylinder is initially rotating with angular velocity ω0. The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases, and the two cylinders rotate at constant rates in opposite directions. Find the angular velocity ω2 of the small cylinder in terms of I1, I2, R1 and R2 and ω0.
Iv2 R2
31. A rod of mass m and length l is held vertically on a smooth horizontal floor. Now it is released from this position, find the speed of its centre of mass when it makes an angle θ with the vertical.
∴
mv1 = mv2 +
ht
or
mg
Iω R According to the problem v2 = ωR mv1 = mv2 +
or
s_
ar m w
Iω R
ht
m(v1 − v2 ) =
2
l 1 ml 2 2vc 1 2 mg (1 − cos θ) = . + mvc 2 2 12 l sin θ 2 or
...(i) ...(ii)
tp
ht
tp s
From the impulse-momentum m(v1 – v2) = F ∆t Iω = F ∆tR From equations (i) and (ii), we get
...(ii)
Substituting the value of ω in equation, we get
w
)
:// w
) Solution
in
l 1 1 mg (1 − cos θ) = I c ω2 + mvc 2 2 2 2
.ja
5
From conservation of energy
fo
co
m
T
an
Y
...(i)
l vc = sin θ ω 2
a.
Y
1.53
) Solution Since, there is no horizontal force and
Rotational Dynamics
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1.54
on a rough horizontal floor so that it starts-off with a purely sliding motion at t = 0. After t0 seconds, it acquires a purely rolling motion as shown in the figure.
Z
,
(i) Calculate the velocity of the centre of mass of the disc at t0. (ii) Assuming the coefficient of friction to be µ, calculate t0. Also calculate the work done by the frictional force as a function of time and the total work done by it over a time much longer than t0.
)
) Solution
5
From angular impulse equation Lz = I1 (ω1 – ω0)
m
Z
co
5
fo
Y
s
ta r
si
er
un
s:
R1R2 I1 ω0 – R ω2 I1 = R1 I2 ω2
or
ω2 (R12 I2 + R22 I1) = R1 R2 I1 ω0
or
RR Iω ω2 = 2 1 2 1 20 (R1 I 2 + R 2 I1 )
2
33. A uniform disc of mass m and radius R is projected horizontally with velocity v0
2 fr mR
...(iii)
So, angular velocity after time t0
1 mR 2 . α 2
α=
ω = α t0 =
2
∴
or
ht
R1I 2 ω2 I1 (ω0 R1 − ω2 R 2 ) 2 2
Torque about the centre of the disc f .R = I. α =
=
I 2 ω2 ωR I1 ω0 − 2 2 R1
R2 = R1
tp
s:
//t
ht
.m
tp
e/
According to the problem, after impact since they rotate at the constant rate, therefore ω1R1 = ω2 R2 ...(iv) From equations (iii) and (iv), we get
io
//t .m e/
tp
(i) By law of motion fr = ma ...(i) Acceleration of cylinder in the negative direction vx = v0 – at0 ...(ii)
un
...(ii)
...(iii)
ht
)
iv
//w
s:
R2 L 2 ω2 = R1 I1 (ω0 − ω1 )
) Solution
er si
w
For small body f . ∆t R2 = I2 ω2 From equations (i) and (ii), we get
ta
m ...(i)
w
.ja
For large body – f . ∆t R1 = I1 (ω1 – ω0)
rio
ar
s_
an
5
in
a.
Z
5
,
iv
Rotational Dynamics
2 f t0 mR
According to the problem, when the disc is in pure rolling motion, then vx = ωR
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Rotational Dynamics
or
or v0 −
2 f t .R f t0 = r 0 m mR
&0
3f t v0 = r 0 m
Y&
)
v0 2v0 = 3 3
(ii) From equation (iv)
ta
s
ta rio
un e/ .m
O
:// t
,
tp s
Z
Y&
1
ht
ft = −µmg v0t0 − m 2
er si
iv
s: //
ht tp t
1 f 0 = − f r v0t − . r t 2 2 m 0 2 r 0
fo
iv un e/ t
0 f Total work done = − f r ∫ v0 − r t dt m 0
l ml 2 cos θ ∆t = ω−0 2 12 where, m is the mass of the rod N
t.m
ht tp
= – fr vx dt Integrating, we have
s_
an ar m
s: //
w
Work done = fr dx
er si
v0 3µg
=
mv0 3µmg
w .ja
t0 =
w
∴
where, fr = μmg
in
a. co
m
mv0 mf r
t0 =
impulsive force acting on the rod is the normal reaction N acting at the end A of the rod. Since, torque of the impulsive normal reaction N about the centre of mass of the rod gives the clockwise rotation to the rod, the end A of the rod starts to move horizontally towards left just after impact. Further as the impact force N at A is vertical, the centre of mass will continue to move vertically after impact but with some new velo vc′ (say). For the rod from angular impulse momentum theorem about the axis passing through its centre of mass and perpendicular to the plane of figure (say z-axis) in the frame of the centre of mass of the rod: τz ∆t = ∆Lz
rio
vx = v0 −
) Solution During the course of impact, the only
..(iv)
From equation (ii)
T
$
v f ∴ r t0 = 0 m 3
1.55
N
l ml 2 cos θ ∆t = ω 2 12
or
...(i)
∴
From the linear impulse-momentum theorem, the translational motion of the rod is in the vertically downward direction.
or
Fy ∆t = ∆Py
– N ∆t = m v′c – mv
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34. A slender rod of length l forming an angle θ with the horizontal strikes a frictionless floor at A with its centre of mass velocity vc and no angular velocity. Assuming that the impact at A is perfectly elastic, find the angular velocity of the rod immediately after the impact.
$
1 = −µmg v0 t0 + µ 2 mg 2 t0 2 2
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...(ii)
Rotational Dynamics
Since, the point A moves towards left and the centre of mass moves vertically downward, the intersection point I of the normal drawn on the velocity of point A and centre of mass is the instantaneous centre of rotation of the rod just after impact. Hence, kinetic relation between vc′ and ω becomes
m v1 ...(i) 2 where, v′ and v1 are the velocities of the ball and sphere 1 respectively after collision (Remember that collision is head on). As the collision is perfectly elastic, from the definition of coefficient of restitution
co m
o s_ in f
a.
rio si ta
s rio rs ita
P
iv e
From equations (i) and (ii), we get 4v0 , directed towards right. 3 In the centre of mass frame of spheres 1 and 2.
e/
un
v1 =
→
→
P1 = − P2
s: //t .m
tp ht
O
iv er
un
e/
s: //t .m
tp
ht
P
Y
P
an ar
m
.ja
w
w
s: //w
tp
ht
35. A ball of mass m moving with velocity v0 experience a head-on elastic collision with one of the spheres of a stationary rigid dumb-bell as shown in the figure. The mass of each sphere equals m and the distance between them is l. 2 Disregarding the size of the spheres, find the proper angular momentum L of the dumb-bell after the collision, i.e., the angular momentum in the reference frame moving translationally and fixed to the dumb-bell’s centre of inertia. P
...(ii)
...(iv)
6vc cos θ l 1 + 3cos 2 θ
Y
v1 – v′ = v0
or
Solving equations (i) and (iv), we get ω=
v1 − v′ v0 − 0
1=
...(iii)
ωl N ∆t = m vc − cos θ 2
ωl cos θ 2 From equations (ii) and (iii), we get v′c =
mv0 = mv′ +
and
→
→
| P1 | = | P2 | = m | v1 − v2 |
1.56
→
→
→
→
Also, r1c = − r2 c
)
) Solution From conservation of linear
momentum along the direction of incident ball for the system consisting of colliding ball and sphere,
→
→ 1 m / 2 4v0 nˆ therefore, L = 2 2 2 3 where, nˆ is the unit vector in the sense of →
→
r1c × P1 Hence,
P
→
r1c ⊥ P1
As,
→
L = 2[r1c × P1 ]
thus
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→
L=
mv0l 3
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1.57
36. A small body of mass m tied to a nonstretchable thread moves over a smooth horizontal plane. The other end of the thread is being drawn into a hole O shown in the figure with a constant velocity. Find the thread tension as a function of the distance r between the body and the hole if r = r0. The angular velocity of the thread is equal to ω0.
Now, from second law of motion for m
T = F = mω2r Hence, the tension 2
37. One fourth length of a uniform rod of length 2l and mass m is placed on a horizontal table and the rod is held horizontal. The rod is released from rest. Find the normal reaction on the rod as soon as rod is released.
mω04 r04 ω r F = T = m 02 0 r = r3 r
co m
fo rio s_ ta
)
iv er
3 edge τ = mg 4
un
s: //
3 I0 α = mg 4
//t
s:
...(i)
3 I0 = I g + m 4
ht
tp
where,
ht
.m
tp
e/
⇒
=
2
4ml 2 9 2 43ml 2 + ml = 12 16 48 O
t.m
ht
si
) Solution Torque applied by gravity about the
→ As N = mg , the net torque of these two forces about any fixed point must be equal to zero. Tension T, acting on the mass m is a central force, which is always directed towards the centre O. Hence, the moment of force T is also zero about the point O and therefore, the angular momentum of the particle m is conserved about O. Let the angular velocity of the particle be ω. When the separation between hole and particle m is r, then from conservation of momentum about the point O
rio
ve ni
e/ u
s: tp
7
PJ
)
s
rs ita
w .ja w
//w
1
2
in
a. an ar
)
m
) Solution Forces acting on the mass m are shown in the figure.
O
O
Rotational Dynamics
*
O
m(ω0r0) r0 = m(ωr) r
ω=
From equations (i) and (ii), we get
2 0 0 2
ωr r
α=
or
O
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36 g 43l https://t.me/universitarios
...(ii)
1.58
Rotational Dynamics
rod about COM as shown below.
3l 27 g ×α = 4 43
D D[ %\
PJ
Putting this value in equation (iii), we get
Then, ∑Fx = 0 ∴ ax = 0
27 g 16mg N = mg − m = 43 43
ΣFy
ay =
rio
ita
\
Let ax and ay be the linear accelerations of COM and α the angular acceleration of the
s
rio
si ta
iv er
un e/ .m //t
s:
tp
$
39. A sphere of mass m and radius a rolls on other sphere of mass m′ and radius b which rolls on horizontal plane; friction is sufficient to prevent sliding and the motion is all parallel to the some vertical plane. Set up the equation to determine the motion of each sphere.
[ PJ
&
ht
%
D\
&
$
6 3g 4 mg and α = 13 L 13
/D
%
/
7
...(iii)
rs
iv e
un e/ t.m
)
/
τ TL cos30° 3 3T = = m(2L) 2 I 2mL 12
Now just after the string breaks, acceleration of point A in vertical direction should be zero. i.e., ay = Lα cos 30º Solving equations (i), (ii), (iii) and (iv) we get T=
s: // tp ht
) Solution Drawing free body diagram of rod immediately after the string breaks.
...(ii)
s_
an
ar
m
.ja
w
w
//w
s: ht tp
mg − T m
in f
a.
a=
m
=
o
m
and
co
38. A thin uniform bar of mass m and length 2L is held at an angle 30º with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after spring breaks.
...(i)
1
a=
where,
From newton’s second law of motion mg – N = ma ....(iii)
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1.59
Rotational Dynamics
&
Y
mg sinθ – fs = m (acc′(t) + ac′t) 1
IV
mg cosθ – N = m (acc′(n) + ac(n))
IV
2 d θ dv = m (a + b) − sin θ dt dt For rotational motion of C, τcz = Icαz
s_ i
ta rio
ht
Now for sphere C ac = acc′ + ac′ = acc′ (t) + acc′ (n) + ac′
where, acc1 (t ) = (a + b) gent) 2
d 2θ ut ′ (along the tandt 2
dθ acc ′ (n) = (a + b) un (along the normal) dt
....(vi)
un
e/
2 d ω′ f x b + f s ′ b = m′ b 2 ...(vii) 5 dt These seven equation are sufficient to determine the motion of each sphere or in other words the obtained equation are sufficient to calculate all the kinematical terms. We might alternately have written down the equation of energy for the whole system instead of equations of dynamics because neither friction forces nor normal reactions do work on system. Therefore
.m //t
tp
s:
...(ii)
...(v)
For the rotational motion of sphere C′ τc′z = Ic′ αz
.m
dθ dt
ta
e/
iv er si
N′ – m′g – Ncosθ – fs sinθ = 0
//t
s:
tp
ht
bw’ = – aw + (a + b)
dv dt
From the equation Fy = macy
un
s:
ht
Hence,
− f s ′ − Í sinθ + f s cos θ = m′
iv er si
m
.ja
w
//w
w
tp
vpc′ (t) = vpC′(t) + vcc′ (t)
a.
an
ar
and on the other hand vp = vpc′ + vc′ Also the point of contact P of two spheres have no relative tangential velocity i.e., the tangential velocity, vp(t) must be same for both the spheres. Hence, vpc′(t) + vc′(t) = vpc (t) + vcc′ (t) + vc(t) and
o
2 dω f s a = ma 2 ...(iv) 5 dt For sphere C′ corresponding of its linear motion Fx = macx
their velocities at an arbitrary time t. Let v denote the velocity of the sphere m′, and let C′ C makes an angle θ with the vertical at time t. As the lower sphere is pure rolling, v = bω′ ...(i) Also, vp = vpc + vc = vpc + v’cc’ + vc′
nf
)
) Solution Let C and C′ be the centres and ω, ω′
co m
1
s
P J
I V
From equation Fn = macn
PJ
T 1
d 2 θ dv = m (a + b) 2 + cos θ dt dt
rio
T
&
the centre for each sphere, we have for sphere C, for wave-linear motion: Ft = mact
3
Z
1 2 12 2 2 1 12 mvc + ma ω + m′vc2 + m′b 2 ω2 + mg (a + b) cos 2 2 5 2 2 5
dv i along1x – axis 12 1 12 mvc2 + ma 2 ω2 + m′vc2 + m′b 2 ω2 + mg (a + b) cos θ = constant dt 2 2 5 2 2 5 From the equations of dynamics for each But vc = vcc′ + vc′ ...(viii) sphere; resolving and taking moment about
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and ac =
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1.60
where,
Rotational Dynamics
vcc ′ = (a + b)
dθ and vc ′ = v dt
3VHXGRIRUFH PD
D D
and angle between vcc′ and vc′ equals θ 2
dθ ∴ vc 2 = (a + b) 2 + v 2 dt dθ + 2(a + b) v cos θ dt
I IULFWLRQ
...(ix)
in
io
ta r
ta
si
er
vr = at
4 2 = 2 3 = m/s (leftwards) 3 3
In the same time, truck has acquired a forward velocity. vt = a0t
ht
tp s
://
ht
t.m
tp
e/
s:
un
//t
iv
.m
ht tp
s:
e/ un
iv
rio
s
er
si
w .ja
//w
w
m
ar
s_
an
a.
fo
co
m
Drawing free body diagram of cylinder with respect to truck. ma0 – f = ma f a = a0 − ...(i) or Thus from equations (viii) and (ix), we have m 2 1 2 dθ 2 2f 2 τ 2 2 fR 2 dθ 21 2 m v + (a + b) + 2(a + b) v cos θ + a 2 ωAlso, m′ αv= + =b ω + =mg (a + b) cos θ ...(ii) 1 2 dt dt 5 2 5 I mR 2 mR 2 2 dθ dθ 2 2 2 1 2 2 2 2 Since cylinder rolls without slipping on the + 2(a + b) v cos θ + a ω m′ v + b ω + mg (a + b) cos θ truck, we have dt dt 5 2 5 a = Rα ...(iii) 1 2 2 2 2 Solving equations (i), (ii) and (iii), we get m′ v + b ω + mg (a + b) cos θ = costant ...(x) 2 5 2 2 2 a = a0 = (1) = m/s 2 40. A cylindrical pipe of diameter 1m is kept on 3 3 3 the truck as shown in the figure. If the truck With above acceleration, cylinder covers a now starts moving with constant acceldistance s = 4 m on the truck in time t given 2 eration of 2m/s , the pipe rolls backward by without slipping on the floor of the truck 1 12 and finally falls on the road. If the pipe or s = at 2 = t 2 2 2 3 moves a total length of 4m on the floor of the truck, find how much distance the pipe t2 For s = 4m, 4 = moves on the road before it finally stops. 3 The coefficient of friction between the pipe The linear velocity of the cylinder relative 2 and the road is 0.4 (g = 10m/s ) to truck is
= (1) 2 3 m/s (rightwards)
a0 = 1 m/s2 (towards right) Let ‘a’ be the linear acceleration of centre of mass of cylinder (towards left) with respect to the truck and α be its angular acceleration (anticlockwise) about its centre of mass.
or
)
) Solution Acceleration of truck
Hence, at the moment cylinder leaves the truck, it is moving with a linear velocity v = vr – vt 4 2 = −2 3 = − m/s 3 3
v=
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2 m/s (rightwards) 3 https://t.me/universitarios
Rotational Dynamics
1.61
nf o
$
s
[COM of semicircular ring lies at a dis-
2r from centre] π
ta rio
tance of
1 2 Since, P and M are the centres of mass of the two parts A and B. 1 2 2r 2 MO = PO = = = 0.32 m ∴ π π
) Solution Here, r = m = 0.5m.
e/ un
s: //t .m
Let C be the centre of mass of the whole ring. MP = 2(0.32) = 0.64 m
ht tp
ht
tp s: //
τ µ mgR 2µg 2 × 4 α' = = = = = 16 rad/s 2 I 1 mR 2 R 0.5 2 Linear velocity becomes zero in time t1 given by 2 1 v t1 = = 3 = s a′ 4 2 3
)
t.m
ht tp
and angular retardation,
%
iv er si
un i e/
s: //
w
f = µg = (0.40(10)) = 4 m/s 2 m
w
ve r
w
Friction acts backward. Linear retardation,
si ta
.ja m
I IPD[ PPJ
a' =
ar a
Z
P V
rio s_ i
Y
(c) friction force
na .c
om
Distance moved by cylinder during this time v 4 8 and its angular velocity, ω = r = = rad/s (anticlockwise) 2 R 3 × 0.5 3 v 4 8 2 ω= r = = rad/s(anticlockwise) (anticlockwise) rad/s R 3 × 0.5 3 v 2 3 1 = = m d= ′ 2 a 2(4) 6 Once the cylinder leaves the truck, there is no external torque on it about centre of mass 41. A ring shown in figure is made up of two axis of rotation. Hence, its angular velocity semicircular rings A and B of markes and linear horizontal velocity do not change. 2 kg and 4 kg respectively. The ring has Therefore, when it touches the ground, it does the diameter of 1m. The ring rolls withnot perform pure rolling. out slipping. Find 8 (a) angular acceleration ω= rad/s 3 (b) normal reactions and
and angular velocity becomes zero in time t2 given by 0
8 1 ω t2 = = 3 = s α’ 16 2 3
3
D
4
1 Then, MC = (0.64) = 0.213 m 3
Since, t1 = t2 i.e., linear velocity and angular velocity become zero simultaneously, hence, 1 s. the cylinder stops in time, t = 2 3
&2
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1.62
Rotational Dynamics
Now, (mA + mB) ay = ΣFy
∴
QC = (OQ) 2 + (CO) 2
or
(mA + mB) a sin θ = 60 cos 30º – N
∴ Normal reaction, N = 60 cos 30º – (mA + mB)a sin θ
CO = 0.32 – 0.213 = 0.107 m
∴
2
2
= (60) (0.86) – (2 + 4) (3.5) sin 11.9º
= (0.5) + (0.107) = 0.511 m
Force diagram is shown below:
∴ Frictional force, f = 60 sin 30º – (mA + mB) a cos θ
4
:HLJKW 1
co m
= (60) (0.5) – (2 + 4) (3.5) cos 11.9º n
fo
in
rio s_
m
ita
ita
rs ve un i
P
$
)
Let α be the angular acceleration of ) Solution 2 = 6.84 rad/s the rod. Acceleration of B with respect to A is 1.2 α at an angle of 60º with horizontal as shown in figure: Using sine law, we get aBA aB aA = = sin 45° sin 60° sin 75°
ht
tp s:
1 (2 × 4 + 2 × 2) 2 (0.5) + (60) (0.86) (0.107) 2 = 6.84 rad/s 2 1 2 × 4 + 2 × 2) 2 (b) Acceleration of centre of mass, C: a = (QC) (α) in the direction shown in figure i.e., perpendicular to QC a = (0.511) (6.84) m/s2 = 3.5 m/s2
//t .
2
e/
s:
tp
ht
(60) (0.5) (0.5) + (60) (0.86) (0.107)
rio
s
rs ve un i
e/
(60sin 30°) (QO) + (60cos30°) (CO) (2mA + 2mB )r 2
=
(a) the angular acceleration of rod (b) the reactions at A and B (g = 9.8 m/s2) %
//t .m
ht tp
s:
//w
w
F1 = 60 cos 30° (a) The ring will rotate about point Q. τ angular acceleration, α = Q IQ
=
42. The extremities of a 1.2m rod weighting 25 kg can move freely and with no friction along two straight tracks as shown. If the rod is released with no initial velocity from the position shown, determine
m
.ja w
θ = 11.9. W = 60 newton F1 = 60 sin 30°
a. ar an
∴
∴ = 9.45 ewton
QO 0.5 cos θ = = CQ 0.511
D
:
or (mA + mB)a cos θ = 60 sin 30º – f
1
)
T T
)
(c) (mA + mB)ax = ΣFx
2
&
n
= 47.27 ewton
D%
[
D
2
T T
T
LQT
DV
D$
$
4
or aA = 1.64 α and aB = 1.47 α
VT
R DF
&
*
%
\
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1.63
Rotational Dynamics
aG = aA = aGA
% *
1%
Therefore, acceleration of G can now be obtained by
$
Z 1
1$
D$
I
D%
I
%
D%$ D
NB = 7.37 α
...(iv)
or
\
NB sin 45º = (25) (1.34 α)
or
rio s_ in fo
o D*$ D
\
Now ΣFx = max
[
na .c om
o D$
$
D\
%
D[
D
aG into x and y components, we
Resolving have
[
*
iv uestions
un type
t.m e/
L = mv h
Putting, h=
(d) m 2 gh3
(∵ r sin θ = h)
⇒
where, θ = angle between v and r
s: //
Angular momentum the particle about O L = mvr sin θ
v02 sin 2 θ v02 sin 2 45° v02 = = , we get 2g 2g 4g L=
Putting,
v = 4 gh , we get L = m 2 gh3
Speed of the particle at the top = horizontal component of the speed of projection. v v = v0 cos θ0 = v0 cos 45° = 0 ⇒ 2
mv02 4 2
)
) Explanation
Q
mv 4 2g
ht tp
(c)
er si ta rio s
er s
un iv
e/
.m
ht tp s:
//t
...(iii)
1. Particle of mass m is projected with a velocity v0 making an angle of 45º with horizontal. The magnitude of angular momentum of the projectile about the point of projection at its maximum height is mv 3 (a) Zero (b) 2g 2 0
ita
.ja
w
w
//w
s:
tp
ht
= 0.173 NA – 0.19 NB
SOLVED OBJECTIVE
ar a
m
Also ΣFy = may ax = 1.64 α – 0.6 α cosº = 1.34 α ...(i) or 245 – NA – NB cos 45º = (25) (0.52 α) N and ay = 0.6 α sin 60º = 0.52 α N A...(ii) (0.6cos30°) − B (0.6cos30° + 0.6sin 30°) or ...(v) τ 2NA + 0.707 NB = 245 – 13 α Free body diagram of rod is α = G = 2 (25)equations (1.2) IG Solving (iii), (iv) and (v), we get 12 N α = 2.33 rad/s2 N A (0.6cos30°) − B (0.6cos30° + 0.6sin 30°) τ 2 and NA = 137 N, NB = 110 N α= G = (25) (1.2) 2 IG 12
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1.64
2. A homogenous rod AB of length L and mass M is pivoted at the centre O in such a way that it can rotate freely in a vertical plane. The rod is initially in a horizontal position. An insect S of the same mass falls vertically with speed V at the point C, midway between the points O and B and stick to the rod. Immediately after the insect fall, the angular velocity of the system is
7 D
5
PJ
) Explanation Let a be the acceleration of the )
6
cylinder. Then mg – T = ma If α be the angular acceleration, then
...(i)
mR 2 T. R. = I α = α 2
...(ii)
m
ita
%
/
mg (b) g and 2 g mg (d) and 2 3
3 a up the plane from 2 a from the centre 2 3
(d) at a distance down the plane 1 '
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P1
& 2
%
$
PJVLQD
2g mg and (a) 3 3 g mg (c) and 3 2
3. A string is wrapped several times round a solid cylinder and then the end of the string is held stationary while the cylinder is released from rest with an initial motion. The acceleration of the cylinder and tension in the string will be
3 a up the plane from 2
ht tp
&
(b) at a distance the centre (c) at a distance the centre
s: //
t.m e/
12 V 7 L
2
$
4. A cube in limiting equilibrium on an inclined plane forming an angle of 30º with the horizontal. The line of action of the normal reaction of the plane on the cube is (a) at the centre of the cube
un
//t
ht tp s:
ω=
2 mg g and T = 3 3
iv
.m
ht
∴
e/
s:
tp
)
ML ML MV = + ω 4 3
⇒
un iv
L2
a=
er si ta rio s
er s
w w //w
ML2
L
+ M ω ) Explanation (MV) = 4 12 16
As the string unwinds without slipping a=Rα ...(iii) Solving these equations, we get
3V 4L
(d)
ar a
3 VL 2
(b)
.ja
12V 7L
(c)
1 VL 2
(a)
%
2
$
rio s_ in fo
&
na .c om
9
Rotational Dynamics
D PJ
PJFRVD
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Rotational Dynamics
)
) Explanation At limiting equilibrium for linear
1.65
equilibrium mg sin 30º = µN and mg cos 30º = N 1 or µ = tan 30º = 3
7 D
*
6. A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is
5. A uniform rod of length l and mass M is suspended on two vertical inextensible string as shown in the figure. Calculate tension T in left string at the instant, when right string snaps.
rio ta si
/
er iv
//t
un
)
s:
e/
//t
.m
ht tp
FL = mg
F=
ht
s:
tp
⇒
…(i)
s
e/ .m
tp ht
si ta er un
iv
)
L 2
mg 2
)
2
mg m a =|α = × 2 3 2
a=
∴
mg T= 4
PJ
/ 2
/
(ii)
7. A horizontal disc rotes freely about a vertical axis through its centre. A ring, having the same mass and radius as the
3 g 4
⇒
Again,
(d) mg (1 – µ)
zero.
and the acceleration of the centre of mass of the rod downwards be a.
mg 4
) Explanation Net torque about the point O is
) Explanation Let the tension in the rope be T Then, mg – T = ma
(b)
rio
m
.ja
w
//w s:
mg 8
)
mg 2
(c)
w
(d)
s_
in
a.
an
ar
(a) infinitesimal
(b) mg
mg (a) 2 mg (c) 4
fo
co m
2 3
a
x=
or
D
PJ
Considering rotational equilibrium about the c.m., we have a µN × = N × x 2
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1.66
∴
Ratio of heat produced to initial kinetic E − E2 energy = 1 E1
disc, is gently placed on the disc. After some time, the two rotate with a common angular velocity. (a) Some friction exists between the disc and the ring. (b) The angular momentum of the ‘disc plus ring’ is conserved. 2 rd of (c) The final angular velocity of 3 the initial angular velocity of the disc. 2 rd of the initial kinetic energy (d) 3 changes to heat.
Rotational Dynamics
=
8. A long horizontal rod has bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ and gravity is neglected, then the time after which the bead starts slipping is
ta
s
si ta
(d) infinitesimal
ve r
)
f = µN = µmαL
e/ un i
⇒ or
mv 2 = f L v2 = µαL2 ω2 = µα
α2t2 = µα
∴
t =
or
ω = 1 3
µ α /
$
I P1
1 P
9. A rod of length l falls on two metal pads of same height from a height h. The coefficients of restitution of the metal pads are e1 and e2 (e1 > e2). The angular velocity of the rod after it recoils is
1 Initial kinetic energy, E1 = I1ω12 2 1 Final kinetic energy, E 2 = (I1 + I 2 )ω2 2 2 Heat produced = less in kinetic energy = E1 – E2
t.m tp ht
Iω ω2 = 1 1 I1 + I 2
or
s: //
ht
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Rotational Dynamics
(d)
e1 + 1 2 gh e2 − 1
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10. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform ω (t) will vary with time t as
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1. A semicircular lamina of mass m has radius r and centre C. Its centre of mass is at a distance x from C. Its moment of inertia about on axis through its centre of mass and perpendicular to its plane is 1 1 (a) mr 2 (b) mr 2 2 4 1 1 (c) mr 2 + mx 2 (d) mr 2 − mx 2 2 2
exercise
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2. A uniform rod of length 2 ℓ is placed with one end in contact with the horizontal table and is then inclined at an angle α to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be
unsolved
OBJECTIVE TYPE QUESTIONS
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mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is 1 1 (b) MR 2 (a) MR 2 4 2
3 g sin α (a) ω = 2
(d)
2MR 2
3. A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear velocity v0 and angular velocity ω0 as shown. The disc comes to rest after moving some distance to the right. It follows that (a) 3 v0 = 2ω0 r (b) 2 v0 = ω0 r (c) v0 = ω0 r (d) 2 v0 = 3 ω0 r
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(a)
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6. Two particles A and B are moving as shown in the figure. At this moment of time the angular speed of A with respect to B is
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g sin α (c) ω =
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Rotational Dynamics
5. One quarter sector is cut from a uniform circular disc of radius R. This sector has
9
7. A sphere kept on a rough inclined plane is in equilibrium by a string wrapped over it. If the angle of inclination is θ, the tension in the string will be equal to 2mg (a) mg sin θ (b) 5 mg sin θ (d) none of these (c) 2
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Rotational Dynamics ;
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(a)
13. A stick of length ℓ and mass m lies on a frictionless horizontal surface on which it is free to move in any direction. A ball of same mass m moving with speed v0 perpendicularly to this length and collide at the end of stick. Find the angular velocity when the collision is perfectly inelastic.
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(b)
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(c)
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10. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX’ is ρL3 8π 2
(d) 3
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12. A string is wrapped around a cylinder of radius R. If the cylinder is realased from rest, the velocity of the cylinder, after it has moved h distance will be
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9. As shown in figure, a small block of mass m is at rest with respect to a smooth hemispherical bowl of radius R rotating with angular velocity ω about its axis. Find the radius of the circle (r) over which block is moving as seen from ground.
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11. A particle P is placed inside a hemispherical bowl which rotates about its vertical axis with a constant velocity ω. It is just prevented from sliding down when OP is inclined at 45º with the axis, O being the center of the bowl. Radius of bowl = 10 2 and µ, the coefficient of friction between the particle and bowl is 0.5, g = 10 m/s2. The value of ω is 1 1 (b) (a) 3 3
8. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending.
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18. A solid sphere is rolling on a rough surface, whose centre of mass is at C at a certain instant. Find at that instant it has angular velocity ω. Its radius is R. Find the angular acceleration at that instant mass of sphere is m. (b)
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2 5 d (g + ω R) (c) 2 R2
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15. A spherical body of mass m, radius r and moment of inertia I about its centre moves along the x-axis. Its centre of mass moves with velocity = v0, and it rotates about its centre of mass with angular velocity = ω0.
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17. The surface density of a thin circular disc of radius ‘a’ depends on the distance from the centre as f(r) = A + Br, where A and B are constants. Its moment of inertia about the line perpendicular to the plane of the disc through its center is
14. A hollow sphere of mass 2 kg is kept on a rough horizontal surface. A force of 10 3 N is applied as shown in the figure. Find the minimum value of ‘µ’ so that the sphere starts pure rolling. (Take g = 10m/s2) (a) 0.3 (b) 0.2 (c) 0.1 (d) for no value of µ it will start pure rolling
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16. A sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotav tional velocity about the centre 0 . The 2r rotational velocity when the sphere starts pure rolling is v v (b) 0 (a) 0 3 3 v0 2v0 (c) (d) 5 7
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Rotational Dynamics
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(d) none of these
ω 2
ω 3
22. A uniform circular disc of radius r placed on a rough horizontal surface has initially a velocity v0 and angular velocity ω0 as shown in the figure. The disc comes to rest (neither translates nor rotates) after v moving some distance. Then 0 is rω0 1 (a) 1 (b) 2 3 (d) 2 (c) 2
(c)
(b)
(a) ω
1.71
19. A disc is freely rotating with an angular speed ω on a smooth horizontal plane. If it is hooked at a rigid pag P and rotates without bouncing. Its angular speed after the impact will be equal to
Rotational Dynamics
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21. An this figure a body is moving with constant velocity v = 3m/s along x-axis and also rotating about is centre with angular velocity ω = 4 rod/sec. Find the co-ordinate of the axis of pure rotation when centre of the body is 4 m from the origin.
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23. A cylinder of radius 10 cm rides between two horizontal bars moving in opposite direction as shown in the figure. The location of the instantaneous axis of rotation and the angular velocity of the roller are respectively (There is no slipping at P or Q) (a) 8 cm from Q, 1.25 rad/s2 (b) 8 cm from Q, 2.50 rad/s2 (c) 15 cm from Q, 1.25 rad/s2 (d) 15 cm from Q, 2.50 rad/s2
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24. A solid sphere of radius R is rolled by a force F acting at the top of the sphere as shown in the figure. There is no slipping and initially sphere is in the rest position, then
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20. A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear velocity v0 and angular velocity ω0 as shown. The disc comes to rest after moving some distance to the right. It follows that (a) 3 v0 = 2ω0 r (b) 2v0 = ω0 r (c) v0 = ω0 r (d) 2v0 = 3ω0 r
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3
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(a) H1 = H (b) H1 > H (c) H1 < H (d) can not be decided
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28. Three mass particle A, B and C having masses m, 2m and 3m respectively are rigidly attached to a ring of mass m and radius R which rolls on a horizontal surface without slipping. At a certain instant the velocity of the centre of the ring is v0 as shown in the figure. The kinetic energy of the system is (a) 7mv02 (b) 9mv02 7 (c) mv02 (d) mv0 2 2
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25. A disc of mass M and radius R moves in the x–y plane as shown in the figure. The angular momentum of the disc at the instant shown is
27. A rod of mass M is dropped on a horizontal smooth surface from the position shown in the figure. It collides elastically the surface. After collision with the surface, the center of mass of rod rises to a height H1. Then
(a) work done by force F when the centre of mass moves a distance S is 2FS. (b) speed of the c.m. when c.m. moves a 20 RS distance S is 7 M (c) Work done by the force F when c.m. moves a distance S is FS (d) speed of the c.m. when c.m. moves a 4RS distance S is M
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Rotational Dynamics
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26. A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. A horizontal impulse P is applied to the end B perpendicular to the rod. The time taken by the rod to turn through a right angle is
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29. A triangular block ABC of mass m and sides 2a lies on a smooth horizontal plane as shown. Three point masses of mass m each strikes the block at A, B and C with
2
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Rotational Dynamics
speeds v as shown. After the collision the particle come to rest. Then
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30. A uniform rod is released from its vertical position with its bottom on a smooth surface. If v is the speed of centre of mass of the rod and ω is its angular velocity when it makes an angle θ with the horizontal, then
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2. 10. 18. 26.
(a) (d) (d) (a)
3. 11. 19. 27.
(b) (b) (b) (c)
4. 12. 20. 28.
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ANSWERS
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(a) the centre of mass of ABC remains stationary after collision (b) the centre of mass of ABC moves with a velocity v along x-axis after collision (c) the triangular block rotates with an 2 3mva angular velocity ω = about its I centre of mass (here I is the moment of inertia of triangular block about its centroid axis perpendicular to its plane)
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5. 13. 21. 29.
(a) (b) (c) (b, c)
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6. 14. 22. 30.
(d) (b) (d) (c)
7. (c) 15. (d) 23. (a)
8. (b) 16. (b) 24. (a, b)
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Rotational Dynamics
EXERCISE 2 cylinder is less than 1/3 tan θ, where, θ is the angle of inclination of the plane with the horizontal.
na .c om
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7. A weightless thread with a body of mass m tied to its end is wrapped around a uniform solid cylinder of mass M and radius R shown in the figure At t = 0, the system is set in motion. Find: (a) angular velocity of the cylinder, and (b) kinetic energy of the whole system, as a function of time. There is no friction anywhere.
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4. Three forces are applied at the three corners of a square plate, as shown in figure. Find the resultant force (both in magnitude and direction) and its point of application, if this point is taken on the side BC. )
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3. A uniform rod AB of mass m = 1.0 kg moves linearly with an acceleration a = 2.0 ms2 due to two anti-parallel forces F1 and F2 = 5.0 N. The distance between the points of application of these two forces is equal to l = 0.2 m. Find the length L of the rod. [Ans. 1.0 m]
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2. A solid sphere starting from rest rolls for 5.3 second without slipping along a smooth inclined plane 1 m in length, the angle of inclination being θ = 0.573º. What is the acceleration due to gravity at that place? [Ans. 9.8 m/s2]
6. A disc of radius R/2 is cut from a uniform disc of radius R = 0.2 m, as shown in the figure. The mass of the remaining shaded portion of the disc is m = 7.3 kg. Find the moment of inertia of the shaded portion of the disc about an axis passing through its centre of mass and perpendicular to its plane. [Ans. 0.15 kg-m2]
1. Four particles each of mass m are placed at the four corners of a square of edge a. Find the moment of inertia of the system about an axis perpendicular to the plane of the square and passing through its centre. [Ans. 2ma2]
unsolved
1.74
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5. Show that a cylinder will slip on an inclined plane if the coefficient of static friction between the plane and the
R
5 7 7
[Ans. Resultant will be 2F acting at the middle point of AB and parallel to the diagonal AC]
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Rotational Dynamics
cal sections of the threads, assuming that the thread does not slip over the pulley and the friction is absent.
7
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3 v2 , what is the M.I. of the body? 4 g What shape(s) it may have?
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12. A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16 kg and length is 3 metre. Two particles, each of mass 0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity of 10 m/s and the other with 6 m/s, as shown in the figure. The first particle strikes the bar at the point A and the other at point B. Points A and B are at a distance of 0.5 m from the centre of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. Calculate the loss of kinetic energy of the system in the above collision process. [IIT, 1989]
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11. A square plate of mass M = 120 g and of side a = 5.0 cm is rotated about one of its sides as axis. Find the torque acting on the plate, if its angular acceleration is α = 0.2 rad/s2. [Ans. 2.0 × 10–5 N–m]
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m1 ( 4m2 + M ) Ans. m2 ( 4m1 + M )
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9. A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has an elevated section and a horizontal part. The horizontal part is 1.0 m above the ground level and the top of the track is 2.4 m above the ground. Find the distance on the ground with respect to the point B (which is vertically below the end of the track as shown in the figure) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centre of mass? [IIT, 1987]
7
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If h =
MR 2 Ans. I = 2
5
8. A body of mass M and radius R is rolling on a smooth horizontal plane without slipping with a velocity v. In the course of its rolling, it meets a rising slope and rolls up to a height h before stopping.
gt 1 mg 2t 2 Ans. (a) , (b) M 2 M + 1 R 1 + 2m 2m
[Ans. 1.99 m] 10. As shown in the figure, the mass of the pulley of radius R is M, and the masses of the two blocks are m1 and m2. Find the angular acceleration of the pulley and T the ratio of the tensions, 1 , in the vertiT2
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[Ans. v = 4 m/s ; ∆E = 2.76 joules]
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1.76
13. A uniform cylinder of radius R, rotating about its axis with an angular velocity ω, is placed in a corner, as shown in the figure. The friction coefficient between the corner walls and the cylinder is µ. Find the number of turns the cylinder will complete before it stops.
without slipping on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of cylindrical part of the carpet when its radius reduces to R/2. [IIT, 1990]
Ans.
17. A uniform solid cylinder of mass m = 8.0 kg and of radius r = 1.3 cm as shown in the figure starts falling at time t = 0 due to gravity. Find: (a) tensions in the threads and the angular acceleration of the falling cylinder, and (b) instantaneous power developed by the force of gravity as a function of time. Take mass of the thread negligible.
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Rotational Dynamics
in
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18. A uniform disc of mass M and radius R is initially at rest with its axis fixed through the point O as shown in the figure. A block of mass m initially moving with speed v1 on a frictionless surface passes over the disc to the dotted position. When it first makes contact with the disc, it slips on it. But the friction between the block and the disc is large enough so that slipping ceases before the block loses contact with the disc. Show that the final velocity of the block is given by v1 v2 = M 1+ 2m
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19. A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v, respectively, strike the bar as shown in
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16. A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts rolling
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4 Ans. 5 ω
7
2 2 Ans. (a )13 N, (b) 3 mg t
15. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Another disc M of same dimension but of mass is 4 placed gently on the first disc. Find the angular velocity of the combined system. [IIT, 1986]
3 ω0 R Ans. 4 . µg
7
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14. A uniform disc of radius R rotating with an angular velocity ω0 is carefully placed on a horizontal surface. Find how long will the disc rotate on the surface. The friction coefficient between the disc and the horizontal surface is µ .
Rω02 1 + µ 2 Ans. 8πµg 1 + µ
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1.77
Rotational Dynamics
force constant k of the spring is 3.0 N/m. If the system is released from rest at a position in which the spring is stretched 0.25m, find the translational and rotational kinetic energy of the cylinder as it passes through the equilibrium position.
the figure and stick to the bar after collision. Denoting the angular velocity (about the centre of mass), total energy and centre of mass velocity by ω, E and vc respectively, find the total energy of the system after the collision. P D
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[IIT, 1991]
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22. A uniform solid cylinder of radius R rolls on a horizontal surface that passes into an inclined plane of inclination θ shown in the figure. Find the maximum value of v0 which still allows it to roll on the inclined surface without a jump. Assume that the cylinder rolls without sliding.
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20. A uniform solid cylinder of mass M and radius R rests on two horizontal supports. A thread of negligible mass is wrapped on the cylinder and its hanging end is pulled vertically down with a constant force F as shown in the figure. What is the maximum value of the force that will not bring about any sliding of the cylinder? The friction coefficient between the cylinder and the horizontal support is µ .
3 2 Ans. 5 mv
Ans.
3µmg Ans. 2 − 3µ
21. A solid cylinder is attached to a horizontal massless spring so that it can roll without slipping along a horizontal surface as shown in the figure. The
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23. A homogeneous rod AB of length l = 1.8 m and mass m is pivoted at the centre O in such a way that it can rotate freely in a vertical plane as shown in the figure. The rod is initially in the horizontal position. An insect of the same mass M falls vertically with speed v on the point C, midway between the point O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity ω.
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24. Show that if a spherical body of radius r rolls without slipping along a horizontal plane with velocity vc and reaches the plane inclined at an angle θ with the horizontal, velocity v′ with which it starts up the inclined plane is
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28. A small disc of mass m and a thin uniform rod of length l and of mass mn (n is a number) lie on a smooth horizontal plane. The disc is set into motion with velocity v in the horizontal direction and perpendicular to the length of the rod and it collides elastically with the end of the rod. Find the angular velocity of the rod after the collision.
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2M mMR dv Ans. (a ) m + 2M θ, (b) (m + 2M) . dt
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27. A man of mass M stands on the edge of a horizontal uniform disc of mass m and of radius R, which can rotate freely about a vertical axis passing through its centre. At time t, he starts moving along the edge of the disc, and after moving through an angle θ relative to the disc, he stops. In the process of motion, his velocity with time varies as v(t). Assuming m << M, find: (a) angle through which the disc had turned by the time the man stopped, and (b) moment of the force about the rotation axis with which the man acted on the disc in the process of motion.
(a) Determine the angular velocity ω in terms of v and l. (b) If the insect reaches the end B, when the rod has rotated through an angle of 90º, determine v. [IIT, 1992]
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(a) the force of friction on each disc. (a) the angle θc through which the cylinder rotates before it leaves contact with the (b) the magnitude and the direction of the edge. frictional torque acting on each disc about the centre of mass O of the object. (b) the speed of the centre of mass of the Express the torque in the vector form in cylinder before leaving contact with the edge, and terms of unit vectors iˆ, ˆj and kˆ in the (c) the ratio of the translational to rotax, y and z directions. [IIT, 1997] tional kinetic energies of the cylinder when its centre of mass is in horizontal Ans. (a ) 6Niˆ, (b) 0.6 2 N − m line with the edge. [IIT, 1995] 31. A wedge of mass m and triangular cross6 mgr = BC = CA = 2R) is moving E 4 4 section (AB Ans. (a) θc = cos −1 , (b) v = gR , (c) T = 7 = 6 velocity −viˆ towards 7 E R 1with mgra constant 7 radius R fixed on a smooth a7 sphere of horizontal table as shown in the figure. 6 mgr ET 4 1 4 7 The wedge makes an elastic collision gR , (c) = = 6 , (b) v = 1 7 7 E with the fixed sphere and returns along mgr R 7 the same path without any rotation. Neglect all friction and suppose that the 30. Two thin circular discs of mass 2 kg and wedge remains in contact with the sphere radius 10 cm each are joined by a massfor a very short time ∆t, during which less rod of length 20 cm. The axis of the the sphere exerts a constant force F on rod is along the perpendicular to the the wedge. planes of the discs through their centres as shown in the figure. This object is kept (a) Find the force F and also the normal on a truck in such a way that the axis of force N exerted by the table on the the object is horizontal and perpendicular wedge during the time ∆t. to the direction of the motion of the truck. (b) Let h denote the perpendicular distance Its friction with the floor of the truck is between the centre of mass of the large enough so that the object can roll on wedge and the line of action of F . Find the truck without slipping. Take x-axis as the magnitude of the torque due to the the direction of motion of the truck and normal force N about the centre of the z-axis as the vertically upward direction. wedge, during the internal ∆t. If the truck has an acceleration of 9 m/s2, [IIT, 1998] calculate
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9. Why there are two propellers in a helicopter? 10. You are given two circular discs of equal masses and thickness but made of different materials. Which one will have a larger moment of inertia about its central axis and why?
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12. The cap of the pen can easily be opened with the help of two fingers than with one finger. Why?
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11. How can you distinguish between a hard boiled egg and a raw egg by spinning them on a table top?
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7. How a swimmer jumping from a height, is able to increase the number of loops made in the air?
3. In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1 m respectively and the acceleration of block A is 5 m/s2 in the downward direction, then the acceleration of block B will be
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5. A thin wheel can stay upright on its rim for a considerable length of time when rolled with a considerable velocity, while it falls from its upright position at the slightest disturbance, when stationary. Explain.
4. Why it is more difficult to revolve a stone by tieing it to a longer string than by tieing it to a shorter string?
1. The moment of inertia of a door of mass m, length 2ℓ and width ℓ about its longer side is 11m 2 5m 2 (a) (b) 24 24 2 m (c) (d) None of these 3
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13. Why the speed of whirlwind in a tornado is alarmingly high? ONLY ONE OPTION IS CORRECT
2. Why spokes are fitted in the cycle wheel?
1. A body is rotating. Is it necessarily being acted upon by an external torque?
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5. Consider the following two statements (A) Linear momentum of the system remains constant. (B) Centre of mass of the system remains at rest. (a) A implies B and B implies A (b) A does not imply B and B does not imply A (c) A implies B but B does not imply A (d) B implies A but A does not imply B
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One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is (a) 3 v cos θ (b) 2 v cos θ (c) 3/2 v cos θ (d) ( 3/2 v cos θ)
The coordinates of point of application of forces is given by
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16. In the above question, the distance the sphere will have moved relative to the ground at time t1 is
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12. A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc
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22. In above question find the condition when bobbin moves to right (a) R sin α = r (b) R sin α > r (c) R sin α < r (d) None of these
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21. The free end of a thread wound on a bobbin is passed round a nail A hammered into the wall. The thread is pulled at a constant velocity. Assuming pure rolling of bobbin, find the velocity v0 of the centre of the bobbin at the instant when the thread forms an angle α with the vertical. (a)
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26. The law of conservation of angular momentum states that (a) the angular momentum of a body is always constant (b) the momentum of the body is conserved (c) the angular momentum of the body is conserved as long as no external torque acts on it. (d) when angular momentum is constant, rotational K.E. is constant. 27. Angular momentum is the vector product of (a) linear momentum and radius vector. (b) moment of inertia and angular acceleration. (c) linear momentum and angular velocity (d) linear velocity and radius vector
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Find minimum friction coefficient between floor and ladder for this equilibrium (a) 2/3 (b) 1/2 (c) 3/4 (d) 1/3
24. Two points of a rod move with velocities 3 v and v perpendicular to the rod and in the same direction, separated by a distance ‘r’. Then the angular velocity of the rod is 3v 2v (b) (a) r r 5v 4v (d) (c) r r 25. A uniform disc of radius R lies in the x-y plane, with its centre at origin. Its moment of inertia about z-axis is equal to its moment inertia about the line y = x + c. The value of c will be R (a) –R/2 (b) ± 2 (c) +R/4 (d) –R
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35. A dancer on ice spins faster when she folds her hand. This is due to (a) Increase in energy and increase in angular momentum (b) decrease in friction at the skates (c) constant angular momentum and increase in kinetic energy (d) increase in energy and decrease in angular momentum
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33. A uniform triangular plate ABC of moment of inertia I (about an axis passing through A and perpendicular to plane of the plate) can rotate freely in the vertical plane about point ‘A’ as shown in figure. The plate is released from the position shown in the figure. Line AB is horizontal. The acceleration of centre of mass just after the release of plate is
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42. A rotating body has no angular impulse imparted to it, then (a) its moment of inertia remains same (b) its angular momentum remains same (c) its angular speed remains same (d) neither moment of inertia nor angular momentum remain same
38. Two point masses M1 and M2 are at a distance r part. The moment of inertia of the system about an axis passing through the centre of mass and perpendicular to the line joining the two masses is (a) [M1 M2/(M1 + M2)]r2 (b) [(M1 + M2)/(M1 M2)]r2 (c) [(M1 – M2)/(M1 + M2)]r2 (d) [(M1 + M2)/(M1 – M2)]r2
37. Two discs of same mass m and same thickness t are made from two different materials of densities d1 and d2 respectively. The ratio of the moment of inertia of the two about an axis passing through the centre and perpendicular to the plane of disc is (a) d1 : d2 (b) d2 : d1 (c) (d1/d2) : 1 (d) 1 : d1 d2
41. For the same total mass which of the following will have the largest moment of inertia about an axis passing through the C.G. and perpendicular to the plane of the body (a) a disc of radius a (b) a ring of radius a (c) a square lamina of side 2a (d) four rods forming a square of side 2a
36. A uniform rod of mass 2 kg is 1 metre long. Its radius of gyration about an axis through its one end and perpendicular to its length in SI units is (b) 1/2 3 (a) 1/ 3 (c) 2/3 (d) 2/ 3
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52. The moment of inertia of a body about a given axis of rotation depends upon (a) the distribution of mass (b) distance of the body from the axis of rotation (c) shape of the body (d) all of the above
49. When an explosive shell travelling in a parabolic path under the gravity explodes, the centre of mass of the fragments will continue to move (a) along a hyperbolic path (b) vertically upwards and then vertically downwards (c) along the original parabolic path (d) horizontally and then follows the parabolic path
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46. When a sphere of moment of inertia I about its centre of gravity and mass m rolls from rest down an inclined plane without slipping, its kinetic energy is 1 1 (b) mv 2 (a) Iω2 2 2 1 2 1 2 (c) I ω + mv (d) I ω + mv 2 2
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65. A string is wrapped several times round a solid cylinder and then the end of the string is held stationary while the cylinder is released from rest with no initial motion.
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57. The ratio of kinetic energies of two spheres rolling with equal center of mass velocities is 2 : 1. If their radii are in the ratio 2 : 1, then the ratio of their masses will be (a) 2 : 1 (b) 1 : 8 (c) 1 : 7 (d) 2 2 :1
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56. The moment of inertia of a circular disc of mass 200 gm and radius 5 cm about a tangential axis normal to the plane of disc will be (a) 750 g cm2 (b) 7500 g cm2 2 (c) 75 g cm (d) zero
(b)
60. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The velocity of its center of mass at the bottom of the plane will be:
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55. A long thread is wrapped round a reel. If one end of thread is held in hand and the reel is allowed to fall under gravity, then the acceleration of the reel will be (a) g (b) (2/3)g (c) (3/2) g (d) zero
(c)
54. The moment of inertia of a hollow sphere of mass 1 kg and inner and outer diameters 0.2 and 0.4 meter respectively about its diametric axis will be (a) zero (b) 0.177 kg m2 (c) 0.0177 kg m2 (d) 177 kg m2
(a)
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Rotational Dynamics
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Rotational Dynamics
71. A small block of mass ‘m’ is rigidly attached at ‘P’ to a ring of mass ‘3 m’ and radius ‘r’. The system is released from rest at θ = 90º and rolls without sliding.
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67. A uniform rod of length 2 ℓ is placed with one end in contact with the horizontal table and is then inclined at an angle α to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be 3 g sin α (a) ω = 2
70. A metre stick is held vertically with one end on the floor and is then allowed to fall. The speed of the other end when it hits the floor assuming that the end at floor does not slip, is (g = 9.8 m/sec2) (a) 3.2 m/sec (b) 5.4 m/sec (c) 7.6 m/sec (d) 9.2 m/sec
66. A string is wrapped around a cylinder of mass m and radius R. The string is pulled vertically upward to prevent the centre of mass from falling as the cylinder unwinds the string. The length of the string unwound when the cylinder has reached a speed ω will be R 2 ω2 Rω (b) (a) 4g 4g R 2 ω2 Rω (d) (c) 8g 8g
The acceleration of the cylinder and tension in the string will be 2 mg mg (a) g and (b) g and 3 3 2 g mg g mg and and (d) (c) 2 3 3 2
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68. A mass M is supported by a massless string wound round a uniform cylinder of mass M and radius R. On releasing the system from rest, the acceleration of mass M is (a) g (b) g/2 (c) (2/3) g (d) depends on R
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73. A disc of mass M and radius R can rotate freely about a horizontal shaft at S distant r
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69. Three thin uniform rods each of mass M and length L are placed along the three axis of a cartesian coordinate system with one end of each rod at the origin. The M.I. of the system about z-axis is (a) ML2/3 (b) (2/3)ML2 2 (c) ML /6 (d) ML2
The value of ω for which the force exerted by the hinge on rod is O in this position is
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from the centre of mass of the disc. The disc is released from the position shown in the figure, its angular acceleration will be
76. A force of (2iˆ − 4 ˆj + 2kˆ) Newton acts at a point (3iˆ + 2 ˆj − 4kˆ) metre from the origin.
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77. A thin wire of length ℓ and mass m is bent in the form of a semicircle.
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Rotational Dynamics
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Its moment of inertia about an axis joining its free ends will be (a) zero (b) mℓ2 m m 2 (c) 2 (d) π 2π 2
78. Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be
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74. A block of mass 2 kg hangs from the rim of a wheel of radius 0.5 m.
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On releasing from rest the block falls through 5 m height in 2 s. The moment of inertia of the wheel will be (a) 1 kg m2 (b) 3.2 kg m2 2 (c) 2.5 kg m (d) 1.5 kg m2
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75. If the rotational kinetic energy of a body is increased by 300% then the percentage increase in its angular momentum will be (a) 600% (b) 150% (c) 100% (d) 1500%
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79. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will take same time.
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4. A rod AC of length ℓ and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes rod at point B making angle 37º with the rod. The collision is elastic. After collision
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3. A ring rolls without slipping on the ground. Its centre C moves with a constant speed u. P is any point on the ring. The speed of P with respect to the ground is v (a) 0 ≤ v ≤ 2u (b) v = u, if CP is horizontal (c) v = u, if CP makes an angle of 60º with the horizontal and P is below the horizontal level of C (d) v = 2u , if CP is horizontal
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80. In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will achieve same kinetic energy. ONE OR MORE THAN ONE OPTION MAY BE CORRECT
2. In the figure, the disc D does not slip on the surface S. The pulley P has mass, and the string does not slip on it. The string is wound around the disc.
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Rotational Dynamics
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Rotational Dynamics
of the same mass m. The wedge is free to move on a smooth horizontal surface. The ball rolls without sliding on the wedge. The required height of the wedge are mentioned in the figure. Which of the following statement about the above system is/are true
5. A disc of circumference s is at rest at a point A on a horizontal surface when a constant horizontal force begins to act on its centre.
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9. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with
7. A small ball (uniform solid sphere) of mass m is released from the top of a wedge
y = b, z = 0. Let dA be the area swept out by the position vector from origin to the particle in time dt and L the magnitude of angular momentum of particle about origin at any time t. Then (a) L = constant (b) L ≠ constant dA 2 L dA L = (c) (d) = dt m dt 2m
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8. P is the centre of mass of four point masses A, B, C and D which are coplanar but not collinear. (a) P may or may not coincide with one of the point masses. (b) P must lie within the quadrilateral ABCD (c) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C and D three at a time. (d) P must lie on a line joining two of the point A, B, C and D
6. A particle of mass m is travelling with a constant velocity v = v0iˆ along the line
of the wedge is gh (b) The total kinetic energy of the ball just before it falls on the ground is (31/18) mgh (c) The horizontal separation between the ball and the edge ‘PQ’ of wedge just before the ball falls on the ground is
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Between A and B there is sufficient friction to prevent slipping, and the surface is smooth to the right of B. AB = s. The disc moves from A to B in time T. To the right of B (a) the angular acceleration of the disc will disappear, linear acceleration will remain uncharged (b) linear acceleration of the disc will increase (c) the disc will make one rotation in time T/2 (d) the disc will cover a distance greater than s in further time T
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12. A solid sphere of mass M and radius R is pulled horizontally on a sufficiently rough surface as shown in the figure.
10. A solid cone and a solid sphere is arranged as shown in the figure. The centre of mass is
Choose the correct alternative. (a) The acceleration of the centre of mass is F/M. (b) The acceleration of the centre of mass is 2 F . 3M (c) The friction force on the sphere acts forward. (d) The magnitude of the friction force is F/3.
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(a) at 3R if m1 = m2 = m and R1 = R2 (b) at 2R from the centre of mass of solid cone if ρ1 = ρ2 and R1 = R2 (c) If ρ1 = 2ρ2 then distance from the centre of solid cone is 11R/3 (d) (a) and (b) 11. The diagram shows a uniform solid cylinder of mass M and length 2ℓ.
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Note that B is the common mid-point of the semicircular rings, while O is the centre of the rings. If m is the total mass of the system and R is the radius of the rings. Which of the following statement is true. (a) Moment of inertia of the system about the axis OB is given by mR2. (b) Moment of inertia of the system about the axis OB is given by mR2/2. (c) Radius of gyration of the system about the axis OB is given by R/ 2. (d) Radius of gyration of the system about the axis OB is given by R. 14. A disc of radius R rolls on a horizontal surface with linear velocity V and angular velocity ω.
XY and PQ are axes passing through the centre of the cylinder and parallel to its plane face. (a) If the moment of inertia of the cylinder about the axis XY is I, the moment of inertia about PQ is also I. (b) The moment of inertia of the cylinder about its axis X' Y' is 2I (c) The moment of inertia about a diameter of one end is I + M ℓ2. (d) All of the above
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velocity v and gets caught in the bag. Then for the combined system (bag + bullet) (a) Momentum is mv/(M + m) (b) KE is (1/2) Mv2 (c) Momentum is mv (d) KE is m2v2/2(M + m)
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(c) and axis parallel to the z-axis and pass10 2 ing through a corner is m 3 5 (d) one side is m 2 2 17. A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a common angular velocity. (a) Some friction exists between the disc and the ring. (b) The angular momentum of the ‘disc plus ring’ is conserved. (c) The final common angular velocity is (2/3) rd of the initial angular velocity of the disc. (d) (2/3) rd of the initial kinetic energy changes to heat.
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1. Statement 1: When you lean behind over the hind legs of the chair, the chair falls back after a certain angle. Statement 2: Centre of mass lying outside the system makes the system unstable. 2. Statement 1: A rigid disc rolls without slipping on a fixed rough horizontal surface with uniform angular velocity. Then the acceleration of lowest point on the disc is zero. Statement 2: For a rigid disc rolling without slipping on a fixed rough horizontal surface, the velocity of the lowest point on the disc is always zero.
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16. Four identical rods, each of mass m and length ℓ, are joined to form a rigid square frame. The frame lies in the xy plane, with its centre at the origin and the sides parallel to the x and y axis. Its moment of inertia about 2 (a) the x-axis is m 2 3 4 2 (b) the z-axis is m 3
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(a) Friction force decelerates the disc M1 and accelerates the disc M2. (b) The final angular speeds of the discs are ω R in the ratio 1 = 1 . ω2 R2 (c) The angular momentum of the system (M1 + M2) is conserved about the point of contact. (d) The final rotational energy is less than the initial rotational energy.
Direction: Each question contains statement 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct.
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There is a point P on circumference of disc at angle θ, which has a vertical velocity. Here, θ is equal to π V V (a) π + sin −1 (b) − sin −1 2 Rω Rω −1 V −1 V (d) π + cos (c) π − cos Rω Rω
Rotational Dynamics
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Direction: Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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2. A uniform disc rolls without slipping on a rough horizontal surface with uniform angular velocity. Point O is the centre of disc and P is a point on disc as shown.
6. Statement 1: A wheel moving down a frictionless inclined plane will slip and not roll on the plane. Statement 2: It is the frictional force which provides a torque necessary for a body to roll on a surface.
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Column I Column II (a) If the coefficient (p) Downward of friction is tan θ, along the then linear accelerinclined ation of the disc is plane (b) If coefficient of (q) Zero friction is less than tan θ, then linear acceleration of the disc is (c) Direction of fric- (r) Perpendicution force is lar to plane of disc (d) Angular accelera- (s) Upward tion of disc is along the inclined plane
Under the given condition, the horizontal surface exerts normal reaction nonuniformly on lower surface of the block. Statement 2: For the cubical block given in statement-1, the horizontal force F has tendency to rotate the cube about its centre in clockwise sense. Hence, the lower right edge of cube presses the horizontal surface harder in comparison to the force exerted by lower left edge of cube on horizontal surface.
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7. Statement 1: When a sphere rolls on a horizontal table it slows down and eventually stops. Statement 2: When the sphere rolls on the table, both the sphere and the surface deform near the contact. As a result, the normal force does not pass through the centre and provide an angular deceleration. MATCH THE COLUMN TYPE QUESTIONS
4. Statement 1: If two different axes are at same distance from centre of mass of a rigid body, then moment of inertia of the given rigid body about both axis will always be same. Statement 2: From parallel axis is theorem I = Icm + md2, where all terms have usual meaning. 5. Statement 1: A uniform cubical block (of side a) undergoes translational motion on a smooth horizontal surface under action of horizontal force F as shown.
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3. Statement 1: A uniform thin rod of length L is hinged about one of its end and is free to rotate about the hinge without friction. Neglect the effect of gravity. A force F is applied at a distance x from the hinge on the rod such that force always is perpendicular to the rod. As the value of x is increased from zero to L, the component of reaction by hinge on the rod perpendicular to length of rod increases. Statement 2: Under the conditions given in statement 1 as x is increased from zero to L, the angular acceleration of rod increases.
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(c) Acceleration of point C (d) Acceleration of point D
In each situation of column I a statement is given and the corresponding results are given in column II. Match the statements in column I with the results in column II. 3
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4. In each situation of column I, a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface as shown. At t = 0 (initially) the angular velocity of disc is ω0 and velocity of centre of mass of disc is v0 (in horizontal direction). The relation between v0 and ω0 for each situation and also initial sense of rotation is given for each situation in column I. Match the statements in column I with the corresponding results in column II.
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Column II (p) Change in magnitude with time (b) The acceleration (q) Is always of point P on disc directed from that point (the point on disc given in column I) (c) The tangential ac- (r) is always zero celeration of point P on disc (d) The acceleration (s) is non-zero of point on disc and remains which is in contact constant in with rough horimagnitude zontal surface.
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7. A spherical ball is rolled without slipping on a table of height h. At a particular instant when the ball is in air, the x and y components of the velocity of the C. M. are vx and vy respectively as shown.
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5. Match the column I with column II and select the correct answer using the codes given below Column I (Rigid bodies) (a) Right circular cone of base radius R, mass m and altitude h perpendicular to its axis of symmetry and passing through its apex end. (b) A bent homogenous rod of mass m about the diagonal axis. (c) A regular polygon of mass M and number of sides equal to n and edge length c about any straight line through its centre. (d) A hollow cylinder of inner radius R1, outer radius R2 and length L through mid-point and perpendicular to axis.
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(s) Not conserved
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1. The value of the coefficient ‘s’ (a) 11/17 (b) 17/11 (c) 5/17 (d) 12/17
3. For the next impact to occur Ki must exceed a minimum value Ki; min which may be written in the form Ki, min = δMga where g = 9.81 m/s2 is the acceleration of gravity the coefficient δ in terms of the slope angle θ and the coefficient r 1 (a) δ = (1 − sin(30° − θ)) r 1 (b) δ = (1 − cos(30° − θ)) r
The mass of the prism is M and it is uniformly distributed. The length of each side of the cross-
2. The kinetic energy of the prism just before and after impact is similarly Ki and Kf. We may write Kf = rKi and value of the coefficient r (a) 121/17 (b) 121/289 (c) 12/17 (d) 17/12
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8. A rod of mass m and length ℓ is lying on a table. At some instant velocity of end A is v and angular velocity ω = 4v / 3ℓ. Column I gives angular momentum of the rod about an axis perpendicular of the rod and passing through points and column II the corresponding angular momentum Ix stands for moment of inertia when the axis passes through corresponding point A, B, C or D
(r) Conserved
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sectional hexagon is a. The moment of inertia I of the hexagonal prism about its central axis is I = (5/12) Ma2 The moment of inertia I' about an edge of the 17 prism is I ′ = Ma 2 . 12 The prism is initially at rest with its axis horizontal on an inclined plane which makes a small angle with the horizontal (figure). Assume that the surfaces of the prism are slightly concave so that the prism only touches the plane at its edges. The effect of this concavity on the moment of inertia can be ignored. The prism is now displaced from rest and starts an uneven rolling down the plane. Assume that friction prevents any sliding and that the prism does not lose contact with the plane. The angular velocity just before a given edge hits the plane is ωi while ωf is the angular velocity immediately after the impact. We may write ωf = s ωi.
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10. Where will be the centre of mass after the removal of square 1 (a) In the first quadrant (b) In the second quadrant (c) In the third quadrant (d) In the fourth quadrant
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6. The velocity of centre of mass immediately after the collision is (a) v/3 (b) 2v/3 (c) 3v/2 (d) 2v/5
12. In the above question, where will be the centre of mass after the removal of squares 1, 2 and 3 (a) In the third quadrant (b) In the fourth quadrant (c) In the first quadrant (d) At the origin
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PASSAGE-2 A uniform sphere with mass 2m and radius 2r is at rest in a place where no external forces are applied. P U
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4. If the condition of previous Question is satisfied, the kinetic energy Ki, will approach a fixed value Ki, 0 as the prism rolls down the incline Ki, 0 = K Mg a, K is equal to tan θ cos θ (b) (a) 1− r 1− r sin θ 1− r (c) (d) 1− r sin θ
8. The angular momentum about the center of mass shortly before and after the collision (a) increases (b) decreases (c) remains same (d) none of these
1 (c) δ = (1 − tan(30° − θ)) r 1 (d) δ = (1 − sec(30° − θ)) r
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PASSAGE-4 The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90º.
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13. In the above question, where will be the centre of mass after the removal of squares 1 and 3 (a) At the origin (b) In the first quadrant (c) In the third quadrant (d) In the second quadrant
Rotational Dynamics
PASSAGE-5 Rod AB of length 20 m is moving on an smooth horizontal plane with v0 = 10 m/s and rotating with ω = 2 rad/sec in anti-clock wise direction. $
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(b) 2v0 (d) v0/2
(a) 2 v0 (c) v0
22. Velocity of point P at time t = t0 is
(d) None of these
21. Distance travelled by C.O.M. during time t = 0 to t = t0 is (a) 7.5 π (b) 5 π (c) 2.5 π (d) None of these
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20. The time t0 at which velocity of point P becomes maximum is (At t = 0 velocity of CM is perpendicular to rod) π π (b) (a) 4 2 3π (c) (d) None of these 4
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18. Velocity of ends A and B when velocity of CM is perpendicular to rod are (a) 30 m/s, 10 m/s (b) 10 m/s, 30 m/s (c) 15 m/s, 25 m/s (d) 25 m/s, 15 m/s
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Rotational Dynamics
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8. A uniform solid body consists of a cylinder of radius 2a and height 2h with another cylinder of radius a and height h with one plane face placed centrally on one plane face of the first cylinder. Find the position of the centre of gravity of the solid.
1. Three particles A, B, C, of mass 2, 3, 4 kg are at points (1, 4), (3, 6), (2, 1) in the xy plane. Find the coordinate of their centre of gravity. 2. Three particles of mass 5, 3, 7 kg are at the points A, B, C whose position vectors are iˆ − 2 ˆj , 7iˆ + ˆj , −3iˆ + 5 ˆj. Find the position vector of their centre of mass. Also find the position vector of the centroid of the points A, B, C.
9. A rod of length R and mass M is free to rotate about a horizontal axis passing through hinge P as in figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the rod hits the block B of mass m and stops. Find the ratio of masses such that the block B completes the circle. Neglect any friction.
SUBJECTIVE QUESTIONS
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5. The vertices of a triangle are at the points iˆ + ˆj , 3iˆ − ˆj , 2iˆ + ˆj. Find the position vector of the centre of gravity of the triangle, assuming it to be a uniform lamina.
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3. A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at a distance x from A. Find the normal reactions at A and B.
23. Choose the incorrect statement.
PASSAGE-6 A ring of mass M and radius R sliding with a velocity v0 suddenly enters into rough surface where the coefficient of friction is µ, as shown in figure
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Rotational Dynamics
16. An automobile engine develops 100 kilowatt when rotating at a speed of 1800 rev/ min. What torque does it deliver?
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19. A disc of mass 2 kg and radius 20 cm is free to rotate about an axis through its centre and perpendicular to the disc. If a force of 50 N is applied tagentially, calculate the angular acceleration.
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20. A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 rad/s. If this mass is brought to rest in 10 sec by a brake, what is the torque that is applied?
22. A solid sphere and a solid cylinder having the same mass and radius, roll down the same incline. What is the ratio of their accelerations? 23. The moment of inertia of sphere is 20 kg-m2 about the diameter. What is the moment of inertia about any tangent?
21. If a flywheel of mass 20 kg and diameter 1 m is rotating 300 revolutions per minute, what is its kinetic energy?
24. The moment of inertia of a wheel is 1000 kg m2. At a given instant, its angular
13. A flywheel starting from rest acquires in 10 sec an angular velocity of 240 revolutions per minute. What is its angular acceleration assumed constant?
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12. A vertical post of mass M is to be driven into the ground. A pile-driver of mass m strikes the post vertically with a velocity v. Assuming that the pile-driver does not bounce off the post, find the velocity with which the post enters the ground. If the combined mass comes to rest when the post has beep driven into the ground to a depth h find the constant force with which the ground resists penetration.
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The angle between the two legs is 60º. The fat person sitting on the ladder has a mass of 80 kg. Find the contact force exerted by the floor on each leg and the tension in the crossbar.
15. A wheel starting from rest, is rotating with a constant angular acceleration of 3.0 rad/ sec2. An observer notes that it traces an angle of 120 radians in a 4.0 sec interval. For how long the wheel has been rotated when the observer started his observation?
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11. The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle.
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14. The angular velocity of a wheel increases from 1200 to 4500 rev/min in 10 sec. Compute its angular acceleration and the number of revolutions made during this time.
10. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal, to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. Find the minimum value of F for which the cube begins to tip about an edge. (assume that cube does not slide).
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Rotational Dynamics
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2. An isolated particle of mass m is moving in horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses m/4 and 3 m/4. An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at [1997]
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(b) y = +20 cm (d) y = –20 cm
(a) y = –5 cm (c) y = +5 cm
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This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the co-ordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position. [1996]
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1. A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of large and small spheres are 4 M and M respectively.
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PREVIOUS YEARS' IIT-JEE QUESTIONS
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28. A hoop of radius R rests on a horizontal surface. A similar hoop moves past it at a velocity v. Determine the velocity vA of the upper point of “intersection” of the hoops as a function of the distance d between their centres, assuming that the hoops are thin, and the second hoop is in contact with the first hoop as it moves past the latter.
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27. It was found that the winding radius of a tape on a cassette was reduced by half in a time t1 = 20 min of operation. In what time t2 will the winding radius be reduced by half again?
26. A solid sphere of mass 0.5 kg and diameter 1 meter rolls without slipping with a constant velocity of 5 m/sec along a smooth straight line. Calculate its total energy.
30. A horizontal weightless rod of length 3ℓ is suspended on two vertical strings. Two loads of mass m1 and m2 are in equilibrium at equal distances from each other and from the ends of the string (Fig.). Determine the tension T of the left string at the instant when the right string snaps.
25. A sphere is rolling down without slipping the incline plane from a vertical height h. Find the angular velocity and linear velocity as it reaches the ground if its mass is m and radius is r.
velocity is 10 rad/s. After the wheel rotates through an angle of 100 radians, the wheel’s angular velocity is 100 rad/s. Calculate (i) the torque applied on the wheel (ii) the increase in rotational kinetic energy
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1.104
3. A wedge of mass m and triangular crosssection (AB = BC = CA = 2R) is moving with a constant velocity- viˆ towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that wedge remains in contact with the sphere for a very short time ∆t during which the sphere exerts a constant force F on the wedge. [1998]
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7. A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is
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8. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown.
(b) mg/4 (d) mg (1 – µ)
(a) Infinitesimal (c) mg/2
5. A disc of mass M and radius R is rolling with angular speed ω on a horizontal as shown. The magnitude of angular momentum of the disc about the origin O is [1999]
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4. A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB respectively. Then [1999]
(d) 2 MR2ω
6. A cubical block of side a moving with velocity v on a horizontal smooth plane as shown. It hits a ridge at point O. The angular speed of the block after it hits O is [1999]
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(a) Find the force F and also the normal force N exerted by the table on the wedge during the time ∆t (b) Let h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge during the interval ∆t.
(a) ωA < ωB (c) ωA = ω
1 MR 2 ω 2 3 (c) MR 2 ω 2 (a)
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Rotational Dynamics
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Rotational Dynamics
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The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its centre of mass is 1.2 kg-m2. Two robber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5 m from the line AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacle is 6 N-s. [2001]
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(a) Angular velocity and total energy (kinetic and potential) (b) Total angular momentum and total energy (c) Angular velocity and moment of inertia about the axis of rotation. (d) Total angular momentum and moment of inertia about the axis of rotation.
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9. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down; One along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as beads slides down are [2000]
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11. Two heavy metallic plates are joined together at 90º to each other. A laminar sheet of mass 30 kg is hinged at the line AB joining the two heavy metallic plates. The hinges are frictionless.
3ρL3 8π2
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10. A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in the direction perpendicular to AB. The collision is completely elastic. After the collision the particle comes to rest. (a) Find the ratio m/M (b) A point P on the rod is at rest immediately after the collision. Find the distance AP. (c) Find the linear speed of the point P at a time πL/(3v0) after the collision. [2000]
The moment of inertia of the loop about the axis XX' is [2000]
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12. One quarter section is cut from a uniform circular disc of radius R. This section has a mass M.
The body is set into rotational motion on the table about A with a constant angular velocity ω. (a) Find the magnitude of the horizontal force exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T. [2002]
Rotational Dynamics
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14. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity ω0. When the tortoise move along a chord of the platform with a constant velocity (with respect to the platform). The angular velocity of the platform ω(t) will vary with time t as: [2002]
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Rotational Dynamics
19. A block of mass m is held fixed against a wall by a applying a horizontal force F. Which of the following option is incorrect
Centre of the circle Outside the circle Inside the circle Point on circumference
[2005]
16. Two particles each of mass M are connected by a massless rod of length ℓ. The rod is lying on the smooth surface. If one of the particle is given an impulse MV as shown in the figure then angular velocity of the rod would be [2003]
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18. A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is (a) 2K (b) K/2 (c) K/4 (d) 4K
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8 mR2 4 mR2 (40/9) mR2 (39/7) mR2
21. A particle moves in circular path with decreasing speed. Which of the following is correct [2005]
1 (c) vQ = vP, vC = vP 2
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(a) vQ > vC > vP
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1.108
23. A rod of length L and mass M is hinged at point O. A small bullet of mass m hits the rod as shown in the figure.
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22. A cylinder of mass m and radius R rolls down an inclined plane of inclination θ. Calculate the linear acceleration of the axis of cylinder. [2005]
(b) only direction of L is constant (c) acceleration a is towards the centre (d) it will move in a spiral and finally reach the centre
25. A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis about the edge and perpendicular to the plane is I. The relation between radius R and r. [2006] 2 2 (a) r = R (b) r = R 15 15 2 2 R (d) r = (c) r = R 15 15
(a) L is constant
Rotational Dynamics
PASSAGE BASED QUESTIONS Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an
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Rotational Dynamics
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(c)
31. Statement 1: If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. because Statement 2: The linear momentum of an isolated system remains constant. (a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement-1 (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True [2007]
a. co m
9I ω 2t
29. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is 2I ω (a) 3t
(b)
Iω2 3 Iω2 (d) 6
(b)
Iω2 2 Iω2 (c) 4 (a)
30. The loss of kinetic energy during the above process is
angular velocity ω by a spring having the same spring constant and compressed by a distance x2. Both the discs rotates in the clockwise direction. [2007] 28. The ratio x1/x2 is (a) 2 (b) 1/2 (c) 2 (d) 1/ 2
5. 13. 21. 29. 37. 45. 53. 61. 69. 77.
(d) (a) (a) (b) (b) (b) (b) (a) (b) (d)
6. 14. 22. 30. 38. 46. 54. 62. 70. 78.
(d) (a) (b) (b) (a) (d) (c) (b) (b) (a)
7. 15. 23. 31. 39. 47. 55. 63. 71. 79.
(a) (c) (c) (d) (b) (b) (b) (a) (b) (d)
8. 16. 24. 32. 40. 48. 56. 64. 72. 80.
(a) (a) (a) (c) (c) (c) (b) (c) (b) (b)
ONE OR MORE THAN ONE OPTION MAY BE CORRECT 1. (a, b, d) 2. (a, b, d) 3. (a, c, d) 4. (a, b, c) 5. (b, c, d) 6. (a, d) 9. (c, d) 10. (a, b, c) 11. (a, c) 12. (b, c) 13. (b, c) 14. (c, d)
7. (b, c, d) 8. (a, c) 15. (b, d) 16. (a,b,c,d)
17. (a, b, d)
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1.110
Rotational Dynamics
ASSERTION AND REASON QUESTIONS 1. (c)
2. (d)
3. (d)
4. (d)
5. (c)
6. (a)
7. (a)
MATCH THE COLUMN TYPE QUESTIONS 2. (a) → p, (b) → q, s (c) → p, (d) → q, s
3. (a) → q, (b) → s, (c) → r, (d) → p
4. (a) → p, q, r (b) → p, q, r (c) → p, q
5. (a) → s, (b) → p, (c) → q, (d) → r
6. (a) → p, q, (b) → q, r, (c) → p, q, (d) → p, q
1. (a) → q, (b) → p, (c) → s, (d) → r
(d) → p, q, r 7. (a) → r, (b) → s, (c) → r, (d) → q
8. (a) → q, (b) → s, (c) → p, r, (d) → p
3. (b) 11. (a) 19. (a)
4. (c) 12. (a) 20. (a)
5. (a) 13. (a) 21. (b)
8. (c) 16. (c) 24. (b)
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3. A fly wheel is a large heavy wheel with a long cylindrical axle passing through its centre and its centre of mass lying on its axis of rotation. It is constructed in such a way that most of its mass is concentrated at the rim of fly-wheel. This increases the moment of inertia about the axis of rotation. The function of moment of inertia is to oppose any change in uniform rotatory motion. Therefore, when a fly-wheel of large moment of inertia is coupled to an engine, it assures the running of engine smoother and steadier.
5. Equilibrium of the wheel when it is still and standing up-right on its rim is unstable. Moreover, when the wheel is rolling, it possesses and angular momentum in the horizontal direction which is conserved quantity of the motion. The angular momentum does not changes unless friction slows down the angular velocity. Since, falling results in a change of angular momentum direction, the wheel does not fall, till is loses its angular velocity. 6. The system tends to keep its angular momentum constant. When the person sits near the edge of the platform, the moment of inertia
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4. The torque τ required to cause an angular acceleration α is given by τ = Iα, where, I (= mr2) is the moment of inertia. More is the distance of the stone from the axis of rotation, greater is its moment of inertia and hence, more torque is required to produce same angular acceleration.
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2. The cycle wheel is constructed in such a way so as to increase its moment of inertia with minimum possible mass. This is done to ensure uniform speed.
7. (a) 15. (a) 23. (c)
HINTS AND SOLUTIONS
1. No, torque is required only for angular acceleration.
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CONCEPTUAL QUESTIONS
6. (a) 14. (c) 22. (a)
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(a) (a) (c) (a)
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PASSAGE BASED QUESTIONS
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Rotational Dynamics
13. In a tornado, the air from the surrounding whirls inwards as the air at the centre rises. Since, the moment of inertia decreases, the angular velocity of the inward whirling wind increases according to the law of conservation of momentum.
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7. (a) Initial velocity of centre of mass of given system is zero and net external force is in vertical direction. Since, there is shift of mass downward, the centre of mass has only downward shift. (b) Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts a force on string which has a horizontal component towards right. Hence, centre of mass of system has a rightward shift. (c) Both block and monkey moves up, hence, centre of mass of given system shifts vertically upwards.
12. When two fingers are applied, two equal and opposite forces act on the cap, thereby constituting a couple.
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11. When the raw and the hard boiled eggs are spun together the raw egg will have a greater moment of inertia as the fluid in it will concentrate along its surface. Assuming that the same rotational torque is applied to both the eggs, raw egg will come to rest first as it will have lesser angular velocity.
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10. We know that I = Σmr2 i.e., the moment of inertia of a body not only depends upon the mass but also depends upon the distribution of mass from the axis of rotation. The disc made of less dense material will have larger moment of inertia because it will have greater surface area, i.e., the particles are at the greater distance from the axis of rotation.
5. (a) Since, all forces on disc pass through point of contact with horizontal surface, the angular momentum of disc about point on ground in contact with disc is conserved. Also the angular momentum of disc in all cases is conserved about any point on the line passing through point of contact and parallel to velocity of centre of mass. The K.E. of disc is decreased in all cases due to work done by friction. From calculation of velocity of lowest point on disc, the direction of friction in case a, b and d is towards left and in case c is towards right. The direction of frictional force cannot change in any given case.
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9. If there were only one propeller in the helicopter, the helicopter itself would have turned in opposite direction due to conservation of angular momentum.
3. (a) Speed of point P changes with time. (b) Acceleration of point P is equal to ω2x (ω = angular speed of disc and x = OP). The acceleration is directed from P towards O. (c) The angle between acceleration of P (constant in magnitude) and velocity of P changes with time. Therefore, tangential acceleration of P changes with time. (d) The acceleration of lowest point is directed towards centre of disc and remains constant with time.
8. By raising the hand, the moment of inertia of the system increases and thus angular velocity ω decreases. Now the turn table will slow down as I ω remains constant in the absence of any external torque.
MATCH THE COLUMN TYPE QUESTIONS
7. The swimmer can increase the number of loops by pulling his legs and arms inwards i.e., by decreasing the moment of inertia. By doing so the angular velocity ω increases because I ω remains constant.
of platform increases and hence, its angular velocity decreases. When the person starts moving towards the centre of platform, the moment of inertia decreases and hence, the angular velocity increases.
1.111
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1.112
Rotational Dynamics
(d) Net external force on given system is zero. Hence, centre of mass of given system remains at rest. 8. Taking into the plane –ve (a) L = MvR + Iω > 0 → v f forward
1ˆ j 3
7h from the base of the larger cylinder 6
8.
5. 2iˆ +
9. Minimum velocity required by block m to complete the motion in 5gR
2v (b) L = MvR – Iω = MvR −1 R MvR – 2 MvR < 0 for ring ⇒ v f forward 4 MvR − > 0 MvR for solid sphere ⇒ v f 5 left (c) L = MvR + Iω > 0 ⇒ v f left
Conserving mech. energy
1 2 R I ω = Mg × ⇒ ω = 2 2
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ML3 MR 2 (since L = R) = 3 3
M = 15 m
10. For troppling about edge xx'
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Fmin =
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N1 = mg – N2 = mg –
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mgx d −x = mg d d
NA = w (1 – x/d), NB = wx/d
D
11. The forces acting on different parts are shown in figure. Consider the vertical equi
mgx N2 = d
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Putting I =
1
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Μ gR = mR 5 gR I
MgRI = m2R25gR
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3. As shown in FBD → equation in vertical direction N1 + N2 = mg Taking moments about A mgx = dN2 1
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Conservation of angular momentum with respect to P before and after collision.
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SUBJECTIVE QUESTIONS
1 1 ˆ (5i + 28 ˆj ), (5iˆ + 4 ˆj ) 3 15
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v MvR − MR 2 > 0 ⇒ v f left 2R
2.
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a.
2 v MvR − MR 2 > 0 ⇒ v f left 5 2R
19 10 1. , 10 3
3 Z
m
(d) L = MvR – Iω
MgR I
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Rotational Dynamics
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19. 250 rad/sec2
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20. 0.9 Nm
π − α with the horizontal and 2 v v = is given by v A = 2cos ϕ 2sin α angle ϕ =
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(ii) 0.04 kg-m2
18. 2 %
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17. (i) 0.01 kg-m2
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1800 × 2π ; 60
10000 = 531 N -m 6π
O1 is at rest, the velocity vA of the upper point A of “intersection” of the hoops must be directed along the tangent to the circle with centre O1 at any instant of time (Figure) At any instant of time the segment AB divides the distance d = OO1 between the centres of the hoops into two equal parts, and hence, the horizontal projection v of the velocity vA is always equal to . 2 Consequently, the velocity vA forms an
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15. 8.0 sec.
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22. 15 : 14
23. 70 kg-m2 2
(ii) 4.95 × 106 Joules
24. (i) 4.95 × 104 N-m
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2
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2 gh 2gh 25. ω = 2 and v = 2 2 2 1 + K /R r +K 27. Let the initial winding radius be 4r. Then the decrease in the winding area as a result of the reduction of the radius by half (to 2r) will be S = π (16r2 – 4r2) = 12 π r2.
2
d Since, sin α = 1 − cos 2 α = 1 − 2R the velocity of the upper point of “intersecv tion of the hoops is v A = 2 d 2 1− 2R
26. 8.75 Joule
m
14. 475
τ=
28. Since, the hoop with the centre at point
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13. 2.51 rad/sec2
100 × 103 = τ
a. co
m 2 v 2 + 2 gh( M + m) 2 mv , 2 h ( M + m) M +m
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12.
2 2 = (392 N ) × = 450 N 3 3
or T = N
or N = (40 kg) (9.8 m/s2) = 392 N Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, N (2 m) tan 30º = T (1 m)
Which is equal to the product of the length ℓ1 of the wound tape and its thickness d. The velocity v of the tap is constant during the operation of the tape-recorder, hence ℓ1 = vt1, and we can write 12 π r2 = vt1 d ...(i) When the winding radius of the tape on the cassette is reduced by half again (from 2r to r), the winding area is reduced by π (4r2 – r2) = 3 π r2, i.e., 3 π r2 = vt2 d …(ii) Where t2 is the time during which the winding radius will be reduced in the latter case. Dividing equations (1) and (2) term wise, t we obtain t2 = 1 = 5 4
librium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg)g and the contact force N + N = 2N due to the floor. Thus 2N = (80 kg)g
1.113
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1.114
2
\
29. The forces acting on the hoop-washer system are the force of gravity and the normal reaction of the plane. These forces are directed along the vertical. Consequently, the centre of mass of the system does not move in the horizontal direction. Since, there is no friction between the hoop and the plane, the momentum conservation law, at any instant of time we have Mu + mvx = 0 .........(1) where u and vx are the horizontal components of the velocities of the centre of the hoop and the washer. Since, vx periodically changes its sign, u also changes sing “synchronously”. The general nature of motion of the hoop is as follows: the centre of the hoop moves to the right when the washer is on segments BC and BE, and to the left when the washer is on segments CD and DE (Figure).
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u = m cos ϕ
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2 gr (1 + cos ϕ) ( M + m)( M + m sin ϕ)
30. At the moment of snapping of the right string, the rod is acted upon by the tension T of the left string and the forces N1 and N2 of normal pressure of the loads of mass m1 and m2 (Figure)
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Solving equations (1) and (3) together, we determine the velocity of the centre of the hoop at the instant when the radius vector of the point of location of the washer forms an angle ϕ with the vertical
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The velocity v of the washer and u of the hoop are connected through the energy conservation law: mv 2 Mu 2 mgr (1 + cos ϕ) = + ......... (2) 2 2 The motion of the washer relative to a stationary observer can be represented at any instant as the superposition of two motions; the motion relative to the centre of the hoop at a velocity vt directed along the tangent to the hoop, and the motion together with the hoop at its velocity u having the horizontal direction (Figure). The figure shows that
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Since, the rod is weightless (its mass is zero), the equations of its translatory and rotary motions will have the form – T + N1 – N2 = 0, N1ℓ = 2N2ℓ. The second equation (the condition of equality to zero of the sum of all moments of force about point O) implies that
Rotational Dynamics
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Rotational Dynamics
(c)
2L 3
v0 2 2
11. (a) I = 0.1 m, (b) w' = 1 rad/s (c) laminar sheet will never come to rest
12. (a)
3mω2
(c)
(b) Fx = F/4
3mω2
13. (a)
x=
14. (c)
m 1 = (b) M 4
10. (a)
9. (b)
N1 = 2N2 .....(1) Combining these conditions, we get T = N1 – N2 = N2 .....(2) At the moment of snapping of the right string, the accelerations of the loads of mass m1 and m2 will be vertical (point O is stationary, and the rod is inextensible) and connected through the relation a2 = 2a1. Let us write the equation of motion for the loads at this instant: m1g – N '1 = m1a1, m2g + N '2 = m2a2, Where N'1 and N '2 are the normal reaction of the rod on the loads of mass m1 and m2. Since, N1' = N1 and N2' = N2, we have m1g – 2N2 = m1a1 m2g – N2 = 2 m2a1
16. (a)
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27. (d)
28. (c)
29. (a) 30. (b)
25. (b)
31. (c)
8. (d)
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PREVIOUS YEAR ’ IIT-JEE QUESTIONS 1. (L + 2R, 0)
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Hence, we can find N2 and consequently the m1m2 tension of the string T = N 2 = g m1 + 4m2
1.115
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c h a p t e r
2
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Gravitation
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2.1 NEWTON’S UNIVERSAL LAW OF GRAVITATION
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“Each body in the universe attracts to the another body”, this is inverse square law, this force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Let, the two particles are of mass m1 and m2 and distance between them is r. Therefore m m F∝ 12 2 r G m1 m2 F= r2 where, G = Universal constant = 6.67 × 1011 Nm2/kg2
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2.2 GRAVITATIONAL POTENTIAL ENERGY Let, the two particles are separated by a distance r, mass of the particles are m and M. Hence, the gravitational force between them F = G m2 M r P
0
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2.2
Gravitation
Work done = change in the gravitational potential energy. ∞ ∞ W = ∫ F⋅dr = −GmM ∫ dr2 r r r Negative sign indicates that work done is against the gravitational force ∞
1 W = −G m M − r = − G mr M r
M Hence, gravitational potential energy = − G m r
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Potential energy decreases, when the separation between them increases. Gravitational potential energy is zero at the infinity.
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2.3 ACCELERATION DUE TO GRAVITY
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If the mass of the earth is M, radius of the earth is R and a particle of mass m is at the surface of the earth, thus the attraction force between them is
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F = G m2M R
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This force imports the acceleration of the particle m, acceleration due to gravity. According to the Newton’s second law, F a=m
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a = g = G m2M = GM R2 R m
On the surface of the earth = GM R2
2.4 VARIATION IN ACCELERATION DUE TO GRAVITY (i) Above the surface: Let a particle is above the earth surface at a height h. Hence the gravitional force F′ = G m M2 (R + h) https://t.me/universitarios_infohttps://www.jamarana.com
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Gravitation
2.3
P K 5 2
m
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From previous, g = GM R2 From equations (i) and (ii), we get 2
...(i)
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or g ′ = G M 2 (R + h)
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g′ = g R + R h where, g = accelevation due to gravity at the surface.
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(ii) Below the surface of the earth: For this part, radius of shadded part is (R – h).
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Volume of the shadded part = 4 π (R − h)3 3 Density of the earth =
M 4/3 π R3
Hence, mass of the shadded part M m′ = 4 π(R − h)3 ⋅ 3 4 πR3 3 =
(R − h)3 M R3
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...(ii)
Gravitation
So g ′ = G m
(R − h)3 M R3 (R − h) 2
=G m M
(R − h) R3
∴ g ′ = g 1 − h R
where, g is the acceleration due to gravity. Note: It should be noted that, the value of g decreases if we move above the surface or below of the surface of the earth.
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Here, r is the radius of the circular path followed by m. ⇒ r = R cos θ The resultant force of masses m is given by
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(iii) With latitudes: Consider a point A on the earth surface where the line joining centre O to A makes an angle θ with the equatorial plane as shown. The angle θ is said to be the latitude of the point A´. Let us place a small particle of mass m at A. Analysing the forces of on m, from the reference frame (or relative to) the earth, we find two forcesgravitational force (mg) and pseudo force (mω2r).
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2.4
F = m 2 g 2 + m 2 ω4 r 2 + 2(mg )(mω2 r ) cos(π − θ) The effective gravity at point A, F = g 2 + ω4 r 2 − 2ω2 g r cos θ g′ = m 2
2 2 g ′ = g 1 + ωgR cos 2 θ − 2 ωgR cos 2 θ where, g is the acceleration due to gravity at the surface of the earth neglecting the effect of the rotation.
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Gravitation
2.5
(a) At poles:
θ = 90º g’ = g
⇒
Hence, the rotation of the earth has no effect on the gravity at poles. (b) At equator: θ = 0º 2 g = g 1 − ωgR
2.5 ESCAPE VELOCITY
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It is the minimum velocity with which a body must be projected from the surface of the earth, so that it permanently overcomes and escapes the gravitational field of the earth. We can also say that a body projected with escape velocity will be able to go to point which is at infinite distance from the earth.
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Let us imagine what happens to a body of mass m, if it is thrown from the earth with a velocity ve (escape velocity). As the body moves away from the earth, it slows down (due to gravitational pull of the earth) and, hence the kinetic energy is converted into gravitational potential energy of the mass earth system. Let us imagine that it is just able to reach up to infinity (where gravitational potential energy is zero). Kinetic energy lost by mass m = Gain in gravitational potential energy of the mass earth system 1 mv 2 = (G.P.E.) − (G.P.E.) f i 2 e 1 mv 2 = 0 − − GMm 2 e R ve =
2GM R
or
ve = 2 gR
Subsitituting the value of g = 9.81 m/sec2 and R = 6400 km, we get ve = 11.3 km/sec
Hence, any object thrown with a velocity of 11.3 km/sec or more will escape the gravitational field and will never come back to the earth.
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2.6
Gravitation
2.6 MOTION OF A SATELLITE OF MASS M AROUND THE EARTH Consider a satellite of mass m revolving in a circle around the earth which is located at the centre of its orbit. If the satellite is at the height h above the earth’s surface, the radius of its orbit is r=R+h where R is the radius of the earth. The gravitational force between m and M provides the centripetal force necessary for circular motion. K
S 5
rio s_ in fo
m
P
ita
.ja
P
un iv
er si ta rio s
er s
w w //w
Orbital Velocity
s:
P
(DUWK
ar a
P
na .c om
P
iv
//t
ht
.m
tp
e/
The velocity of a satellite in its orbit is called the orbital velocity. Let v be the orbital velocity of the satellite, then
un t.m e/
v = GM r
Also v =
s: //
or
ht tp s:
GMm = mv 2 r r2
GM R+h
ht tp
Hence, the orbital velocity of a satellite is decided by the radius of its orbit or its height above the earth’s surface. GM v = GM r ≈ R = gR
Time Period The time taken to complete one revolution is called the time period. It is given by r 2π ⋅ r r 2πr T = v = 2πr = GM GM ⇒
4π2 3 T2 = r Gm
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Gravitation
2.7
Note: For a satellite whose time period is 24 h (same as the time period of the earth’s rotation). Radius of orbit, r = 3
GMT 3 (T = 24 hrs) 4π 2
This satellite seems stationary if observed from the surface of the earth. This is also known as Geostationary satellite.
2.7 TOTAL ENERGY OF THE SATELLITE
=
1 GM Gm M m − r 2 r
=
GMm GmM − 2r r
s_ in f si ta
GMm 2r
s
GmM 2r
rio
ni v
e/ u
tp s:
Binding energy = – (Total energy) =
ita
//w w
w
=−
er
.ja m
rio
ar a
na
o
.c o
1 2 Gm M mv + − r 2
=
m
Total energy = kinetic energy + potential energy
er s
iv m e/ un
ht
:// t.m
The energy that must be given to the orbiting satellite to make it escape to infinity is known as binding energy.
tp s
2.8 TRAJECTORY OF A SATELLITE FOR DIFFERENT SPEEDS
//t .
ht tp s:
ht
Let v be the velocity given to satellite. Let νc respresent the orbital velocity and νe be the escape velocity.
vc =
GM 2GM and ve = r r
where, r is the distance of the satellite from centre at the earth.
(i) ν < νc The satellite follows an elliptical path with the centre of the earth as the farther. In this case, if statellite is projected from near the surface of the earth, it will hit the earth’s surface without completing the orbit. (ii) ν = νc The satellite follows a circular orbit with the centre of the earth as the centre of orbit. (iii) νc < ν < νe The satellite follows an elliptical orbit with the centre of the earth as the focus nearer to the point of projection. (iv) ν = νe The satellite escapes the field of the earth along a parabolic trajectory. (v) ν > νe The satellite escapes the field of the earth along a hyperbolic trajectory.
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2.8
Gravitation
2.9 TIME PERIOD OF A SATELLITE AS OBSERVED FROM THE EARTH Consider a satellite in the circular orbit with the time period Ts. The earth also rotates with the time period TE = 24 h. If an observer on the earth sees this satellite, the angular velocity of the satellite relative to the earth will be
7(
7V
→
ωSE = ωS −ωE
rio s_ in fo
→
→
na .c om
(DUWK
m
ar a
Hence, the time for one revolution will appear to be different from Ts if observed from the earth. This time period can be calculated as TSE from the above equation.
er s un iv
w //w
er si ta rio s
⇒ TSE = TSTE TE − TS
w
2π 2π 2π = − TSE TS TE
ita
.ja
(i) If satellite and the earth are rotating in the same direction.
e/
TSTE TS + TE
iv
.m
TSE =
ht tp s:
un
//t
ht
tp
s:
(ii) If satellite and the earth are rotating in opposite directons.
t.m e/
2.10 KEPLER’S LAWS
ht tp
s: //
(i) The Law of the Orbits: Each planet revolves around the sun in an elliptical orbit with the sun at one focus of the ellipse. (ii) The Law of Areas: This law state that the radius vector from the sun to the planet sweeps out equal areas in equal time intervals. Both shaded areas are equal if the time from A to B is equal to the time from P to Q. 3ODQHW
4
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Gravitation
2.9
(iii) The Law of Periods: It states that, the square of the time taken by the planet about the sun is proportional to the planet’s mean distance from the sun. If T be the time period of the planet and r be the mean distance of planet from the sun (average of maximum and minimum distances from the sun), then
⇒
r=−
rmin + rmax 2
T2 is same for all planets. r3
na .c om
rio s_ in fo
x = Separation between two stars. d = Distance between two stars (centre to centre) = 90 × 103 m
ita iv
un
or
v
)
) Solution
First method: m1 = m2 = m = 0.74 × 1029 kg r1 = r2 = r = 11.25 × 103 m
dv 2Gm =− 2 dx x
V
d /2
0
d
∫ vdv = −2Gm ∫
dx x2
or
2. Two neutron stars, each of mass 0.74 × 1029 kg and radius 11.25 km are separated by a centre to centre distance of 90 km. They are initially at rest with respect to each other. (i) How fast are they moving when their saparation has decreased to one-half of its initial value? (ii) How fast are they moving just before they collide? (ignore elastivistic effects)
ht tp
= 1.117 × 104 ms–1
GM x2 Since, the stars approach to each other, therefore 2GM a = 2a1 = 2 x dv 2GM or =− 2 dt x [negative sign indicates that x decreases as velocity increases] a1 = a2 =
t.m e/
= 2 × 9.8 × 6.37 ×106
s: //
v = 2 gR
//t
2GM = 2gR R
ht tp s:
n2 =
∴
.m
tp
ht
∴
D
G
e/
s:
potential energy 1 2 GmM 1 1 mv = −GM − m = or 2 R ∞ R
D
er si ta rio s
un iv
)
//w
w
) Solution Gain in kinetic energy = loss in
er s
w
.ja
m
ar a
1. A body is released at a distance far away from the surface of the earth. Calculate its velocity, when it is near the surface of the earth. g = 9.8 ms–2 and R = 6.37 × 106 m. Assume no resistance.
WORKED OUT EXAMPLES
d /2
v2 1 1 1 − = −2Gm − = 2Gm 2 /2 x d d d 2 v 1 = 2Gm. 2 d
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2.10
∴
∴
3. A hypothetical planet of mass M has three moons each of the equal mass m each revolving in the same circular orbit of radius R. The mass are equally spaced and thus form an equilateral triangle. Find (i) The total potential energy of the system. (ii) The orbital speed of each moon such that they maintain this configuration.
Gm v=2 d v v1 = v2 = 2
Gravitation
Gm 6.67 × 10−11 × 0.74 × 1029 = d 90 × 103 = 7.41 × 106 m/sec
=
2r
v2 1 1 1 = −2Gm − = 2Gm − 2 2r d x d
om
v2 = (6.67 × 10–11)
( 0.74 ×10 ) 22.51×10
v = 1.28 × 107 m/sec
s_ ta si
er iv un
e/
GmM R WB = Work done to place mass m at B =
GM Gm = + m AB R WC = Work done to place mass m at C GM Gm Gm = + + m BC CA R From ∆ ABC, cos 30° =
29
3
−
1 90 × 103
&
W0 = Work done to place mass M at O from infinity = 0 WA = Work done to place mass m at A
or
or
1 1 Gm − = v2 r1 + r2 d
⇒
rio
si ta //t
s:
ht tp
Gm = 7.41 × 106 m/sec. d Gm1m2 Gm1m2 1 1 − = m1v12 + m2 v 22 (ii) r1 + r2 d 2 2 v=
or
in fo
a. c .m
tp s
ht
∴
2
%
:// t.m
ht
GM = v2 d
)
∴
BC/2 R
BC = 3 R = BA = AC
And AO = BO = CO = R
tp s
e/ un
iv
w
://
Second method: (i) Using the energy conservation Loss in potential energy = gain in kinetic energy Gm1 m2 Gm1 m2 1 1 or − = m1v 12 + m2 v 22 d /2 d 2 2 But m1 = m2 and v1 = v2
)
s
w
3 × 6.67 × 10 × 0.74 × 10 90 × 103 = 1.28 × 107 m/sec
=
29
rio
−11
$ )
er
w
.ja
or
P
) Solution )
v 3 Gm v1 = v 2 = = 2 90 × 103
0 5
P
an
3 Gm 90 × 103
v=2
ar
∴
P
m
1 1 = 2Gm − 3 3 22.5 × 10 90 × 10 2 Gm × 3 = 90 × 103
(ii)
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Gravitation
Let the distance x from the larger block the force is zero. Then GmM 16 MGm = (10a − x) 2 x2
3Gm m M+ R 3
or x2 = 16 (10a – x) or x = 4 (10a – x) or x = 8a From the energy conservation at point P
GMm along AO R2 Force on A due to mass m at Gmm C, F2 = AC 2
Gm 2 along AC 3R2
or
∴ Net force on A = F1 + F2 sin 60° + F2 sin 60°
D
in
s_
s
rio
ta
3 5GM 2 a
si er
e/
un
iv
5. Taking the earth to be a uniform sphere of radius 6400 km and the value of g at the surface to be 10 m/sec2, calculate the total energy needed to raised a satellite of mass 2000 kg to a height 800 m above the earths’s surface and to set it into circular orbit at that altitude.
.m
) Solution )
//t
s:
tp
ht
D
[
GmM 45 a 8 Gm 45 v2 = ⋅ a 4 =
v=
e/ .m
ht
)
) Solution
D±[
∴
//t
s:
tp
4. Distance between the centre of two stars is 10a. The mass of these stars M and 16 M and their radii a and 2a respectively. A body of mass m is fired staright from the surface of the larger star towards the smaller star. What should be its minimum intial speed to reach the surface of the smaller star? Obtain the expression in terms of G, M and a.
D
rio
si er un
iv
//w
s:
tp
ht ∴
w
or
w
Gm Gm v= + R 3R
ta
.ja
or
For circular motion at m1, mv 2 Gm m = 2 M + R R 3
1 2 GmM 64 − 16 + 1 − 4 mv = 2 a 8
an
ar
or
m
Gm2 3 GMm = + 2 2 2 R 3R 2 Gm m m+ 2 R 3
1 2 GmM 16 16 1 1 mv = − + − 2 a 2 8 8 2
a.
1 2 GmM Gm(16M) mv − − 2 8a 2a GmM Gm(16M) =− − 2a 8a
co m
=
=
fo
(ii) Force on A due to mass M at O, F1
(i) Potential energy of the system = – [W0 + WA + WB + WC] =−
2.11
GMm R where, m = mass of the satellite and M = mass of the earth GmM Final energy = (R + h)
Initial energy =
Change in energy = Final energy – Initial energy GmM GmM GmM GmM = −− − = R+h R R R+ h
GM Rh h = GmM = 2 .m R(R + h ) R (R + h)
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2.12
s
rio
si ta
er
un
e/ )
t.m :// tp s
ht
rmin + rmax r + R = 2 2
ht
iv
//t
tp
s:
r=
Now, T 2 =
∴
3
T=π
( r + R )3 2GM
8. A planet A moves along elliptical orbit around the sun. At the moment when it
4π 2 r 3 GM
r+R 4π 2 2 3 2 = π (r + R) = GM 2GM
9.8m −2 × ( 6.4 × 106 m )2 = 7.8 × 106 m ) ( = 7174 ms–1 (iii) Let t = time taken between two successive times at which the satellite is overhead to an observer fixed at a position on the equator. θ1 = Angle turned through by the rotating earth during t seconds.
s_
rio
ita rs
iv e un
e/
.m
tp
ht
1/2
= 7417 sec
7. A planet of mass M moves around the sun along the ellipse, so that its minium distance from the sun is equal to r and the maximum distance is equal to R. Using Kepler’s laws, find its period of revolution around the sun. ) Solution According to the Kepler’s Laws,
1/2
in fo
ar an
m
w .ja
//w
s:
1/2
4π 2 r 3 (i) Period of the satellite, T = GM 1/2 2 4π = 2 r3 gR
1/2
2π Here, TS = ω = Period of the satellite S = 6831 sec. 2π and TE = ω = Period of rotation of E
the earth = 24 × 3600 sec. T ⋅T 24 × 3600 × 6861 t= S E = sec TE − TS 24 × 3600 − 6861 ∴ T ⋅T 24 × 3600 × 6861 t= S E = sec TE − TS 24 × 3600 − 6861
w
)
Radius of the satellite orbit, r = R + h = (640 + 1400) km = 7.8 × 106 m
4π2 × (7.8 × 106 m)3 = −2 6 9.8ms × (6.4 × 10 m) = 6831 sec (ii) Circular orbital speed,
a. co m
= 1.6 × 106 J 6. Assuming the radius of the earth to be 6400 km, calcluate the period of revolution of a satellite which is describing an equatorial orbit at 1400 km above the surface and its speed in its orbit. If the satellite is travelling in the same direction as the rotation of earth, i.e., west to east, what is the interval between two successive times at which it will appear vertical overhead to an observer at a fixed point on the equator?
GM R2 vc = = g r r
6400 × 103 × 800 = 10 × 2000 3 6400 × 10 + 800
) Solution
θ2 = Angle turned through by the revoling satellite during t seconds. θ2 2π θ1 = + Then, θ2 = 2π + θ1 and t t t θ2 = Angular speed of the But, t satellite = ωS θ1 = Angular speed of the earth = ωE t 2π = ωS − ωE ∴ t 1 ω ω 1 1 ∴ = S − E = − t 2π 2π TS TE
Rh Energy needed = g . m R+h GM where, g= 2 R ∴ Energy needed
Gravitation
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2.13
Gravitation
was at the distance r0 from the sun its velocity was equal to v0 and the angle between the radius vector r0 and the velocity vector v0 was equal to α. Find the maximum and minimum distance that will separate this planet from the sun during its orbital motion.
angular momentum L of this planet to the centre of the sun. )
) Solution From the conservation of momentum
1 2 Gs ms m 1 2 Gs ms m mv1 − = mv2 − 2 r1 2 r2
)
) Solution From the conservation of angular
momentum about the sun mv0r0 sin α = mv1r1 = mv2r2 or v1r1 = v2r2 = v0r0 sin α
in
ta r
(r2 + r1 ) 1 = 2Gs ⋅ ms r22 r1 ⋅ r2
er si ta rio
s
v12
iv
v12 = 2Gs ⋅ ms
.m
or
e/ un
//t .
s:
ht tp
2Gs ms v12 r12 2Gs ms = 2 − r1 r2 r2
r2 1 1 v12 1 − 12 = 2Gs ms − r2 r1 r2
si
m
or
e/ un
iv
or
er
w
w
:// w
tp s
2G m −2Gs ms ± 4G s2 ms2 + 4 ( v02 r02 sin 2 α ) v02 − s s r0 r1 = 2G m 2 v02 − s s r0
ht
...(iii)
io s_
m
.ja
.c om
an a
v12 −
or
2 2G m 2 2 2 or v0 − s s r1 + 2G s ms − v0 r0 sin α = 0 r 0 or
mv1 r1 v1 r1 = m r2 r2
From equations (ii) and (iii), we get
ar
v02 Gs ms v02 r02 sin 2 α Gms − = − 2 r0 2r12 r1
From equation (i), v2 =
...(ii)
fo
From equation (i)
2Gs ms 2G m = v22 − s s r1 r2
1 2 G s ms m 1 2 G s ms m mv0 − = mv1 − 2 r0 2 r1
v12 −
or
...(i)
For the conservation of energy
v1 = 2Gs ms
r2 r1 ( r1 + r2 )
r2 r1 (r1 + r2 )
s: //t
From equation (i) Angular momentum = mv1 r1
ht tp
= m 2Gs ms
r1r2 (r1 + r2)
where,
η=
v02 r0 Gs ms
) Solution )
Case I When r < R Let us divide the sphere in two parts by imagining a concentric spherical surface passing through P.
9. A planet of mass m moves along an ellipse around the sun, so that its maximum and minimum distance from the sun are equal to r1 and r2 respectively. Find the
m = mass of sun; and Gs = gravitational constant of sun.
10. There is uniform sphere of mass M and radius R. Find the strength E and potenial V of the gravitational field of this sphere as a function or the distance r from its centre (with r < R and r > R). Draw the approximate plates of the function E (r) and V (r).
1 ± 1 − (2 − η)η sin 2 α = r0 (2 − η)
2 v2 1 ± 1 − v02 r02 sin α − 0 r0 rms = 2 v0 − r0 Gs ms
...(i)
mv1 r1 = mv2 r2 From conservation of energy,
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Gravitation
The inner part of mass 5
GM − R3 r for r ≤ R g= GM − r for r ≥ R R3
Hence, 3 U
G[
4 Mr 3 × πr 3 = 3 4 3 3 R πR 3
M
The potential V due to this inner part GM′ V1 = − r M 3M dm = 4πx 2 dx = 3 x 2dx 4/3πR3 R Gdm 3M d V2 = − = − 3 xdx x R
m
rio si ta
er
Let v1 and v2 be the linear speeds with respect to the planets S1 and S2 and R1, R2 their radii respectively. From the Kepler’s law the square of the time period is proportional to the cube of the radius.
iv
un e/ or
R23 = η2 R13 R2 = η2/3 R1
or
2
R 2 ηT = R1 T
As the time period of S1 is T, therefore
...(i)
2πR1 T 2πR 2 Similarly, v2 = T v1 =
tp
s:
.m
3
∴
2πη2/3 R1 ...(ii) T Clearly, at the closest separation, they are moving in the opposite directions. Hence, the speed of S2 with respect to S1 becomes =
When r > R G dm G dV = − = − ∫ dm r R GM =− r GM gr = 3 r r
Case II
ht
GM = − 3 r R
//t
ht
GM for r ≤ R R
dV 3 2r GM = − 0 − 2 dr 2 3R R
6
s
er
//t
s:
tp
3 r = − 1 − 2 2 3R gr =
) Solution
.m
ht
GMr 2 3GM 2 − (R − r 2 ) R3 2 R3 GM = − 3 ( 3R2 − r 2 ) 2R =−
6
5
iv
un
e/
V = V1 + V2
2
5
si ta
.ja
w
w
...(ii)
tp s:
//w
3 GM 2 2 R − r =− 2 R2
o
s_
R
GM 3GM x2 V2 = ∫ − 3 xdx = − 3 R R 2 r r
)
R
rio
m
ar an
in f
a.
co
...(i)
11. Two satellite S1 and S2 revolve around a planet in coplaner circular orbits in the opposite sense. The period of revolutions are T and ηT respectively. Find the angular speed of S2 as observed by an astronaut in S1 when they are closest to each other.
M′ =
3 r 2 Gm − − 1 for r ≤ R 2 3R3 R V= GM − R for r ≥ R
[
2.14
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2.15
Gravitation
Hence, total field at point P2 g = g1 + g 2
2πR1 ...(iii) (1 + η2/3 ) T At the closest separation, the distance between S1 and S2 is
=
T ( η2/ 3 − 1)
si
ta
rio
( 6a )
2
(ii)
GM 36 a 2
=
At point P2, due to the thin sphere GM g2 = 2 ( 5a )
si
GM e T 2 = 2 4π
1/3
gr 2 T = e2 4π
1/3
10 × ( 6400 × 103 )2 ( 8.64 × 104 )2 = 2 4 3 . 14 ( ) 1/3 GM 10 × ( 6400 × 103 )2 (8.64 × 104 )2 = ≅ 4.28 × 107 m = 6.7 r 25a 2 = 2 e 4 ( 3.14 ) https://t.me/universitarios_infohttps://www.jamarana.com https://t.me/universitarios
Mm M = −Gm e + R − r r r1 = Altitude of geosynchronous orbit 1/3
GM
g1 =
At point P2, due to inner sphere.
er
iv
GM 16 a 2
Here, re = radius of the earth R = distance from the earth to the moon = 60 re M = 11,000 kg, rm = radius of the moon Potential energy of the body = Potential energy of the body due to the earth + Potential energy of the body due to the moon GM e m GM m m =− − r R −r
un
e/
ht tp
Since thin sphere inside the field is zero, so at point P1, gravitational field g=
ta
)
GM GM = 2 (4a ) 16a 2
rio
s
er
iv
ht
g1 =
s: //t .m
)
tp s
) Solution Field at point P1 due to the inner sphere,
(i)
un
) Solution
:// t
ht
.m
2
3
e/
D
3
s_ i
an
ar
m
.ja
w w
:// w
D
tp s
D
nf o
a. co m
12. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a shown in the figure. The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P1 and P2 shown in the figure.
2π ( η2/ 3 + 1)
=
v 2 +v1 R 2 − R1
ω=
Hence, the relative angular speed
GM GM 61 GM + = 36 a 2 25a 2 900 a 2
13. (i) Find an expression for the gravitational potential energy of mass m located at a distance r from the earth’s centre along a line joining the earth and the moon. (ii) Find the work done in boosting the body of mass 11,000 kg from geosynchronous orbit to the moon surface. (iii) Hence or otherwise, find an expression for the net gravitational force and find a point where the force vanishes. Mass of earth = 81 × mass of moon Radius of earth = 3.68 × radius of moon = 6400 km Acceleration due to gravity near the surface of the earth = 10 ms–2 Distance of the moon from the earth = 60 × radius of the earth
...(iv)
R2 – R1 = R1(η2/3 – 1)
v1 + v2 =
≅ 4.28 × 107 m
2.16
U′geo = Potential energy of system at geosynchronous orbit
M Mm = −Gm e + r R − r1 1 GM m m GM e m − R − rm rm
Umoon = −
1 = Work needed to lift the body from geosynchronous orbit to the moon, W = Umoon – Ugeo M Mm M Me = Gm e + − e− r R − r r R − rm 1 m 1
14. A spherical hollow is made in a lead sphere of radius r, such that its surface touches the outside surface of the lead sphere and passed through its centre. The mass of the sphere before hollowing was M. With what force according to the Law of Universal Gravitation will the hollow lead sphere attract small sphere of mass m, which lies at a distance d from the centre of lead sphere on the straight line connecting the centres of the spheres and of the hollow?
Gravitation
a. co
m
G
s
rio
iv
er
si
ta
un
e/
iv
.m
un
//t
e/
s:
or
∴
Mm M = e 2 r2 (R − r )
r M m = (R − r ) me r=
R Me Me + Mm
(60re ) 81 = = 54 rr = 3.46 × 108 m 51 + 1
) Solution Let the mass A is at a distance r from
Force is zero, when
15. A tunnel is dug through the earth along one of its diameters. Show that the body dropped in this tunnel will execute simple harmonic motion about the centre of the earth. Assume that the earth is a sphere of uniform density. )
GM e m GM m m + r2 (R − r ) 2
=−
ht
tp
s:
//t
.m
tp
ht
ta
si
er
w
w
//w
s:
tp
ht
rio
)
.ja
m
ar
s_
an
in
fo
2 M P M M M 5 e e e e = Gm + − − r r1 81( 60re − r1 ) 81 re 60r − e e 3.68 3.68 ) Solution Mass of the hollow sphere 3 4 R π Me Me Me − − M 3 2 = ×M = 81( 60re − r1 ) re 60r − re 4 81 8 e × πR 3 3.68 3.68 3 Force of attract = Due to the bigger sphere – 1 1 1 1 = M g re + − − due to the smaller sphere 6.7 81(53.3) 22 59.7 GMm GmM = − ≈ 6.14 × 1010 J. 2 2 d R 8 d − (iii) Gravitational force lies along the earth2 moon line dU ∴ F=− 1 1 dr = GmM 2 − 2 d d Me Mm R = Gm + 8 d − dr r R −r 2
the centre of sphere
M 4 3 Gm ⋅ πr 3 GmM 4/3πR 3 = r F= 2 R3 r
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Gravitation
Hence it is S.H.M. F = Kr GmM K= R3
$ P
⇒
5
Time period, T = 2π
m K
mR3 R3 = 2π GMm GM R 6400 × 103 = 2π = 2 × 3.14 × g 9.8
The magnitude of this force is proportional to r and its direction is towards centre. Hence, this acts as a restroing force towards the equilibrium position at centre O.
2.17
= 2π
= 84.6 min
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3. P is a point at a distance r from the centre of a solid sphere of radius a. The gravitational potential at P is V. If V is plotted as a function of r, which is the correct curve?
2. A simple pendulum of length one metre is taken to a height R, which is the radius of earth. If the acceleration due to gravity on the surface of earth is π2 m/s2, what is the period of the pendulum at this height?
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2
(a)
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Putting r = 6.4 × 10 m and g0 = 9.8 m/sec , we get T = 84 min 6
l . g where g is acceleration due to gravity on the surface of earth and is given as π2 m/s2. At height R, the acceleration due to gravity, T = 2π
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we get ω =
(b) 2 s (d) 4 s
) Explanation Time period of simple pendulum,
) Explanation Putting ge = 0, in the expression ge = g0 – rω2
q
(a) 1 s (c) 3 s
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1. The period of rotation of the earth so as to make any object weightless on its equator is (a) 84 minutes (b) 74 minutes (c) 64 minutes (d) 54 minutes where M are r are the mass and radius of earth.
m
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2.18
5. Three particles each of mass m are placed at the corners of an equilateral triangle of side d. The potential energy of the systems is
U D
(c) −
3Gm 2 d
Gm 2 d
(d) none of these
) Explanation For the system of two particles, )
(c)
9
(b)
U
2
3Gm2 d
(a)
(b) 9
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2
Gravitation
gravitational energy is given as Gm1m2 r UA = U2 + U23 + U31
U
U=−
U A = −3
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Hence (c) is the correct. –ve sign resembles that the particles are bounded by their mutual gravitational field.
6. Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west and B, from west to east. Which train will exert greater force on the tracks? (a) A (b) B (c) They will exert equal force (d) The mass and the speed of each train must be known to reach a conclusion.
Given g = g′
G 4 3 G 4 πR ρ = 2 × πR′3 (2ρ) R2 3 R′ 3 R R′ = or 2
∴
4 3 3 πR (2ρ)
) Explanation Let, v = speed of each train relative
G (R′) 2
to the earth’s surface, vE = speed of earth’s surface relative to the earth’s axis, vA, vE = speeds of A and B relative to the earth’s axis.
For the planet, g ′ =
)
where ρ is the density of the earth.
GM G 4 3 = πR ρ R 2 R 2 3
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tp
g=
G
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) Explanation We have
P
P
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tp
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4. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of earth R will be R R (b) (a) 2 4 R R (d) (c) 3 8
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−GM 2 2 ) Explanation V = 3 (3a − r ) inside, and 2a −GM = outside the sphere. r Hence (c) is the correct.
GMm 3Gm 2 =− d d
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a.
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2
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Gravitation
Then, vA = vE – v, vB = vE + v
)
) Explanation At infinity, the total energy of the
body is zero. Therefore the total energy of the body just before hitting the planet of P will be zero according to the conservation of energy ⇒ Fp = E∞ = 0
v NA = mg − m R
2 A
⇒
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8. A body of mass m is approaching towards the centre of a hypothetical hollow planet of mass M and radius R. The speed of the body when it passes the centre of the planet through its diametrical chute is
3GMm 8r 2
(d)
(b)
(c)
GMm 8r 2
(a)
GMm 4r 2
GMm 8r 3
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2GM R (d) none of these (b)
GM R (c) Zero (a)
mgr h 2 1 + r
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9. A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is at a distance 3 r from the centre of the sphere as shown in the figure. The gravitational force exerted by the sphere on the ring will be
ve r
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v=
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⇒
h
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=−
na
)
) Explanation Energy required to raise it to a heigh h E1 = change in potential energy
−
Since the force imparted on a particle inside spherical shell is zero, therefore the veloctiy of the particle inside the spherical shell remain constant. Therefore, the body passes the centre 2GM of the planet with same speed v = . R
⇒
Up + Kp = 0
7. The ratio of the energy required to raise a satellite up to a height h above the earth to that required to put it into orbit there is (a) h : R (b) R : h (c) 2h : R (d) R : 2h
⇒
v2 N B = mg − m B R NA > N B
∴
2.19
at a distance
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2.20
Gravitation
GMm 3r r 2 +
=
2 3/2
( 3r )
\
3 Gm 8r 2
E=
3 GMm 8r 2
10. A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of co-ordinates. Two spheres of equal radii 1 unit with their centers at A (–2, 0, 0) and B (2, 0, 0) respectively are taken out of the solid leaving behind spherical cavities as shown in the figure. (a) The gravitational field due to this object at the origin is zero (b) The gravitational potential is the same at all points of the circle y2 + z2 = 36 (c) The gravitational potential is the same at all points on the circle y2 + z2 = 4 (d) All of these
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)
) Explanation Here cavities and mass of the
m
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2. The rotation of the earth about its axis speeds up such that a man on the equator becomes weightless. In such a situation. What would be the duration of one day?
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4. A small body of super dense material, whose mass is twice the mass of the earth but whose size is very small compared to the size of the earth, starts from rest at a
,
(b)
R g
1 R 2π g 1 (c) 2π Rg (d) Rg 2π 3. The energy required to remove a body of mass m from earth’s surface is/are equal to (a) 2mgR (b) mgR (c) –mgR (d) zero
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sphere is symmetrically situated about the origin, therefore gravitational field at the origin of this object is zero. The circle y2 + z2 = 36 has a radius 6 and all points on it are at a distance 6 units from the centre where whole mass of the sphere can be supposed to be connected. Circle is outside the sphere. Situation is similar in the case with y2 + z2 = 4.
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[
OLVED OBJECTIVE TYPE UESTIONS
1. In a gravitational force field a particle is taken from A to B along different paths as shown in the figure. Then (a) work done along path I is more (b) work done along path III is less (c) work done along path IV is more (d) work done along all paths is the same
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∴ Force = M × E =
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Gravitation
F1 r22 = if r1 > R and r2 > R F2 r12
(c)
F1 r1 = if r1 > R and r2 > R F2 r2
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12. If the radius of the earth shrinks by 1%, its mass remaining same, the accelera
11. The ratio of the radius of the earth to that of the moon is 10. The ratio of the acceleration due to gravity on the earth to that on the moon is 6. The ration of the escape velocity from the earth’s surface to that from the moon is (a) 4 (b) 6 (c) 8 (d) 12
8. A spaceship moves in the absence of external force with constant velocity. To change the direction of motion, a jet engine is switched on. If it starts ejecting gas relative to the spaceship and directed at right angle to the spaceship’s motion, then (a) angular velocity will be constant (b) the magnitude of velocity will be change (c) it will follow circular path (d) none of these
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10. A point p lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from and reaches C under gravitational attraction only, if speed at C will be
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7. If E be the energy require to raise a satellite to a height h above the surface of earth and E′ be the energy require to put the same satellite into orbit at that E height, then is equal to E′ 2R 3h (a) (b) h 2R R 2h (c) (d) h R
m
(d)
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6. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be (a) 1/2 h (b) 1 h (c) 2 h (d) 4 h
(b)
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5. Let the acceleration due to gravity be g1 at a height h above the earth’s surface, and g2 at a depth d below the earth’s surface. If g1 = g2, h < < R and d < < R, then (a) h = d (b) h = 2d (c) 2h = d (d) It is not possible for g1 to be equal to g2
4H 3g
(d)
9. The magnitude of the gravitational field at distance r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then F r (a) 1 = 1 if r1 < R and r2 < R F2 r2
H g
2H 3g
(b)
(c)
2H g
(a)
height H < < R above the earth’s surface, and reaches the earth’s surface in time t. Then t is equal to
2.21
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2.22
Gravitation
tion due to gravity on the surface of the earth will (a) decrease by 2% (b) decrease by 0.5% (c) increase by 2% (d) decrease by 0.5% 13. A particle of mass m is placed inside a spherical shell, away from its centre. The mass of the shell is M. The particle will (a) move towards the centre. (b) move away from the centre, towards the nearest wall. (c) move towards the centre if m < M, and away from the centre if m > M. (d) remain stationary
d will increase, v will increase d will increase, v will decrease d will decrease, v will decrease d will decrease, v will increase
17. A sphere of mass M and radius b has a concentric cavity of radius a as shown in figure. The graph which shown variation of gravitational potential V with distance r from the center of sphere is 9
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16. A satellite going round the earth in a circular orbit loses some energy due to a collision. Its speed is V and distance from the earth is d. Then
9 D
(d)
E
U
D
18. A body of mass m is raised to a height h (h << R) from the surface of the earth
E
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tp s:
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E
9
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(b)
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(d) none of these
15. A planet is moving around a sun in circular orbit of radius R. If we increase the velocity of the planet 2 times. Find the path of the planet. (a) ellipse (b) circular (c) parabola (d) hyperbola
V1r12 r2
U
14. A planet is revolving in an elliptical orbit around the sun as shown in figure. What will be the areal velocity? rV Vr (a) 1 1 (b) 2 1 3 q (c)
E
(a)
a. c
(a) (b) (c) (d)
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2.23
Gravitation
rio s_ in fo
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7. (c) 15. (c)
EXERCISE 1
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6. (c) 14. (a)
ita er s
w //w s: tp ht
(
5. (c) 13. (d)
1. Four identical particles, each of mass m, move along the circumference of a circle of radius r under the action of their mutual gravitational attraction. Determine the speed of each particle. Gm 2 2 +1 Ans. 4r
25 9 5 (d) 27 (b)
20. A satellite is revolving around a planet in an elliptical orbit under its gravitational field. It is seen that the linear momentum of the satellite varies with the radius vector as R–1/2. Then the angular momentum of the satellite is proportional to (a) R1/2 (b) R 3/2 (c) R (d) R0
na .c om ar a
4. (a) 12. (c) 20. (d)
m
3. (b) 11. (c) 19. (d)
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2. (a) 10. (b) 18. (c)
w
1. (d) 9. (b) 17. (c)
19. Two satellites A and B of masses m1 and m2 respectively are moving in the same circular orbit of radius R, in the opposite m 5 directions, where, 1 = . Just after a m2 3 perfectly elastic collision between them, the ratio of their kinetic energies will be
ANSWERS
2. If a double star (a system of two stars moving around the centre of inertia of the system due to gravitation) has a mass M and its period of revolution is T, then find the distance between the components of the double star.
&0
3 5 9 (c) 25 (a)
where the acceleration due to gravity is g. The loss in weight is mgh mgh (a) (b) R 2R 2mgh mgh (c) (d) R 4R
1/3 GmT 2 Ans. 2 4π
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P
2.24
8. Find the gravitational attraction due to a semicircular wire of length L and of mass M on a particle of mass m placed at the centre of the semicircular wire.
3. Find the gravitational potential energy of a particle of mass m and a thin uniform rod of mass M and of length L, if they are situated at a distance l apart. Find the force of interaction between them.
2πGMm Ans. L2
GMm GMm L Ans. U= L log e 1 + l ;F = l (L + l )
rio s_
10. A planet P moves in an elliptical orbit round the sun. At the instant when its distance from the sun was r0, its velocity was v0 and the angle between the radius vector r0 and the velocity vector v0 was equal to θ. Find the maximum and the minimum distances of the planet from the sun during its motion.
e/
si
ta
un i
rio
ve
s
rs
ita
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w
w
//w
s:
//t
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ș U
3
6
9
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//t
ht
.m
e/
s:
tp
6. Using the conservation laws, show that the total mechanical energy of a planet of mass m moving in an elliptical orbit round the sun of mass M depends only on its semi-major axis a. Find this energy as a function of a.
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er
9
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tp
ht
2GMr1r2 r1 + r2
Ans.
in fo
co m
an a.
m
ar
5. A planet of mass m moves in an elliptical orbit round the sun of mass M so that its farthest and the nearest distances from the sun are r1 and r2 respectively. Find the angular momentum of the planet about the centre of the sun.
9. A satellite is put in an orbit just above the earth’s atmosphere with a velocity 1.5 times the velocity for a circular orbit at that height. The initial velocity is parallel to the surface. What should be the maximum distance of the satellite from the earth, when it is in the orbit? [Roorkee, 1980] [Ans. 2R + 3h]
4. Consider an earth satellited position so that it appears stationary to an observer on earth and serves the purpose of a fixed relay station for intercontinental transmission of television and other communication. What should be the height at which the satellite should be positioned and what would be its direction of motion? (Radius of earth R = 6400 km) [IIT, 1973] 6 [Ans. 35.94 × 10 m]
Gravitation
tp s:
9
ht
r0 r0 2 rmax = 2 − η 1 + 1 − η(2 − η)sin θ ; rmin = 2 − η 1 − 1 − η( GMm Ans. E = − 2a , where, 2a = r1 + r2 = major axis of the ellipse. r0 r0 2 2 rmax = 2 − η 1 + 1 − η(2 − η)sin θ ; rmin = 2 − η 1 − 1 − η(2 − η)sin θ m , where, 2a = r1 + r2 = major axis ofthe ellipse. 11. What will be the acceleration due to 7. If the radius of the earth were to shrink gravity on the surface of the moon if by one pernocent, its mass remaining the 1 its radius were th the radius of the same, the acceleration due to gravity on 4 1 the earth’s surface would decrease by earth and its mass is th of the mass what amount? [IIT, 1981] 80 of the earth? [Ans. 2%]
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R Ans. 2 gR 2 − 1 v0
17. The astronauts in the satellite orbiting the earth feel weightless. Does the weightlessness depend upon the distance of the satellite from the earth? If so, how? Explain your answer. [Roorkee, 1985] [Ans. The satellite does not exert any force on the astronaut and hence the astronauts in a satellite are in a state of weightlessness. This state of weightlessness does not depend upon the distance of the satellite from the earth.]
12. A planet of mass m moves towards the sun with a velocity v0 (when far away from the sun) and directed along a line whose perpendicular distance from the sun is d. Assuming the mass of the sun to be M, find the minimum distance of the planet from the sun.
2.25
R and acceleration due to gravity on its surface is g.
What will be the escape velocity on the surface of the moon if it is 11.2 km/s on the surface of the earth? [Roorkee, 1983] [Ans. 2.5 km/s]
Gravitation
19. A satellite of the earth has to move in an equatorial plane close to the surface of the earth either in the direction of rotation of the earth or against it. Calculate by how many times the kinetic energy of the satellite in the latter case exceeds that in the former. [Ans. 1.27] 20. Two satellites S1 and S2 revolve round a planet in coplanar circular orbit in the same sense. Their periods of revolution are 1 hour and 8 hour respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1, find: [IIT, 1986] (a) the speed of S2 relative to S1 (b) the angular speed of S2 actually observed by an astronaut in S1. [Ans. (a) 3.14 × 104 km/hour. (b) – 2.907 × 10–4 rad/sec]
21. A solid sphere has a mass M and radius R. A spherical cavity is made in it in such
16. On the north pole of the earth a body is thrown vertically upward with a velocity v0. Find the maximum height attained by the body if the radius of the earth is
18. Find the rate at which the earth should rotate about its axis so that the apparent value of acceleration due to gravity at the equator becomes zero. What will be the length of the day in this case? [Ans. 1.41 hour]
15. If a planet of mass m be suddenly stopped in its circular orbit of radius r, show that it would fall into the sun in a time ( 2 / 8), times the period of revolution of the planet.
ht
ht
14. Imagine a planet having mass and radius half of the corresponding quantities for earth. The surface temperature of this planet reaches upto 800 K. Estimate whether there is a possibility of presence of oxygen molecule in the planet’s atmosphere. Given escape velocity from earth = 11.2 km/second, Boltzman constant k = 1.38 × 10–23 J/K, mass of a molecule of oxygen = 5.3 × 10–26 kg. [Roorkee, 1984] [Ans. There is possibility of oxygen molecule on the surface of the planet as the molecule cannot escape from its surface.]
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tp s
3 Gm 2 Ans. 2 a
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13. Three identical particles, each of mass m, are placed at the three vertices of an equilateral triangle ABC of side a. Find the work required to be done to increase the distance between the particles from a to 2a.
v4d 2 GM Ans. 2 1 + 02 2 − 1 v0 G M
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2.26
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iv
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un
t.m e/
ht tp s:
//t
ht
er si ta rio s
un iv
w
//w
tp
s:
22. An artificial satellite of moon revolves in a circular orbit whose radius exceeds the radius of the moon n times. Assuming that the satellite experiences a resistive force due to cosmic dust that depends upon the velocity v of the satellite as F = kv2, where k is a constant, Find how long the satellite will stay in the orbit until it falls on to the moon’s surface?
26. Three particles, each of mass m are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle move in a circle while maintaining the original separation a. Find the initial velocity that should be given to each particle and also the time period of circular motion. [IIT, 1988; MNR, 1990]
er s
w
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m
1 Ans. F = GMm 1 − 2 r2 R 8 1 − 2r
rio s_ in fo
na .c om ar a
U
25. Find the distance of a point from the centre of the moon at which the strength of the earth’s and the moon’s gravitational fields are equal and opposite and the resultant field is equal to zero. Take the mass of the earth to be η = 81 times greater than that of the moon and the distance between the centres of these planets to be n = 60 times greater than the radius of the earth R. [Ans. 3.84 × 104 km]
3
2c
2
24. A space-ship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the space-ship in the orbit to overcome the gravitational pull? (Radius of the earth = 6400 km, g = 9.8 m/s2) [Roorkee, 1988] [Ans. 3.2804 km/s]
a way that the boundary of the cavity touches the surface of the sphere and passes through its centre, as shown in the figure. Find an expression for the gravitational force on a particle of mass m at a distance r from O along the line joining the centres.
ita
Gravitation
[Ans. 7.826 × 10–4 rad/sec]
28. The mass M of a planet-earth is uniformly distributed over a spherical volume of radius R. Calculate the energy
much as present. Take the equatorial radius as 6400 km. [Roorkee, 1988]
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s: //
27. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape m ( n − 1) moon’s from sufacethe earth; Ans. Time after which the satellite will fall on to thevelocity k height gR of the satellite (a) Determine the above the earth’s surface. m ( n − 1) which the satellite will fall on to the moon’s suface (b) If the satellite is stopped suddenly in k gR its orbit and allowed to fall freely into the earth, find the speed with which it 23. Determine the speed with which the earth hits the surface of the earth. has to rotate on its axes so that a person [IIT, 1990] 3 on the equator would weigh th as [Ans. (a) h = 6400 km, (b) v = 7.919 km/s] 5
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Gravitation
astronaut fall on a straight-line from the earth’s centre. The radius of the earth is 6400 km. [Roorkee, 1998] –2 [Ans. 3 × 10 N] 30. A body is projected vertically upwards from the bottom of a crater of Moon R , where R is the radius of of depth 100 Moon with a velocity equal to the escape velocity on the surface of the Moon. Calculate maximum height attained by the body from the surface of the Moon. [Ans. 99.50 R]
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s:
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s:
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.m
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s:
un
iv
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w w
er si
ta
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m
rio
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s_
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in
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fo
co
m
29. A cord of length 64 m is used to connect a 100 kg astronaut to a space-ship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth’s surface. Also assume that the space-ship and the
needed to deassemble the planet against the gravitational pull amongst its constituent particles. Given, MR = 2.5 × 1031 kg. m and g = 10 m/s2. [IIT, 1992] [Ans. 1.5 × 1032 joules]
2.27
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Gravitation
UESTION BANK
1. Can a satellite coast in a stable orbit in a plane not passing through the earth’s centre? Explain your answer.
6. Does the escape speed of a body from the earth depend on (a) the mass of the body (b) the location from where it is projected (c) the direction of projection (d) the height of the location from where the body is launched?
Q
CONCEPTUAL UESTIONS
Q
2.28
2
7. An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 ms–2. What is the acceleration of the astronaut, the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.)
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5. A satellite in low earth orbit experiences a small drag force from the earth’s atmosphere. What happens to its speed because of this drag force?
3. Explain why the moon has no atmosphere.
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(c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
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9. The escape velocity of a planet is Ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be (a) ve (b) ve / 2 (c) ve/2 (d) zero
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10. A (non-rotating) star collapses onto itself from an initial radius Ri with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration ag on the surface of the star as a function of the radius of the star during the collapse? DJ
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(b) 1.5 E0 (d) E0
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23. A particle is projected vertically upward from the surface of the earth with a speed 3 gR, R being the radius of the earth of 2 and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance) (a) 3R/4 (b) R (c) 2R (d) 3R
24. A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit of radius r + ∆ r (∆r << r) with time period T + ∆ T then, ∆T 2 ∆r ∆T 3 ∆r = = (b) (a) T 3 r T 2 r ∆T ∆r ∆T ∆r = =− (d) (c) T r T r
19. A satellite can be in a geostationary orbit around earth at a distance r from the centre. If the angular velocity of earth about its axis doubles, a satellite can now be in a
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21. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be (a) 1/2 h (b) 1 h (c) 2 h (d) 4 h
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17. Three identical stars, each of mass M, form an equilateral triangle (stars are positioned at the corners) that rotates around the centre of the triangle. The system is isolated and edge length of the triangle is L. The amount of work done, that is required to dismantle the system is 3GM 2 3GM 2 (a) (b) 2L L 2 3GM GM 2 (c) (d) 4L 2L
20. A planet of mass m is in an elliptical orbit about the sun (m << Msun) with an orbit period T. If A be the area of orbit, then its angular momentum would be (a) 2 mA/T (b) mAT (c) mA/2T (d) 2 mAT
16. A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be (a) zero 3GM (b) R
15. Two point masses of mass 4 m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be (a) 1 : 4 (b) 1 : 5 (c) 1 : 1 (d) 1 : 2
geostationary orbit around earth if its distance from the centre is r r (b) (a) 2 2 2 r r (c) (d) 1/3 (4) (2)1/3
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25. A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is – (opposite to the centre of gravity) [Here g = GM/R2, where M is the mass of the sphere] mg 3mg (a) (b) 2 8 mg (c) (d) None of these 16
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1. Assuming the earth to be a sphere of uniform density the acceleration due to gravity (a) at a point outside the earth is inversely proportional to the square of its distance from the centre (b) at a point outside the earth is inversely proportional to its distance from the centre (c) at a point inside is zero (d) at a point inside is proportional to its distance from the centre.
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3. A double star is a system of two stars of masses m and 2 m, rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars then their time
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30. The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is M s r2 (a) × 100 M m r1
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26. A satellite is launched in the equatorial plane in such a way that it can transmit signals upto 60º latitude on the earth. The angular velocity of the satellite is GM GM (a) (b) 8R3 2 R3 (c)
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Gravitation
earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall and the acceleration of the particle varies with x (distance of the particle from the centre) according to
4. Mark the correct statement/s (a) Gravitational potential at curvature centre of a thin hemispherical shell of radius R and mass M is equal to GM/R. (b) Gravitational field strength at a point lying on the axis of a thin, uniform circu
period of rotation about their centre of mass will be proportional to (a) r3/2 (b) r (c) m1/2 (d) m–1/2
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9. Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start under their mutual gravitational attraction (a) acceleration of m1 is more than that of m2. (b) acceleration of m2 is more than that of m1. (c) centre of mass of system will remain at rest in all the reference frame (d) total energy of system remains constant
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
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Direction: Each question contains statement 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c), and (d) out of which only one is correct.
1. Statement 1: Space rockets are usually launched in the equatorial line from west to east.
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14. Inside a uniform spherical shell (a) the gravitational potential is zero (b) the gravitational field is zero (c) the gravitational potential is same everywhere (d) the gravitational field is same everywhere
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13. A planet is revolving round the sun. Its distance from the sun at Apogee is rA and that at Perigee is rp. The mass of planet and sun is m and M respectively, vA and vP is the velocity of planet at Apogee and Perigee respectively and T is the time period of revolution of planet round the sun. π2 (a) T 2 = (rA + rP )3 2Gm
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(c) its angular momentum about the earth decreases (d) its period of revolution around the earth increases
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Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II. 1. Two identical spacecrafts, each with a mass of 3250 kg, are in the same circular orbit at a height of 270 km above the earth’s surface. Spacecraft A leads spacecraft B by 105 s, that is, A arrives at any fixed point 105 s before B. At a particular point P, the pilot of B fires a short rocket burst in the forward direction, reducing the speed of B by 0.95%. The orbital parameters of B will change after the “burn”.
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7. Statement 1: Two satellites A and B are in the same orbit around the earth, B being behind A. B cannot overtake A by increasing its speed. Statement 2: It will then turn into a different orbit.
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6. Statement 1: Comets move around the Sun in elliptical orbits. The gravitational force on the comet due to Sun is not normal to the comet’s velocity, but the work done by the gravitational force over every complete orbit of the comet is zero. Statement 2: Gravitational force is a conservative force and the work done by a conservative force over a closed path is always zero.
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11. Statement 1: Escape velocity is independent of the angle of projection.
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5. Statement 1: If an object is projected from earth’s surface with escape velocity, path of object will be parabola. Statement 2: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity. Path of object will be ellipse.
9. Statement 1: For the planets orbiting around the sun, angular speed, linear speed and K.E. changes with time, but angular momentum remains constant. Statement 2: No torque is acting on the rotating planet. So its angular momentum is constant.
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4. Statement 1: Consider a planet in an elliptic orbit around sun. Its kinetic energy changes with time and angular momentum remains constant. Statement 2: Gravitational force is a central force. As a result, no torque acts on the planet about the position of sun and so angular momentum remains conserved.
3. Statement 1: If earth suddenly stops rotating about its axis, then the value of acceleration due to gravity will become same at all the places. Statement 2: The value of acceleration due to gravity is independent of rotation of the earth.
8. Statement 1: The escape speed does not depend on the direction in which the projectile is fired. Statement 2: Attaining the escape speed is easier if a projectile is fired in the direction, the launch site is moving as the earth rotates about its axis.
2. Statement 1: In a free fall, weight of a body becomes effectively zero. Statement 2: Acceleration due to gravity acting on a body having free fall is zero.
Statement 2: The acceleration due to gravity is minimum at the equator.
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3. Assume v0 be the velocity of projection of a body from the surface of the earth enables it to become a satellite of the earth close to the earth surface. If R be the radius of GM the earth then v0 = gR where g = 2 ; R G – universal gravitational constant and M is the mass of the earth. If ve be the escape v 1 . If v be velocity on the earth then 0 = ve 2 the velocity of projection of body on the surface of earth then, match the following
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4. A satellite of mass m is orbiting the Earth at a height h from its surface. [M is mass of the earth and its radius is R] Column I Column II −GMm (a) Kinetic energy of the (p) ( R + h) satellite 1 GMm (b) Potential energy of (q) 2 ( R + h) the satellite −GMm (c) Total energy of the (r) 2( R + h) satellite GMm (d) The energy must be (s) ( Rr + h) spent to pull the satellite out of the earth’s gravitational field.
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2. A satellite is revolving round the earth in a circular orbit of radius ‘a’ with velocity v0. A particle is projected from the satellite in forward direction with relative velocity 5 v= − 1 v0 . During subsequent motion 4 of particle, match the following Column I
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Column I Time period of B before the burn (in sec) Total energy of B after the burn (in J) Semi major axis of B after the burn (in km) Velocity of B before the burn (in m/s)
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(d) A satellite of mass m is revolving around planet of mass M its total energy is given GMm as − then it 2r0 may follow
The ratio of the gravitational force on A and B is the ratio of their gravitational masses; that is, FAC m′A = FBC mB′
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The word mass has been used in two quite different experimental situations. For example, if we try to push a block that is at rest on a horizontal frictionless surface, we notice that it requires some effort to move it. The block seems to be inert and tends to stay at rest, or if it is moving it tends to keep moving. Gravity does not enter here at all. It would take the same effort to accelerate the block in gravity free space. It is the mass of the block which makes it necessary to exert a force to change its motion. This is the mass occurring in F = ma in our original experiments in dynamics. We call this mass m the inertial mass. Now, there is a different situation which involves the mass of the block. For example, it requires effort just to hold the block up in the air at rest above the earth. If we do not support it, the block will fall to the earth with accelerated motion. The force required to hold up the block is equal in magnitude to the force of gravitational attraction between it and the earth. Here inertia plays no role whatever; the property of material bodies, that they are attracted to other objects such as the earth, does play a role. m′M e The force is given by: F = G , Re2 When m’ is the gravitational mass of the block. Consider two particles A and B of gravitational masses m'A and m'B acted on by a third particle C of gravitational mass m'C. Let the third particle be at an equal distance r from the other two. Then, the gravitational force exerted on A by C
Therefore, the law of universal gravitation contains within it the result that the weights of various bodies, at the same place on the earth, are exactly proportional to their gravitational masses. Now, suppose we measure the inertial masses mA and mB of the particles A and B by dynamical experiments. Having done this, we then let these particles fall to the earth from a given place and measure their accelerations. We find experimentally that objects of different inertial masses all fall with the same acceleration g arising from the earth’s gravitational pull. But the earth’s gravitational pulls on these bodies are their weights, so that using the second law of motion we obtain. W m WA = mA g , WB = mB g or A = A WB mB
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3. If inertial mass m and gravitational mass m' of body are not same then equation of motion for freely falling bodies will have to be modified. Suppose a body falls with initial velocity zero by a height h then speed of it at any instant will be given by V = k 2gh here k is a constant which will be equal to
4. For a spring mass system if time period of oscillation on a horizontal frictionless surface be T then the same for the oscillation in vertical plane of the aforementioned system will be (a) ( m′/m′ )T (b) T ( m /m′ ) (c) T (d) T (m'/m)
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7. If the two have same mass (a) A would have more potential energy than B while on their initial circular paths (b) A would have more kinetic energy than B while on their initial circular paths (c) Relative to Earth’s centre, angular momentum of A when it is in elliptical path would be less than angular momentum of B (d) During the whole process angular momentum of B would be more than angular momentum of A
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While we’re on the subject of space travel, let’s talk about free fall and something that is misleadingly called “weightlessness”. As you are reading this, you can probably feel your chair pushing upwards on you with a force of several hundred newtons. If your feet are not touching the ground this is an upwards force equal in magnitude to your weight (a downwards force). My weight is 680 N downwards, so I know that the force from the chair is about this much, upwards, You can also feel your abdominal muscles holding your abdominal organs in place. �����������������������������������������
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Two satellites A and B are revolving around the earth in circular orbits of radius r1 and r2 respectively with r1 < r2. Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. Focal lengths of the elliptical path are r1 and r2 respectively. At position 2, A is given another impulse so that velocities of A and B at 2 become equal and the two move together. For any elliptical path of the satellite time period of revolution is given by Kepler’s planetary law as T2 ∝ r3 where, ‘a’ is semi major axis of the r +r ellipse which is 1 2 in this case. 2
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Also angular momentum of any satellite revolving around the Earth will remain a constant about Earth’s centre as force of gravity on the satellite which keeps it in elliptical path is along its position vector relative to the earth centre.
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11. This passage refers to weight that is felt by an observe if he himself moves under different circumstances. In this context of
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In an orbiting spacecraft, the spacecraft and the cosmonaut are both accelerating towards the centre of the earth at the same rate (their centripetal acceleration is ac = v2/r, where v is the orbital speed and r the radius). Their weight is what keeps them in orbit: W = mac. Because they are both accelerating towards the centre of the earth at the same rate, there is on average no force between the cosmonaut and the spacecraft. This absence of forces from seat, floor, abdominal wall etc. is what is commonly but misleadingly called ‘weightlessness’: the cosmonauts in the space station are not without weight, in fact they have (almost) their usual weight. It’s just that they donot feel the force of chairs on their bums and they donot feel their abdomens holding in their organs. If the cable of a lift (elevator) has broken, both the lift and its passenger are in free fall, accelerating downwards at g. The passenger no longer feels the force on his abdominal muscles, and so, like the cosmonaut, he might say that he feels ‘weightless’. He is not, of course, without weights. His weight is still W = mg. A NASA airplane (nicknamed the ‘vomit comet’) cuts the power in its engines and, for about 25 seconds, travels in a trajectory that is nearly parabolic. Both the plane and its occupants accelerate towards the Earth at g: all are in free fall. Astronauts are thus exposed to free fall and ob-
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tain brief periods of experience in working in this conditions. Some physicists tend to use the word ‘weightless’ in scare quotes to make it clear that they are not talking about a situation in which there is no weight. Many physicists prefer to avoid the word altogether and talk instead about free fall. There are some similarities between the passenger (mass m) in the lift (let’s put it at the equator) and a cosmonaut (mass m) in low Earth orbit. The weight of each is about mg. Both accelerate towards the centre of the Earth at approximately g. The difference is that the spacecraft makes a circle around the Earth in about 90 minutes, whereas, the lift makes a circle around the Earth in about 24 hours. The acceleration g is just enough to keep an object in low Earth orbit with a period of 90 minutes. It is far too great for the ‘orbit’ of the helpless passenger in the lift. If a satellite loses speed, it gradually spirals in towards the Earth. The horizontal speed of the passenger in the lift is so low that his ‘spiral’ towards the centre of the Earth is almost a straight line.
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These force and some others give you the sensation of having weight. You do not readily sense your weight directly very much, because it is applied homogeneously over your whole body. When the forces from the chair or on your abdominal wall are reduced or zero, you may feel ‘weightless’ the feeling you get when a lift starts to accelerated rapidly downwards, or when you go quickly over a peak on a roller coaster.
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12. In case II in which man is in free fall with the lift, suddenly makes a jump upward with velocity 5 m/s relative to the lift which is of height 10 m and then collide with the ceiling elastically and again hits the bottom of the ceiling and sticks to it. For masses of the lift and man being same, how much time he spent in the air (a) 4 seconds (b) 2 seconds (c) 8 seconds (d) None of these
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14. Two commercial satellites, Seeker I and Seeker II, have circular orbits of R and 2 R, respectively, about the same planet. What is the orbital velocity of Seeker II if the orbital velocity of Seeker I is v? (a) v/2 (b) v/(2)1/2 1/2 (c) v(2) (d) 2v 15. Two boxes of negligible mass are placed 2 metres apart on the surface of Planet X. Initially 15 identical steel ball bearings are placed in each box. Which statement is most accurate if 10 of the ball bearings are transferred from box 1 to box 2? (a) The force of gravity between the two boxes remains the same
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(a) Newton’s second law, F = ma, applies (b) Newton’s law of universal gravitation states that every body attracts every other body with a force of magnitude Fgravity = GMm/r2, where M and m are the masses of two interacting bodies, r is the distance between
An enterprizing physics student has developed a software package that calculates gravity-based parameters for actual and theoretical planets and satellites. The program is based on the following simple principles and approximations
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Since, the more massive planet is stationary, EK depends only on the motion of the satellite. The potential gravitational energy increases from a large negative value to zero as the distance increases to infinity. The speed needed for a projectile to completely escape the pull of the planet is called the escape velocity. A projectile is completely beyond the gravity of a planet has a total mechanical energy of zero with respect to that planet. Conservation of energy requires that the total mechanical energy of the projectile when it was at the planet’s surface must also equal to zero.
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13. In the case of vomit comet, the airplane remains in the parabolic path for 25 seconds and again comes to the same altitude. At the new position it will be (a) Moving with speed with its nose down as against initial situation. (b) Moving with same speed with its nose up as in the initial case irrespective of the movement of the passengers inside the plane. (c) Orientation of the plane at t = 25 s will depend upon movement of the passengers inside the plane. (d) None of these
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their centers of mass, and G is the gravitational constants. For a pair of interacting bodies if one is significantly more massive than the other, the heavier body is considered stationary with the smaller body orbiting around it. The orbit of the smaller body is a circle. The smaller body experiences a centripetal force. Fcentripetal = mv2/r, directed towards the centre of the more massive body. For bodies on or near the surface of a planet, r is simply the planet’s radius, R. Planets are spheres with volume, V = (4/3) πR3. The total mechanical energy of a satellite is the sum of its kinetic and gravitational potential energy E = EK + EP = mv2/2 – GMm/r = constant
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weight. If he is moving in a circle whose radius is 10 m with a constant speed of 10 m/s and plane of the circle is horizontal, how much weight will he feel if his mass is 50 kg [Take g = 10 m/s2] (a) 500 N (b) 1000 N 500 (c) 500 2 N (d) N 2
Gravitation
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= 5.4 × 109 kg distributed uniformly over its circumference. A highly dense particle of mass m2 = 6 × 108 kg is placed on the axis of the ring at a distance x0 = 3 m from the centre. Neglecting all other forces, except mutual gravitational interaction of the two, calculate (a) displacement of the ring when particle is closest to it (b) speed of the particle at that instant.
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6. A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is M and radius of the earth is R.
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7. A small body is dropped into a narrow channel drilled along the Earth axis. Considering the Earth to be a homogeneous sphere and disregarding air resistance. Find (a) the magnitude of the body’s acceleration a(r) as a function of r, (b) the magnitude of the body’s speed v(r) as a function of r, (c) the body’s speed v(0) at the instant when it reaches the centre of the Earth. Compare v (0) with the orbital velocity and (d) the time T after which the body returns to the initial point.
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1. Given a thin homogeneous disk of radius a and mass m1. A particle of mass m2 is placed at a distance ℓ from the disk on its axis of symmetry. Initially both are motionless in free space but they ultimately collide because of gravitational attraction. Find the relative velocity at the time of collision. Assume a << ℓ.
4. A pendulum beats seconds on the surface of the Earth. Calculate as to how much it loses or gains per day if it is taken to: (a) a mine 8 km below, (b) a point 8 km above, the surface. (Radius of the Earth = 6.4 × 106 m)
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19. What is the minimum velocity required to launch a rocket from the surface of Planet Z ? The planet has a mass of M and a radius of R? (a) (2GM/R2)1/2 (b) (2GM/R)1/2 (c) (GM/R2)1/2 (d) (GM/R)1/2
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18. Which expression is an accurate description of the total mechanical energy for a satellite of mass m in a circular orbit of radius r around a planet of mass M? (a) GMm/r (b) –GMm/r (c) GMm/2r (d) –GMm/2r2
SUBJECTIVE UESTIONS
3. A particle is fired vertically upwards from the surface of Earth and reaches a height 6400 km. Find the initial velocity of the particle if radius of the Earth R = 6400 km and g at the surface is 10 m/s2.
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17. Which expression is an accurate description of the kinetic energy for a satellite of mass m in a circular orbit of radius r around a planet of mass M? (a) 2GMm/r (b) GMm/r (c) GMm/2 r (d) GM/r2
16. Planet Y has density D, and surface gravitational acceleration, g. The radius of Planet Y is suddenly doubled while its density remains the same. Compared to the original g, the new value of the surface gravitational acceleration would be, which of the following? (a) 4 g (b) 2 g (c) g (d) g/2
(b) The force of gravity between the two boxes remains the same (c) The force of gravity between the two boxes increases (d) Any change in Fgravity between the boxes cannot be determined without knowing the mass of the steel ball bearings
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Gravitation
8. A body on the equator of a planet weights half its weight at the pole. The density of matter of the planet is 3 g/cm2, Determine the period of rotation of the planet about its axis. 9. Find the potential energy of the gravitational interaction of a mass point of mass m and a thin uniform rod of mass M and length ℓ, if they are located along a straight line at a distance a from each other, also find the force of their interaction.
16. The masses of two stars are m1 and m2, and their separation is ℓ. Determine the period T of their revolution in circular orbits about a common centre.
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18. The radius of Neptune’s orbit is 30 times the radius of the Earth’s orbit. Determine the period TN of revolution of Neptune around the sun.
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12. A satellite of mass Ms is orbiting the earth in a circular orbit of radius r. It starts losing energy at a constant rate C due to friction. Show that the satellite falls to the earth in GM e M s 1 1 a time t given by t = − 2C Re r where Re and Me are the radius and mass of earth.
17. A meteorite approaching a planet of mass M (in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in the circular trajectory of radius R. The mass of the station is ten times as large as the mass of the meteorite. As a result of collision, the meteorite sticks in the station which goes over to a new orbit with the minimum distance R/2 from the planet. Determine the velocity u of the meteorite before the collision.
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11. A synchronous satellite is one whose velocity in its circular orbit allows it to remain above the same position on the rotating earth’s surface. Calculate the required distance H above the earth’s surface to the satellite. Locate the position of the orbital plane of the satellite and calculate the angular range β of longitude on the surface of the earth for which there is a direct line of sight to the satellite.
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10. What is the minimum work that has to be performed to bring a spaceship of mass m = 2 × 103 kg from the surface of the Earth to the Moon?
15. A body is projected vertically upwards from the surface of earth with a velocity, sufficient to carry it to infinity. Find n if the time taken by it to reach a height of thrice the radius of earth (Take radius of earth = R and acceleration due to gravity as g) is (n/3) ( 2 R /g ).
the satellite as F = αv2, where, α is a constant, find how long the satellite will stay in orbit until it falls onto the Moon’s surface.
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(b) 129 (d) 730
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2. Distance between the centres of two stars is 10a. The masses of these stars are M and 16 M and their radii a and 2a respectively. A body of mass m is fired at night from the surface of the larger star towards the smaller star. What should be its minimum initial speed to reach the surface of the
14. An artificial satellite of the Moon revolves in a circular orbit whose radius exceeds the radius of the Moon η times. In the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming the resistance force to depend on the velocity of
1. If the distance between the earth and the sun were half its present value, the number of days in a year would have been [1996]
13. A planet of mass M moves around the Sun along an ellipse so that its minimum distance from the Sun is equal to r and the maximum distance to R. Making use of Kepler’s laws, find its period of revolution and angular momentum around the Sun.
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PREVIOUS YEARS’ IIT-JEE UESTIONS
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3. An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) E0. Its potential energy is [1997] (a) –E0 (b) 1.5 E0 (c) 2E0 (d) E0
(b) TA > TB (if mA > mB) 2
T r (c) A = A TB rB (d) TA = TB
8. A spherically symmetric gravitational system of particles has a mass density ρ for r ≤ R ρ= 0 where ρ0 is a con 0 for r > R stant. A test mass can undergo circular motion under the influence o the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by [2008]
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(b)
(a) W1 = W2 = W3 (c) W1 = W2 > W3
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7. A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then [2006] (a) TA > TB (if rA > rB)
4. A cord of length 64 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. [1998]
smaller star? Obtain the expression in terms of G, M and a. [1996]
Gravitation
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9. Statement 1: An astronaut in an orbiting space station above the earth experiences weightlessness. because Statement 2: An object moving the earth under the influence of Earth’s gravitational force is in a state of “free-fall”. [2008] (a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. [2008]
Gravitation
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7. (b, c) 14. (b, c, d)
5. (a, b, c, d) 6. (a, c) 12. (a, c) 13. (b, c, d)
4 (b, c) 11 (b, c, d)
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2. (c) 10. (a)
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Gravitation
2.45
Q
MATCH THE COLUMN TYPE UESTIONS 1. (a) → s, (b) → r, (c) → r, (d) → q
2. (a) → p, (b) → s, (c) → r
3. (a) → p, (b) → q, (c) → r, (d) → s
4. (a) → q, (b) → p, (c) → r, (d) → q
5. (a) → p, q (b) → p, q, r, (c) → r, s (d) → r, s
Q
PASSAGE BASED UESTIONS 2. (c) 10. (c) 18. (d)
3. (b) 11. (c) 19. (d)
4. (c) 12. (a)
5. (b) 13. (c)
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6. (b) 14. (b)
7. (b, c) 15. (c)
8. (c) 16. (b)
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5. As due to air drag some mechanical energy of the satellite will be converted into heat energy, there will be loss of ME of the satellite, i.e., E = – (GMm/2 r) will decrease.
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This in turn implies that the radius of the orbit will decrease and so the satellite will follow a spiral path towards the earth as shown in figure. Now as in case of satellite motion, GMm GMm PE = and KE = r 2r So with decrease in r, PE will decrease but KE and hence, speed will increase. Increase in speed (or KE) inspite of decrease in mechanical energy is accounted by the fact that the rate at which PE decreases (due to decrease in r) is more than the rate at which mechanical energy decreases.
7. Since, there are no nearby stars to exert gravitational force on him and the small
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1. In case of satellite motion centripetal force is provided by gravitation between the earth and the satellite. If earth is not at the centre of the orbit, i.e., if the plane of the orbit is not passing through the centre of earth, the gravitational force F will not be directed towards the centre of the orbit O’ and hence, the orbit will not be stable. The component of gravitational force perpendicular to the plane of orbit will pull the satellite till its plane passes through the centre of the earth as shown in figure.
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CONCEPTUAL UESTIONS
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2.46
Gravitation
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V M RA = VA R RB
=
2
σ4π( RA2 + RB2 ) RA = 2 2 1/2 ( RA + RB ) σ4πRA2
ωR d 2. (a) g 1 − = g − ω2 R; d = g R
3. (d) The acceleration due to gravity at earth’s surface is g and at a distance R from earth’s surface it is g/4. Hence,
=
7. (a) V = ωR g = g0 – ω2R [g = at equator, g0 = at poles] g0 = g 0 − ω2 R; 2 g g R ω2 R = 0 ; V 2 = 0 2 2 Ve = 2 g 0 R = 4V 2 = 2V
4 Dd πR 3 GM 3 4. (b) g = 2 = [d = density] R R2 4 d = Gd πR 3
T1 = 2 [ T = 2π 1/g ] T2
RA2 + RB2 5 = RA 3
s: //t .m
%
[
σ4πR 2 = σ4π( RA2 + RB2 )
ht tp
2 U
s
si iv
un
e/
s: tp ht GT
T
2
VA RA 3 4 = = then RB = RA VB RB 4 3 for new shell of mass M and radius R M = M A + M B = σ4πRA2 + σ4πRB2
GP
(
σ4πRA2 RB RA = σ4πRB2 RA RB
Given
.m //t
ht
\
=
er
w
w
2Gλ ˆ 2GM ˆ 2Gmπ ˆ j= j= j (along y-axis) r r 2
tp
=
π π Gλ ˆ ˆj sin θ i cos θ + ∫ ∫ r 0 0
s: //w
E=
B
s_
A
m
Gλrd 0 ∫0 dE = ∫0 r 2 (iˆ cos θ + ˆj sin θ)
an
π
ar
π
in
a.
co m
Since, g is constant and R increased by factor 5, d decreases by factor 1/5. M G m GMn 5. (a) Change in force of gravity = 2 − 32 R R M 8. (a) path 1 decreased E (more negative) GMn G m 3 (only due to mass M/3 due to gives decreased semi major axis “a”; = 2 − R R2 (b) less decreased “a” gives decreased T. shell gravitational field is zero (inside the shell)) ONLY ONE OPTION IS CORRECT 2GMm m = 1. (d) Let mass per unit light of wire, λ = R2 6. (c) M A = σ4πRA2 , M B = σ4πRB2 , where, σ and πr = , r = . mass of element, dm = π is surface density Gdm −GMA GMB λrdθ then dE = 2 VA = , VB = r R R
spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.
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2.47
fo
tp s:
ht
P
2
G 3m Gm + cos30° 3d 2 4d
G &0
P 2
1 d 1 4mω2 4× KEm 2 5 = 25 = 1 = 2 16 4 KEm 1 4d mω2 25 2 5
16. (d) Gravitational potential at P,
G 3m 2 cos 60° 12d 2
GMm = 2 E0 . r
15. (a) Due to mutual force of attraction they will revolve about centre of mass with same ω.
In vertical direction,
+
io s
ta r
si
er
iv
un e/
.m
Therefore, P.E. = −
Gm 2 Gm 2 = − =0 8d 2 8d 2
2
GMm GMm GMm − − 2r r 2r
G
Gm 2 cos 60° 4d 2
Net force =
...(ii)
ta si er
iv un
e/
E0 =
//t
−
G 2mm cos30° 12d 2
1 2 GMm mv = 2 2r
or
1 2 mv in equation 2 (i) from equation (ii), we get
//t
s:
tp
ht
Net force =
mv 2 GMm = 2 r r
Substituting the value of
.m
tp
ht
12. (c) In horizontal direction
in
s_
rio
ar
w .ja
//w w
s:
Further;
If the body is to escape, the velocity at surface of earth is 2 gR . If E’ is the kinetic energy corresponding to this velocity then 1 E ′ = m( 2 gR ) 2 ⇒ E ′ = 2 E 2
14. (c) Total energy = kinetic energy + Potential energy 1 GMm E0 = mv 2 − ...(i) 2 r
m
11. (b) Orbital velocity close to surface 1 of earth is gR . So, E = m( gR ) 2 2 1 ⇒ E = mgR 2
m
1 R2 R decreasing g increase curve b represent variation.
10. (b) g ∝
3Gm 2 1 + 8 + 3 3Gm 2 along SQ. d 2 24 2d 2
13. (a) In figure A force due to 5M, 2M cancel and similarly in figure B force due to 2M, 2M cancel, hence, FA = FB. In fig. C also force due to 5M cancel but net force will be more in comparison to figure A or figure B as man at the centre is double similarly think for figure D.
a. co
v R 2π = gR = e 2 π R /g 2
an
=
=
GMm 1 2 GMm = 2 gR mve = ; ve = R 2 R
In tunnel body will perform SHM at centre Vmax = Aω (see chapter on SHM)
3Gm 2 3Gm 2 3Gm 2 + + 2 2 24d 3d 8d 2
9. (b)
4 = R 2Gd π as Ve ∝ R for same density, 3 VA =2 VB
=
4 2Gd πR 3 2GM 3 8. (a) Ve = 2 gR = R= R R2 R2
Gravitation
vp =
−GM 5R
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2.48
05
5
P
2
18. (d) TE in the orbit at a height equal to radius of earth 1 GMm −1 GMm = mv 2 − = 2 2R 4 R
−GM R
Gravitational potential at O, vO =
Gravitation
mv 2 GMm = 2 2 R (2 R)
3
Change in energy = −
s
rio
rs ita
ve
e/ un i t.m
3/2
ht tp
GM 2 mv 2 = 2 L ( L / 3)
22. (a) Remember Keplers law are applicable for satellite also. dA J 1 = = vr Areal velocity = dt 2m 2
tp s:
T2 6400 = T1 36000
or
3/2
0
3
A =J T
21. (c) We know that
Integrating, 2m
s: // Therefore,
[J = angular momentum]
6400 or T2 = × 24 ≈ 2 h 36000
q
0
dA J = dt 2m
T2 ∝ R3 or (T2/T1) = (R2/R1)3/2
ht &20
un iv
.m e/
//t
ht
0
20. (a) Using Kepler’s 2nd law
w
w
s: //
tp
GMM GM cos30° = 3 2 L2 L This acts as centripetal force. ⇒ 2
2
T2 ∝ r'3 r r ′ = 1/3 4
er
2 L × L cos30° = 3 3 Force on each star M due to the other two
si ta rio s_ in fo
ar m
w
.ja
17. (b) Stars move around COM. Distance of each star from
=
GM to dismantle the system, L energy equal to total energy of the system must be provided.
⇒ v=
1 GM mv 2 GMm r = 2 2 r r r
23. (d)
GM 2 1 3 GM 2 i.e., 3 × MV 2 + 3 × =− 2 L 2 L
Areal velocity ∝ velocity = 1 : 2
2GM 1 (1 − ) R 5
COM is
19. (c) Angular speed of earth = angular speed of geostationary satellite. If ω is double, time period become half using
an a. co m
GM GM 1 2 m − = mv or 5R 2 R
3 GMm 3 = mgR 4 R 4
=
work KE theorem W = ∆K ⇒ m[vp – vO] 1 = mv 2 2
1 GMm GMm −− 4 R R
r . Hence ratio of areal
GM 1 2 GMm GMm mv0 − =− and g = 2 R 2 R R+h
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Gravitation
1 3 R2 g × × gR − Rg = − 2 2 R+h We get h = 3R
GMm = m(OB )ω2 (OB) 2
Hence,
$
24. (a) Since, T2 = kr3 5
Differentiating the above equation ∆T 3 ∆r ∆T ∆r = =3 ⇒ T 2 r T r
o
s_
ar
rio
ita rs
rio
si ta
un iv
GMm 2rs ...(ii) The angular momentum L is given by
e/
.m //t
1/2
GM L = ms vs rs = ms rs
tp s: ht
28. (c) By conservation of energy
−
GMm 1 m GM GMm + =− RE 2 N 2 2 RE H
⇒ H=
R ⇒ OB = 2 R OB Here gravitational force will provide the required centripetal force.
26. (a) In ΔAOB: cos 60º =
M 1 & ε0 = 3 (4 / 3)πR 4πG
rs = (GMms2rs)1/2
..........(iii) From eqs. (ii) and (iii), we get L = (2Emsrs2)1/2
5
&0
3mg then F = 8
GMms rs
er
P.E. = −
mρR Net force on m → F = mE = 8ε 0 Here, ρ =
s
1 1 GM K.E. = ms vs 2 = ms ; 2 2 rs
e/ un iv e
tp ht 5
...(i)
∴ Total energy E = K.E. + P.E. = −
s: //
ht
t.m
tp
s:
//w
Net gravitational field ρR ρR E = E1 + E2 = − 6ε 0 24ε 0 ρR = 8ε 0
P
GM GMms ms vs2 = ∴ vs = 2 rs rs rs
.ja m w
(−ρ) R −ρR =− 24ε0 R 2 24ε0
w
=−
in f
27. (a) The velocity vs of the satellite is given by
an a.
Gravitational field at mass m due to cavity (–ρ) (−ρ)( R /2)3 r ρa 3 E2 = ........ using E = 3ε0 R 2 3ε 0 r 2 3
%
GM GM GM ⇒ ω= = 3 3 (OB) (2 R) 8R3
⇒ ω=
co m
0
q
(DUWK
0
q
2
25. (b) Gravitational field at mass m due to full solid sphere ρr ρR 1 E1 = = ........ ε 0 = 4πG 3ε 6ε
⇒ 2
Therefore,
2.49
N 2 RE N 2 −1
Altitude = H – R =
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RE N 2 −1
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2.50
3. (a, d).
2π 2π = π 2 2π ω1 + ω2 + T1 T2
29. (c) Given, 8 −
Gravitation
U
T1 = 24 hours for earth. ⇒ T2 = 12 hours (T2 being the time period of satellite, it will remain same as the distance from the centre of the earth remains constant).
m co
fo
in
rio
si ta
s
rio
er
iv
un e/ .m
//t
s:
tp
ht
2GM m r12 ∆a × 100 = × × 100 aav r22 GM s
7. (b, c) Net force towards centre of earth mgx = mg ′ = R
si ta
iv
un
//t
s: tp
Fav GM s = 2 Me r1 Percentage change in acceleration Average force on earth, Fav =
=
Thus pressing force N =
Mg constant and independent of x. 2 Hence, (b). Tangential force F = ma = mg' cos θ
2
1 PJ T 5 [
ONE OR MORE THAN ONE OPTION MAY BE CORRECT
g=
GM r R3
mgx R R 2x
N=
r Mm = 2 1 × 100 r2 M s
1. (a, d) Outside earth g =
32π2 r 3 27GM
Normal force N = mg' sin θ
ht
∆F 2GM m = Me r22
e/
.m
ht
Change in acceleration of earth ∆a =
2GM m M e r22
w
//w
s:
tp
Change in force, ∆F = F1 − F2 =
T22 =
6. (a, c) Due to air resistance satellite will experience tangential force due to which it will not remain in its orbit and approach towards earth hence r decreases and v increases. Thus KE increases. Due to tangential force angular momentum decreases.
er
GM s M e GM m M e − r12 r22
4π2 r23 GM
T2 ∝ r3/2; T2 ∝ m–1/2
a.
an
ar
w
.ja
m
F2 =
T22 =
s_
GM s M e GM m M e + r12 r22 When moon goes on the opposite side of earth. Effective force of attraction, F1 =
P
2mr 2r = m + 2m 3
r2 =
2π 2π = = 23 hours ω2 = ω1 2π − 2π T2 T1
30. (c) During total eclipse Total attraction due to sun and moon,
⇒ T=
&0
U
P
U
GM , Inside earth r2
Q = g ′ cos θ =
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gx R
R2 − x2 4 x
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Gravitation
gx R2 − 4x2 R
GM e m 1 At any distance ‘r’ T.E. = mu 2 − r 2 but angular momentum conservation,
T.E. at any distance ‘r’
1 5 GM e a GM e m = m − but through conr 2 4 r2 servation of total energy
3GM e m 1 5 GM e a GM e m = m = =− r a 2 4 r2
ita rio s_ in
ar
2. Angular momentum of particle = m(v0 + v)
GM e 5 mv0 a.....v0 = 4 a Total energy of particle
a=
GM e m 1 = m(v0 + v) 2 − 2 a GM e m 1 5 = × mv02 − a 2 4
=
3GM e m 5 GM e m GM e m − =− 8 a a a
s
rs
e/ un i
8. (c) In the continuation with same comprehension
(r2 + r2 /3) = 2r2 /3 2 Let time period of A in elliptical path is T0 r=
2
T r ∴ 0 = 2 T r
3
Q
MATCH THE COLUMN TYPE UESTIONS
ht tp s:
//t
.m
//t
tp s:
ht
ve
.m
ht
R R+h Curved area AB on earth = 2πR2(1 – sinθ) Area on earth escaped from satellite = 4πR2 – 2πR2(1 – sin θ) = 2πR2(1 + sin θ) Minimum colatitude, sin θ =
Q
ita
ni e/ u
tp s: %
GMm −GMm for and K.E. = 2r r same m, If r1 > r2, KE1 < KE2 and U1 > U2 Also, angular momentum of A before reaching the position 2 (during elliptical path) was less than that of B. Since, some impulse is needed (at position 2) for A in direction of its motion to equal its speed and angular momentum equal to that of B.
7. (b, c) U = −
∴ T0 =
T (2 2) 3 3
∴ Time taken by A to move from 1 to 2,
K
//w
w
5
PASSAGE BASED UESTIONS
T
w
T
ve rs
.ja
m
$
On solving, 3r2 – 8ar + 5a2 = 0 ⇒ (r – a) (3r – 5a) = 0 ⇒ r = a, r = 5a/3 minimum distance = a, maximum distance = 5a/3
fo
co
an a.
12. (a, c) From figure
T
m
11. (b, c, d) Inside shell gravitational field GM intensity = 0 and potential = − R
5
5GM e 5 GM e a a⇒u = 4a 4 r2
mur = m
9. (a, d) Same force acts on both masses 1 (F = ma) Hence a ∝ m In absence of external force (remember mutual gravitational force is an internal force for the system) total energy remains constant.
R Curve is parabolic and at x = , a = 0. 2 Hence (C).
rio
a=
2.51
t = T0/2 =
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2.52
Gravitation
SUBJECTIVE UESTIONS
Q
or v 2 = ve2 +
3. If a body is projected from the surface of Earth with a velocity v and reaches a heights h, by conservation of energy. Kinetic Energy = Change in potential Energy
i.e., i.e.,
dr 2R =R dt r 1
∫ dt = t=
an a.
R
3/2 1 2 R h 1 + − 1 as h = 3R; 3 g R
s_ i
7 2R ;n=7 3 g
rio
ita
s
rio
ta
er si
ni e/ u
..........(1)
un iv
.m //t
m e/ //t .
ht tp s:
s:
ht tp
11. H = 35800 km, β = 162.6º 13. T = π (r + R)3 /2 γM
G (m1 + m2 ) and the required period of 1 revolution of these stars is
14. t ≈ ( η − 1)m /α gR 15. If at a distance r from the centre of earth, the body has velocity v, by conservation of mechanical energy,
T = 2π
G (m1 + m2 )
17. Let v1 be the velocity of the station before the collision v2 the velocity of the station and the meteorite immediately after the col
1 2 −GMm 1 2 −GMm mv + = mve + 2 r 2 R
10. W = 1.3 × 105 kJ
γmM ln 1 + , F = − a ( a + ) a
mM 9. U = −γ
r1/2 dr
Here ω1 are the angular velocities of rotation of the stars ℓ1 and ℓ2 are the radii of their orbits, F is the force of interaction between the stars, equal to Gm1 m2/ℓ2 where ℓ is the separation between the stars, and G is the gravitational constant. By the definition of the centre of mass. m1ℓ1 = m2ℓ2, ℓ1 + ℓ2 = ℓ ..........(2) solving equation (1) and (2) together, we obtain G (m1 + m2 ) ω1 = ω2 = = −1 13
8. 3.16 × 10–4 rad/sec
∫
m1ω12 1 = F , m2 ω22 1 = F
(d) T = 2π R /g
R+h
16. Since, the system is closed, the stars will rotate about their common centre of mass in concentric circles. The equations of motion of the stars have the form
ve rs
w
:// w
g 0 R = v1
ht
tp s
(b) v(r ) = g 0 ( R 2 − r 2 )/R (c) v(0)
t=
ar
w .ja m
7. (a) α(a) = g0(r/R)
6. d = 2R
(b) 107.9 sec
5. m(g ± 2ωv)
v2 = gh, i.e., v = 10 × 6400 × 103 = 8 km/s 4. (a) 54 sec
1 R 2g
nf o
Here, h = R = 6400 km and g = 10 m/s so,
i.e.,
co m
0
2
GM R2
2 gR 2 R v2 = 2gR + 2gR − 1 ⇒ v = r r
(b) 18 m/sec
2. (a) 0.3 m
1 2 mgh mv = 2 [1 + (h /R)]
R − 1 r
But as ve = 2 gR and g =
1/2
2 1 1. 2G (m1 + m2 ) − a
2GM R
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Gravitation
velocity v2 of the station immediately after the collision through the relation
VR = v2 x R. 2
Solving equations (1) = (4) together and considering that v1 = GM /R , we determine the velocity of the meteorite before the collision: u =
rio s_ in fo
ar a
Here, mN, mE, ωN, ωE, RN and RE are the masses, angular velocities, and orbital radii of Neptune and the Earth respectively, and M is the mass of the sun. We now take into account the relation between the angular velocity and the period of revolution around 2π 2π the sun ωN = , ωE = TN TE
ita
un iv
e/
iv
.m
un
//t
Here TN and TE are the periods of revolution of Neptune and the Earth. As a result, we find that the period of revolution of Nep-
t.m e/
ht tp s:
5 0
er s
w
.ja
m
w
//w
s:
tp ht
X
RN3 ≅ 165 RE3 years. A similar result is obtained for elliptical orbits from Kepler’s third law. tune around the sun is TN = TE
ht tp
s: //
Y
11mM 11m 2 + (v2 x + v22 y ) R 2
Q
1. (b)
2. Vmin =
3 5GM 2 a
3. (c)
4. T = 3 × 10–2N 5. (a)
Where V is the velocity of the station at the moment of the closest proximity to the planet. Here we have used the formula for the potential energy of gravitational interaction of two bodies (of mass m1 and m2): Wp = –Gm1 m2/r. According to kepler’s second law, the velocity V is connected to the
.........(3)
11mM 11m 2 + v , R /2 2
PREVIOUS YEARS’ IIT-JEE UESTIONS
= −G
−G
6. h = 99R
\
GMmE GMmN , mE ω2E RE = RN2 RE2
er si ta rio s
[
mN ω2N RN =
........(1)
and mu = 11 mv2y .........(2) After the collision, the station goes over to an elliptical orbit. The energy of the station with the meteorite stuck in it remains constant during the motion in the elliptical orbit. Consequently,
58GM R
18. We shall write the equation of motion for Neptune and the Earth around the sun (For the sake of simplicity, we assume that the orbits are circular)
na .c om
Hence, v1 = GM /R . In accordance with the momentum conservation law, the velocities u, v1, and v2 are connected through the following relation mu + 10 mv1 = 11 mv2 We shall write the momentum conservation law in projections on the x– and y-axes
10mv12 10mM =G R R2
10 mv1 = 11 mv2x
.........(4)
lision, m the mass of the meteorite, and 10 m the mass of the station. Before the collision, the station moved around a planet in a circular orbit of radius R. Therefore, the velocity v1 of the station can be found from the equation
2.53
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2.54
7. (d)
G (4/3)πρ0 R 3 v 2 1 = r ≤ R. Hence, v ∝ r2 r r
4 v 8. (c) G πρ0 r = , r ≤ R. Hence, v ∝ r 3 r
s_ in
fo
a. c
om
9. (a)
an
si ta rio
ar .ja m
s
er
w w ht
tp
s:
//t
ht
.m
tp
e/ u
s:
ni
//t
ve
.m
rs
e/
ita
un
iv
//w s: tp ht
2
rio
Gravitation
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c h a p t e r
ar a
3.1 DEFORMING FORCE
3
rio s_ in fo
na .c om
Elasticity
iv
un
t.m e/
ht tp s:
3.2 ELASTICITY
er si ta rio s
un iv
.m
//t
ht
e/
s:
The force acting on the spring (F1) is the deforming force. The force exerted by the spring to oppose the deforming action is the restoring force (F2). By Newton’s third law, we can say that F1 and F2 equal and opposite forces turns forming an Action-Reaction pair.
tp
//w
w
er s
w
ita
.ja
m
It is the external force applied to a body which tends to change the size or shape of the body. Under the action of deforming force, a body opposes any change in its shape and size due to net effect of internal (molecular) forces. The resulting force which opposes the deformation is known as restoring force. When one tries to stretch a spring:
ht tp
s: //
The property of a body due to which it opposes the action of the deforming forces is called as elasticity. A material is said to be elastic if it returns back to its original shape or size, when the deforming forces are removed. On the other hand, plastic material remain parmanently distorted when the deforming forces are removed. There are a number of elastic modulii (Young’s modulus, bulk modulus, shear modulus) that measure of response of an elastic solid which is acted upon by the deforming force.
The deforming force is measured (described) in terms of a physical quantity, the stress that is developed in the body. The deformation of the solid is described in terms of physical quantity the strain, that is created in the body as results of deformation force.
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3.2
Elasticity
3.3 STRAIN AND STRESS When an elastic body deforms, under deforming forces, its length dimension, volume dimension and/or its shape may change. Then Strain =
change in dimension original dimension
Tensile strian = Bulk strain =
change inlength ∆l = original length l
change in volume ∆V = original volume V
rio s_ in fo
deforming force area of application
er si ta rio s
un iv
//w
w
er s
w
.ja
Hence at equilibrium, stress =
ita
m
ar a
na .c om
(i) External deforming forces cause deformation (strain) in the body. (ii) Strain cause internal restoring forces, called elastic forces to come into action. (iii) Internal elastic force per unit area of their application is called stress. (iv) As the strain increases, stress also increases. (v) At steady equilibrium state elastic force = deforming force
s:
3.4 LONGITUDINAL STRESS AND LONGITUDENAL STRAIN
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
Consider a rod (or a string) of length l and cross-sectional area A. Suppose that the a force F is applied at the ends parallel to the length so as to stretch the rod by ∆l. The rod is said to be under tension. The tensile stress is defined as the ratio of the force of tension to the cross-section area. F Tensile stress = A The resulting tensile strain is defined as the ratio of the change in length to the original length. ∆l l If the ends of a rod are subjected to equal pushes, then rod is said to be under compression. F Compressive stress = A Tensile strain =
l + ∆l l
Compressive strain =
∆l l
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Elasticity
3.3
Tensile and compressive stresses and strains are also known as longitudinal stress and longitudinal strain. Example A bob of mass m is suspended from a light steel wire of length l and radius r. Assuming mass of wire to be negligible, what is the longitudinal stress and strain?
) Solution Let T be the tension developed in the steel wire. As the bob is stationary T = mg
T
e/
tp
s:
3.5 VOLUME STRESS AND VOLUME STRAIN
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
ar a
Longitudinal strain ∆l = l
l+∆ ∆ll
m
rio s_ in fo
na .c om
Longitudinal stress mg = 2 πr
l
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
Sometimes, the deforming force acting on the body tends to change its volume. Let the deforming force changes the volume of a body from V to V + ∆V (∆V is negative, if volume is decreases). We define volume strain as the ratio of change in V to the original volume.
V + ∆V
∆V V The stress associated with the volume strain is called as volume stress. It equals the deforming forces acting on a unit area. It is usually called as pressure (P). F Volume stress = P = A Volume strain =
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3.4
Elasticity
Example A cube is subjected to pressure of 5 × 105 N/m2. Each edge of the cube is shortened by 1%. Determine volume strain. Given: Volume stress = P = 5 × 105 N/m2
) Solution Let l be the length of cube (undeformed) ∴
99 Distroted volume, V f = l 100 ∆V = Vf – Vi
3
3
na .c om
∴
99 = l − l3 100 3 ∆V 99 Volume strain = = − 1 ≅ – 0.03 Vi 100
rio s_ in fo
3.6 SHEAR STRESS AND SHEAR STRAIN
e/
B
un t.m e/
Q
P
ht tp
s: //
ht tp s:
C
iv
.m
A
//t
ht
tp
s:
un iv
//w
D
er si ta rio s
w
er s
w
ita
.ja
m
ar a
If the deforming force tries to change the shape of a body, a shear stress is developed in the body. In the diagram we see a force F parallel to the surface ABCD of area A, deforms the rectangular area ABPQ into a parallelogram.
Let AA′ = ∆x and AQ = y Hence, due to F, shape of the body is altered. Shear stress =
shearing force F = cross-section of ABCD A ∆x
A
A'
B
B'' B
y θ
Q
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P
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Elasticity
3.5
Shear strain is defined as: ∆x Shear strain = = tan θ (θ is very small) y
rio s_ in fo
un iv e/ .m
tp
er si ta rio s
er s
w //w
s:
3.7 HOOK’S LAW
Motion
ita
µmg
w
.ja
m
ar a
na .c om
Example A block of weight 15 N slides on a horizontal table, the coefficient of sliding friction = 0.4, the area of the block in contact with table is 0.05 m2. What is the shear stress? µ mg ) Solution Shear stress = A 0.4 × 15 = = 120 N/m 2 0.05
iv
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
If the deforming forces are within a limit (known as elastic limit), the stress created in the body is proportional to the resulting strain. i.e., stress ∝ strain stress The ratio is known as modulus of elasticity. strain According to various type of stresses, we have three modulii of elasticity. (i) Young’s modulus of elasticity (Y): Y=
longitudinal stress F/A = longitudinal strain ∆l /l
Fl A∆l The following forms of the result must be carefully noted. Y=
∆l = ∆l =
Fl ∆l and F = AY AY l
Fl ∆l and F = AY AY l
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3.6
Elasticity
(ii) Bulk modulus (K): volume stress p =− volume strain ∆V/V pV ⇒ K=− ∆V ‘–’ sign indicates decrease in volume The reciprocal of K is known as modulus of compressibility. K=
na .c om
(iii) Shear modulus (η): shear stress F/A η= = shear strain θ F η= Aθ n is called as modulus of rigidity
rio s_ in fo
un iv
//w
w
3.8 ELASTIC ENERGY
er si ta rio s
er s
w
ita
.ja
m
3 1 1 = + Y η 3K
ar a
Relation between Y, K and η The three modulus of elasticity are related as
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
When an elastic body is deformed, work is done by applied force. This work is stored as elastic potential energy and is released when the body returns back to its original shape or size. We have already come across an example of elastic potential energy in case of a compressed or stretched spring. The general expression for the elastic potential energy in a unit volume of a deformed body is given by 1 Elastic energy stored per unit volume = (stress) (strain) 2 1 = (modulus of elasticity) (strain)2 2 In case of a longitudinal stress (compressive or tensile)
l
x
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Energy stored 1 ∆l = Y Volume 2 l
Elasticity
3.7
2
1 YA (∆l ) 2 2 l 1 Total energy = F ∆l 2
Total energy =
Work done = ∫ F.dx F/A x /l
Y=
co
m
3.9 THERMAL STRESS
s
rio
si ta
iv er
un
s: //
ht
t.m
tp
s:
e/ un
iv
//w
w
er
w
si
ta
.ja
m
rio
ar
s_
an
in f
a.
o
If the ends of the rod are rigidly fixed as to prevent expansion or contraction and the temperature of the rod is changed, elongation or compressive stress, called thermal stress, will be set up in the rod. This strain may become very large, so as to strain the rod beyond its elastic limit or even beyond its breaking strength. Hence, in the design of any structure which is subject to change in temperature, some provision must be made for expansion. In a long steam pipe this is accomplished by the insertion of expansion joints or a section of pipe in the form of a · V. In bridge, one end may be rigidly fastened to its abutment while the other rests on rollers. It is easy to compute the thermal stress set up in a rod which is not free to expand or contract. Suppose, that a rod at a temperature t has its ends rigidly fartened and that while they are thus held, the temperature is reduced to a lower value t0. The fractional change in length if the rod were free to contract would be
Hence,
.m
ht
tp
s: //t
ht tp
e/
∆L = α (t − t0 ) = α ∆ t L Since, the rod is not free to contract, the tension must increase by a sufficient amount to prevent the same fractional change in length. But from the definition of Young’s modulus. F/A Y= ∆L/L F = AY
Substituting for
∆L , we get L
F = AYα ∆ t
∴
∆L L
Stress in the rod =
F = Y α∆t A
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3.8
Elasticity
3.10 TORSION Consider a cylinder whose upper ends is rigidly fixed and the other end is twisted through an angle θ about the axis of the cylinder. The twisted cylinder exerts a restoring torque given by τ=
1 πηr 4 θ 2 l
Fixed twisted trhough an angle θ
rio s_ in fo
na .c om
l
ar a
where, n is the modulus of rigidity, r is the radius and l is the length of the cylinder. If we express torque as τ = Cθ
er s
w
un iv
//w s:
t.m e/
s: //
ht tp
Stress = Force per unit area mg 2000 × 980 = πr 2 π(0.05) 2
= 2.496 × 108 dyne cm–2 change in length Linear strain = original length =
0.24 × 10−3 = 12 × 10−5 2
iv
//t
ht tp s:
) Solution By the formula
Young’s modulus of elasticity =
un
ht
.m
tp
e/
WORKED OUT EXAMPLES
1. A wire 2 m long, 1 mm in diameter when stretched by a weight of 2 kg increases in length by 0.24 mm. Find the stress, strain and Young’s modulus of the material of the wire. Assume that the change in diameter is negligible.
=
er si ta rio s
ita
.ja
m
πηr 4 θ , is known as the torsional rigidity. 2l
w
then, C =
=
stress strain
2.496 × 108 dyne cm −2 12 × 10−5
= 20.8 × 1011 dyne cm–2 2. If the young’s modulus of the material of wire is 2.2 × 1012 dyne cm–2. What mass must be suspended from the wire fixed at one end to stretch it by 1 mm? Length of the wire is 2.5 m and diameter is 1.0 mm.
) Solution Stress =
weight of body mg = 2 area of wire πr
Strain =
change in length ∆l = original length l
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ta
si
rio
ita
rs
un i e/
//t
s:
tp
= –1.024 × 10–8 m2 = 10.24 mm3
4. What would be greatest length of a steel wire, which when fixed at one end can hung freely without breaking? Density of steel = 7.8 gm–3 and breaking stress for steel is equal to 7.8 × 109 cm–2
) Solution Let the greatest length of steel is l. )
−P ∆V/V
7 3 − PV − ( 2 × 10 ) (0.04) ∆V = = K 1.25 × 1011
ht
∆l 0.035 = = 4.67 ×10−4 l 75 stress 5.9 × 107 Y= = strain 4.67 × 10−4
∆V ∆V = V (0.04)3
.m
ht )
= 5.9 ×107 N/m 2
Strain =
Volume strain =
Bulk modulus, K =
π ( 6.5 × 10
−4 2
=
force F = area A 8 × 9.8
ve
.m
tp s:
//t
ht
(iv) Volume stress, P = 2 × 107 N/m2
)
(i) Stress =
s
er
un iv
e/
s:
tp
face displaced 4.0 mm relative to the bottom surface. What are the shearing stress, shearing strain and shear modulus? (iv) Compute the volume change of a solid copper cube, 44 mm on each edge, when subjected to a pressure of 2 × 107 N/m2.
) Solution
fo
in
s_
rio
.ja
w
//w
w
an a.
m ar
co m
3.9 mg stress πr 2 mgl Young’s modulus of elasticity = = 2 (ii) Young’s modulus, Y = ∆l strain πr ∆ l stress mg l mg = = 2. l stress πr 2 mgl strain πr ∆l Young’s modulus, Y = = = 2 2 ∆ l strain πr ∆ l Yπr ∆l ∴ Required mass, m = l gl mgl ∆l = 2 2.2 × 1012 × 3.14 × 25 × 10−4 × 1× 10−1 πr Y = 980 × 250 80,000 × 9.8 × 4 = = 7049 g π(4.5) 2 × 10−4 × 1.9 × 1011 = 7.049 kg. = 2.6 × 10–3 m = 2.6 mm 3. (i) A metal wire 75 cm long and 0.13 cm tangential force is diameter stretches 0.035 cm, (iii) Shear stress = area of face when a load of 8 kg is hung on its 0.5 ends. Find the stress, strain and = = 333 N/m 2 Young’s modulus. 15 × 10−4 (ii) A solid cylinder steel column is 4 m ∆x (displacement) Shear strain, θ = = long and 9 cm in diameter. What y height will be decrease in its length when 0.4 carrying a load 80,000 kg? Given: = = 0.133 11 2 Y = 1.9 × 10 N/m 3 (iii) A box shaped piece of gelatine shear stress Shear modulus, η = dessert has a top area of 15 cm2 shear strain and height of 3 cm. When a shear333 ing force of 0.5 N is applied to = = 2500 N/m 2 0.133 the upper surface, the upper surElasticity
2
Then, weight of steel = ρAlg
= 1.27 × 10 N/m 11
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3.10
stress 7.8 × 109 = ρg 7.8 × 980
Young’s modulus, Y =
5. A wire of radius r stretched without tension along a straight line is tightly fixed at A and B. A mass m is suspended from the mid point of the wire. Due to the weight of mass, the wire is pulled into the shape ACB. Find the depression ‘d’ in the wire. The length of the wire is 2l and its Young’s moduls is Y.
=
θ
∆l l
2
+ l2 ) − l
o
nf
s_ i
s
rio
er si ta
iv
un
e/ ∴
Strain =
=
T mg = A 2A sin θ
(d
)
mg 2sin θ
Stress =
T=
T l × πr 2 ∆l
T=
Y × πr 2 ∆l l 2
l
0.315 −2 2 × 102 × π× × 10 × 0.1 2 = 100 = 1.559 × 106 dyne
1/2
d2 d2 = 1 + 2 − 1 ≈ 2 l 2l
stress strain
T T l A = = × ∆l A ∆l l
ht
C m
⇒
t.m
T
when the extension is 1 mm.
Young’s modulus of elasticity =
s: //
θ
θ
B
tp
T
6. A sphere of mass 50 g is attached to one end of the steel wire, 0.315 mm diameter and one metre long. In order to form a conical pendulum, the other end is attached to a vertical shaft which is set rotating about its axis. Calculate the number of revolutions necessary to extend the wire by 1 mm. Young’s modulus of elasticity of steel = 2 × 1012 dyne cm–2.
) Solution Let T be the tension in the wire,
s: //
ht
θ
tp
2l
mgl 2l 2 . 2πr 2 d d 2
rio
iv un e/ t.m
)
ht
) Solution Let the T be tension of wire A
stress strain
1/ 3
er
w w
s: tp
m
2 T sin θ = mg
stress strain
si ta
.ja
B
//w
C
(Neglecting d2,
mg d = 2 2πr Y
m
2l
A
ar an
a.
∴
co m
= 1.02 × 106 cm.
d + l2
Young’s modulus, Y =
⇒
l=
stress 7.8 × 109 = ρg 7.8 × 980
2
because d2 << l2) mgl Stress = 2πr 2 d 2
l=
d
sin θ =
∴
As
weight ρAlg = Area A (where ρ = density of the steel) Now, Stress =
Elasticity
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Elasticity
Frequency of revolution =
∴
θ
1 0.3561 = 2.808 revs–1 = 168.5 rev mm–1
T
T cos θ
7. A truncated solid cone of length l radius of its ends equal to r1 and r2 (r1 < r2). The bigger end is fixed. A force F is applied at the other end distributed uniformly over its area of cross-section. Find the total elongation in the cone. Young’s modulus of cone is Y.
θ
T sin θ mg
m
l = lm
m
ita
.ja
rio ita rs
ni
e/ u
where, r is the radius of circle described. r = l sin θ = 100 × 1 = 100 cm
dl =
⇒
∴
From equation (i), rT sin θ m
r −r Yπ r1 + 2 1 x l
2
r
F 2 dx 2 Y ∫r1 (r2 − r1 ) π r1 + x l
Let,
∴
100 × 1.559 × 106 × 1 = 1766cms −1 50 2πr Period of revolution = v 2π× 100.1 = = 0.3561 sec. 1766 =
F dx
∫ dl =
v=
(r2 − r1 ) x l
F stress A = Young’s modulus = strain dl dx
//t
.m
Thus, r = r1 +
s:
2
ve
)
from the lower radius is r.
ht tp
mv r
s
rs iv e t.m
) Solution Let the radius of cone at distance x
s: //
tp
ht
sin θ = 1 − cos 2 θ = 0.9998 ≈ 1 T sin θ =
dx
x
un e/
s:
tp
ht
∴
rio
ar
r1
mg 50 × 980 cos θ = = = 0.0314 T 1.55 × 106
∴
r2
r
s_ i
an a.
nf o
co
l
F
w w //w
r
From the figure T cos θ = mg
1 period
=
T
θ
3.11
(r2 − r1 ) x = z lim of z , r → r2 l (r2 − r1 ) dx = dz l
r1 +
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3.12
Elasticity
wires. One wire is of steel and the other two are of copper. All the wires have the same areas of cross-section. Consider the modulus of elasticity of steel to be double than that of copper.
r
r
=
1 1 Fl − πY(r2 − r1 ) r1 r2
=
Fl π Y r1 r2
8. A solid uniform rod of length l is hanging vertically from its upper end. Find the total elongation of the rod due to its own weight if ρ is the density and Y the young’s modulus of the materials of the rod.
c
s
c
)
m
s
si
ta
rio
s Ts
c Tc
er
m
iv
//t
According to problem YS = 2YC ∴ TS = 2TC From equations (i) and (ii), we get Mg Mg TC = and TS = 4 2
...(ii)
ht
tp
s:
TC TS = YC YS
.m
tp
e/
un
//t s:
or
ht (l–x)
ta si er un
iv
c Tc
.m e/
tp ht x
fo
in
s_
.ja
w w
//w
ρAlg dx lA Y
s:
dl = (l − x)
an m
ar
∴
For the equilibrium, mg = 2TC + TS ...(i) Since elongation of all wires is same, therefore T /A T /A dl = C = S YCl YSl
rio
)
T T dx A Young’s modulus, Y= = dl A dl dx
C
) Solution
a.
W ) Solution Tension,T = (l − x) l where, W = ρ lg A
m B
A
co
2 2 Fl dz Fl 1 l= = − 2 ∫ πY(r2 − r1 ) r1 z πY(r2 − r1 ) z r1
l
9. A homogenous block with a mass m hungs on three vertical wires of equal length arranged symmetrically. Find tensions of
ρg x2 1 ρ g l2 = lx − = Y 2 0 2 Y
10. Two rods made of different materials are placed between massive walls. The cross-section of the rod is A and their respective lengths are l1 and l2. The rods are heated by t degrees. Find the force with which rods act on each other if their coefficient of linear thermal expansion are α1 and α2 and modulii of elasticity of their materials Y1 and Y2 are known.
l
ρg (l − x)dx Y ∫0
Integrating, ∫ dl =
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Elasticity
Y1
Y2
α1
α2
l1
l2
B
) Solution )
C
Y1
A
Y2 F
α2
) Solution Since, the blocks moves with uniform )
l2
velocity (i.e., acceleration = 0)
ar a
Expansion of the wire due to temperature is balanced by the contraction of the compressed wire. Fl dl1 = 1 …(i) Y1A
µN
m
un
mA.g
From block B T – µmB g = 0 There is no friction force acting on A, since the normal reaction exerted by C on A is zero. mA ·g = T mA 2 = = 10 kg µ 0.2
mB =
Stretching force on the spring, T = mAg = 2 × 9.8 N Tension of the spring l= spring constant 2 × 9.8 = = 10−2 m 1960 Potential energy stored in the spring 1 = (2 × 9.8) (10−2 ) = 9.8 × 10–2 J 2
ht tp
s: //
t.m e/
ht tp s:
A
11. Two blocks A and B are connected to each other by a string and a spring, the string passes over a frictionless pulley as shown in the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surface of block is 0.2. Force constant of the spring is 1960 Nm–1. If the mass of block A is 2 kg, calculate the mass of block B and the energy stored in the spring.
T
iv
…(iii)
Fl Fl + 2 = (l1α1 + l2 α 2 ) t Y1A Y2 A (l α + l α ) t A F= 1 1 2 2 l1 l2 + Y1 Y2
ita
un iv e/ .m
//t
ht
∴
tp
Also dl1 + dl2 = l1 α1 t + l2 α2t = (l1α1 + l2 α2) t
er s
w //w
…(ii)
s:
Fl2 Y2 A
T
B
.ja
w
And dl2 =
N
er si ta rio s
l1
rio s_ in fo
F
na .c om
α1
⇒
3.13
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3.14
∴
Putting the value of x in equation (iii), we have YA mg l (mg ) l E= = 2YA 2l YA
If the energy is converted to heat when the bob get snapped, we have E = Heat absorbed by the wire
( mg )
0
( mg )
2
2YA 2ρc
ita
.ja e/
Hence
.m
{(100kg) (10ms )}
t.m e/
2
−3
= 0.00457 K
13. A street shaft of radius 10 cm and of length 2 m is transmitting power of 80 kW at 300 r.p.m. If n = 8 × 1010 Nm–2, calculate the twist between the two ends. )
) Solution Power, P = torque × angular velocity = Tω P ∴ T= ω ⇒
T=
1 πr 2 P θ= 2 l ω
∴
θ=
2Pl πηr 4 ω
mg mg l where Y = A = x Ax l
…(iii)
x
YA YA 2 E = ∫ x dx = x l 2l 0
From equations (i) and (ii)
2
( 7860 kgm )
s: // ht tp
−1
(2) ( 2.1× 1011 Pa ) (1.256 × 10−3 m 2 )
un
//t
iv
un iv
er si ta rio s
Area of cross-section of the wire, A = πr2 = (3.14) (2 × 10–3 m)2 = 1.256 × 10–5 m2
er s
w ht tp s:
…(ii)
= (ALρ) c ∆ T
rio s_ in fo
ar a
∆T =
w //w s: tp ht
YA F= x l
∴
L
2AY
∆T =
m
Fl = Ax
2
where, m is mass of the wire. Rearranging the above expression, we get
m
F stress A = Young’s modulus, Y = strain x l
i.e.,
…(i)
na .c om
Energy stored due to rod = ∫ F.dx
)
x
2
2
) Solution
mg l x= YA
12. A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature at the wire snapped, and change in temperature of the wire ignoring radiation losses. (For the steel wire, Young’s modulus = 2.1 × 1011 Pa, Density = 7860 kg/m3, specific heat capacity = 420 J/kg– K)
Elasticity
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Elasticity
η = 1010
)
) Solution
Torsional torque, T = Force × radius of cylinder
N/m–2, r = 0.1 m 300 × 2π rad/sec. 60 = 10π rad s −1
ω = 300 rev min −1 =
T = Fr =
2
1 π ( 6 × 10 = 2
10
= 4.05 × 10–4 rad.
14. A cylinder of length 2m and radius 8 mm is camped at one end and a force tangential to its cross-sectional boundary and in the plane of the cross-section is applied at the other end to produce one radian twist in the cylinder. If the modulus of rigidity of the material of the cylinder is 6 × 1010 Nm–1, calculate the force applied.
) (8 ×10 )
−3 3
×1
2 F
= 2.41 × 104 N.
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
ar a
rio s_ in fo
na .c om
∴
2 × 80 × 103 × 2 4 = × 10−3 π × 8 × 1010 × 10−4 π2
θ=
∴
1 πηr 4 θ 2 l 1 πηr 3θ F= 2 l
⇒
Here, P = 80 × 103 W, l = 2m,
3.15
iv
un
t.m e/
Hence
(d)
d1 Y = 1 d2 Y1
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(b)
(c)
d1 Y1 = d 2 Y2
(a)
L is maximum in case of (a). A
3. Two wires of different materials support a massive object suspended from the centre of a uniform horizontal bar with its ends connected to two vertical wires of equal length. What must be the ratio of the diameters of the wires so that the bar may remain horizontal when the object is removed? [Young’s modulus of the wires are Y1 and Y2 respectively].
2. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied to all? (a) L = 50 cm; D = 0.05 mm (b) L = 100 cm; D = 1 mm (c) L = 200 cm; D = 2 mm (d) L = 300 cm; D = 3 mm
L
(∵ F and Y are common)
s: //
∆L1 F1 L1 A2 40 × 1 22 = × = × =1 ∆L2 F2 L2 A1 80 × 2 12
⇒
ht tp
)
) Explanation Young’s modulus, F /A F /A Y= 1 1 = 2 2 ∆L1 / L1 ∆L2 / L2
FL
) Explanation ∆L = AY ∝ A )
e/
ht tp s:
//t
ht
.m
tp
1. Two steel wires of length 1 m and 2 m have diameters 1 mm respectively. If they are stretched by forces of 40 N and 80 N respectively their elongation will be in the ratio of (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 1 : 8
s:
SOLVED OBJECTIVE TYPE QUESTIONS
d1 Y2 = d 2 Y1 d1 Y = 2 d2 Y1
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3.16
)
om
s_
rio
ita
∆l = l
Mg 3KA
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ht
(c)
(d)
⇒
F 1 = Lρα A 2
8. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless pistion of area A floats on the surface of the liquid. When a mass m is placed on the pistion to compress the liquid, the fractional change in the radius δR of the sphere, is R Mg Mg (a) (b) KA 2 KA
1 1 = l ∆l
rio
ita
e/
Stress =
⇒
⇒
F 2F 2 2 πr = 4πr ∆ L 3L
L Aρ.α 2
F=
.m
//t
∴
s:
F /A
( ∆l /l )
er s
iv
un
)
the mid point of the rod is
tp
)
) Explanation Y =
(d) 4l
(b) l
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7. A uniform rod of length L and density ρ is being pulled along a smooth floor with a horizontal acceleration α. The magnitude of the stress at the transverse cross-section through the mid-point of the rod is Lρα (a) Lρα (b) 2 2 (d) none of these (c) Lρα 3
) Explanation Force at the cross-section throught
5. A wire of length L, radius r, when stretched with a force F, changes in length by l. What will be the change in length in a wire of same material having length 2L, radius 2r, radius 2r and stretched by a forca 2 F?
∆V Change in pressure = V Bulk modulus
3 × 107 − 105 299 = × 10−5 × 100% 8 × 1010 8 = 0.0374% ≈ 0.04%
=
Y1 α1 3 = = Y2 α 2 2
∴
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) Explanation
) Explanation Thermal stress = Yα∆t
4. By what fraction is the volume of an aluminium sphere be reduced as it is lowered from the surface to the depth of 3 km in ocean? Bulk modulus of Al = 8 × 1010 N/m2. At the surface of ocan P = 105 N/m2. At a depth 3 km, P = 300 atmospheres. (a) 20% (b) 10% (c) 1% (d) 0.04%
d1 Y = 2 d2 Y1
or
(∵ A ∝ d)
)
d12Y1 = d22 Y2
or
6. Two rods of different materials having coefficients of thermal expansion α1, α2 and Young’s modulii Y1, Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo then same increase in temperature. There is no bending of the rod. If α1 : α2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9
Fl AY For the bar to remain horizontal, ∆l1 = ∆l2 ∴ A1Y1 = A2Y2 (∵ F and l are common)
) Explanation ∆l =
Elasticity
Mg 4 KA
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Elasticity
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7. For a given wire, the Young’s modulus is 2.4 times that of its rigidity. The Poisson’s ratio is (a) 0.5 (b) 0.2 (c) – 0.5 (d) –1.
8. Two blocks of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is 3.18 × 1010N/m2, then the radius of the wire is
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4. One end of a steel wire of length 1 m and cross-sectional area 1 mm2 is clamped at rigid support. The increase in its length when 1 kg-wt is suspended from it will be (if Y = 20 × 1010 pascal) (a) 5 cm (b) 0.5 mm (c) 0.25 mm (d) 0.05 mm
6. A solid sphere of radius R made of material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the suface of the liquid. When a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is Mg Mg (a) (b) MA 3BA 3Mg Mg (c) (d) 4 BA 4 BAR
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3. A uniform pressure P is exerted on all sides of a solid cube. It is heated through ∆tº in order to bring its volume back to the value it had before the application of pressure. Then P B (a) ∆t = (b) ∆t = Bγ Pγ Bγ (c) ∆t = BγP (d) ∆t = P where γ is the cubical coefficient of expansion.
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2. The length of a wire increased by 1% by a load of 10 kg suspended from it. If the same load is suspended from the wire of same material but of double length and double radius, by what percentage is the length increased? (a) 0.25% (b) 0.5% (c) 2% (d) 4%
is twisted by an angle of 30º. The angle of shear is (a) 12 (b) 1.2 (c) 0.12 (d) 0.012
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1. A force of 400 kg weight can break a wire. The force required to break a wire of double the cross-section will be (a) 800 kg-wt (b) 200 kg-wt (c) 1600 kg-wt (d) 100 kg-wt
OBJECTIVE TYPE QUESTIONS
δR Mg = R 3KA
5. The upper end of a wire 1 m long and 4 mm radius is clamped. The lower end
δV 3δR = V R
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∴
∴
δV = 4πR2δR
∴
Mg / A δV / V δV Mg = V KA 4 2 V = πR 3
) Explanation K =
3.17
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1 kg
4 kg
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3.18
10. The correct graph verifying Hooke’s law is
(d)
K ρP ρK . P
13. Two capillaries of same diameter are dipped into liquids of specific gravity 0.4 and 0.8 respectively. If their surface tensions are in the ratio 6 : 5 then the ratio of heights of liquids in them will be (a) 12 : 25 (b) 5 : 12 (c) 4 : 8 (d) 8 : 4
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14. A stone of 0.5 kg mass is attached to one end of a 0.8 m long aluminium wire of 0.7 mm diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such that the wire makes an angle of 85º with the vertical. Find the increase in the length of the wire (Y for aluminium = 7 × 1010 N/m2, sin 85º = 0.9962 and cos 85º = 0.0872) (a) 0.7 mm (b) 1.7 mm
(d)
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Extension
Load
ρP K
(b)
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(c)
P ρK
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Extension
Load
(a)
(c)
m
Load
12. A material has normal density ρ and bulk modulus K. The increase in the density of the material, when it is subjected to an external pressure P from all sides is
a. c
Extension
(a)
(b)
(d) 2 YA α2 Lt2
1 YA α 2 Lt 2 4
1 YA α 2 Lt 2 2
(d) 17 cm
(c) 1.7 cm
Load
(c)
(b)
(a) YA α2 Lt2
9. A steel rod of length L, area A, Young’s moduls Y and coefficient of linear expansion α is heated to tºC. The work that can be performed by the rod when heated is
11. A wire 3 m in length and 1 mm in diameter at 30ºC is kept in a low temperature at –170ºC and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is [Y = 2 × 1011 N/m2, g = 10 m/s2 and α = 1.2 × 10–5/ºC] (a) 5.2 mm (b) 2.5 mm (c) 52 mm (d) 25 mm
(b) 2 × 10–5 m (d) 4 × 10–5 m
(a) 1 × 10–5 m (c) 3 × 10–5 m
Elasticity
ANSWERS 1. (a) 9. (b)
2. (a) 10. (c)
3. (a) 11. (a)
4. (d) 12. (c)
5. (c) 13. (a)
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6. (b) 14. (b)
7. (b)
8. (d)
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Elasticity
of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section 10–3 m2 and the other is of brass of cross section 2 × 10–3 m2. Find out the position along the rod at which a weight may be hung to produce (a) equal stresses in both wires, (b) equal strains in both wires.
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l1, α1, γ1
A
4. A horizontally oriented rod of length 2L = 1.0 m is being rotated about vertical axis passing through its centre. Find the minimum number of revolutions at which this rod will rupture. The tensile strength of copper is σm = 0.30 GPa and its density is ρ = 8.9 g/cm3. [Ans. 0.8 × 102 r.p.s.]
B
C
Ans. A (l1α1 + l2 α 2 ).t l1 l2 + Y1 Y2
5. A light rod of length 2.00 m is suspended from the ceiling horizontally by means
7. A force of 106 N/m2 is required for breaking a material. If the density of material is 3 × 103 kg/m3, then what should be the
6. Two rods of different metals having the same area of cross-section A are placed between two massive walls as shown in the figure. The first rod has a length l1, coefficient of linear expansion α1 and Young’s modulus Y1. The corresponding quantities for the second rod are l2, α2 and Y2. The temperature of both the rods are now raised by tº C. Find the force with which the rods act on each other (at higher temperature) in terms of given quantities. [IIT, 1975]
Young’s modulus of steel = 20 × 1010 N/m2 Unstretched length of wire = 500 cm Radius of wire = 0.05 cm [IIT, 1972] [Ans. 3.76 m/s]
Young’s modulus of steel = 2 × 1011 N/m2 [IIT, 1974] [Ans. (a) 1.33 m, (b) 1.00 m]
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3. A sphere of radius 10 cm and mass 25 kg is attached to the lower end of a steel wire which is suspended from the ceiling of the room. The point of support is 521 cm above floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate velocity of the ball at its lowest position.
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Given, α for steel = 1.2 × 10–5/º C. [IIT, 1971] [Ans. 1.92 × 103 N]
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2. A steel rod 25 cm long has a crosssectional area of 0.8 cm2. What force would be required to stretch this rod by the same amount as expansion produced by heating it through 10º C?
Given: Young’s modulus of brass = 1011 N/m2
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[IIT, 1970]
practice
EXERCISE 1
1. A steel wire of diameter 0.8 mm and length 1m is clamped firmly at two points A and B which are 1m apart and in the same horizontal plane. A body is hung from the middle point of the wire such that the middle point sags 1 cm lower from the original position. Calculate the mass of the body. Given, Y = 2 × 1011 N/m2 [Ans. 8.2 × 10–2 kg]
3.19
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3.20
Elasticity
culate the displacements of the points B, C and D using the data on Young moduli given below (neglect the weight of the rod). YAB = 2.5 × 1010 N/m2, YBC = 4.0 × 1010 N/ m2, YCD = 1.0 × 1010 N/m2. [Roorkee, 1986] [Ans. Displacement of point B, = 3.92 × 10–6 m] Displacement of point C = 8.82 × 10–6 m Displacement of point D = 23.52 × 10–6m]
13. A simple pendulum is made by attaching a 1 kg bob to 5 m long copper wire of diameter 0.08 cm and it has certain time period of oscillation. Next a 10 kg bob is substituted for the 1 kg bob. Calculate the change in time period if any (Young’s modulus of copper = 12.4 × 1010 N/m2). [Roorkee, 1988] [Ans. 0.0035 s]
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Total extension in the entire rod,
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15. A copper wire of negligible mass, 1 m length and cross-sectional area 10–6 m2 is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotating with an angular velocity of 20 rad/sec. If the elongation in the wire is 10–3 m, obtain the Young’s modulus. If on increasing the angular velocity to 100 rad/sec, the wire breaks down, obtain the breaking stress. [Roorkee, 1992] [Y = 4.00 × 1011 N/m2].
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12. A rod AD, consisting of three segments AB, BC and CD joined together, is hanging vertically from a fixed support at A. The lengths of the three segments are respectively 0.1 m, 0.2 m and 0.15 m. The crosssection of the rod is uniformly 10–4 m2. A weight of 10 kg is hung from D. Cal-
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mω2l r 2 Ans. T = 1 − ; 2 l2
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11. A ring made of lead, of radius r = 25 cm is rotated about a vertical axis passing through its centre and perpendicular to its plane. Find the number of revolutions per second at which this ring will rupture. The density of lead ρ= 11.3 g/cm2 and its tensile strength σ = 0.015 GPa. [Ans. 23 rps.]
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[Roorkee, 1982]
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10. A metallic wire is stretched by suspended weight to it. If α is the longitudinal strain and Y, its Young’s modulus of elasticity, show that the elastic potential energy per 1 unit volume is given by Yα2. 2
[Ans. 2.0 × 10–5/º C; 1.1 × 1011 N/m2]
15. A uniform iron rod of length l and of mass m is rotated with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Find an expression for the tension in the rod as a function of the distance r from the axis of rotation. What is the elongation of the rod?
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9. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of the constituents do not change with rise in temperature. Find the Young’s modulus and coefficient of linear expansion of the second rod. Given: αcu = 1.7 × 10–5/º C, Ycu = 1.3 × 1011 N/m2 [IIT, 1979]
8. A composite rod is made by joining a copper rod end to end with a second rod of different material, but of same cross section. At 25º C, the composite rod is 1 m in length, of which the length of the copper rod is 30 cm. At 125º C, length of the composite rod increases by 1.91 mm.
length of the wire made of material so that it breaks by its own weight? [Roorkee, 1979] [Ans. 34.01 m]
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Elasticity
3.21
QUESTION BANK equal to the jumper’s weight? Explain why or why not.
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7. A bungee jumper leaps from a bridge and comes to a stop a few centimeters above the surface of the water below. At that lowest point, is the tension in the bungee cord
1. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W1 is suspended from its lower end. If S is the area of crosssection of the wire, the stress in the wire at a height (3L/4) from its lower end is (a) W1/S (b) [W1 + (W/4)]/S (c) [W1 + (3W/4)]/S (d) (W1 + W)/S
6. A cylindrical steel bar is compressed by the application of forces of magnitude F at each end. What magnitude forces would be required to compress by the same amount (a) a steel bar of the same cross-sectional area but one half the length? (b) a steel bar of the same length ut one half the radius?
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ONLY ONE OPTION IS CORRECT
2. Two rods of different materials having coefficients of thermal expansion α1 and α2 and Young’s moduli Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 and α2 are in the
5. The columns built by the ancient Greeks and Romans to support temples and other structures are tapered, they are thicker at the bottom than at the top. This certainly has an aesthetic purpose, but is there an engineering purpose as well? What might it be?
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4. Does it take more force to break a longer rope or a shorter rope? Assume the ropes are identical except for their lengths and are ideal there are no weak points. Does it take more energy to break the long rope or the short rope? Explain.
9. Read each of the following statements carefully and state, with reason if it is true or false. (a) The modulus of elasticity of rubber is greater than that of steel. (b) The stretching of a coil is determined by its shear modulus. (c) When a material is under tensile stress, the restoring forces are caused by interatomic attraction. While under compressional stress the restoring force is due to interatomic repulsion. (d) Elastic forces are strictly conservative when Hooke’s law is obeyed. (e) A piece of rubber under an ordinary stress can display 100% strain, yet when unloaded it returns to its original length. This shows that the restoring force in the rubber piece are strictly conservative.
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3. A karate student hits downward on a stack of concrete blocks supported at both ends. A block breaks. Explain where it starts to break first, at the bottom or at the top. (The block experiences shear, compressive and tensile stresses. Recall that concrete has much less tensile strength than comprehensive strength. What part of the block is stretched and which is compressed when the block bends in the middle?
2. What are the advantages of using the concepts of stress and strain to describe deformations?
8. Are you more likely to find steel rods in a horizontal concrete beam or in a vertical concrete column? Is concrete more in need of reinforcement under tensile or compressive stress?
1. Young’s modulus for diamond is about 20 times as large as that of glass. Does that tell you which is stronger? If not, what does it tell you?
CONCEPTUAL QUESTIONS
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3.22
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13. When temperature of a gas is 20ºC and pressure is changed from p1 = 1.01 × 105 Pa to p2 = 1.165 × 105 Pa then the volume changed by 10%. The bulk modulus is (a) 1.55 × 105 Pa (b) 0.115 × 105 Pa (c) 1.4 × 105 Pa (d) 1.01 × 105 Pa
12. A uniform rod of mass m, length L, area of cross-section A and Young’s modulus Y hangs from a rigid support. Its elongation under its own weight will be (a) zero (b) mgL/2YA (c) mgL/YA (d) 2 mgL/YA
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11. A load W produces an extension of 1 m in a thread of radius r. Now if the load is made 4 W and radius is made 2 r all other things remaining same, the extension will become (a) 4 mm (b) 16 mm (c) 1 mm (d) 0.25 mm
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8. The length of elastic string, obeying Hooke’s law is ℓ1 metres when the tension 4 N and ℓ2 metres when the tension is 5 N. The length in metres when the tension is 9 N is (a) 5ℓ1 – 4ℓ2 (b) 5ℓ2 – 4ℓ1 (c) 9ℓ1 – 8ℓ2 (d) 9ℓ2 – 8ℓ1
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7. If the work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of the same material but double the radius and half the length y 1 mm is (a) 16 J (b) 8 J (c) 4 J (d) (1/4) J
(b) YA = YB (d) YA = 3YB
10. A steal wire 1 m log and 1 mm2 cross section area is hung from rigid end. When weight of 1 kg is hung from it then change in length will be (given Y = 2 × 1011 N/m2) (a) 0.5 mm (b) 0.25 mm (c) 0.05 mm (d) 5 mm
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6. Work done in increasing the length of wire 1 m long and of cross section area 1 mm2 by 1 mm by 20 kg-wt will be (a) 0.1 Joule (b) 1 Joule (c) 10 Joule (d) 5 Joule
;
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(a) YB = 2YA (c) YB = 3YA
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5. When load of 5 kg is hung on a wire then extension of 3 m takes place, then work done will be (a) 75 Joule (b) 60 Joule (c) 50 Joule (d) 100 Joule
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4. How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? [Young’s modulus for brass = 0.9 × 1011 N/m2] (a) Nearly 17 N (b) Nearly 34 N (c) Nearly 51 N (d) Nearly 68 N
%
14. Two bodies are suspended from two rods of same length and same area of cross-section.
(d)
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(c)
strain 2 Y× volume stress × strain × volume strain 2 × volume 2Y 1 Y × stress × strain × vol. 2
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3. The elastic energy stored in a wire of Young’s modulus Y is (a)
9. The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and YB are the Young’s moduli of the materials, then
ratio 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9
Elasticity
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(a) (b) (c) (d)
15. The Young’s modulus of a metal is 1.2 × 1011 Nm–2 and its inter atomic force constant is 3.6 × 10–9 N/m, then the average distance between atoms is (a) 15 Å (b) 3 × 10–16 m (c) 3 × 10–20 m (d) None of these
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21. The work required to be done in producing an extension of 1.0 mm in a wire of length 50 cm and area of cross-section 1 mm2 will be (Y = 2 × 1010 N/m2) (a) 6 × 10–2 J (b) 4 × 10–2 J –2 (c) 2 × 10 J (d) 1 × 10–2 J
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22. The work done in increasing the length of a wire of area of cross-section 0.1 mm2 by 1% will be (Y = 9 × 1011 Pa) (a) 2 × 102 J (b) 4.5 × 102 J 2 (c) 3 × 10 J (d) 6 × 102 J
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(d) 9 : 4
(c) 3 3 : 2 2
(b) 3 : 2
3: 2
(a)
19. Find the bulk modulus of elasticity of helium at 1 atmospheric pressure if the variation of pressure and volume is adiabatic in nature.
24. Two springs of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 2 : 3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of
18. A metallic rod of Young’s modulus 2 × 1011 Nm–2 undergoes a strain of 0.05%. The energy stored per unit volume of the rod will be (a) 2.5 × 108 Jm–3 (b) 0.5 × 104 Jm–3 (c) 2.5 × 104 Jm–3 (d) 0.5 × 108 Jm–3
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23. The following four wire (length L and diameter D) are made of the same material. Which of these will have the largest extension when the same tension is applied? (a) L = 50 cm, D = 0.5 mm (b) L = 100 cm, D = 1 mm (c) L = 200 cm, D = 2 mm (d) L = 300 cm, D = 0.5 mm
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17. An iron rod of length 2 m and cross sectional area 50 mm2 stretches by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of the iron rod is (a) 19.6 × 1010 N/m2 (b) 19.6 × 1015 N/m2 (c) 19.6 × 1018 N/m2 (d) 19.6 × 1020 N/m2
20. A steel wire 1.5 m long and of radius 1 mm is attached with a load 3 kg at one end the other end of the wire is fixed. It is whirled in a vertical circle with a frequency 2 Hz. Find the elongation of the wire when the weight is at the lowest position. [Y = 2 × 1011 N/m2, g = 10 ms–2] (a) 1.77 × 10–3 m (b) 7.17 × 10–3 m (c) 3.17 × 10–7 m (d) 1.37 × 10–7 m
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16. A steel wire of uniform cross-sectional area 2 mm2 is heated up to 50ºC and is stretched by clamping its two ends rigidly. The change in tension when the temperature falls from 50ºC to 30ºC is given by – (α = 1.1 × 10–5ºC, Y = 2.0 × 1011 N/m2) (a) 88 N (b) 5 N 10 (c) 1.5 × 2.0 N (d) 2.5 × 2.010 N
7.31 × 105 Pa 1.73 × 105 Pa 5.73 × 107 Pa None of the above
Their Young’s modulii are in the ratio 2 : 3. Then (a) Their lengths will increase in the ratio 3:4 (b) They will be subjected to stresses which are in the ratio 2 : 3 (c) When bodies are displaced vertically and released, they will oscillate with time periods in the ratio 3 : 2 (d) When bodies are displaced and released, time periods will be in the ratio √3 : √2
3.23
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3.24
4. For a substance undergoing extension without any change in its volume (a) σ shows drop in radius with increase in length (b) The poisson’s ratio is σ = 0.5 (c) σ can be –ve (d) σ can be +10
1. A ring made by cutting directly from a metal plate has radius R. The arc ABC makes angle θ at the centre as shown in the figure. If the ring gets heated
5. A wire of radius 1 mm and length 2 m is twisted through 90º. If the modulus of rigidity is 5 × 1010 N/m2, then (a) Angle of shear is 7.85 × 10–4 radians at a point midway. (b) Angle of shear is 7.85 × 10–4 radians on the surface. (c) Couple per unit twist is 3.92 × 10–2 N-m/rad (d) Torsional couple is 6.16 × 10–2 N/m
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
Elasticity
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7. Choose the correct information. 1 (a) Work done in stretching = × 2Y (stress)2 (b) For constant volume poisson’s ratio σ = 0.5 (c) Stress caused by heating is independent of length (d) normal stress causes shear strain always.
8. Arrangement of three metallic rods of equal cross sectional area riveted to each other is as shown in figure. α Al is the coefficient of linear expansion for aluminium and αS (where α Al > α S ) for steel. The arrangement is heated, then
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3. A light rod of 2 m length is suspended from the ceiling horizontally using two vertical wires of equal length tied to its ends. One of the wire is made of steel of cross-sectional area 0.1 cm2 and young modulus 20 × 1011 dyne/cm2, while the second is made of brass of C.S. area 0.2 cm2. and Young’s modulus 10 × 1011 dyne/cm2. Then (a) For stress to be same, a mass can hang at 0.6 m from brass wire end. (b) For stress to be same, a mass can hang at 0.6 m from steel end. (c) For equal strain, the mass should be hung at mid-point of the rod. (d) For equal force, the mass should be at 1/4th from steel end.
6. A beam of light of infinity at an angle 60º with the normal falls on a partially reflecting surface of reflectivity 0.8 there will be (a) tangential stress on the surface (b) normal pressure on the surface (c) only normal pressure on the surface (d) normal pressure will be 3 times the tangential stress on the surface
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(a) Tensile stress is developed in the aluminium rod. (b) Compressive stress is developed in the aluminium rod (c) No stress is developed in the aluminium rod (d) Stress developed in aluminium rod is twice that in steel rod
mg where ℓ is the incre2 ment in length of wire.
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Statement 2: In the above statement, loss in gravitational energy is mgℓ while the loss mg . in energy to surrounding is 2
3. Statement 1: Young’s modulus for a perfectly plastic body is zero. Statement 2: For a perfectly plastic body, restoring force is zero.
ASSERTION AND REASON QUESTIONS
4. Statement 1: Identical springs of steel and copper are equally stretched. More work will be done on the steel spring. Statement 2: Steel is more elastic than copper.
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7. Statement 1: Water is more elastic than air. Statement 2: Air is more compressible than water.
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6. Statement 1: Steel is more elastic than rubber. Statement 2: When same deformation is produced in two identical bodies of these material, greater restoring force develops in the steel body.
(d) Statement 1 is False, Statement 2 is True.
1. Statement 1: The stress-strain graphs are shown in the figure for two materials A and B are shown in figure. Young’s modulus of A is greater than that of B.
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Direction: Each question contains state-ment 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct.
3.25
1. Match the following:
(b) Normal stress
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2. Statement 1: If a metal wire is attached to the ceiling of a room and mass m is attached to another end, the energy stored in the
Column II (p) Micrometer with a spirit level (q) Bulk modulus
Column I (a) Hooke’s law
Statement 2: The Young’s modules for stress small strain is, Y = = slope of linear strain portion, of graph; and slope of A is more than slope that of B.
Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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MATCH THE COLUMN TYPE QUESTIONS
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3.26
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5. Match Column I (Elastic properties of an isotropic elastic material) with Column II (nature of strain produced) and select the correct answer using the codes given below the lists.
1. For which range, during unloading the above curve will be retraced? (a) up to OA only (b) up to OB (c) up to C (d) Never retraced its path
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Column II (p) Depends on temperature (q) Depends on length (r) Depends on area of crosssection (s) Depends on the nature of material
It is observed that for a substantial range of stresses, the stress-strain relation is linear, and the specimen recovers to its original dimensions when the load is removed.
Column I (a) Young’s modulus of a substance (b) Bulk modulus of a substance (c) Modulus of rigidity of a substance
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(d) Position’s ratio
PASSAGE-1 Figure shows qualitatively the relation between the stress and the strain as the deformation gradually increases.
PASSAGE BASED QUESTIONS
(s) energy
(c) Compressibility
Column II (p) pressure (q) energy density (r) angle
Column I (a) Stress (b) Strain (c) Modulus of elasticity (d) Torque
2. Match the following:
Column I Column II (a) Young’s modulus (p) Shear strain (b) Modulus of rigidity (q) Normal strain (c) Bulk modulus (r) Transverse strain (d) Poisson’s ratio (s) Volumetric strain
(r) Change in volume depends on normal stress (d) Searle’s method (s) Elastic limit
(c) Velocity of sound
Elasticity
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4. Two different types of rubber are found to have the stress-strain curve as shown (i) to absorb vibrations one would prefer rubber A (ii) For manufacturing car tyre one would prefer B
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PASSAGE-2 In Searle’s apparatus for determining the Young’s modulus for a wire, a student is taking measurements of different parameters. Length of the wire, L = 30 cm measured with a millimeter scale Diameter of the wire d = 1 mm measured with a screw gauge Least count of the screw gauge = 0.01 mm Least count of the spherometer attached to the apparatus frame = 0.005 mm Spherometer measured the extension (e) of the wire W = weight of the loads.
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3.28
8. The load vs extension graph should be (if the extension is in elastic limit) (a) straight line (b) curved line (c) broken line (d) any of the above
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1. A human bone is subjected to a compressive force of 5.0 × 105 N/m2. The bone is 25 cm long and has an approximate cross sectional area 4.0 cm2. If the ultimate compressive strength of the bone is 1.70 × 108 N/m2, will the bone be compressed or will it break under this force?
(a) Which material has greater Young’s modulus? (b) Which material is more ductile? (c) Which is more brittle? (d) Which of the two is stronger material?
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6. A rod AD consisting of three segments AB, BC and CD joined together is hanging vertically from a fixed support at A. The lengths of the segments are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly 10–4 m2. A weight of 10 kg is hung from D. Calculate the displacements of points B, C and D using the data on Young’s moduli given below (neglect the weight of the rod). YAB = 2.5 × 1010 N/m2, YBC = 4.0 × 1010 N/m2 and YCD = 1.0 × 1010
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5. A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m stretches under a load by 0.7 mm. Calculate the load, given that the Young’s modulus for copper is 1.1 × 1011 Pa and that for steel is 2.0 × 1011 Pa (1 Pa = 1 Nm–2)
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Flat faces at the narrow end of the anvil have a diameter of 0.5 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil? 3. The stress versus strain graph for two materials A and B are shown in figure. The graphs are on the same scale.
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7. The radii of two rods of the same length and same material are in the ratio r1 : r2. If these rods are twisted by applying the same torsional torque, then the ratio of the angle of twist produced in the two rods will be 8. A steel wire of length 4.7 m and crosssection 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-section 4.0 × 10–5 m2 under a
4. What is the density of ocean water at a depth, where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3? Compressibility of water = 45.8 × 10–11 N–1 m2]
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2. Anvils made of single crystals of diamond, with the shape as shown in figure, are used to investigate behaviour of materials under very high pressures.
SUBJECTIVE QUESTIONS
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Elasticity
with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance r from the rotation axis. Find the elongation of the rod. Density of rod is ρ.
9. A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section 10–3 m2 and the other is of brass of cross section 2 × 10–3 m2. Find out the position along the rod at which a weight may be hung to produce, (a) equal stresses in both wires (b) equal strains on both wires Young’s modulus for steel is 2 × 1011 N/m2 and for brass is 1011 N/m2.
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The wires A and C are of copper 1 mm2 in cross-section and B of iron 2 mm2 in crosssection. If the Y for copper and iron is 1 × 1011 and 2 × 1011 N/m2 respectively, find the tension in each wire.
(b) 2 × 10–11 N/m2 (d) 2 × 1013 N/m2
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3. In Searle’s experiment, which is used to find Young’s modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer aof least count 0.001 cm). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error. [2004]
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2. The adjacent graph shows the extension (∆ℓ) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10–6 m2, calculate the Young’s modulus of the material of the wire. [2003]
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1. A thin rod of negligible mass and area of cross-section 4 × 10–6 m2, suspended vertically from one end, has a length of 0.5 m at 100ºC. The rod is cooled to 0ºC, but prevented from contracting by attaching a mass at the lower end. Find (i) its mass and (ii) the energy stored in the rod, given for the rod. Young’s modulus = 1011 N/m2, coefficient of linear expansion 10–5 K–1 and g = 10 m/s2. [1997]
PREVIOUS YEARS’ IIT-JEE QUESTIONS
given load. What is the ratio of the Young’s modulus of steel to that of copper?
3.29
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3.30
4. The pressure of a medium is changed from 1.01 × 105 Pa to 1.165 × 105 Pa and change in volume is 10% keeping temperature constant. The bulk modulus of the medium is
lar reading, the student measures the extension in the length in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young’s modulus obtained from the reading is [2007]
Elasticity
[2005] 204.8 × 105 Pa 102.4 × 105 Pa 51.2 × 105 Pa 1.55 × 105 Pa
(a) (b) (c) (d)
ASSERTION AND REASON QUESTIONS 2. (a)
3. (a)
4. (a)
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5. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particu
(a) (b) (c) (d)
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MATCH THE COLUMN TYPE QUESTIONS 1. (a) → s (b) → q, r (c) → q (d) → p
2. (a) → p, q (b) → r (c) → p, q (d) → s
3. (a) → q (b) → s (c) → p (d) → r
4. (a) → p, s (b) → s (c) → p (d) → p
5. (a) → q (b) → p, (c) → s, (d) → r
PASSAGE BASED QUESTIONS 1. (b)
2. (b)
3. (b)
4. (d)
5. (a)
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6. (b)
7. (c)
8. (d)
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Elasticity
3.31
HINTS AND SOLUTIONS 5 0 9 0 4 0 = = A ( − ) A( 3 − 0 ) A(1 − 0 ) 2 0
Y=
9. (a) False: For a given load, Young’s modulus is inversely proportional to strain which is more for rubber. So elasticity of rubber is less than that of steel. (b) True: Under the application of deforming force, stretching of coil changes its shape without any change in the length of wire used. (c) True: When a material is subjected to tensile stress, the interatomic separation increases and becomes more than r0, due to which interatomic force of attraction comes into play. In case of compressional stress, interatomic separation decreases and becomes less than r0, due to force for repulsion comes into play. (d) False: Elastic forces are conservative as long as loading and unloading curves are identical even if the curves are not linear. (e) False: In case of rubber, elastic hysteresis takes place which signifies nonconservative forces.
These give 4 5 = 1 − 0 2 − 0
CONCEPTUAL QUESTIONS
8. (b) Let ℓ0 be the unstretched length and ℓ3 be the length under a tension of 9 N. Then
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F 5 × 105 = = 12.5 × 108 N/m2 A 4 × 10−4
Since this stress exceeds the ultimate compressive stress of the lone, the bone will break. 2. Here F = 50,000 N, d = 0.5 mm = 5 × 10–4 m
Work necessary in the second case = 8 × 2 = 16 J
F L . A
SUBJECTIVE QUESTIONS
∴
T = 2π
∴ F2 = 8F1 ⇒ W2 = 8W1
YA = tan 60° = 3 YA =3 ⇒ YB = tan 30° = 1/ 3 YB
F YA = 1 L Where ℓ is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by
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ONLY ONE OPTION IS CORRECT 7. (a) Work W =
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3.32
7. Torisional torque required for θ twist is ηπr 4 θ 2 Thus for same τ, ℓ and η, τ = Cθ =
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9. (a) 1.33 m, (b) 1 m mg 10. ∆ = 2AY
4
r 2 1 ∴ θ1 : θ2 = 2 = = 16 :1 4 r η 1
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F 50000 × 4 = = 2.5 × 1011 Pa A 3.14 × (5 × 10−4 ) 2 3. (a) For a given strain, stress is more for material A. stress Young’s modulus = strain Hence, Young’s modulus is greater for A than that of B. (b) For material A, there is a larger stress strain variation beyond elastic point, therefore A is more ductile, than B. (c) For material B, there is a small stressstrain variation beyond elastic point, therefore, B is more brittle than A. (d) The strength of a material, is the measure of the amount of stress required to cause fracture, therefore A is stronger than B.
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Elasticity
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Hydrostatics and Hydrodynamics
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Hydrostatics and Hydrodynamics
The Vertical forces Acting on this fluid are (i) F1, vertically upward by the fluid below it. (ii) F2, vertically downward by the fluid above it. (iii) Weight W, vertically downward. Let the pressure at the surface A be p and the pressure at B be p + dp. Then, F1 = p ∆S And F2 = (p + dp) ∆S The volume of the fluid considered is (∆S) (dz). If the density of the fluid at A is ρ, the mass of the fluid considered is ρ (∆S) (dz) and, then its weight
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4.2 pASCAL’S LAW If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude. https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
4.3
Simply, if the pressure at one point is increased then all other points pressure is also increased because pressure difference in two points is equal. As an example, suppose a flask fitted with a piston is filled with a liquid as shown in the figure. Let an external force F be applied on the piston. If the cross-section area of the piston is A, the pressure just below the piston is increased by F/A. By Pascal’s law, the pressure at any point B will also increase by the same amount F/A. ) $
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Principle of Archimedes states that when a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to weight of the displaced fluid. According to Archimedes’ principle B = mg = weight of fluid displaced where, B = buoyant force B = Vρg where, V = volume occupied by the solid inside the liquid.
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Hydrostatics and Hydrodynamics
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Note: 1. When a body is in air, the net downward force on it is due to earth’s gravity only. This net force = mg = Vdg. This force is known as weight of the body. 2. When the same body is immersed in a liquid of density ρ, the net downward force on it = mg – B where, B = buoyant force ⇒ Fd = Vdg – vρg This downward force is also known as apparant weight in the liquid. 3. When a body is immersed in a liquid, it feels lighter due to buoyant force. Hence, the buoyant force is also known as loss in weight (= vρg)
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Example 90 g of sulphuric acid of specific gravity 1.8 is mixed with 90 g of water. If the specific gravity of the mixture is 1.5, find the contraction when the liquid are mixed. Density of water = 1 gm–3.
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Density of water = 1 gm–3 ∴ Density of sulphuric acid = 1.8 times density of water = 1.8 g cm–3 Volume of 90 g of water = 90 cm3 90 Volume of 90 g of sulphuric acid = = 50 cm3 1.8 ∴ Total volume before mixing = 140 cm3 After mixing, total mass is 180 g and specific gravity of the mixture is 1.5, i.e., density of the mixture is 1.5 g cm–3. 180 = 120 cm3 1.5 1 Hence the mixture contracts by 20 cm3 i.e., of its original volume. 7 Volume of the mixture =
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Example A vertical U tube of uniform inner cross-section contains mercury in both of its arms. A glycerine (density 1.3 g cm3) column of length 10 cm is introduced into one of the arms. Oil of density 0.8 g cm–3 is poured in the other arm until the upper surface of the oil and glycerine are in the same horizontal level. Find the length of oil column. Density of mercury is 13.6 g cm–3. )
) Solution
Let the height of x in the column is poured. When the horizontal line is same of the oil and glycerine. For our previous concept, pressure in horizontal line is equal to each point. So, the pressure at B and E is equal.
4.4
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Hydrostatics and Hydrodynamics
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4.5
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Let the atmospheric pressure p0. Then, pressure at B = p0 + ρ1g × 10 where, ρ1 = density of glycerine Pressure at, E = p0 + ρ2 gx + p3 (10 – x) g where, ρ2 = density of liquid and p3 = density of mercury Hence, p0 + ρ1 g (10) = p0 + ρ2gx + ρ3 g (10 – x) or 1.3 × 10 = 0.8 × x + 13.6 × (10 – x) or 13 = 0.8x + 136 – 13.6 x or 12.8x = 123
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Example A stone of density 2.5 g cm–3 completely immersed in sea water is allowed to sink from rest. Calculate the depth to which the stone would sink in two seconds. Neglect the effect of friction. Specific gravity of sea water is 1.025 and acceleration due to gravity is 980 cms–2. % 9SJ
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Hydrostatics and Hydrodynamics
) Solution
Let the volume of the stone is V. Then weight of stone = Vdg where, d is density of stone. Buoyant force of sea water, B = Vρg Thus, net downward force,
or
Vdg – Vρg = Vda
or
a=
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(2.5 − 1.025) × 980 cm/s 2 2.5 = 578.2 cm/s2 1 s = ut + at 2 2 1 = 0 + × 578.2 × 4 2
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F = Vdg – Vρg From Newton’s Law F = ma
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Example A tank of square cross-section (2m × 2m) is filled with water upto a height of 2.5 m. Find the thrust experienced by the vertical and bottom face of the tank. (g = 10 m/s2).
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) Solution Here, the pressure is uniformly distributed over the bottom surface. Thrust at the bottom surface. = Pressure × Area = (2.5 ρg) × A = 2.5 × 1 × 103 × 10 × 2 × 2 = 1 × 105 N
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Hydrostatics and Hydrodynamics
4.7
On the vertical surface in contact with liquid the pressure is not same at all points. Points at greater depth experience large pressures and hence greater thrust. Consider a vertical surface ABCD (l × h) in contact with liquid of height h. Divide the vertical surface ABCD into rectangular strips of height dx and area ldx. h ρg l h 2 F = ∫ pd A = ∫ ( xρg ) l dx = 2 0 Expressing F as F = [Pressure at C.G. of vertical surface] × [area of surface] h = ρg hl 2
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Example The figure shows a cross-sectional view of a mercury dam whose length perpendicular to the diagram in 30 m. The depth of water behind the dam is 10 m. The masonary of which the dam is constructed has a density of 3000 kg/m3.
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(i) Find the dimensions x and 2x if the weight of the dam is to be 10 times the horizontal force extended on it by water. (ii) Check the stability of the dam with respect to its overturning about the edge through point O.
) Solution (i) Weight of the dam = volume × density
1 = ( x + 2 x)12 × 30 3000 kg 2 = 1620000 x kgf https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
Horizontal thrust due to water in kg f (on a vertical surface) = Pressure at G × Area = 5 × 103 × (30 × 10) = 15 × 105 kg f According to problem
1620000 x = 10 (15 × 105) ∴
x=
2x = 18.52 m
and
1500 = 9.26m 162
Anti-clockwise torque exerted by weight of the dam
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11 2 x g 9 = 169.78 × 106 g Clockwise torque due to liquid thrust
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(ii) Let G be the centre of gravity of the dam. Taking O as origin and left as + x-axis
4.8
= 5 × 106 g As the torque due to liquid thrust is much less than that due to weight of the dam, hence the dam is stable.
Example A cube of wood supporting a 200 gm mass just float in the water. When the mass is removed, the cube rised by 2 cm. What is the size of cube?
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Hydrostatics and Hydrodynamics
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Vertical force = mg = Weight of the cube or 200 g = Weight of the mass B = l3ρg upward where, l = side of the cube At equilibrium (m + 200) g = l3 ρg From figure (b), mg = l2 (l – 2) ρg From equations (i) and (ii), we get 200 = ρ [l3 – (l – 2) l2] Where ρ = 1 g/cc of water Solving, we get l = 10 cm
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…(i) …(ii)
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4.10
Hydrostatics and Hydrodynamics
Example An expansible balloon filled with air floats on the surface of a lake with 2/3 of its volume submerged. How deep must it be sunk in the water so that it is just in equilibrium, neither sinking further nor rising? It is assumed that the temperature of the water is constant and height of the water barometer is 10 metres. )
) Solution
p1V1 = p2V2
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Let V1 be the volume of the balloon, when floating Mass of the balloon = Mass of water displaced 2 = × V1 × 1g 3 2 = V1 gf 3 Let V2 be the volume of the balloon when it is sunk in the water. Since, mass of the balloon is constant, therefore 2 V1 = V2 3 2 −10 × V1 = p2 × V1 3 ∴ p2 = 15 m of water At the surface of the lake pressure is 10 m of water Hence the balloon must be sunk at a depth of (15 – 10) = 5 m.
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When liquid velocity at any point is constant in magnitude and direction at all times, the motion of the liquid is said to be steady. A flow-line in a velocity field is an imaginary line in that space, where the tangent to the line at any point on the line gives the direction of the velocity at that point. A flow line is also called a velocity line or a stream line. The flow of a liquid can be classified as:
4.5 HYD ODYNAMICS
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(i) Stream-line flow In stream-line flow, the stream lines in the flow space remains steady as time progresses. (ii) A Turbulent-flow In turbulent-flow, the stream lines in the flow space shift their positions as the time progresses. In a stream-line flow, a group of stream lines from a tubular volume of the flow space, the surface of which is tangential to the stream lines, forming the lateral boundary of that tubular volume. Such a tubular volume in the flow space is called a tube of flow.
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Hydrostatics and Hydrodynamics
4.11
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(iii) Steady state flow It is the flow in which the liquid proporties at any point in the velocity field do not change with time.
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4.6 EQUATION Of CONTINUITY
Let us consider two cross-sections of a tube of flow at the point A and B. Let the area of cross-section at A be A1, and that at B be A2. Let the speed of the fluid be v1 at A and v2 at B. dM = ρ Av = constant. dt If the fluid is incompressible, density is constant at all points, hence equation of continuity is Av = constant A1 v1 = A1 v2
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4.12
Hydrostatics and Hydrodynamics
The product of the area of cross-section and the speed remains the same at all points of a tube of flow. The equation of continuity is law of conservation of mass in fluid dynamics.
4.7 BErNOULLI’S EQUATION Suppose, the area of cross-section at A = A1 The area of cross-section at B = A2
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Speed at A = v1 Speed at B = v2 Pressure at A = p1 Pressure at B = p2 and, Density of liquid = ρ PQ = horizontal line. Let, δ l1 = length of the fluid flowing into the tube at A1 per second. δ l2 = length of the fluid flowing out of the tube at A2 per second. Since, the fluid is incompressible, therefore A1 δ l1 = A2 δ l2 ...(i) Total energy of the fluid at any cross-section = pressure potential energy + gravitational potential energy + kinetic energy 1 Total energy at A = p1 (A1 δ l1) + (A1 δ l1 ρ) v12 + (A1 δl ρ) gh1 2 where, ρ is the density of the liquid. Total energy at B = p2 (A2 δ l2) + Then, p1 (A1 δ l1) +
1 (A2 δ l2 ρ) v22 + (A2 δ l2 ρ) gh2 2
1 (A1 δ l1 ρ) v12 + (A1 δ l1 ρ) gh1 2
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Hydrostatics and Hydrodynamics
= p2 (A2 δ l2) +
1 (A2 δ l2 ρ) v2 + (A2 δ l2 ρ) gh2 2
4.13 ...(ii)
Dividing both sides by ρ A1 δl1 = ρ A2 δ l2 = rate of the mass flow, we get p1 v12 p ν2 + + gh1 = 2 + 2 + gh2 ρ 2 ρ 2 Generalizing and removing suffixes, we get p ν2 + + gh = constant ρ 2
4.8 VENTUrIMETEr
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This equation is called Bernoulli’s equation for steady non-viscous and incompressible fluid flow.
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This is an instrument for measuring the rate of flow of liquids and gases [fluids]. The instrument is connected horizontally in the tube in which the rate of flow is to be measured. If pA is pressure at A and pB is pressure at B, then pA – pB = h ρg where, h = difference of liquids of density ρ in vertical tubes. Applications of Bernoulli’s equation at A and B in two cases (i.e., liquid, gas).
2 2 ν 2 2 − ν12 = ( pA − pB ) = hρg ρ ρ
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ν 2 2 − ν12 = 2hg
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4.14
Hydrostatics and Hydrodynamics
or
Q2 Q2 − = 2 gh A 22 A12
∴
Q = A1 A 2
(since, Q = Aν)
2hg A12 − A 22
for Gases Let d be the density of the gas pA – pB = hρg
2hρg d (A12 − A 22
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4.9 pITOT TUBE
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Pitot tube is used to measure the speed of flow of a fluid. The cross-section of tube at B is perpendicular to the direction of flow and tube is at rest, so that at B, velocity of fluid becomes zero, i.e., vB = 0 Let v be the velocity of fluid at A and pA and pB are pressures at A and B respectively. ⇒ pB – pA = h ρg where, ρ = density of liquid in vertical tubes.
for Liquids If points A and B are in same horizontal line and area of cross-section is same at A and B, gravitational head is same and the pressure difference is due to difference in velocity. https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
Apply Bernoulli’s theorem at A and B
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Let ρ be the density of the liquid in the vertical tube
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4.15
4.16
Hydrostatics and Hydrodynamics
4.10 VELOCITY Of EffLUX Let us find the velocity with which liquid comes out of a hole at a depth h below the liquid surface. Using Bernoulli’s Theorem
A
B hole
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1 1 pA + ρ vA 2 + ρghA = pB + ρ vB 2 + ρghB 2 2
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1 1 Patm + ρ vA 2 + ρgh = patm + v 2 + 0s 2 2
or
v=
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≈ 2 gh
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Example Water is in stream line flow through a horizontal tube. If the velocity at a crosssection of 4 cm2 is 100 cms–1, calculate. (i) The rate of laminar at that section. (ii) The velocity of the flow at another cross-section of 1 cm3. (iii) Pressure difference between the two sections. (ρ = density of water = 1 g cm–3)
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Hydrostatics and Hydrodynamics
4.17
) Solution (i) Rate of flow = (cross-section) (velocity) = (4 cm3) (100 cms–1) = 400 cm–3 s–1 400 cm −3s −1 (ii) Velocity of 1cm 2 = 1 cm 2 = 400 cm s −1 1 1 p1 + ρv12 = p2 + ρv2 2 2
(iii) We know,
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Example Water shoots out of a pipe and nozzle as shown in the figure. The cross-sectional area for the tube at point A is four times that of the nozzle. The pressure of water at point A is 40.78 × 103 Nm–2 (guage). Find the height h above the nozzle to which water jet will shoot. Neglect all the losses occurred in the above process. K OO P
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1 1 pA + ρvA 2 = ρvN 2 + ρgl 2 2
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…(i)
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Hydrostatics and Hydrodynamics
Using continuity equation AA vA = AN vN vN =
∴
AA vA = 4 vA AN
…(ii)
From equation (i) and (ii), we get pA = ρgl + 7.5 v A 2 =
15 2 ρvA 2
(40.78)103 − (9.8)(1.1) 103
vA = 2 ms–1
⇒
…(iii)
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Applying Bernoulli’s equation between point A and the water jet, we have
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Example Water stands at a depth H in a tank whose sides are vertical. A hole is made in one of the walls at a depth h, below the water surface, calculate
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4.18
KROH %6WUHDP
) Solution (i) Vertical distance travel by steam = (H – h) When path of the stream is parabolic, then, the time taken by the stream to reach the bottom surface is
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Hydrostatics and Hydrodynamics
2(H − h) g
…(i)
t=
In horizontal direction there is no acceleration
Velocity of stream at B = 2gh 2(H − h) g
R = x = v × t = 2 gh 2(H − h) g
R = x = v × t = 2 gh
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R2 = 4 hH –4h2
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(ii) Horizontal Range,
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1. A wooden plank of length 1 m and uniform cross-section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. (exclude the case θ = 0º)
P
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WO KED OUT EXAM LES
4.19
ș
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4.20
Hydrostatics and Hydrodynamics
volume by the pressure and the gravity forces as the fluid flows from point P to point Q.
)
) Solution
Weight of the plank = Aldg where, d = density of the plank. B = (Volume of submerged part) × ρg Let V′ = submerged volume = A (0.5) sec θ where, (0.5) see θ is submerged length. Taking moment at hinged point, we have
4
3 P P
) Solution )
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From equation (i) 1 p2 − p1 = ρ (v12 − v2 2 ) + ρg (h1 − h2 ) 2 1 = × 1000(1 − 0.25) + 1000 × 9.8(2 − 5) 2 1 = × 1000 × 0.75 + (−29400) 2 = 29025 Jm–3 Work done by gravity = ρg (h2 – h1) = 1000 × 9.8 (5 – 2) = 2.94 × 104 Jm–3
3. A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous
2. A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points P and Q at height of 2 metres and 5 metres are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work-done per unit
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l 1 W × sin θ = B ⋅ (0.5sec θ) ⋅ sin θ 2 2 l Al dg sin θ = A × 0.5 sec θ ρg × 0.5 or 2 1 sec θ ⋅ sin θ 2 or d = 0.5 × 0.5 ρ sec2 θ Here, d = 0.5 ρ ∴ 0.5 ρ = 0.5 × 0.5 ρ sec2 θ or sec2 θ = 2 1 or cos 0 θ = 2 1 cos θ = or 2 ∴ θ = 45°
…(i)
Given that h1 = 2 m, h2 = 5 m A1 = 4 × 10–3 m2 A2 = 8 × 10–3 m2 v1 = 1 m/s Applying continuity equation, we have ρ1 A1 v1 = ρ2 A2 v2 But ρ is constant, therefore A1 v1 = A2 v2
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ș
1 1 p1 + ρv12 + ρgh1 = p2 + ρv2 2 + ρgh2 2 2
s_
an
9GJ
m
Applying Bernoulli’s theorem
&*
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4.21
Hydrostatics and Hydrodynamics
where,
and incompressible liquids of density d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure p0.
of liquid of density d.
and
$
Hence from equation (i) D =
G
(ii) Total pressure at the bottom P = patm + pressure due to liquid of density d + pressure of density of liquid 2d + pressure of the solid cylinder.
%
na .c om
K
rio s_ in fo
un iv
er si ta rio s
er s
ita
L = dg . 4
3
iv
.m
un
s: //
This pressure is due to the cylinder. Hence, total pressure p = p0 + dg
ht tp
3
(b) (i) Using Bernoullis equation
)
A A 3L A L LDg = . dg + . 2dg 5 5 4 5 4 …(i)
H H L + 2dg + dg 2 2 4
3H L = p0 + + dg 2 4
) Solution
(a) (i) Weight of solid cylinder is balanced by buoyant force of two liquids
G G
t.m e/
ht tp s:
L 3L or p1 − p2 = d ⋅ g − 2d ⋅ g 4 4
e/
s:
//t
ht
tp
//w
w
w
.ja
m
ar a
(a) A homogeneous solid cylinder of length H A L L < , cross-sectional area is 2 5 immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid. Determine: (i) density D of the solid; and (ii) total pressure at the bottom of the container. (b) The cylinder is removed and the original arrangement is restored. A tiny hole of area s (s << A) is punched on the vertical side of the container at a height h (h < H/2). Determine: (i) The initial speed of efflux of the liquid at the hole. (ii) The horizontal distance x travelled by the liquid initially. (iii) The height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm (Neglect the air resistance in these calculations).
H H + 2dg + pressure 2 2 due to cylinder
or p = p0 + dg
;
5d 4
1 pA + ρ1 vA 2 + ρ1 ghA 2
1 = pB + ρ2 vB 2 + ρ2 ghB 2 Here, ρ1 = d and ρ2 = 2d
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G
+
AL 2dg = Buoyant force of 54
liquid of density 2d.
+
A 3L dg = Buoyant force 5 4
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…(i)
4.22
one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ (> ρ). )
) Solution
The centre of mass composite body. (Wooden stick + small piece of metal) can be obtained by using the expression.
pA = pB because we have opened the liquid to atmosphere and also sv Since, vA = B , therefore A vA = 0 According to the problem given that (s << A), hence from equation (i) H H 1 + 2d − h = 2d vB2 2 2 2
xcm =
dg
3 vB 2 = H − 2h g 2
3 vB = H − 2h g 2 (ii) After liquid injected from the hole vertical velocity is zero. So its free fall 1 h = gt 2 2
o nf
a. rio
ar
3H 3H 3 3H − 4 × = H 8 8 4
4. A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its
rio
i.e.,
ita
rs
iv e
un e/ t.m
tp
ht
dx 2 = 3H − 8 h = 0 dh 3H ∴ h= 8
If l is the length of wooden stick dipped in the liquid, the centre of mass of the buoyant force will be obtained at l xb = 2 For the composite body floating vertically in equilibrium, we will have xcm = xb
ML l = 2(M + m) 2
or
M(L/2) + m(0) M+m ML = 2(M + m)
xcm =
s: //
or x2 = h (3H – 4h) dx For the maximum value = 0, dh d 2x so is also zero. dh
or xm =
s
iv er un
.m
//t
x = h (3H − 4h)
(iii)
ht tp s:
= h (3H − 4h)
where, the subscripts A and B represents wooden stick and piece of metal respectively. Taking the origin at the small piece of metal, we get
e/
s:
ht
tp
2h 3 = H − 2h × g 2
si ta
m
.ja
w
//w
w
Horizontal distance, x = v × t
O
s_ i
an
2h g
∴ t=
/
/
co
∴
MA + MB
m
or
M A xA.cm + M B xB.cm
l=
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M L M+m
Hydrostatics and Hydrodynamics
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…(i)
4.23
Hydrostatics and Hydrodynamics
Now, from the mass of displaced liquid we write (M + m) = (πR2) lσ Substituting l from equation (i), we get
)
ita
un
t.m e/
s: //
ht tp
er si ta rio s
Ah ρB g = ma = A (hA + hB + h) d a
iv
.m
//t
ht tp s:
er s
un iv
e/
s:
tp
ht
1.8 − 1.6 = 0.25 cm 0.8 (iii) When the cylinder dipped in the liquid completely, top of the cylinder is top of the liquid. It means the equilibrium condition is destored. If the volume is more than the cylinder volume dipped in liquid, this extra volume is buoyant upward force. Hence or h =
or
m
//w
w
w
.ja
3
a=
0.25 × 1.2 g g = 2.25 × 0.8 6
6. Two identical cylinderical vessels with their bases at the same level contain a liquid of density ρ. The area of either base is A, but in one vessel the liquid height is h1 and the other h2. Find the work done by gravity in equalizing the levels when the two vessels are connected.
5. Uniform solid cylinder at density 0.8 g/cm floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm3 and 1.2 g/cm3 respectively. The height of liquid A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm (i) Find the total force exerted by liquid A on the cylinder. (ii) Find h, the length of the part of the cylinder in air. (iii) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.
= πR 2 L [ ρσ − ρ]
ar a
rio s_ in fo
m = πR 2 L ρσ − M
or (1.2 + 0.8 + h) × 0.8 = 1.9 × 0.7 + 1.2 × 0.8 or (2.0 + h) × 0.8 = 0.84 + 0.96
M + m = (πR 2 L) ρσ
or
na .c om
or (M + m)2 = πR2 MLσ The mass M of the wooden stick is given by M = πR2 Lρ Substituting this on the right side of equation (ii), we get (M + m)2 = (π R2 L)2 ρσ or
(i) Liquid exerts only horizontal force but this force is balance to each other. Buoyancy force is applied in liquid B. (ii) At equilibrium, weight of the cylinder is balanced by the buoyant force of the liquid. Let the area of the cylinder is A. Then, Volume of the cylinder, V = A (hA + hB + h). From equation of equilibrium A (hA + hB + h) dg = A hA ρAg + A hB ρB g
…(ii)
M (M + m) = πR 2 Lσ M+m
) Solution
K DLU $
3
K$ K
%
3
KD
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K
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Hydrostatics and Hydrodynamics
) Solution )
) Solution )
Obviously, the work done by gravity in equalising the levels is equal to the decrement of the potential energy of the liquid mass. Taking the reference level of potential energy at the common level of the base, the initial height of the C.G. of the system.
Z
\
\
Ah1ρgh1 Ah2ρgh2 + 2 2 = Ah1ρgh1 + Ah2ρgh2
2
fo
in
a.
s_
an
ta
si
rio
iv
un
ρgy2
∫
e/
s:
.m
tp
dp =
x =l
∫ ρω x dx
g ( y2 − y1 ) =
ω2l 2 2
∴
y2 − y1 =
ω2l 2 2g
s:
or
8. A cylindrical vessel with a liquid moves linearly with acceleration a. Find the shape of the surface of the liquid in the vessel.
) Solution )
7. The manometer shown below contain a liquid of density ρ. Find the difference in the levels when the manometer rotates at a constant angular velocity ω about one of its vertical limbs.
ht
tp
h 2 + h22 (h + h ) = A ρg (h1 + h2 ) 1 − 1 2 4 g (h1 + h2 )
1 = A ρg ( h1 − h2 )2 4
2
0
ρgy1
//t
ht
W =A ρg (h1 + h2) (H1 – H2)
ta
h1 + h2 4 Hence, the required work done =
s
er
//t
ht
.m
tp
e/
mg (h1 + h2 ) / 4 + mg (h1 + h2 ) / 4 mg + mg
un
s:
iv
//w w
w
.ja
m
rio
ar
Finally, when the levels are equalized, each cylinder holds the same mass (say m) of liquid on account of the quality of the base area, (h + h ) and the level in each is at a height 1 2 2 from the common base level. Therefore, final height of the C.G. of the system
H2 =
%
m …(i)
h +h 2(h1 + h2 )
;
Let A be the cross-section of the lower limb. The element of liquid of length dx has the mass dm = ρ (A dx) and its acceleration an = ω2x, normally towards the axis of rotation. If the pressure at the end of the mass element are p and p + dp at the distance x and x + dx respectively from the point O, we have Fn = man (p + dp) A – pA = ρ (A dx) ω2x or dp = ρω2 dx …(i) But at point O, p0 = ρgy1 and at B, p = ρg y2
co
2 2
=
2 1
G[
si
m1 gh1 m2 gh2 + 2 H1 = 2 m1 + m2
er
4.24
Let us consider a liquid element of crosssectional area A and of length l as shown in
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4.25
Hydrostatics and Hydrodynamics
h1 − h2 a = = constant l g
9. A rod of length 6 m has a mass of 12 kg. It is hinged at one end at a distance of 3 m below the water surface. (i) What weight must be attached to the other end of the rod so that 5 m of the rod are submerged? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material is 0.5.
the figure. Using Newton’s second law of motion for the liquid element F x = m ax or p1 A – p2 A = (A lρ) a ...(i) As the liquid has no vertical acceleration, therefore p1 – p2 = (h1 – h2) ρg ...(ii) From equation (i) and (ii), we get
)
) Solution
in f
m
ita ta si er ni v
AG = 3 m = 300 cm Density of the material of the rod
e/ u
.m
//t
= 0.5 g cm–3 Volume of the submerged part
s:
…(ii)
10 kg 0.5g cm −3
= 2 × 104 cm3 Force of buoyancy FB = Weight of water displaced = Vdg = 2 × 104 × 1.0 gf
Using equations (ii) and (iii), we have dp = – ρa dx – ρg dy But on the free surface dp = 0, thus from equation (iii) dy a =− dx y Thus, the sought shape is a straight line where a slope is − . g
=
mass of 5m of rod Density of rod
ht
tp
V=
But,
And
rio s
rs
.m
s:
tp
ht
∂p = ρa ∂x ∂p − − ρg = 0 ∂x ∂p ∂p dp = dx + dy ∂x ∂y
−
W
θ
AG' = 2.5 m = 250 cm
//t
Let us indicate the positive x-axis in the horizontal direction directed toward right and positive y-axis in upward direction. Then, we have ax = 0 ay = 0 and gy = – g, gx = 0 …(i)
W = Mg
G'
R φ
e/ un i
//w
ht tp s: →
−∇ p + ρg = ρa
FB G
L
ve
w
w
.ja
That’s why the shape of the liquid in the vessel is a straight line as shown in the figure. Another Method From Euler’s equation:
B
Weight of the rod, W = 12 kg f = 12 × 103 gf (i) Let R be the reaction at the hinge. Taking the moments of all forces acting on the rod about A, for equilibrium of the rod, the algebraic sum of moments of the forces must be zero. Therefore
l
So,
o
om
ar
rio s_
2
1
an a
.c
h2 h1
In the figure, G is the centre of gravity of the rod and G' is the central gravity of the submerged part of the rod. Hence, G' is also the centre of buoyancy through which the force of buoyancy FB acts vertically upwards.
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4.26
A ⇒ v2 = 1 (v1 ) = 4 v1 A2
m
ht tp s:
s
rio
si ta
ni
//t .m
e/ u
s: //
ht tp
= 280 × 103 + (– 108) × 103
= 172 ×103 Nm–2 (ii) Let Fx and Fy be the external forces on the pipe bend along x-direction and y-direction respectively. Equation of fluid is
t.m
ht tp
10. Oil enters the bend of a pipe in the horizontal plane with velocity 4 ms–1 and pressure 280 × 103 Nm–2 as shown in the figure. Find (i) The pressure of oil at the point Q. (ii) The force required to hold the bend in the place. (Take specific gravity of oil as 0.9 and sin 37º = 0.6)
of oil = 0.9 × 103 kg m–3]
φ = 0, R is vertical and negative sign just shows that it is directed vertically downwards (θ = 180º).
nf o
s_ i
si ta
ve r
e/ un i
s: //w
w
17 7 = + 12 − 20 kgf = − kgf 3 3
rio
m
w
.ja
∴ R = w + W – FB
∴ φ = 0 or 180º
Also, (R cos φ + FB) = (w + W)
1 1 p1 + ρ v12 = p2 + ρ v2 2 2 2 (Pressure in gravity does not include because plane is horizontal.) 1 p2 = p1 + ρ (v12 − v2 2 ) 2 1 = p1 + × 0.9[16 − 256] × 103 [density 2
co
ar an a.
⇒ R sin φ = 0
Applying Bernoulli’s equation between points P and Q, we have
7 or w = × 103 g = 2.33 kgf 3 Hence, a weight of 2.33 kg should be suspended from the end B.
p2 = A2
+y
V2 Q
+x p1A1 V1 37° P
Q
V2
A2 = A1 4
A1= 0.2 m V1= 4 ms-1
2
37° P
p1 A1 + p2 A2 cos 37° + Fx = [ρQ] (– v2 cos 37° – v1) A 4 ⇒ p1 A1 + p2 1 × + Fx 4 5 4 = ρ A1 v1 −(4 v1 ) − v1 5
)
or w (600) = (2 × 104 × 250) – (12 × 103 × 300)
w · AB = FB AG´ – W × AG
(ii) Resolving R into horizontal and vertical components for translation equilibrium of the rod, we must have
(i) Let A1 and A2 be the cross-sectional area of the pipe at points P and Q respectively. Let v1 and v2 be the velocities of oil at the points P and Q respectively. By conservation of mass. Q = A1v1 = A2v2
∴
(Taking clockwise moments to be negative)
) Solution
Hence, –W · AG sin θ – w · AB sin θ + FB AG´ sin θ = 0
ve r
Hydrostatics and Hydrodynamics
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4.27
Hydrostatics and Hydrodynamics
Fx p 16 = p1 + 2 + (ρ v12 ) 1 + A1 5 5
upward at the back wall of the vessel. To prevent water from flowing out of the hole H, the acceleration of the vessel should have such a 172 21 3value that it occupies a face area DBH and a Fx = −(0.2) 280 + + 0.9 × 16 × × 10 5 5 A width of the vessel given by L 172 21 Fx = −(0.2) 280 + + 0.9 × 16 × × 103 1 5 5 Area of ∆DBH = bc 2 ≈ – 7.5 × 104 N (to the left) D L Along y-direction – p2 A2 sin 37° + Fy ρQ (v2 sin 37°) −
o
ita rs
w
∴
rio
s
1 A Volume of liquid retained = bc × 2 L Let ρ be the density of the water. bc Aρ Mass of the water retained = 2L
er iv
un
//t
Mass of cart and water = M +
bc Aρ 2L
m e/
s: tp
b
ht
tp
s:
//t .
ht c
.m e/
L
ht
tp
s:
un
//w
iv e
w
F = Fx2 + Fy2 = 76 × 103 N.
F
H
rio
ar
m
.ja
= 12.1 × 103 N
s_
B
12 3 × 172 3 ⇒ Fy = (0.2) 0.9 × 16 × + × 10 5 20 ∴
in f
a.
12 3P = ρ v12 + 2 5 20
A1
b
si ta
Fy
⇒
A1 3 3 × + Fy = ρ A1 v1 4 v1 4 4 5
co m
⇒ − p2
an
∴
F
R
b mg
a
11. A vessel with a hole in its bottom is fastened on the cart. The mass of the vessel and the cart is M. With what force F should the cart be pulled so that a maximum amount of water remains in the vessel. The dimensions of the vessel are as shown in the figure.
As the cart is drawn by a force F, the water in the vessel takes up a slant position rising
)
) Solution
Let us consider a small element of the water of mass m. Now, the element is in equilibrium under the action of following forces: (i) Its weight mg acting downwards. (ii) The inertial force ma acting in a direction opposite to the acceleration a of the cart. (iii) The normal reaction, R.
θ
c
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4.28
Hydrostatics and Hydrodynamics
If θ be the inclination of the water surface with the horizontal, therefore
2 1
ma mg
tan θ =
b c Required force, F = mass × acceleration
or
a = g tan θ = g ×
Rate of flow Q = A1 v1
om
in
.m e/
si
ta
un iv
er
ni v
//t
m
e/ u
13. A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional are A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contain a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate: (i) The acceleration of the container. (ii) Its velocity when 75% of the liquid has drained out.
) Solution )
A2 − A2 1 p1 − p2 = ρ v12 1 2 2 2 A2
ht
2
//t .
…(ii)
1 1 vA p1 + ρ v12 = p2 + ρ 1 1 2 2 A2
2 × 8 × 10−2 × 9.8 × 13.6 × 103 103 × π2 (54 − 34 ) × 10−6
= 13991 × 10–5 m3/sec = 139.9 cc/sec
Putting the value of v2 in equation (i), we get
or
s_
rio
ta
Q = π (3 ×10–2)2 π (5 × 10–2)2
tp s:
tp s:
v1A1 A2
ht
v2 =
ht A 2 v2 A1
v1 =
Substituting the numerical values, we get
er si
w .ja
tp s
://
w w
1 1 p1 + ρ v12 = p2 + ρ v2 2 …(i) 2 2 where, v1 and v2 are the velocities at these points respectively A1 v1 A2 v2
or
fo
an a. c
ar
)
In the Venturimeter, if p1 is the pressure at 1 and p2 the pressure at part 2, we have
2( p1 − p2 ) ρ (A12 − A 22 )
where, (p1 – p2) is the pressure difference and A1 and A2 are the area of cross-sections of pipeline and throat.
m
) Solution
= A1 A 2
s
12. The diameter of the throat of a venturimeter is 6 cm. When it is inserted in a horizontal pipe line of diameter 10 cm, the pressure difference between the pipe and throat is 8 cm of mercury. Calculate the rate of flow of the water.
rio
bc A ρ gb = M + 2 L c
h
∴
v1 = A 2
2( p1 − p2 ) ρ (A12 − A 22
Let the height of the liquid column in the container during its motion at an arbitrary instant
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4.29
Hydrostatics and Hydrodynamics
of time t is z (t) and the speed of container at that moment is v(t) as shown in the figure.
∴
t=
200 1/2 1/2 ( z0 − z ) 2g
Initially, m0 = ρA z0 1 and Finally m0 = ρAz 4
ρA −ρ A dz = 2 gzdt 100
s_ in fo
dt =
−100 dz ⋅ 1/2 2g z
s
rio
ita
rs
ve
e/ un i
ht
dV dh =A dt dt
…(ii)
or dV = A dh or A dh = a u dt From equations (i) and (iii), we get A dh dt = a 2 gh
Now at t = 0, v = 0 g v= t ∴ 50 For the liquid in the container
or
u = 2 gh On the time interval dt, volumn of the water ejected through orifice dV = au dt ...(i) Volume of water in the vessel at time t, V = Ah ...(ii) Differentiating equation (ii) with respect to time, we get
or
t
or
dv dm = Fext + vrcl dt dt dv g = = 19.6 × 10−2 m/s 2 dt 50
Let at any time t, height of water in the vessel be h. Thus, speed of flow of the water through the orifice at that moment
∫ dt = 0
∴
m
tp
s:
//t
ht
.m
tp s:
ρA 2 gzdt …(i) 100 [This equation (i) can also be derived by using the well known equation for a variable mass system (in this case container with liquid)], ρ Az dv =
)
e/
.m
//t
ht tp s:
un iv
//w
time dt. So, from the conservation of linear momentum for the container with remaining liquid + ejected mass of liquid system along horizontal direction gives
) Solution
w
w
ρA A 2 gz 2 gz = 2gzdt in mentum ρ 100 100
14. A cylindrical vessel of height H and base area A is filled with water. An orifice of area a is opened in the bottom of the vessel. Neglecting the viscosity of the water, determine how soon all the water will pour out of the vessel?
er si ta rio
m
2gz towards right, taking out mo-
.ja
velocity
ar a
na .
co
Let us choose an inertial frame in which the container is at rest at time t. In this frame the velocity of the container at time t + dt is dv to the left, while the liquid moves out with
200 2g
m
z(t)
t=
∴
m 1/2 m 1/2 0 − 0 …(iii) ρA 4 ρA From equation (iii) and (ii), we get the speed of container as m 1/2 m 1/2 v = 2 2 g 0 − 0 ρA 4 ρA
V(t)
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t=
A a 2g
0
∫
H
dh h
A 2H a g https://t.me/universitarios
…(iii)
4.30
s_ in f
o
ita rio
ita
ve rs
)
) Solution
e/ un i
∴ t=
2h g
Total time back to the original position, 2h = t1 (given in the problem) g
1 ∴ h = g t12 8
2t = 2
From equations (i) and (ii), we have T – T0 = Vda or T = T0 + Vda
1 2 gt 2
h
ht tp
s: //t .m
ht tp
rio s
ni ve rs e/ u t.m
s: //
)
(i) If d < dL, obtain an expression (in terms of d, t1 and dL) for the time t2, the ball takes to come back to the position from which it was released. (ii) Is the motion of the ball simple harmonic? (iii) If d = dL, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.
(i) Equation of motion, h =
When the block is at rest, from Newton’s law B – mg + T0 = 0 ...(i) where, B = Buoyancy force, and mg = Weight of block. Let the volume of the block is V. It is given that in the problem density of the liquid is greater than the solid block so, let the density of liquid is ρ and the density of solid is d. ∴ Vρg – Vdg + T0 = 0 ...(i) The block is in rest with respect to the vessel which is accelerated with respect to the ground. Hence, the acceleration of the block with respect to ground is ‘a’ upward. T + Vρg – Vdg = Vda ...(ii)
a = T0 1 + g
an a. m ar .ja w w w tp s: // ht ) Solution
T0 a g
16. A ball of density d is dropped on a horizontal solid surface. It bounces elastically from the surface and return to its original position in a time t1. Next the ball is released and its falls through the same height before striking the surface of a liquid of density dL.
co m
a given byT0 1 + . g
or
T = T0 +
or
15. The tension in a string holding a solid block below the surface of a liquid (of density greater than solid) is T0, when the containing vessel, as shown in the figure is at rest. Show that tension T, where the vessel has an upward acceleration a, is
Hydrostatics and Hydrodynamics
…(i)
In the second case, body released the some height when it is striking the surface it falls in liquid.
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Hydrostatics and Hydrodynamics
(ii) No
Hence, t2 = Time taken, Fall height h + time taken when the body goes in liquid in zero velocity + come back to the original position where the block is released.
(d L − d ) g =0 d Thus, the upward acceleration is zero. So, thus the ball continuous goes down in the liquid.
(iii) a =
where, t3 = Time taken when velocity of solid is zero.
t4 = Return back to the surface. t5 = From surface to original position. Upward acceleration of body in liquid,
rio s
ta si er iv un
m //t .
ht tp
(d L − d ) g d 2h × = 2 gh d (d L − d ) g
) Solution
At extreme position, Weight of the rod = S L d1 g Buoyancying force B = S L d2 g
=
s:
t1 d d 2h = (d L − d ) g 2(d L − d )
d2
e/
tp s
ht
2h3 2hd ⋅ d = a (d L − d ).g (d L − d )
v4 = at4
in
s_
rs ve ni
e/ u
t.m
://
So, t4 = =
d1
v2 2 gh d hd = = 2a 2.g (d L − d ) (d L − d )
ht
h3 =
ita
w .ja m
//w
2 gh d t1 d v = = a g (d L − d ) 2(d L − d )
tp s:
t3 =
w
∴
rio
Velocity of the block before strike the liquid v = 2 gh
fo
om a. c an
g (d L − d ) d
)
=
V d L g − Vdg Vd
ar
a=
17. A thin rod of length L and area of crosssection S is pivoted at its lowest point P inside a stationary, homogeneous and nonviscous liquid. The rod is free to rotates in a vertical plane about a horizontal axis passing through P. The density d1 of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by a small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic motion and determine its angular frequency in terms of the given parameters.
2h + t 3 + t 4 + t5 g
t2 =
2h t1 = g 2
t5 =
Hence, t2 =
4.31
t1 t1 d t1 d t + + + 1 2 2(d L − d ) 2(d L − d ) 2 o
dL d = t1 1 + = t1 (d L − d ) dL − d
θ
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4.32
Hydrostatics and Hydrodynamics
I = Moment of intertia at the rod about O. 1 = (S L d1 ) L2 3
Taking torque about a point O, we have L τ0 = (S L d2 g – S L d1 g) sin θ 2 L2 = Sg (d2 – d1) θ 2 (∵ θ is very small then sin θ ≈ θ) τ0 = Iα SL2 g (d 2 − d1 ) θ 2I Hence, angular acceleration is directly proportional to the angle so that the motion is simple harmonic. ∴
α=
3 d 2 − d1 g 2 d1 L
=
rio s_ in fo
na .c om
SL2 g (d 2 − d1 ) 1 2. (SL d1 ) L2 3
ω=
P
(c) the balance A will read less than 2 kg and B will read more than 5 kg (d) the balance A and B will read 2 kg and 5 kg respectively
=
Wa W Wa = a = =4 Wa − WW ∆W Wa / 4
er si ta rio s
un
m
[since weight of the body decreases by 25% of its original weight Wa (weight in air)] 2. The spring balance A read 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid as shown. In this situation (a) the balance A will read more than 2 kg (b) the balance B will read more than 5 kg
B
Ans. (b) and (c)
) Explanation When the mass is submerged )
A
t.m e/ s: //
Weight of the body in air Weight of the body in air − Weight of the body water
ht tp
=
ht tp s:
weight
iv
.m
//t
ht
)
) Explanation Relative density = Specific gravity
ita
e/
un iv
er s
w
s:
tp
//w
w
.ja
m
1. A body weights W in air and it loses its weight by 25% in water. The relative density of the body is 1 (a) 4 (b) 4 3 4 (c) (d) 4 3 Ans. (a)
ar a
SOLVED OBJECTIVE TY E QUESTIONS
in the liquid, a force of buoyancy act on it which reduces the reading of balance A. The reaction of this buoyancy is experienced by the liquid therefore the reading of balance B increases.
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3. A cube of mass m and density D is suspended form the point P by a spring of stiffness k. The system kept inside a beaker filled with a liquid of density d. The elongation in the spring, assuming D > d, is
⇒
A
a
ita rio s_
h
(b) lρa (d) hρg + lρa
ta
rio
s
er s
(a) hρg (c) hρg – lρa Ans. (d)
un iv
si
)
er
of width dx and area of cross-section ‘a’, at a distance x from the pont of the tank Mass of element = dm = (αdx) ρ Net force to the right on the element = (P + dP) α – Pα = αdP ∴ αdP = (Pαdx) α
iv
.m //t tp s:
.m e/
un
//t s:
tp
ht
or
C
C
A
A
∫ dP = ∫ Padx
ht
(i) spring force k x, where x = elongation of the spring, (ii) gravitational force w, weight of the cube = mg (iii) buoyant force Fb (or upward thrust) imparted by the liquid on the cube given as Fb = Vdg here V = volume of the immersed portion of the cube. For complete immersion, V = volume of the cube For equilibrium of the cube, kx + Fb = mg
dx
x
C
) Explanation Consider an element of the liquid
e/
s:
ht tp
mg
l
in fo
na .c
ar a
m
.ja
w
w
//w
)
α P
or PC – PA = Pal Also PB – PC = Pgh or PB – (PA + Pal) = Pgh or PB – PA = hPg + lPa
Fb
4. A sealed tank containing a liquid of density ρ moves with a horizontal acceleration a as shown in the figure. The difference in pressure between the points A and B is
B
) Explanation The cube is in equilibrium under
m
mg d 1 − k D
om
R
kx
x=
(d) none of these P
three forces
mg − Fb mg − Vdg = , k k
⇒
mg D 1 − k d
. . . . ... .. . . . ... . .. . . . . .. . . . . . . . . . . . . ... . .. . . . .. . ... . .. .. . . . ... . .. . . . . .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . .... . .. . . . . . . .. . . . . .. . .. . . . . . . . . . . . . . . ... . . .. . . . . .... ... . ..... .. . . . ... . .. . . . . .. . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . .. .. . . . . . . ... . .... ... . . ..... ... . . ..... .. . . . .... .. . . . . . . . . . . . . . .. .. . . . . . . .. . . . . .... .. . . . . . . . . . . .. . . . . .. .. . . . . .. .. . . . .. . . . . . . . . . .... ... . ..... .. . . m . . . . . . . . .. . . . . .. ... . . .. . . . . .... ... . . ..... .... . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .... . . .. . .. . . . . . . . . . . . .. . . .. . .. . . . . . .. .. . . . .. ... . . .. . . . . .... ... . . ..... .... . . . . . . .. .
4.33
m where V = D
mg d 1 + k D
(c)
(b)
mg d 1 − k D
(a)
x=
Hydrostatics and Hydrodynamics
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4.34
)
of mass, M = 1000 kg
∴
m
ta
h
si er iv
(a) l decreases and h increases (b) l increases and h decreases (c) both l and h increase (d) both l and h decrease Ans. (d)
decrease As density of coin is more than that of water, h will decrease
= (1.013 + 0.049) × 105 N = 1.062 × 105 N 6. A vessel contains oil (density = 0.8 g/cm2) over mercury (density = 13.6 g/cm3). A homogeneous sphere floats with half its volume immersed in mercury and the
8. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The column of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density ρ where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is
103 × 9.8 × 1 2 = 1.013 × 105 (1) N 2
) Explanation Due to decrease in weight, l will )
ht
tp
s: //
t.m
e/
un
//t
tp s:
ht
Pressure at C = PC =
ρh 2 F = Patm + ( ) 2
rio s
er e/
.m
tp
ht
PA + PB 2 = average pressure over the side wall ρg = Patm + 2 The force exerted on the side wall is
l
iv un
s: //
Given: Volume of the tank, V = 1m3 and Side of the tank = b = 1m Pressure at A = PA = Patm Pressure at B = PB = Patm + ρgh = Patm + ρgl ∴
Coin
si
.ja w w
w
B
o
ta rio
ar a
s_ i
nf
na .
A C
d1 + d 2 0.8 + 13.6 = = 7.2 g /cm 2 2 2
co m
1000 kg = = 1m3 1000 kg /m 2
h
d=
7. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there. After some time the coin falls into the water. Then
m ρw
Volume of water V =
)
4 3 2 2 πr dg = πr 3d1 g + πr 2 d 2 g 3 3 3
) Explanation The cubical tank contains water ⇒
) Explanation From the principle of floation
other half in oil. The density of the material of the sphere in g/cm3 is (a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 Ans. (c)
5. A cubical water tank is completely filled with water of mass m = 1000 kg. The force exerted by the water on the side wall of the tank, is (a) 1.062 × 105 N (b) 1.62 × 105 N 5 (c) 0.049 × 10 N (d) None of these Ans. (a)
Hydrostatics and Hydrodynamics
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Hydrostatics and Hydrodynamics
4.35
)
) Explanation Applying Bernoulli’s Principle
between the points 1 and 2, we have 1 1 ρgh1 + P1 + ρv12 = ρgh2 + P2 + ρv22 2 2 where, P1 = P0 + ρgh, P2 = P0, h1 = h2 = h1, P0 is the atmospheric pressure.
h
p 2R
h
m
in fo
a. co
(b) Mg – Vρg (d) ρg(V + πR2h)
(a) Mg (c) Mg + πR2 hρg Ans. (d)
h'
an
ta rio
ar
.ja
si
ta r
iv
si er
t.m
:// t.m
R = v2 × t t = time of fall can be given by
1 2 2h1 gt ⇒ t = 2 g
⇒ R = v2
ht tp s
h1 =
iv
Range where
e/ un
://
tp s
ht
9 AV 2 a <
1 ρgh = ρv22 , v2 = 2 gh (say ) 2
un
e/
tp s
ht
R
io s
er
w
w
:// w
Since,
m
and
9. A small hole is made at a height of 1 h′ = m from the bottom of a cylin 2 drical water tank and at a depth of h = 2 m from the upper level of water in the tank. The distance where the water emerging from the hole strikes the ground is
2h1 ;Putting v2 2 gh , we get g
R = 2 hh′
1 m and h = 2 m, 2 we get R = 2m
(c) 2 m Ans. (c)
(b) 1 m (d) none of these
(a) 2 2 m
R
10. In a cylindrical water tank there are two small holes Q and P on the wall at a depth of h1 from upper level of water and at a height of h2 from the lower end of the tank respectively as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of h1 is and h2 is
Putting h′ =
s_
cylinder – force at the top of cylinder = buoyant force on the cylinder i.e., F – πR2 ρgh = Vρg ⇒ F = ρg (V + πR2h)
)
) Explanation The Force on the bottom of the
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4.36
R1 = R 2 = R ⇒ v1t1 = v2t2 where, v2 = velocity of the efflux at
Q
Hydrostatics and Hydrodynamics
....(i)
2 = 2 g (H − h2 ) P
t1 = time of fall of the water stream through 2 (H − h1 ) g
(1) =
t2 = time of fall of the water stream through
2h2 g Putting these values in equation (i), we get (2) =
)
na .c om
) Explanation Since two streams strike at the
(H – h1) h1 = (H – h2) h2
or
ar a
v1
ita
h1
e/
un t.m e/ P
olved
ht tp s:
//t
iv
.m
tp ht un
OBJECTIVE TY E QUESTIONS of both the limbs are equal, what will be the difference in heights h of the liquid in the two limbs of U-tube, when the system is given an acceleration a in horizontal direction towards right as shown?
s: //
1. A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ0 (ρ0 rel="nofollow"> ρ0). The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle θ with the horizontal. Then
1 ρ0 2 ρ
(d) sin θ =
ρ0 ρ
ρ ρ0
(b) sin θ =
(c) sin θ =
1 ρ0 2 ρ
(a) sin θ =
ht tp
S
h1 =1 h2
a H
2. A liquid stands at the plane level in the U-tube when at rest. If area of cross-section
or
er si ta rio s
er s un iv
v2
s:
//w
w
P
w
.ja
H
[H – (h1 + h2)] [h1 – h2] = 0
Here, H – h1 + h2 is irrelevant because the holes are at two different heights ∴ h1 = h2
m
Q
h2
same point on the ground, therefore
rio s_ in fo
(b) 2 (d) < 1
(a) 1 (c) > 1 Ans. (a)
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L
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Hydrostatics and Hydrodynamics
La H
h
Hg (d) a
L2 a (c) H g
(b)
g L2 aH
(a)
4.37
si rio
s
er
iv
er
iv
un
8. An open tank containing non viscous liquid to a height of 5 m is placed over the ground. A heavy spherical ball falls from height 40 m over the ground in the tank. Ignoring air resistance, find the height to which ball will go back. Collision between ball and bottom of tank is elastic having e = 0.5. (a) 45 m (b) 35 m (c) 40 m (d) 20 m
9. A metallic sphere floats in immiscible mixture of water (ρw = 103 kg/m2) and a liquid (ρL = 13.5 × 103 kg/m3) such that
e/
.m
ht
tp
s:
//t
ht
(b) 20 cm (d) 100 cm
si
ta
un
e/
.m
tp
s:
//t
ht
s_ rio ta
w
//w w
s:
tp
(a) 10 cm (c) 50 cm
6. A conical block, floats on water with 90% height immersed in it. Height h of the block is equal to diameter, i.e., 20 cm. The mass to be kept on the block, so that the block just floats at the surface of water, is (ρwater = 1 g/cc)
fo
in
a.
an
ar
m
.ja
g −a (d) 2 g ρg g+a
5. The weight of a ballon is W1 when empty and W2 when filled with air. Both are weight in air by the some sensitive spring balance and under identical conditions. (a) W1 = W2 as the weight of air in the ballon is offset by the force of bouyancy on it (b) W2 < W1 due to the force of bouyancy acting on the filled balloon (c) W2 > W1 as the air inside is at a greater pressure and force has greater density than the air outside (d) W2 = W1 + weight of the air inside it
m
co
7. A block of mass 10 kg connected to another hollow block of same size and negligible mass, by a spring of spring constant 500 N/m, floats in water as shown in the figure. The compression in the spring is (ρwater = 1 × 103 kg/m3, g = 10 m/s2)
4. A beaker containing a liquid of density ρ moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is (a) hρg (b) hρ (g + a) (c) hρ (g – a)
(b) 980 g (d) 196 g
(a) 568 g (c) 112 g
3. There is a hole at the bottom of a large open vessel. If water is filled up to a height hm, its flow rate is ν. If water is filled to a height 4h, its flow rate will be (a) 4v (b) 3v (c) 4v (d) v
where, L = length of horizontal portion of the tube, H = initial height of liquid in both limbs
1 4 portion is in water and th 5 5 portion of the sphere is in the liquid. The density of the metal in kg/m3 is
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4.38
o
nf
s_ i
s
3g 8
si
ta
rio
rio
si ta
(d)
er iv un e/
.m //t
A
B
s:
16. A conical vessel is completely filled with a liquid of density s as shown in the figure. The force exerted by the liquid on the curved surface is
12. A wooden cube just floats inside water when a 200 g mass is placed on it. When the mass is removed, the upper surface of the cube is 2 cm above the water level. The size of the cube is
F
O
15. Air stream flows horizontally past an aeroplane wing of surface area 4m2. The speed of air over the top surface is 60 m/s and under the bottom surface is 40 m/s. The force of lift on the wing is (density of air = 1 kg/m3) (a) 800 N (b) 1000 N (c) 4000 N (d) 3200 N
ht
R
tp
P
m
er
un
e/
.m
//t
s:
tp
3 ρgR 3 2 (d) 0
(b)
ht
3g 4
(c)
iv
//w
s:
tp
ht
9 ρgR 3 2 (c) ρgR3
(a)
14. A unifrom wooden rod of length L and specific gravity S is free to rotate in vertical plane about the end a. The rod is released in the horizontal position inside liquid of specific gravity 1.5 S as shown in the figure. The instantaneous acceleration of the end B is g (a) g (b) 4
co a.
an ar m .ja
w w
11. The figure shows a semi-cylindrical massless gate (of width R) pivoted at the point O holding a stationary liquid of density ρ. A horizontal force F is applied at its lowest position to keep it stationary. The magnitude of the force is
P
D
P
13. A boat carrying iron balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water? (a) It will remain unchanged (b) It will rise (c) It will fall (d) First it will rise and then fall
10. The minimum horizontal acceleration of the container so that the pressure at the point A of the container becomes atmospheric is 3 4 (a) g (b) g 2 3 3 (c) g (d) g 4
(b) 10 cm (d) 20 cm
(a) 5 cm (c) 15 cm
(b) 4.0 × 103 (d) 3.0 × 103
(a) 4.5 × 103 (c) 3.5 × 103
Hydrostatics and Hydrodynamics
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4.39
Hydrostatics and Hydrodynamics
R
H2
H1
R
(d) 10 W
W + ρgHS 10
iv
.m
W −ρgHS 10
(b)
W
s: //
t.m e/
un
//t
ht tp s:
W 10
er si ta rio s
(c)
un iv
(a)
e/
s:
tp
ht
Liquid of density p
H
18. Water flows from a large reservoir through a pipe of cross sectional area S into a big stagnant lake. The level of water in the lake is at a height H1, above the end of the pipe as shown in the figure. The flow rate of water under steady state is
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(d) S 2 g (H1 − H 2 )
(c) S 2 g (H 2 − H1 )
20. A cylindrical drum open at the top contains 30 litre of water. It drains out through a small opening at the bottom 10 litres of water comes out in time t, the next 10 litres in further time t2 and the last 10 litres in further time t3. Then (a) t1 = t2 = t3 (b) t1 > t2 > t3 (c) t1 < t2 < t3 (d) t2 > t1 = t3
(b) S 2 gH 2
F
(a) S 2 gH1
ht tp
rio s_ in fo
ita
w w
//w
17. A piece of wood floats in water kept in a beaker. If the beaker moves with a vertical acceleration a, the wood will (a) sink deeper in the liquid if a is upward (b) sink deeper in the liquid if a is downward, with a < g (c) come out more from the liquid if a is downward with a > g (d) remain in the same position relative to the water
m .ja
πR 2 σg (↓) 3
(d) 2
19. A heavy load W is supported on a platform of area S by applying a force F on a small piston of area S/10. The value of F for equilibrium is
πR 2 σg (c) (↑) 3
er s
πR 2 σg (↓) 3
Lake
na .c om
(b)
ar a
πR 2 σg (↑) 3
(a)
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4.40
Hydrostatics and Hydrodynamics
ANSWERS 4. (b) 12. (d) 19. (b)
1. A cubical block of iron 5 cm on each side is floating on mercury in a vessel. (a) What is the height of the block above mercury level? (b) Water is poured into the vessel so that it just covers the iron block. What is the height of the water column? Density of mercury = 13.6 gm/cm3, density of iron = 7.2 gm/cm3. [IIT, 1973] [Ans. (a) 2.35 cm, (b) 2.54 cm]
8. (d)
4. An ornament, which weighs 36 g in air, weighs only 34 g when immersed in water. It is known that some copper is required to be mixed with the gold to prepare the ornament, find the amount of copper added. Take the density of gold to be 19.3 g/cm3 and that of copper to be 8.9 g/cm3. [Ans. m = 2.2 g]
er si ta rio s
ita
5. A tube bent at right angle is lowered into a water stream, as shown in the figure. The velocity of the stream relative to the tube is v = 3 m/s. The closed upper end of the tube situated at a height h0 = 15 cm from the water surface has a small orifice. Find the height h upto which the water jet will spurt. [Ans. 30.9 cm]
iv
un iv
e/
.m
V
h1
[Ans. 3.00 m/s]
7. The density of air in atmosphere decrease with height and can be expressed by the relation d = d0 e–αh
3. A large block of ice 5 m thick has a vertical hole drilled through it and is floating in the middle of the lake. What is the
6. A solid cylinder of radius R = 10 cm and of mass M = 2 kg floats in water with its axis vertical. Show that if it is slightly depressed and released, it will execute simple harmonic motion and find its period. [Ans. T = 0.5 s]
ht tp
s: //
t.m e/
un
//t
ht tp s:
minimum length of the rope required to scoop up a bucket full of water through the hole? [IIT, 1975] [Ans. h = 0.5 m.]
er s
w
//w
s:
tp
ht
7. (a) 15. (c)
m
w
.ja
ar a
na .c om
2. A wide vessel with a small orifice in its bottom is filled with water and kerosene of densities ρ1 = 1 g/cc and ρ2 = 0.8 g/cc, respectively. Find the velocity of water flow if the height of the water layer is h1 = 30 cm and that of the kerosene layer is h2 = 20 cm. Neglect viscosity.
h2
6. (a) 14. (c)
EXE CISE 1
pr
5. (c) 13. (c) 20. (c)
rio s_ in fo
3. (c) 11. (d) 18. (b)
R
2. (b) 10. (c) 17. (d)
actice
1. (a) 9. (c) 16. (c)
where d0 is the density at sea level, α is a constant, and h is the height. Calculate
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Hydrostatics and Hydrodynamics
na .c om
8. A thin tube, sealed at both ends, is 1.0 m long. It lies horizontally, the middle 0.1 m containing mercury and the two equal ends containing air at standard atmospheric pressure. If the tube is now turned to vertical position, by what amount will mercury be displaced? Standard atmospheric pressure = 0.76 m of Hg.
B
m
ita er s
un iv
.m
W
un
t.m e/
ht tp s:
12. Two solid uniform spheres, each of radius 5 cm are connected by a light string and is totally immersed in a tank of water. If the specific gravities of the spheres are 0.5 and 2, find the tension in the string and the pressure between the bottom of the tank and the heavier sphere. [Ans. R = 2.566 N] 13. The side wall of a wide vertical cylindrical vessel of height h = 60 cm has a narrow vertical slit running all the way down to the bottom of the vessel. The length of the slit is l = 40 cm and the width b = 0.1 cm.
ht tp
s: //
[Ans. θ = 45º]
10. Two identical cylindrical vessels with their bases at the same level, each containing a liquid of density ρ. The height of the liquid in one vessel is h1 and that in the other vessel is h2. The area of either base is A. What is the work done by gravity in equalizing the level when the two vessels are connected? [IIT, 1981] 2 h − h2 Ans. Aρg 1 2 11. A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water upto a height of 0.5 m. The specific gravity
0.5 m
iv
θ
//t
ht
D
B′
C
e/
s:
tp
//w
w
w
.ja
F
er si ta rio s
ar a
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[IIT, 1978]
of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. [Exclude the case θ = 0] [IIT, 1984]
the atmospheric pressure at sea level? Assume g to be constant. The numerical values of the constants are g = 9.8 m/s2, do = 1.3 kg m–3, α = 1.2 × 10–4 m–1 [Roorkee, 1977] [Ans. 1.06 × 105 N/m2]
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16. Calculate the rate of flow of glycerine of density 1.25 × 103 kg/m3 through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 newton/m2. [Roorkee, 1991] –4 3 [Ans. Q = 6.44 × 10 m /s]
With the slit closed, the vessel is filled with water. What is the resultant force of reaction of the water flowing out of the vessel immediately after the slit is opened?
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18. A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross section of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. [IIT, 1997]
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[Ans. F = 3.14 N]
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17. A cylindrical tank 1 m in radius rests on a platform 5 m high. Initially, the tank is filled with water to a height of 5 m. A plug, whose area is 10–4 m2, is removed from an orifice on the side of the tank at the bottom. Calculate: (a) initial speed with which the water flows from the orifice, (b) initial speed with which water strikes the ground, (c) the time taken to empty the tank to half its original value. Given, g = 10 m/s2 [Roorkee, 1995]
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15. A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev. per sec, find the difference in height of the liquid at the centre of the vessel and at its sides. [Roorkee, 1987] [Ans. 0.01m]
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2. A large tank is filled with two immiscible liquids of densities ρ1 and ρ2 as shown in figure. A small hole is made at the bottom of tank from which liquid of density ρ2 starts flowing out. Bernoulli’s equation can be applied between points
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1. A cylindrical vessel filled with water is released on an inclined surface of angle θ as shown in figure. The friction coefficient of surface with vessel is µ (< tan θ). Then the constant angle made by the surface of water with the incline will be
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4. Atmospheric pressure at a height of about 6 km decreases to nearly half the value at the sea level, through the height of the atmosphere is more than 100 km. Explain.
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3. The blood pressure in humans is greater at the feet than at the brain. Explain.
12. The size of the needle of a syringe controls flow better than the thumb pressure exerted by a doctor while administering an injection. Why?
15. Does it matter if one used gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
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2. The force required by a man to move his limbs immersed in water is smaller than the force for the same movement in air. Explain.
1. A balloon filled with helium does not rise in air indefinitely but halts after a certain height. Explain.
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3. Shape of the meniscus formed by two liquids when capillaries are dipped in them are shown. In I, it is hemispherical whereas in II, it is flat. Pick correct statement regarding contact angle formed by the liquids in both situations.
6. A square box of water has a small hole located in one of the bottom corner. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed v0, as shown in figure. When the box is half empty, it is tilted by 45º so that the hole is at the lowest point. Now the water will flow out with a speed of
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F = 100 N F = 50 N F = 25 N F, as applied, cannot balance the effect of the force on the first piston.
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(a) the pressure in the water is higher at lower points in the stream (b) the pressure in the water is lower at lower points in the stream (c) the pressure in the water is the same at all points in the stream (d) pressure variation will depend upon density and exit speed of the water.
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20. A rain drop of radius r has a terminal velocity v m/s in air. The viscosity of air is η poise. The viscous force on it is F. If the radius of drop be 2r and the drop falls with terminal velocity in same air, the viscous force on it will be (a) F (b) F/2 (c) 4 F (d) 8 F
23. A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by
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Hydrostatics and Hydrodynamics
to pull a thin wire ring up (refer figure) if it is initially resting on a horizontal water surface. The circumference of the ring is 20 cm and its weight is 0.1 N
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7. A beaker is filled in with water is accelerated a m/s2 in + x direction. The surface of water shall make an angle (a) tan–1 (a/g) backwards (b) tan–1 (a/g) forwards (c) cot–1 (g/a) backwards (d) cot–1 (g/a) forwards
3. Water is flowing through a long horizontal tube. Let PA and PB be the pressures at two points A and B of the tube (a) PA may be equal to PB (b) PA may be greater than PB (c) PA may be smaller than PB (d) PA = PB only if the cross-sectional area at A and B are equal
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10. In a peculiarly shaped tube as shown in the diagram the lower part is a quarter circle having radius of curvature R. The vertical limbs of the tube are separated by a distance R. A liquid of density ρ is poured into the tube of two height R and 2R in left and right limbs respectively. A valve fitted in the circular part maintains this liquid height difference. The valve is suddenly opened. If the gauge pressure at the lowest point is P and acceleration of the liquid is ‘a’ just after opening the valve, choose the correct choice (a) P = 2ρgR 4 + 2π (b) P = ρgR 4+π (c) P = πρgR 2g (d) a = 4+π
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1. An arrangement of the pipes is shown in the figure. The flow of water (incompressible and nonviscous) through the pipes is steady in nature. Three sections of the pipe are marked in which section 1 and section 2 are at same horizontal level, while being at a greater height than section 3. Correctly match order of the different physical parameter with the options given. 5
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2. Bernoulli’s equation can be written in the following different forms (column I). Column II lists certain units each of which pertains to one of the possible forms of the equation. Match the unit associated with each of the equations.
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3. Match the column I with column II.
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4. Match the column I with column II.
2 (b) Time to acquire (q) V = 2 r g (ρ − σ) 9 η terminal speed (c) Radius of drops (r) Less viscous force of water < 0.01 mm (d) Greater veloc- (s) Independent of ity of flow of a the density of liquid liquid.
Column I Column II (a) The net force exert- (p) 2hρgA2 ed by liquid on side walls is (b) The force exerted by (q) zero the liquid on the base of the vessel is (c) The weight of the (r) hρg (A2 – liquid is A1) on the liquid (d) The walls of the (s) Less then vessel at the level X 2hρgA2 exert a downward force is
Column II (p) Drops in mist
Column I (a) Larger Reynold’s number
Hydrostatics and Hydrodynamics
ASSAGE BASED QUESTIONS
PASSAGE-1 A liquid of density ρ is filled in a beaker of cross section A to a height H and a cylinder of ice of mass m base are ‘a’ is floating in it as shown in the figure.
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10. If the thin plate moves such that the plate is at a distance of 2 cm from one of the plates and 4 cm from the other, now the force exerted by the liquid on the plate is (a) 0.6 N (b) 0.4/3 N (c) 1.6/3 N (d) None
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4. The angle of the free surface with the horizontal is equal to (a) 30º (b) tan–1 (2/3) –1 (c) tan (1/3) (d) 45º
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PASSAGE-2 If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquids gets changed. In case of horizontally accelerated liqa uid (ax), the free surface has the slope x . In case g of vertically accelerated liquid (ay) for calculation of pressure, effective g is used. A closed box with horizontal base 6 m by 6 m and a height 2 m is half filled with liquid. It is given a constant horizontal acceleration g/2 and vertical downward acceleration g/2.
7. Maximum value of water pressure in the box is equal to (a) 1.4 MPa (b) 0.14 MPa (c) 0.104 MPa (d) 0.014 MPa
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1. Pressure due to the weight of ice will be maximum at (a) 1 (b) 2 (c) both 1 and 2 (d) neither 1 or 2
Take atmospheric pressure as P0.
6. Water pressure at the bottom of centre of box is equal to (atmospheric pressure = 105 N/m2, density of water = 1000 kg/m3, g = 10 m/sec2) (a) 1.1 MPa (b) 0.11 MPa (c) 0.101 MPa (d) 0.011 MPa
5. Length of exposed portion of top of box is equal to (a) 2 m (b) 3 m (c) 4 m (d) 2.5 m
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14. A penguin floats first in a liquid of density ρ0 then in a fluid of density 0.95ρ0 and then in a fluid of density 1.1ρ0. Rank the densities according to the magnitude of buoyant force on the penguin greatest first. (a) 1.1 ρ0, ρ0, 0.95 ρ0 (b) 0.95 ρ0, ρ0, 1.1 ρ0 (c) ρ0, 0.95 ρ0, 1.1 ρ0 (d) All tie
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13. Figure shows four containers of same oil. Rank them according to pressure at depth h, greatest first. (a) A, C, B (b) B, A, C (c) C, A, B (d) All tie
Hydrostatics and Hydrodynamics
1. A cube with a mass m = 20 g floats on the surface of water. Each face of the cube a is 3 cm long. What is the distance between the lower face of the cube and the surface of the water? (S.T. of water = 70 dyn/cm]
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3. A 50 kg girl wearing high steel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
2. The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
4. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The
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PASSAGE-4 The pressure at a point in a fluid in static equilibrium depends on the depth of that point but not on any horizontal dimension of the fluid or its container. When a body floats in a fluid, the magnitude Fg of the gravitational force on the body is equal to weight mg of the fluid that has been displaced by the body.
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SUBJECTIVE QUESTIONS
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9. A manometer reads the pressure of a gas in an enclosure as shown in figure (a) When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers in mercury and the atmospheric pressure is 76 cm of mercury. (a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury. (b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
5. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 Nm–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kgm–3 (g = 9.8 ms–2).
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10. Mercury is completely filled in a rectangular tank of height 72 cm. The atmospheric pressure at the place is 72 cm of Hg. Find the distance in cm of point of application from bottom of tank of the net force, on the inner surface of side vertical wall of tank.
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8. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 ºC) is 2.50 × 10–2 Nm–1 If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
11. A long, square wooden block is pivoted along one edge as shown. The block is in equilibrium when immersed in water to the depth shown. Assuming friction in the pivot is negligible, evaluate the density of the wood (in SI unit, up to two significant digit). /
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6. What is the minimum volume of a block of wood (density = 850 kg/m3) if it is to hold a 50 kg women entirely about the water when she stands on it?
length of the slider is 30 cm. What is the surface tension of the film?
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Hydrostatics and Hydrodynamics
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2. A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in figure. The lower density liquid is open to the atmosphere having pressure P0. [1995] (a) A homogeneous solid cylinder of length L (L < H/2), cross-sectional area A/5 is immersed such that if floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid. Determine
(b) 4d/5 (d) d/5
(a) 5d/4 (c) 4d
12. In the incompressible flow through the device shown in figure, velocities may be considered uniform over the inlet and outlet sections.
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(i) the density D of the solid. (ii) the total pressure at the bottom of the container. (b) the cylinder is removed and the original arrangements is restored. A tiny hole of area s(s << A) is punched on the vertical side of the container at a height h (h < H/2). Determine (i) the initial speed of efflux of the liquid at the hole. (ii) the horizontal distance x travelled by the liquid initially and (iii) the height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm. (Neglect the air resistance in these calculations)
1. A homogeneous solid cylinder of length L (L < H/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure P0. Then, density D of solid is given by [1995]
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13. The 3 m diameter log (sp. gr = 0.8) shown in figure is 6 m long into paper. Determine the net vertical and horizontal reaction at point A.
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If the fluid flowing is water, obtain an expression for the mass flow rate at section 3. The following conditions are known A1 = 0.1 m2, A2 = 0.2 m2, A3 = 0.15 m2, V1 = 5 m/s and V2 = 10 + 5 cos(4πt) m/s.
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Hydrostatics and Hydrodynamics
tube is kept inclined in the vertical plane as shown in the figure. The area of crosssection of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. [1997]
string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is [2001]
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5. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to [2000]
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7. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance ℓ and h are shown there. After sometime the coin falls into the water. Then [2002]
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(b) Mg – Vρg (d) ρg(V + πR2h)
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4. Water from a tap emerges vertically downwards with an initial speed of 1 ms–1. The cross-sectional area of tap is 10–4 m2. Assume that the pressure is constant throughout the steam of water and that the flow is steady, the cross-sectional area of stream 0.15 m below the tap is [1998]
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8. A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm3 and 1.2 g/cm3, respectively. The height of liquid
(b) 2πL (d) L/2π
6. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and mass M. It is suspended by a
(a) L / 2π (c) L
(a) (b) (c) (d)
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A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm.
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The capillary has outer radius 0.002 m and inner radius a. When pressure P is applied at the top of the tank volume flow rate of the liquid is 8 × 10–6 m3/s and if capillary tube is detached, the liquid comes out from the tank with a velocity 10 m/s. Determine the coefficient of viscosity of the liquid. [2003]
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11. A container of width 2a is filled with a liquid. A thin wire of weight per unit length λ is gently placed over the liquid surface in the middle of the surface as shown in the figure. As a result, the liquid surface is depressed by a distance y (y << a). Determine the surface tension of the liquid. [2004]
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12. A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity. [2004] 13. Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger is 8 mm and the diameter of the nozzle is 2 mm. The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range of water stream on the ground. (g = 10 m/s2) [2004]
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10. A liquid of density 900 kg/m is filled in a cylindrical tank of upper radius 0.9 m and lower radius 0.3 m. A capillary tube of length ℓ is attached at the bottom of the tank as shown in the figure. 3
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9. A soap bubble is being blown at the end of very narrow tube of radius b. Air (density ρ) moves with a velocity v inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is T. After sometime the bubble, having grown to radius r separates from the tube. Find the value of r. Assume the r >> b so, that you can consider the air to be falling normally on the bubble’s surface. [2003]
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[2002] (a) Find the total force exerted by liquid A on the cylinder. (b) Find h, the length of the part of the cylinder in air (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.
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Hydrostatics and Hydrodynamics
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Hydrostatics and Hydrodynamics
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PASSAGE-1 The density of block is ρ/3 and is used to block a hole H.
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18. If water level is lowered below h2, then (a) the block will not rise (b) the block will start rising again at h2 = h/3 (c) the block will start rising again at h2 = h/2 (d) the block will start rising again at h2 = h/5 [2006]
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15. Water is filled in a cylindrical container to a height of 3 m. The ratio of the crosssectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s2) 2
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17. If the water level is lowered such that it comes below the top of the block keeping the block constant at which height h2 (h2 is depth of the upper surface of water level from the top of the block) will the block be at a stable condition? 4h h (a) h2 = (b) h2 = 9 3 2h 2h (c) h2 = (d) h2 = 9 3
14. A U-shaped tube contains a liquid of density ρ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column. [2005]
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Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude. [2007] 2 (a) |2P0Rh + πR ρgh – 2RT| (b) |2P0Rh + Rρgh2 – 2RT|
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Hydrostatics and Hydrodynamics
(c) |P0πR2 + Rρgh2 – 2RT| (d) |P0πR2 + Rρgh2 – 2RT| 20. Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. [2008] and Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. (a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1 (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True
ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has subhemispherical soap bubble as shown in figure. Just after opening the valve,
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2. (a) → q, (b) → r, (c) → p
3. (a) → r, (b) → s, (c) → p, q, (d) → r
4. (a) → r, (b) → q, (c) → p, (d) → s
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ASSAGE BASED QUESTIONS 2. (a) 10. (a)
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7. Surface tension of the liquid is the force per unit length of an imaginary line drawn in the free surface of liquid. So, it does not depend on the area of the surface.
4. The density of air is maximum near the surface of earth and decreases rapidly with height up to 6 km. and becomes nearly half of its value at the sea level at a height of 6 km. So, atmospheric pressure at a height of 6 km. decrease to nearly half of its value at sea level. Beyond 6 km height density of air changes very slowly.
8. Detergents having small angle of contact dissolve easily in water. 9. In the absence of external forces the liquid drop experiences only surface tension force and tries to minimise its surface area. For a given volume the minimum surface area is of a sphere.
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10. If we blow under the paper, the pressure on the lower side will decreases. The pressure above the paper is atmospheric pressure which will bend the paper downward. 11. When we try to close a water tap with our fingers, it reduces the area of outlet of water
6. For water-glass surface the angle of contact is acute. To achieve this acute valve of angle of contact the water spreads out an a clean glass surface. On the other hand for mercury-glass, angle of contact is obtuse. In order to achieve this obtuse valve of angle of contact, the mercury form drops.
5. The pressure is transmitted equally in all directions. In other words, there is no fixed direction for the pressure due to liquid. Hence, hydrostatic pressure is a scalar quantity.
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2. The density of water is much greater than air, so the upthrust on the limbs of a man is more. As a result of which the effective weight of limbs of a man decreases in water. So, it is easier for him to move his limbs in water than in air.
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1. Initially, the weight of the displaced air (upthrust) is greater than the weight of helium filled balloon. We know that the density of air and acceleration due to gravity decreases with height. Due to which upthrust also decreases. Therefore, the balloon halts at a height where weight of balloon just equals to the weight of air displaced.
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SUBJECTIVE QUESTIONS 1. Downward force = Buoyant force
1 12. Bernoulli’s equation P + ρv 2 + ρgh = con2 stant, shows that P occurs with power one and v occurs with power two. Therefore, influence of velocity is more than pressure. Moreover needle of syringe has very small cross sectional area compared to the thumb. That is why a needle controls the flow rate better.
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13. The fluid flowing out of a small hole acquires a large velocity and hence, large momentum. As no external force is acting on the vessel, according to principle of conservation of momentum, the vessel gets a backward momentum.
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3. 6.2 × 106 Nm–2
4. 2.5 × 10–2 Nm–1
5. For the first bore, pressure difference (between the concave and convex side) = 2 × 7.3 × 10–2/3 × 10–3 = 48.7 Pa. Similarly for the second bore, pressure difference = 97.3 Pa. Consequently, the level difference in the two bores is [48.7/(103 × 9.8)] m = 5 mm. The level in the narrower bore is higher. 6. v = 0.33 m3
ht
tp
s:
//t
.m
s:
tp
er
//t .m
P
ht
ta r
un
0.2 + 0.0084 0.2084 = = 0.02315 m 9 9 = 2.315 cm.
e/
tp s
ht
1 2 mv ρv 2 2 is is energy per unit weight = 2 (m /ρ) 1 2 mv v2 2 = is energy per unit volume and m 2 energy per unit mass
20 × 10−3 × 10 + 4 × 70 × 10−3 × 3 × 10−2 9 × 10−4 × 103 × 10
si
iv e
rs
w
w
:// w
=
=
1. Use equation of continuity and concept “pressure is greater at lower and broader section.” 1 2 mv 1 2 v 2 2 = 2. Kinetic energy is mv ∴ 2 2g (mg )
Mg + 4Ta a 2ρg
x=
.ja
15. No, one can use gauge pressure instead of absolute pressure in applying Bernoulis’ theorem, if the pressure at the two points where Bernouli’s equation is applied are significantly different.
m
rio
ar
14. No, Bernouli’s theorem is used for streamline flow only.
MATCH THE COLUMN TY E QUESTIONS
[
7
jet and hence, velocity of water increase. (av = constant)
4.62
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7. 48.2º or tan–1(1.118)
R
PR
1. (a)
2. (i)
h(3H − 3h)
nf
o
5. (a)
s_ i
6. (d)
rio
7. (d)
si ta
8. (a) zero (b) 0.25 cm. (c) g/6 9. 4T/ρv2
13. Horizontal 81 × 104, Vertical 33.9 × 104
s
dQ ∝ r5 dt
un
12.
13. 2 m
14. H =
ω2 L2 2g
15. (a)
16. (b)
17. (a)
18. (a)
19. (b) 20. (a) As the stream falls down, its speed will increase and cross-section area will decrease. Thus it will become narrow. Similarly as
si ta
iv
λa 2y
er
11.
12. 2500 + 999 cos(4πt)kg/s
360 B + 4320 B – 8.64 B ρ = 0 4680 ρ= = 540 kg/m3 8.64
rio
10. (1/720) Ns/m2
t.m e/
s: //
tp
ht
∑τ = 0
m
un
e/
Torque due to upthrust force = (0.6 × 1.2 × B) × 103 × B × 10 × 0.6 = 4320 B Torque due to weight = ρ × (1.2)2 × B × q × 0.6 = 8.64 Bρ
3H 3 , H 8 4
4. (c)
.m
//t
s:
× 10 × (x) (0.6 – x)B.dx = 360B
0
tp
3
ht
∫ 10
(iii) At h =
iv er
w
//w
s:
tp
ht
(ii)
g 2
3. 29025 J/m3, 29400 J/m3
co
a.
an
ar
m
w
.ja
2
H H H ρgb × × 2 2 3 = 4 H = 32 cm. = 2 H 9 ρgbH 2 + ρgb 2 11. Torque due to liquid pressure 0.6
(3H − 4h)
(b) (i)
10. Force due to atmospheric pressure = ρ0A = ρgH × bH F1 = ρgbH2 This force acts at a height H1 = H/2 from bottom Force due to mercury on side wall, ρgbH 2 F2 = 2 This force acts at a height H2 = H/3 from bottom. ∴ distance of net force from bottom F H + F2 H 2 = 1 2 F1 + F2 ρgbH 2 ×
5d dg (6 H + L) (ii) P = P0 + 4 4
9. (a) absolute pressure = 96 cm of Hg, gauge pressure = 20 cm of Hg for (a) absolute pressure = 58 cm of Hg, gauge pressure = 18 cm of Hg for (b) mercury would rise in the left limb such that the difference in its levels in the two limbs becomes 19 cm.
8. 20.0 Nm–2, 1.06 × 105 Nm–2.
4.63
EVIOUS YEA S’ IIT-JEE QUESTIONS
Hydrostatics and Hydrodynamics
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4.64
Hydrostatics and Hydrodynamics
P2 = pressure just inside the bubble at the end 1 = P0 +
21. (b) P1 = pressure just inside the bubble at
⇒ Air will flow from end 1 to end 2
4T R
s rio ve
ht
tp s:
//t
.m
ht tp
e/ u
s:
ni
//t
ht
.m
tp
e/
rs i
ta
un
iv
s: //w
w
er s
w
ita
.ja m
rio s_
ar an
in
a.
fo
co m
the end 2 = P0 +
4T R
R > r ⇒ P2 < P1
the stream will go up, speed will decrease and cross-section area will increase. Thus it will become broader. Hence statement 1 is correct and statement 2 is correct explanation also.
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c h a p t e r
5
m
ar a
5.1 SIMPLE HARMONIC MOTION
rio s_ in fo
na .c om
Simple Harmonic Motion
er si ta rio s
un iv
2
%
iv
.m
un
//t
ht
tp
s:
$
e/
//w
w
er s
w
ita
.ja
Simple harmonic motion is defined as a type of motion in which a particle moves to and fro about fixed point in such a way that its acceleration is directly proportional to displacement and it is always directed toward mean position.
t.m e/
ht tp s:
A and B is extremes position O is the mean position.
ht tp
s: //
Example Motion of simple pendulum is simple harmonic motion. Motion of large mass droped in the tunnel passing through the centre of the earth is simple harmonic motion.
A
B
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C
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5.2
Simple Harmonic Motion
5.2 ANALYSIS OF SIMPLE HARMONIC MOTION Consider a particle of mass m executing simple harmonic motion along x-axis with mean position at origin. From above definition, if displacement of particle is positive x-direction then the acceleration is negative x-direction.
rio s_ in fo
ar a
na .c om
O
er si ta rio s
un iv
e/
ma = – k x
.m
ht
tp
s:
//w
w
er s
w
ita
.ja
m
So a∝–x or F ∝ –x Thus, a is directly proportional to x then force is also directly proportional to x. F=–kx ...(i) where, k is constant From equation (i), if mass of the particle is m, then
s: //
t.m e/
ht tp s:
un
//t
iv
d 2 x -k = x ...(ii) dt 2 m This equation is called as the differential equation of simple harmonic motion The general expression for x (t) satisfying the equation is x (t) = A sin (ωt + φ) ...(iii) k [from equations (ii) and (iii)] m
ht tp
where, ω =
Amplitude The amplitude of a particle executing simple harmonic motion is its maximum displacement. From equation (iii), the amplitude of particle is A. x (t) = A sin (ωt + φ) Since maximum value of sin θ = 1, therefore – A ≤ x ≤ A
Time Period Time period of the particle in simple harmonic motion is the time taken by the particle to complete one cycle of motion and is denoted by T. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.3
Consider the equation of simple harmonic motion, i.e., x (t) = A sin (ωt + φ) 2π , then ω
If t is increased by
2π x (t ) = A sin ω t + +φ ω = A sin (ωt + 2π + φ) = A sin (2π + ωt + φ) = A sin (ωt + φ)
Thus, the function x (t) repeats itself after on interval of
om s_ i
nf
o
.c
m
Frequency
rio
k 2π m ..... since ω = = 2π m ω k
an a
T=
⇒
2π ω
ar
Time period, T=
2π . Therefore, ω
ve
ni
...(iv)
e/ u
s: //
ht
tp s:
//t
ht
.m
tp
rs ita
e/
t.m
ht tp
s: //
ω 1 k = 2π 2π m For solution of equation of simple harmonic motion a = – ω2 x dv ∴ = -ω2 x dt dv dx or × = -ω2 x dx dt vdv or = -ω2 x dx v(nu) =
rio s
un iv
w w
er
w
si ta
.ja
The frequency of a particle executing simple harmonic motion is equal to the number of oscillations completed in one second. It is measured in cycle per second (Hertz) and is denoted by
∫ vdv = -ω ∫ x dx 2
or
v2 = – ω2x2 + C …(v) 2 2 2 v =C–ω x It is clear that v increase as x decrease and at a particle, when v = 0 and x = A (amplitude) or
v2 = ω2A2 – ω2x2
v = ω A2 - x2
or
Again
…(vi)
C = ω 2 A2
then,
dx = ω A2 - x2 dt
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Simple Harmonic Motion
Integrating,
or or
sin -1
dx A2 - x2
= ∫ ω dt
x = ωt A
x = sin (ωt + φ) A x = A sin (ωt + φ)
…(vii)
∴
dx v= = ωA cos (ωt + φ) dt
and
a=
na .c om
dv = -ω2 A sin (ωt + φ) = -ω2 x dt
5.3 CONCEPT OF PHASE
rio s_ in fo
or
∫
ar a
Phase means state of motion which described by velocity displacement and acceleration.
ita er si ta rio s
er s
un iv
//w
w
w
.ja
m
x = A sin (ωt + φ) dx v= = Aω cos (ωt + φ) dt a = – ω2 A sin (ωt + φ)
e/
iv
.m
Let or or
ht tp
Initial Condition
t.m e/
ht tp s:
un
//t
ht
tp
s:
v, x and a depends on (ωt + φ), i.e., (ωt + φ) is controller of phase. Hence, (ωt + φ) is phase angle on simple pendulum. θ = ωt + φ At t = 0, θ = φ known as initial phase or epoch. φ represent initial condition and actual equation of simple harmonic motion depends emit a condition.
s: //
5.4
x = 0 at t = 0 x = A sin (ωt + φ) 0 = A sin φ sin φ = 0 φ = 0 and π φ=0 x = 0+ t = 0 φ=π x = 0– t = 0 A
B
Actual equation of simple harmonic motion: For x = 0+ at t = 0 https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
Let or or
x = A sin ωt for φ = 0 x = A and at t = 0 x = A sin (ωt + φ) A = A sin (ωt + φ) sin φ = 1
π 2 Actual equation ∴
φ=
π x = A sin ωt + = A cos ωt 2
na .c om
Concept of Reference Circle
rio s_ in fo er si ta rio s t
Pt=0 x Reference line
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp ht x = A cos (ωtt + φ) y = A sin (ωt + φ)
P t = t
O
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
Projection of a particle in uniform circular motion perform simple harmonic motion.
ht tp
dx = A ω sin (ωt + φ) dt aθ = – ω2 A cos (ωt + φ) = – ω2x
vx =
Physical Signification of Phase Physically phase represents the mood. Mood of the particle: (i) Equation of projective wave x = A sin (ωt – kx) indicate that mood (phase) of particle is function of time. (ii) Equation of stationary wave x = 2a cos kx × sin ωt indicate that mood (phase) depends on time only. https://t.me/universitarios_infohttps://www.jamarana.com
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5.5
5.6
Simple Harmonic Motion
5.4 ENERGY CONSIDERATION IN S. H. M. Simple harmonic motion’s defined by the equation F = – kx The work done by the force F during a displacement from x to x + dx is dW = Fdx = – kx dx The work done in a displacement from x = 0 to x is x
W = ∫ (-kx) dx
0
in fo
an a. c
om
1 = - kx 2 . 2 Let U(x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force.
rio si
ta
un .m
iv er
e/
k m
iv
//w
ω=
//t
ht
As
tp
s:
1 and U( x) = kx 2 . 2
s
er s
w
w
ita
.ja
m
rio
ar
s_
1 U( x) - U(0) = - W = kx 2 2 Let us choose the potential energy to be zero when the particle is at the centre of oscillation x = 0 Then, U(0) = 0
s: //
ht
tp
t.m e/
s:
un
1 k = mω2 - U( x) = m ω2 x 2 2 The displacement and the velocity of a particle exceeding a simple harmonic motion are given below. x = A sin (ωt + φ)
tp
v = Aω cos (ωt + φ)
ht
and
Thus potential energy at time t is 1 U = mω2 x 2 2 1 = mω2 A 2 sin 2 (ωt + φ) 2
And the kinetic energy at time t is
1 K = mv 2 2
1 = mA 2 ω2 cos 2 (ωt + φ) 2
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Simple Harmonic Motion
5.7
The total mechanical energy at time t is E=U+K 1 = m ω2 A 2 2 [sin2 (ωt + φ) + cos2 (ωt + φ)] 1 = m ω2 A 2 2 We see that, the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected. Hence, force is conserved as mechanical energy is conserved. Graph of potential energy
na .c om
1 U = mω2 A 2 2
e/ s: //
t.m e/
ht tp s:
un
//t
iv
.m
er si ta rio s
un iv
//w s: tp ht
rio s_ in fo
ita
er s
w
v2 x2 + =1 ω2 A 2 A 2
w
⇒
.ja
m
ar a
Graph between velocity vs displacement v2 = ω2 (A2 – x2) ⇒ v2 = ω2A2 – ω2x2 ⇒ v2 + ω2x2 = ω2A2
ht tp
x=A
u=0 x=0
x=+A
This is the equation of ellipse.
5.5 CHARACTERISTICS OF SIMPLE HARMONIC MOTION (i) It is a periodic motion, i.e., motion repeats itself after second interval of time. All periodic motions are not simple harmonic motion but all simple harmonic motions are periodic. e.g., motion of the earth, round is periodic but not simple harmonic motion (ii) It is an oscillatory motion, i.e., to and fro motion (i.e., vibrating motion). All oscillatory motions are not simple harmonic motions but all simple harmonic motions are oscillatory motions.
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Simple Harmonic Motion
Example A ball of density (d) is dropped in the liquid of density ρ (ρ > d) from height h. What type of motion the ball will perform, is it simple harmonic motion? Find time period of motion (viscous force on resistance has to be negligible).
) Solution
Weight of liquid, W = Vρg Displaced liquid = B Weight of the body = Vdg = Mg
Hence, acceleration inside liquid is upward
ita
v = 2gh
er s
w
.ja
h
C
un
//t
iv
.m
tp
e/
s:
un iv
//w
w
B
V ρg
er si ta rio s
m
A
rio s_ in fo
g (ρ - d ) d
ar a
=
B - Mg Vρg - Vdg = M Vd
na .c om
a=
ht
The motion of ball is to and fro, i.e., oscillatory but not simple harmonic motion because acceleration is not directly proportional to displacement. Time period of motion:
∴ t1 =
1 2 g t1 2
s: //
h=
ht tp
Time from A to B
t.m e/
ht tp s:
5.8
2h g
Time from B to C vc = vB – at2 = 0 ∴ t2 =
2 gh VB = a a
Total time = 2 (t2 + t1) 2 gh 2h = 2 + a g https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.9
Example Find time period of motion of ball on the given inclined plane.
) Solution
Motion of ball on the inclined plane is oscillatory but not simple harmonic motion because a ∝ x. C
m
ar a
α
un iv
e/
1 g sin α t12 2
iv
.m
2h cosec α 2h = cosec α g sin α g
ht tp
s: //
t.m e/
ht tp s:
un
//t
⇒ t1 =
B
er s
w
w
//w ht
tp
s:
h cosec α =
β
ita
.ja
Time from A to B 1 s = ut + at 2 2
h
er si ta rio s
h
rio s_ in fo
na .c om
A
elongation = d
Block in equilibrium
Similarly, t2 =
2h cosec β g
Hence, time period of oscillation = 2 (t1 + t2) (iii) Acceleration is directly proportional to displacement. (iv) Acceleration is always opposite to displacement from mean position which is always considered as initial position for the sake of measuring displacement. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
Note: Direction of displacement is not distance by direction of velocity of simple harmonic motion Here, a∝–x a = – ω2 x 2π ω= = 2πn T
⇒ ⇒
∆K + ∆U = 0
er s
w
ita
.ja
Example
rio s_ in fo
na .c om
m
ar a
∆K = – ∆U F · ∆x = ∆U -∆U ⇒ F= ∆x Change in kinetic energy, ∆K = work-done = F ∆x ∂u f = ∂x
(v) The force acting on the particle is conservative i.e., (K + U) = constant
)
//t
er si ta rio s
) Solution
iv
ht
.m
tp
e/
s:
un iv
//w
w
The equation of a particle exceeding simple harmonic motion is x = (5 m) sin π −1 ( π s )t + 3 . Write down the amplitude time period and maximum speed. Also find the velocity at t = 1s.
t.m e/
ht tp s:
un
Comparing with equation, x = A sin (ω t + φ) We see that amplitude = 5 m 2π 2π = = 2sec. ω πs -1 Maximum speed = Aω = 5 m × πs–1 = 5π m/sec. dx Velocity at time t = = Aω cos (ωt + φ) dt
ht tp
At t = 1s
s: //
And, time period =
π −5π m/sec v = (5 m) (πs -1 ) cos π + = m / sec. 3 2
Example A particle of mass 0.2 kg undergoes simple harmonic motion according to the equation
(i) What is the time period of oscillation? (ii) What are initial condition?
π x (t) = 3 sin π t + 4
5.10
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Simple Harmonic Motion
5.11
(iii) What is the total energy of the particle if potential energy is zero at mean position? (iv) What are the kinetic and potential energies of particle at time at t = 1s? (v) At what time instants is the particle’s energy purely kinetic? )
) Solution
Comparing the given equation with x (t) = A sin (ωt + φ)
m
fo io iv
.m
er
si
ta
iv
un
e/
From the position x = 1.5 2 m
tp
e/
s:
un
//t
1 With the velocity = (3π ) m/sec. 2
.m
1 (iii) Kinetic energy = mω2 A 2 2 1 = (0.2) (π) 2 (3) 2 2 = 0.9 π2
ht
π 3 m (iv) At t = 1s, x(t ) = 3sin π + = 4 2 π v(t ) = 3π cos π + 4 3π =m/sec. 2
tp
s:
//t
ht
s
w
Hence, the particle starts moving at t = 0
s:
tp ht
si
3π m/sec. 2
⇒ v(0) =
//w
er
w
.ja
m
dx π = v(t ) = 3π cos πt + dt 4
ta r
ar
s_
= 1.5 2 m
rio
an
in
π 4
(ii) x (0) = 3sin
a. co
we get A = 3 m ω = π rad/sec. π φ= 4 2π = 2 sec. (i) T = ω
1 Kinetic energy = mv 2 2
9π 2 1 = (0.2) = 4.44 J 2 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
1 U = kx 2 2 1 = m w2 x 2 2 1 9 = (0.2)(π) 2 2 2 = 4.44 J (v) Energy is purely kinetic at mean position i.e., when x = 0 π x (t ) = 3sin πt + 4
π = 0, π, 2π, 3π ...... 4
ar a
πt +
ita
m
1 3 7 11 t=- , , , 4 4 4 4
⇒
.ja
w
rio s_ in fo
π 0 = 3sin πt + 4
we have
na .c om
Using
er si ta rio s
un iv
//w
w
er s
At these time instants, particle crossing origin and hence its energy is purely kinetic.
e/
ht tp s:
un
//t
iv
.m
tp
s:
(i) If particle starts at t = 0 from mean position, its equation is x(t) = A sin ωt, if initial velocity is positive. x(t) = – A sin ωt, if initial velocity is negative. (ii) If the particle starts at t = 0 from right extremes, its equation is
ht
t.m e/
x(t) = A cos ωt.
s: //
(iii) If the particle starts at t = 0 from left extremes, its equation is x(t) = – A cos ωt.
ht tp
5.12
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Simple Harmonic Motion
5.13
Example The friction coefficient between the two blocks shown in the figure is µ and the horizontal plane is smooth. (i) If the system is slightly displaced and released, find the time period. (ii) Find the magnitude of frictional force between the blocks when the displacement from the mean position is x. (iii) What can be the maximum amplitude if the upper block does slip relative to the lower block?
ar a
) Solution
m M
rio s_ in fo
na .c om
k
er si ta rio s
.m
M+m k
iv
(ii) Acceleration of the blocks at displacement x from the mean position is -kx a = -ω 2 x = M+m
t.m e/
ht tp s:
un
//t
ht
er s
un iv
e/
And, time period, T = 2π
tp
s:
//w
w
w
k M+m
ω=
ita
.ja
m
(i) For small amplitude the two blocks oscillate together. The angular frequency is
Thus resultant force on the upper block is
s: //
-m kx M+m
ht tp
ma =
This force is provided by the friction in the lower block. Hence, magnitude of the frictional force =
mk | x | . M+m
(iii) Maximum force of friction required for simple harmonic motion of the upper mk A block is at the extreme positions. But the maximum frictional force can M+m mk A only be µ mg. Hence, = µ mg M+m or A =
µ ( M + m)g k
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Simple Harmonic Motion
Example A particle is moving in a straight line with simple harmonic motion. Its velocity has the values 3 m/sec and 2 m/sec when its distance from mean positions are 1 m and 2 m respectively. Find the length of its path and period of its motion. )
) Solution
At x = 2 m,
v1 = 3 m/sec. v2 = 2 m/sec.
At x = 1 m,
From the formula
v2 = ω2 (A2 – x2) ...(i)
9 = ω2 (A2 – 1)
w .ja
s
iv
un
si ta
5 = 1.29 rad/s 3
ω=
t.m
tp
And
e/
s: //
w
w
er si
A = 6.4 = 2.53m
∴
un
2π = 4.875 ω
ht
t.m
tp
Time period, T =
e/
s: //
iv
Hence, length of the path = 2 A = 5.06 m
er
5 A2 = 32
or
ht
ta rio
9 A2 – 36 = 4 A2 – 4
m
or
rio
ar
an
9 A2 -1 = 4 A2 - 4
s_ in fo
co m
...(ii)
a.
and 4 = ω2 (A2 – 4) From equations (i) and (ii), we have
s: //
Example
ht
tp
A particle of mass m is located in a undimensional potential field where the potential energy of the particle depends on the co-ordinate x as U(x) = U0 (1 – cos ax); U0 and a is constants. Find the period of small oscillations that the particle performs about the equilibrium position.
) Solution )
∴
As U(x) = U0 (1 – cos ax)
Fx = -
dU dx ...(i)
= – U0 a sin ax
or
Fx = – U0 a · ax
(because for small angle of oscillations sin ax ≅ ax) or
5.14
Fx = – U0 a2x
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Simple Harmonic Motion
5.15
But for small oscillation Fx = – mω2x U0a2 m
ω=a
or
U0 m
Hence, time period, T =
2π m a U0
na .c om
=
2π ω
= 2π
m a U0 2
rio s_ in fo
∴ ω2 =
er si ta rio s
er s
w
w
) Solution
ita
.ja
m
ar a
Example Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.
e/
d 2x = mg - 2 F θ dt 2
un t.m e/
ht tp s:
m
iv
.m
By Newton’s Second law
//t
ht
tp
s:
un iv
//w
At that position, the force on the ball = mg – 2 F sin θ
s: //
(because θ is very small then sin θ ≅ θ) θ
ht tp
θ
F
x
F
mg
or
m
d 2x x = mg - 2F dt 2 l/2 4 Fx = mg l
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5.16
Simple Harmonic Motion
4 Fx l
a=g-
Thus,
4F mgl x ml 4F
=-
x′ = x -
Putting,
mgl , we get 4F
-4 F x′ ml 4F ω2 = ml
a=
Thus,
2
m
ml F = 0.2 sec.
T=π
ta si rio
s
er
//w w
w
.ja m
or
in fo
ar
rio s_
an
a.
A π2 4 F = T2 ml
or
co
or
4F 2π = T ml
ita
un iv
tp s:
5.6 GENERAL APPROACH FOR ANALYSING SIMPLE HARMONIC MOTION
ve
ht
.m
rs
e/
The following steps must be carefully followed while analysing simple harmonic motion for a particle:
e/ un i
//t
.m
tp
ht
tp s:
//t
ht
s:
(i) Displace the particle mean position (equilibrium position) by a distance x along the line of simple harmonic motion. (ii) Analyse the forces on the particle and try to show that (a) The magnitude of net force is proportional to x. (b) The force is directed towards the mean position, i.e., it has a restoring nature. (iii) Find the value of proportionality constant from the expression of the magnitude of force. m (iv) Find the time period using, T = 2π k
5.7 SIMPLE PENDULUM A simple pendulum is an idealized body consisting a point mass, suspended by a light inextensible cord. When pulled to one side of its equilibrium position and then released, the pendulum oscillates. In figure, the string of a simple pendulum of length l attached to a bob of mass m is shown making an angle θ with the vertical.
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Simple Harmonic Motion
θ
5.17
l
T
na .c om
mg sinθ
θ mg sinθ mg
rio s_ in fo
ar a
The forces acting on m are mg or W (gravitational pull of the earth) and T (Tension in the cord).
er si ta rio s
ht
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
⇒ Choose the axis along tangent to the arc of motion and along the radius. Resolve the forces along the two axes. ⇒ The radial component of magnitude mg cos θ balances the force and tension in the string. ⇒ The tangential component of magnitude mg sin θ is the restoring force acting on m to return it to the equilibrium state, so the magnitude of restoring force is F = mg sin θ
arc ..... θ = radius mg x So F = mg θ = …(i) l ⇒ F∝x So the simple pendulum executes simple harmonic motion. The relation between magnitudes of force and displacement is F = kx ...(ii) From equations (i) and (ii), we get mg ⇒ k= l ⇒ θ=
x l
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
This force is acting towards mean position. If the angle θ is very small, then sin θ is very close to θ (sin θ → θ) in radians. So sin θ ≅ θ or F ≅ mg θ (here θ is angular displacement) Now, the magnitude of displacement along the arc is x (but for small θ, are can be taken as a straight line segment).
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5.18
Simple Harmonic Motion
So, the time period of a simple pendulum, when its amplitude is small, is given as m k m mg / l
= 2π
l g
Also, Frequency, v = =
1 k 2π m 1 2π
g l
ar a
5.8 LIQUID OSCILLATING IN A U-TUBE
rio s_ in fo
= 2π
na .c om
T = 2π
//t
er si ta rio s iv
.m
tp ht
D
2x
A
B
Displaced liquid
t.m e/
ht tp s:
un
In equilibrium
C
x
x
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
Consider a liquid column of total length 2l inside a U-tube as shown:
ht tp
s: //
In the U-tube, liquid is displaced from state of equilibrium by distance x. It is clear from diagram that the total difference in liquid levels between limbs is 2x. Consider equal cross-section A on both sides (tube is uniform throughout). The restoring force acting on the liquid is the thrust of excess pressure due to the column ABCD. The magnitude of restoring force is F = pressure × area = (2x) ρgA F = (2 A ρg) x or
...(i)
F∝x
Hence the liquid executes simple harmonic motion in the tube Comparing equation (i) with F = kx, we get k = 2Aρg Hence, the time period is https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
T = 2π = 2π
5.19
m k m 2A ρg
T = 2π
2A lρ 2A ρg
or
T = 2π
l g
na .c om
∴
rio s_ in fo
But mass of oscillating liquid = m = A (2l) ρ
ar a
5.9 OSCILLATION OF A CYLINDER FLOATING IN LIQUID
iv
er si ta rio s
e/
L
.m ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Consider a cylinder of mass m and density d floating on the surface of a liquid of density ρ. The total length of a cylinder is L and the length immersed in the liquid is l. Taking A as the area of cross-section, we apply the law of floatation on the cylinder.
Cylinder in equilibrium
Cylinder displaced downward by a distance x https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
B
mg
Weight of cylinder = Buoyant force mg = Vρg
or
Ld = lρ
rio s_ in fo
(AL) dg = (Al) ρg
ar a
∴
na .c om
where, V is the volume inside the liquid. ...(i)
un iv
e/
.m
tp
s:
//w
F = B – mg = (l + x) A ρg – LA dg = Aρgx + l A ρg – LA dg
er si ta rio s
w
er s
w
ita
.ja
m
If the cylinder is displaced from this position, it oscillates up and down. Let, it displace through a distance x and we analyse the net force acting on it. As the volume occupied inside the liquid increases, the buoyant force B becomes greater than weight and hence, net force acts upwards. Magnitude of net force is
ht
un
//t
iv
From equation (i), F = (Aρg) x Hence, the magnitude of the net force is proportional to x and the direction of force is towards the equilibrium or mean position. Thus, the cylinder will executes simple harmonic motion Computing with F = kx, we get k = A ρg
ht tp
s: //
t.m e/
ht tp s:
5.20
Hence, time period, T = 2π = 2π
m k m Aρg
Using m = (AL) d and Ld = lρ, we can further simplify for T as T = 2π
ALd Alρ = 2π Aρg Aρg
T = 2π
l g
Hence, the time period depends only on the length of the immersed portion. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.21
5.10 OSCILLATIONS OF A BLOCK CONNECTING TO A SPRING (i) Horizontal spring
na .c om
Elongation = d +x
rio s_ in fo
Block displaced downwards by x
er si ta rio s
.m
e/
F = ksx
iv
i.e., F is proportional to x. Block executes simple harmonic motion with k = ks m ks
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
Let a block of mass m be placed on a smooth horizontal surface and rigidly connected to a spring of force constant ks whose other end is permanently fixed. When the spring is in its underformed condition, block is in equilibrium. Let it displace the block by a distance x as shown. The net force acting on the block has a magnitude ksx and is directed towards its mean position. Hence, magnitude of the force is
s: //
Time period, T = 2π
ht tp
(ii) Vertical spring If the spring is suspended vertically from a fixed point and carries the block at its other end as shown, the block will oscillate along the vertical line.
ksd elongation = d
Undeformed spring
Block in equilibrium
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mg d = mg ksd=
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Simple Harmonic Motion
In its equilibrium position, the block elongates the springs by a distance d. In this position the weight of block and elastic force of spring balance each other. If the block displaced downwards by a distance x, the elongation in springs becomes (d + x) and the elastic force becomes greater than the weight of the block. The net force now acts upwards and has a magnitude.
er si ta rio s
er s
un iv
e/
ks((d+x)
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
s:
From equation (i), F = ksx
ita
.ja
//w
w
w
F = ks (d + x) – mg F = ks d + ks x – mg
rio s_ in fo
Block displaced downwards by x
m
ar a
na .c om
Elongation =d+x
ht
5.22
net force upwards
mg
Hence, the magnitude of net force is proportional to x and its direction is towards the mean position (equilibrium position). The block executes simple harmonic motion and k = ks. Time period = 2π
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m ks
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Simple Harmonic Motion
5.23
5.11 COMBINATION OF SPRINGS (i) Springs in series: When two springs of force constants k1 and k2 are connected in series as shown, they are equivalent to a single spring of force constant k which is given by 1 1 1 = + k k1 k2 k1k2 k1 + k2
na .c om ar a
≡
er s
un iv
k1k2 k 1 + k2
un
t.m e/
ht tp s:
//t
Block displaced downwards by x
k=
iv
.m
e/
s: tp ht
ita
m .ja //w
w
w
k2
k
rio s_ in fo
k1
er si ta rio s
k=
ht tp
s: //
(ii) Springs in parallel: For a parallel combination as shown, the effective spring constant is
k1
k = k1 + k2
k2
≡
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k = k1 + k2
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5.24
Simple Harmonic Motion
k1
k2
k1+k2
) Solution
rio s_ in fo
ar a
na .c om
Example A block of mass M, at rest on a horizontal, frictionless table, is attached to a rigid support by a spring of force constant k. A bullet of mass m and speed v strikes the block as shown in the figure. The bullet remains embedded in the block. Determine the amplitude of the resulting simple harmonic motion in ter ms of m, M, v and k.
0
N
ht tp
s: //
t.m e/
Y
un
iv
er si ta rio s
e/
ht tp s:
//t
ht
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Velocity of block and bullet after collision. mv = (m + M) v′ mv ∴ v′ = (m + M)
For the maximum amplitude 1 1 1 (m + M)v′2 = k x 2 = k A 2 2 2 2
…(i)
From equation(i), m+M A= v′ k = =
(m + M) mv 1 . = mv k (m + M) k(m + M )
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Simple Harmonic Motion
5.25
Example A massless spring of force constant 3.60 N/cm is cut into halves.
ar a
ita
er s
w
.ja
m
(i) Force constant = k1 + k2 = 3.60 N/cm + 3.60 N/cm = 7.20 N/cm
un iv
e/
.m
M k
iv
1 1 k = f 2π M
or
M= =
k f2 4π 2
t.m e/
1 1 k = ⋅ f 2 4π 2 M
ht tp
or
k
s: //
where, f = frequency
un
//t
or
T = 2π
ht tp s:
ht
tp
s:
//w
w
(ii) Time period of single spring,
er si ta rio s
) Solution
M
rio s_ in fo
na .c om
(i) What is the force constant of each half? (ii) The two halves, suspended separately support a block of mass M shown in the figure. The system vibrates at a frequency of 2.87 Hz. Find the value of the mass M.
7.20 × (2.87) 2 = 1.48 kg. 4 × (3.14) 2
Mg
Example A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is a0 and the length of the pendulum is l, find the time period of small oscillations about the mean position.
) Solution
For mean position, the acceleration of the bob with respect to the car should be zero. If θ be the angle made by the string with the vertical, the tension, weight and the pseudo force will add to zero in this position.
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Simple Harmonic Motion
α
T
θ
ma0 (θ + α) mg
na .c om
Suppose, at some instant during oscillation, the string is further deflected by an angle α so that the displacement of the bob is x. Taking the components perpendicular to the string,
m
ar a
rio s_ in fo
Component of T = 0 Component of mg = mg sin (α + θ) Component of ma0 = – ma0 cos (α + θ). Thus, the resultant component
ita
.ja
F = m [g sin (α + θ) – a0 cos (α + θ)]
er si ta rio s
un iv
//w
w
er s
w
Expanding the sine and cosine and putting cos α ≈ 1, sin α ≈ α ≈
x , we get l
tp
e/
s:
x F = m ρg sin θ - a0 cos θ + g cos θ + a0 sin θ l
…(i)
iv
.m
At x = 0, the force F on the bob should be zero, as this is the mean position. Thus, by equation (i) 0 = m [g sin θ – a0 cos θ] ...(ii) a Solving, we get tan θ = 0 g
cos θ =
s: //
a0
…(iii)
ht tp
Thus sin θ =
t.m e/
ht tp s:
un
//t
ht
5.26
(a0 ) 2 + g 2 a0
…(iv)
(a0 ) 2 + g 2
Putting equations (ii), (iii) and (iv) in equation (i), we get F = m g 2 + a02 where,
ω=
x l
or
F = mω2x
g 2 + a02 l
This is an equation of simple harmonic motion with the time period t=
2π l = 2 ω ( g + a02 )1/4
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Simple Harmonic Motion
5.27
mao
mg
An easy working rule may be found out as follows. In the mean position, the tension, the weight and the pseudo force balance. From figure, the tension is
rio s_ in fo
T = a02 + g 2 m
ar a
or
na .c om
T = (m a0 ) 2 + (mg ) 2
l l = 2π 2 T/m [ g + a02 ]1/4
er si ta rio s
ita
er s
w
w
.ja
t = 2π
m
This play the role of effective g. Thus, the time period is
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
Example Determine the period of oscillations of mercury of mass m = 200 g poured into a bent tube seen the figure whose right arm forms an angle θ = 30º with the vertical. The cross-sectional area of the tube is A = 0.50 cm2. The viscosity of mercury is to be neglected.
θ Equilibrium position
θ x
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5.28
Simple Harmonic Motion
) Solution
Total displaced liquid = x (1 + cos θ) Weight of liquid = ρAx (1 + cos θ) g ...(i) If the mercury rises in the left arm by x, it must fall by a slanting length equal to x in the other arm. Total pressure difference in the two arms will then be given by equation (i) This will give rise to a restoring force – ρg Ax (1 + cos θ)
...(ii)
or ma + ρg A x (1 + cos θ) = 0 ρg A (1 + cos θ) x m This equation in simple harmonic motion
m ρgA (1 + cos θ)
er si ta rio s
er s
un iv
//w
w
w
Putting the value, we get T = 2.94 sec.
ita
.ja
m
∴ T = 2π
2
2π ρ g A or = (1 + cos θ) m T
rio s_ in fo
ρg A (1 + cos θ) m
ar a
ω2 =
na .c om
or a = -
e/
tp
s:
5.12 ANGULAR ACCELERATION OF A RIGID BODY
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
Linear simple harmonic motion is the right-left (or up-down) oscillations of the particles. A rigid body can also execute clockwise and anti-clockwise oscillations. These oscillations are called as angular oscillations. A very simple example of angular oscillations is a thin disc suspended by a wire attached to the centre of the disc. If the disc is rotated in the horizontal plane, the wire will be twisted. The twisted wire will exerts a torque on the disc tending to return it to its original position. This is a restoring torque (compare it with a restoring force in linear oscillations). For small twist angles θ, the restoring torque is proportional to the angular displacement θ (Hook’s law).
Wire
Disc
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Simple Harmonic Motion
5.29
Hence τ = – C θ ...(i) Here negative sign indicates that if θ is anti-clockwise, torque is clockwise and vice versa i.e., torque is opposite to θ. The equation (i) is the condition for angular simple harmonic motion. It can easily be compared to F = – kx Torque = – Cθ Iα = – Cθ or
I
d 2θ = - Cθ dt 2 d 2θ C =- θ dt 2 I
⇒
ar a
d 2θ k =- x dt 2 m
rio s_ in fo
na .c om
This differential equation for angular simple harmonic motion can also be compared with the same equation for linear simple harmonic motion
m
Following the same comparison we can see that
I and the variation of θ with time C θ (t) = θmax sin (ωt + φ)
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
Time period, T = 2π
e/
.m
tp
s:
where, θmax is the angular amplitude or the maximum angular displacement from the mean position.
iv un
//t
ht
5.13 PHYSICAL PENDULUM
ht tp
s: //
t.m e/
ht tp s:
A rigid body is suspended from a fixed point O. It can perform angular oscillations in a vertical plane about the horizontal axis through O perpendicular to the plane of oscillations. This rigid body can be called as a physical pendulum.
O θ
d C
mg
Let C be the centre of gravity of such a rigid body suspended from O. The distance of C from point of suspension O is d. If we displaced the body anti-clockwise through a small angle θ, the weight of the body creates a clockwise restoring torque about axis through O. Magnitude of restoring torque is https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
τ = mg (d sin θ) = mg dθ (if θ is small) Comparing with τ = C θ, we have C = mgd Let I be the moment of inertia of the body about the axis through O. Therefore
Time period = 2π
I mgd
co m
Example A body makes angular simple harmonic motion of amplitude π/10 rad. and time period 0.05 sec. If the body is at a displacement θ = π/10 rad. at t = 0, write the equation giving the angular displacement as a function of time. θ = θm sin (ωt + φ)
ar an a.
m
s
ni v
//t s: ht tp
sin φ = 1
π 2
e/ u
φ=
.m
or
or
ht tp s: //t
.m
e/
s:
At t = 0, θ =
…(i)
er
//w
w
π θ = sin[(40πs -1 )t + φ] 10
π rad. 10 Putting in equation (i), we get π π = sin φ 10 10
ht tp
ita
un iv er s
w
∴
2π 2π = = 40πs -1 T .05
.ja
ω=
π rad 10
si ta rio
Here θm = amplitude =
rio s_ in fo
)
) Solution Let the required equation be
π π π By equation (i), θ = sin (40π s -1 )t + = cos [(40πs −1 ) t ] 2 10 10 Example
A solid cylinder is attached to a mass less spring with constant k = 3 Nm–1 and it rolls without slipping. If the system is stretched by 0.25 m and released. (i) Show that under these condition the centre of mass of cylinder executes 3m simple harmonic motion with period T = 2 π 2k (ii) Find the translational kinetic energy and rotational kinetic energy as it passes through equilibrium position.
5.30
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na .c om
Simple Harmonic Motion
) Solution
5.31
er s
e/
…(i)
un
//t
iv
.m
er si ta rio s
un iv
//w
w
1 where, v = ωR and I = mR 2 for solid cylinder 2 1 mω2 R 2 + mω2 R 2 + kx 2 = constant or 2 Differentiating equation (i) with respect to t, we get
s: tp ht
rio s_ in fo
ita
.ja
1 2 1 2 1 2 mv + Iω + kx = constant 2 2 2
w
i.e.,
m
ar a
(i) First method: From the conservation of energy Total energy of cylinder and spring system is constant
ht tp
s: //
t.m e/
ht tp s:
dω dx dω mR 2 2ω +ω + 2kx = 0 dt dt dt dx Here, = v = ωR dt or x = Rθ 3α m + 2 kθ = 0 2k α=θ or 3m This equation is the simple harmonic equation because angular acceleration is directly proportional to the angular displacement. T = 2π
3m 2k
Second Method: Let the cylinder rotates in antic-lockwise direction. In this rotation if angle rotates is θ, then the spring shifts atleast a distance Rθ from mean position. The friction force opposed this force.
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Simple Harmonic Motion
Rθ θ Mean position
na .c om
Thus, net force = (k Rθ – f) towards mean position. Applying τ = Icm α, and F = ma
rio s_ in fo
f R = I α, and k Rθ – f = ma
∴ k Rθ – f = m α R f = (k Rθ – m α R)
w
er s
w
or
un iv
kR 2 θ Icm + mR 2
kRθ
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
e/
⇒ α=
tp
s:
//w
or (k Rθ – m α R) R = Icm α
er si ta rio s
ita
.ja
m
a = αR
ar a
where, a is the linear acceleration and α is the angular acceleration. For rolling without slipping.
ht
5.32
f
The magnitude of the angular acceleration is proportional to the angle θ. Hence, the cylinder performs simple harmonic motion ω2 =
kR 2 Icm + mR 2
2
k R2 2π or = 1 T m R 2 + mR 2 2 or
T = 2π
3m 2k
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5.33
1 1 (ii) Translational kinetic energy = mv 2 = m (Aω) 2 2 2 1 2π A = (0.25) = m A 2 ⋅ 2 T
Here,
Simple Harmonic Motion
2
1 (2π) 2 2k = m A2 ⋅ ⋅ 2 (2π) 2 3m 1 3N × 2 = × 0.25 × 0.25 × = 2 3 = 6.25 × 10–2 J
na .c om
ar a
m
11 = mv 2 2 2
er si ta rio s
ita
er s
un iv
e/
s:
The rotational energy is half of the translational kinetic energy.
iv
ht tp s:
un
//t
ht
.m
tp
//w
w
w
.ja
1 = × 6.25 × 10-2 J 2 = 3.1 × 10−2 J
rio s_ in fo
1 1 1 = Iω2 = ⋅ mR 2 ω2 2 2 2
Rotational kinetic energy
t.m e/
)
s: //
ht tp
) Solution
Obviously for small β, the ball excutes parts of simple harmonic motion. Due to the perfectly elastic collision, the velocity of ball simply reversed. As the ball is in simple harmonic motion (| θ | < α on the left). Its motion law is ma + mg sin θ = 0
1. A ball is suspended by a thread of length l at the point O on the wall, forming an angle α with the vertical. Then the thread with the ball was deviated through a small angle β (β > α) and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum.
WORKED OUT EXAMPLES
O αβ
θ l
mg sinθ
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mg cosθ mg
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5.34
Simple Harmonic Motion
) Solution
θ θ = – g sin θ = - g l = – ω2 θ ……. (i) If we assume that the ball is released from the extreme position θ = β at t = 0, the solution of differential equation would be taken in the form θ = β cos ωt
)
or
When the hydrometer is displaced, then the net upward force
F = ρπ r2 (h + x) g – mg
...(ii)
From equations (i) and (ii) F = ρπ r2 xg ρπ r 2 xg ∴ a= x = ω2 x m
……. (ii)
g t l
= β cos
...(i)
mg = ρ π r2hg
Let at equilibrium, h is length in the liquid. Then at equilibrium
o
nf s_ i io ta r si rio
s
er
er si ta
iv
m ρπ r 2 g
T = 2π
e/ un
tp
t.m
ht
4πm ρg r 2
3. A cylinder of mass m and radius r rolls without slipping on a circular surface of radius R as shown in the figure. Find the frequency of oscillation when the cylinder is displaced slightly from its equilibrium position.
tp
2. Calculate the period of small oscillations of a hydrometer shown in the figure, which was slightly pushed down in the vertical direction. Mass of the hydrometer is m = 50 g, the radius of its tube is r = 3.2 mm, the density of the liquid is ρ = 1.00 g/cm3. The resistance of the liquid is assumed to be negligible.
=
s: //
ht
m
a. iv e/
s:
//t
l π -1 α + sin g2 β
π -1 -1 becausesin x + cos x = 2
(h + x)
Hence acceleration is directly proportional to the displacement
.m
tp
ht
=2
un
-1 α π - cos β
s:
l g
//w
time period, T = 2t′ =2
a∝x
an
-1 α π - cos β
∴
m .ja
l g
=
or
ar
-α l cos -1 g β
t′ =
w w
or
co
g -α = β cos t′ l
If t′ be the time taken by the ball to go from the extreme position θ = β to the wall, i.e., θ = – α, then from equation (ii)
R m
r
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Potential energy = mgh
)
) Solution
v = velocity of centre of mass of the cylinder, and ω = angular velocity of rotation of the cylinder For pure rolling,
= mg (R – r) (1 – cos θ) E = Total energy of the system at B
1 1 = mv 2 + Iω2 + mgh 2 2 2
co m
fo
3 d θ m(R - r ) 2 + mg (R - r ) (sin θ) = 0 2 dt
in
s_ io
ta r
io
s
er iv
ta r
si
un
iv
d 2θ 2 g =θ dt 2 3 R -r
er
e/
un
//t .
e/
Hence the motion is simple harmonic motion. Hence angular frequency of oscillation,
1 I ωc 2 2 2
1 R - r 2 = I θ 2 r
ω=
2 g . 3 R -r
ht tp
s:
//t .
m
s:
ht tp
….. [For small θ, sin θ. θ] ⇒
m
Reference position
=
d 2θ 2 g + θ=0 dt 2 3 R - r
si
.ja w w ht tp
o
A
The total energy of the system consists of kinetic energy (rotational and translational) and potential energy should remain the same all the time. 1 Translational kinetic energy = mv 2 2 1 = m(R - r ) 2 θ 2 2 1 Rotation kinetic energy = I ωc 2 2
⇒ ∴
R
dE =0 dt
an m
(R – r) cos θ
ar
θ
s: //w
B
Now,
a.
C R –r
3 = m(R - r ) 2 θ 2 + mg (R – r) (1 – cos θ) 4
…(ii)
V R -r ωc = = θ r r
⇒
mg (R – r) (1 – cos θ)
…(i)
v = (R - r ) θ
∴
1 11 R - r 2 = m(R - r ) 2 θ 2 + mr 2 θ + 2 2 2 r
Let θ be the angular displacement of the line OC as the cylinder passes through the position B.
4. A certain mass of a perfect gas is enclosed in a cylinder of volume V0 fitted with a smooth heavy piston of mass m and area A. The piston is displaced through a small distance downwards so as to compress the gas isothermally, and then it is let go. Show that the resultant motion is simple harmonic motion and find its period. Take the atmospheric pressure as Patm.
v = r ωc
Let
5.35
Simple Harmonic Motion
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5.36
weight and hence magnitude of net upward force is
)
) Solution
F = PA – mg – Patm A
PatmA
xA F = P0 1 + A - mg - Patm A V0 From equation (i),
⇒ PO A
mg
Po
F=
Hence, magnitude of the force is proportional to x and its direction is towards equilibrium position. Comparing with F = kx, we get
co m
Piston in equilibrium
mg
in
s
iv
si
ta
un
er
e/
//t
tp s:
ht
⇒
T = 2π
= 2π
mV0 mg 2 Patm + A A
mV0 Patm A 2 + mg A
5. A uniform rod is placed on two spinning wheels as shown in the figure. The axis of the wheels are separated by a distance l = 20 cm the coefficient of friction between the rod and the wheels is µ = 0.18. Demonstrate that in this case the rod perform harmonic oscillations. Find the period of these oscillations.
.m
e/ u
//t
s:
tp
ht
ni v
.m
ht
-1
m V0 P0 A 2
xA P = P0 1 V0
m k
rio
w //w s: tp
Let A be the area of croass-section of the cylinder. When, the piston is in equilibrium, forces on it must balance. ∴ P0 A = Patm A + mg ...(i) Let us now displace the piston through a small distance x in downward direction. As the compression of the gas is isothermal, the final pressure P is given by P0 V0 = P (V0 – xA) PV P= 0 0 V0 - xA
= 2π
er s
w
P
ita rio s_
.ja m
PA
P0 A 2 V0
Time period, T = 2π
ar
x
k=
an a.
PatmA
P0 A 2 x V0
fo
Simple Harmonic Motion
xA = P0 1 + V0 [∵ (1 + t)n ≈ 1 + nt for small t.]
Considering the forces on the displaced piston, thrust of the gas pressure now exceeds its https://t.me/universitarios_infohttps://www.jamarana.com
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From equations (i), (ii) and (iii), we get
)
) Solution
In the equilibrium position, the centre of mass of the rod lies midway between the two rotating wheels. Let us displaced the rod horizontally by some small distance and then released it. Let us depict the forces acting on the rod when its centre of mass is at distance x from its equilibrium position. Since, there is no net vertical force acting on the rod, Newton’s second law gives
d 2x 2g =-µ x 2 dt l Hence, time period, T = 2π
fo
co m
in
a.
s_
an
rio
ar m
si
s rio ta si iv er
mg x
e/
...(i)
)
The physical system consists with a pulley and the block. Earth and spring are considered as external object. Here the physical phenomenon consists in the oscillatory motion of system. Choosing on inertial frame, let us direct the x-axis as shown in the figure. Initially the system is in equilibrium position. Now from the condition the translation equilibrium for the block, T0 = mg ...(i) Similarly, for the rotational equilibrium of the pulley k ∆l R = T0 R or
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T0 = k ∆l
…(iii)
l + 2x l - 2x N1 = N2 2 2
As the rod experience no net torque about an axis perpendicular to the plane of the figure through the centre of mass of the rod
…(ii)
ht
tp
For the translational motion of the rod from the equation
v
un
x
s: //t .m
ht N1 + N2 = mg
Fx = max md 2x µN1 - µN 2 = dt 2
x
) Solution First Method: Shown in figure (b).
er iv un
e/
µN2
mg
O
:// t.m
x
T
T
ta
.ja w c
tp s
ht tp
s:
µN1
//w
o
k∆ l
T 0`
N2
w
N1
2l µg
6. Find the time period of the system shown in the figure. The radius of the pulley is R, its moment of inertia relative to the rotation axis is I, the mass of the body is m and the spring constant is k.
l 2 c
l 2µg
T=π
l 2
5.37
Simple Harmonic Motion
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...(ii)
5.38
econd method: The mechanical energy of the system (pulley + block) is the joint field of spring and gravity is conserved.
s
From equations (i) and (ii) mg ∆l = ...(iii) k Now let us disturb the equilibrium of the system, no matter in which way to analysis its motion at an arbitrary shown in the figure. Now from Newton’s second law of motion for the block Fx = max d 2x mg - T = m 2 ...(iv) dt
Therefore, ∆T + ∆U = 0
2 2 1 x 1 2 1 v0 1 2 m x + I mv + ( ) I 0 2 R 2 2 R 2
1 1 + k (∆l + x) 2 - k (∆l ) 2 - mgx = 0 2 2
Similarly, for the pulley for its rotational motion from the equation. τz = I αz (where z-axis is the axis of rotation) d 2θ TR - k (∆l + x) R = I 2 ...(v) dt
m
Here, v0 = 0 at x = 0 which is constant.
in
rio
ta
s
rio
ta
iv
er si
un e/
Hence,
.m s:
I m+ 2 R 2π ω
= 2π
or
tp
k
Hence, the time period T =
ht
ω2 =
k x = - m + I2 k
//t
ht
d x k x = - 2 dt m+ I R 2 2
I x + kx = 0 m + 2 R
[because at equilibrium position kAl = mg]
//t
s:
tp
I d 2x -kRx = mR + 2 R dt
or
x x (Al + x) - mg = 0 R R
er si or
Solving equations (iv) and (vii) and using the initial condition of the problem, we get
or
1 I x 1 x+ k 2 m + 2 2 2 R R 2
iv un
...(vii)
e/
I d 2x R dt 2
.m
ht
tp
TR - k (∆l + x) =
s:
From equations (v) and (vi)
s_
ar an
w
...(vi)
//w w
.ja
2
or
Differentiating with respect to time, we get
m
a = αR d x d 2θ = R dt 2 dt 2
fo
co
a.
But
2
1 1 x 1 m( x ) 2 + I + k (∆l + x) 2 - mgx = 0 2 2 R 2
m+ k
I R2
T = 2π
x
m+ k
I R2
7. Two springs, each with unstretched length 0.20 m but having different force constants k1 and k2 are attached to opposite ends of a block with mass m on a level, frictionless surface. The other ends of the springs are now attached to two points P1 and P2, 0.10 m from the original position of the ends of the spring. Let k1 = 1.0 N/m, k2 = 3.0 N/m and m = 0.20 kg.
Simple Harmonic Motion
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Simple Harmonic Motion
5.39
Determine the length of each spring when the block is in its new equilibrium position after the springs have been attached to the pins.
R M
C θ α
P2
P1
A
)
) Solution
fo
s
rio
si ta
er
iv
un
mA gR – R mg sin α = 0 mA = m sin α
e/
s: //t .m
ht tp
∴ or
//t .
s:
ht tp
| F1′ | = | F2′ | k1 (0.1 + x) = k2 (0.1 – x) 0.1 + x = 0.3 – x 4.0 x = 0.2 m x = 0.05 m length of spring = 0.35 m and 0.25 m
8. A uniform cylindrical pulley of mass M and radius R can freely rotate about the horizontal axis C. The free end of thread tightly wound on the pulley carries a dead weight A. At a certain angle α if counter balances a point mass m fixed at the rim of the pulley, find the angular frequency of small oscillation of the arrangement.
si ta
e/
m
ht tp
Since, | F1 | ≠ | F2 |, the mass will move towards the right so as to attain equilibrium position at which these forces become identical. If x is the distance moved by the mass towards right side, we have
i.e., or or or
er
un
| F2 | = k2 (0.1 m) = (3.0 N/m) (0.1) = 0.3 N
iv
| F1 | = k1 (0.1 m) = (1.0 N/m) (0.1 m) = 0.1 N
s: //w w
w
.ja
m
rio
ar a
When the springs are attached to the pins, each spring will under go elongation by 0.1 m. The elongation will exerts forces on the mass which are given by
s_
in
na
)
) Solution
If we get the system in motion, no matter in which way, from the conservation of mechanical energy of oscillation at an arbitrary position (as shown in the figure). 1 1 1 MR 2 2 mA (θ ) 2 R 2 + m(θ ) R 2 + (θ) + mg 2 2 2 2 [R cos (α + θ) – R cos α] – mA g Rθ = constant
R2 M 2 m sin α + m + (θ) + mg R[cos 2 2 (α + θ) – m cos α] - m sin α Rθ = constant or
0.2 m
In this problem, the weight of A is restoring force for half oscillation and for the other half oscillation the weight of point mass m is restoring. Hence, the solution of this problem from energy conservation is more meaningful. It is also because the restoring and conservative forces are responsible for the conservation of mechanical energy of oscillation. Before writting the equation of energy, let us find the mass of the dead weight A. From the condition of the problem, at the equilibrium position: τC2 = 0
om
0.1 m
.c
0.1 m
0.2m
Differentiating with respect to time, we get R2 M m sin α + m + 2 ⋅ θθ + mg R (α + θ) 2 2 θ – mg R sin α = 0
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5.40
or
M R m sin α + m + θ + mg 2 [sin α cos θ + cos α sin θ] – mg sin α = 0
or
m in rio
iv
)
un
) Solution
e/
(i)
un e/
s:
X1
ht
tp
//t
ht
.m
tp
s:
//t
iv
.m
tp
ht
s
er si
ta
.ja
w
s: //w w
9. A rod AB of mass M is attached as shown below to a spring of constant k. A small block of mass m is placed on the rod at its free end A. (i) If end A is moved down through a small distance d and released, determine the period of oscillation. (ii) Determine the largest allowable value of d if the block m is to remain at all times in contact with the rod.
si
2mg cos α [M + 2m(1 + sin α)]R
er
ω=
rio
2mg cos α θ [M + 2m(1 + sin α)]R
s_
an
ar
θ=-
m
∴
fo
M R m sin α + m + θ + mg cos α θ = 0 2
ta
a. co
[for small θ, cos θ ≃ 1 and sin θ ≃ 0]
or
10. In the given figure, the spring has a force constant 5000 N/m. The pulley is light and smooth. The spring and the string are light, the suspended block is of mass 1 kg. If the block is slightly displaced vertically down from its equilibrium position and released, find the period of its vertical oscillations.
M R m sin α + m + θ + mg 2 [sin α ⋅ cos θ + cos α ⋅ sin θ] – mg sin α = 0
or
M R 2 m sin α + m + θ + mg R 2 sin (α + θ) – mg R sin α = 0
Simple Harmonic Motion
At equilibrium X2 T
T B
or
)
) Solution Try yourself.
l
or
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b
m
A
Initially, there is no elongation of spring. When it goes in equilibrium position 2 T = Kx1 T = mg 2 mg x1 = k
k
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...(i)
5.41
and x2 = 2x1 4 mg = k mg x2 = k/4
Simple Harmonic Motion
O
…(i)
(ii) Similarly, 2T = mg T = k x1
l
m k/4
Hence, time period, T = 2π
)
co m
) Solution
in
si
s
rio
ta
si
er iv
e/
1 ml 2 2 l 1 (θ) + mg (1 - cos θ) + k (l θ) 2 = 2 3 2 2 constant
un
//t
e/
s:
m
tp
//t .
Hence, time period, T = 2π
1 2 l 1 I(θ) + mg (1 - cos θ) + k (l θ) 2 = 2 2 2 constant
Differentiating with respect to time, we get 1 ml 3 l 1 g θθ + mg sin θθ + kl 2 2θθ = 0 2 3 2 2
m 4k
θ θ
11. Find the frequency of small oscillation of a thin uniform vertical rod of mass m and length l hinged at the point O shown in the figure. The combined stiffness of the springs is equal to k. The mass of the springs in negligible.
s:
mg 4k
tp
x2 =
ht
T 2k
ht
=
or
.m
tp
x x2 = 1 2
or
ht
mg
or
un
iv
//w
s:
2T
w
er
w
At equilibrium
X2
1 2 1 Iω + mgh + kx 2 = constant 2 2
ta
.ja
m
T
rio
ar
s_
an
X1
fo
Let us locate the rod at the position when it makes angle θ from the vertical and in this problem both, gravity and spring forces are restoring conservative force. From the energy conservation
a.
T
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Simple Harmonic Motion
Thus, for very small θ
)
) Solution The electro-magnetic field induced in the loop
3g kl θ = - 1 + θ 2l mg
ε = l (v × B) = v . Bl
...(i)
5.42
L
12. A loop is formed by two parallel conductors connected by a solenoid with inductance L and a conducting rod of mass m which can freely (without friction) slide over a conductor. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction B. The distance between the conductors is equal to l. At the moment t = 0 the rod is imparted an initial velocity v0 direction to the right. Find the law of motion x(t) if the electric resistance of the loop is negligible.
di - iR = ε - vBl dt
Applying Kirchhoff’s law
3g kl 1 + 2l mg
ω=
…(ii)
L
…(iv)
rio s_ in
fo
ar an
ta
-
…(v)
rio
si ta
d 2 v B2 l 2 + v=0 dt 2 mL
iv er
v = A sin ω t
un
∴
v (t) = v0 sin ω t
e/ where,
ω=
Hence,
x(t ) =
Bl ml v0 cos ωt ω
13. A body of mass m fell from a height h onto the pan of a spring balance. The masses of the pan and the spring are negligible, the stiffness of the later is k. Having struck to the pan, the body starts performing harmonic oscillations in the vertical direction. Find the amplitude and the energy of these oscillations.
ht
.m //t s: tp
Lm d 2 v = vBl Bl dt 2
s
er si iv
e/
un
or
//t l
…(iii)
From equations (iii) and (v), we get
s: tp B
dv dt
co m
a.
i Bl = - m
di m d 2v =dt Bl dt 2
or
ht L
di = vBl dt
Ampere’s force oppose the velocity
+z
.m
l
B
R=0
But
∴
m
.ja
w w //w
Vo
s: tp
L
ht
Hence,
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Simple Harmonic Motion
5.43
because at the extreme position kinetic energy is zero. Although the weight of body m is conservative force, it is not restoring in this problem, hence Uextreme is only concerned with the spring force. Therefore
m
h
)
m
) Solution
14. A body A of mass m1 = 1.00 kg and a body B of mass m2 = 4.10 kg are interconnected by a spring as shown in the figure. The body A perform free vertical hurmonic motion with amplitude a = 1.6 cm and frequency ω = 25 s–1. Neglecting the mass of springs, find the maximum and minimum values of force that this system exerts on the bearing surface.
1 m2 g 2 E = U extreme = k a 2 = mgh + 2 2k
ta si er iv
tp
mg k
ht
∆x =
e/ un
.m
s:
While the body A is at its upper extreme position, the spring is obviously elongated by the mg amount a - 1 k If we indicate y-axis in vertically downward direction, then by Newton’s second law of motion mg m1 g + k a - 1 = m1 ω 2 a k
The mechanical energy of oscillation which is conserved equals, E = Uextreme’
or
m g k a - 1 = m1 (ω 2 a - g ) k
mg 2 hk 1+ k mg
a=
2
mg 2 mgh a = x - ∆x = + k k
Hence, amplitude,
) Solution
//t
ht
∴
rio
iv un
//t
s:
At the equilibrium position, there is no force. Thus mg = k ∆x
B
)
2
mg mg 2 mgh + + k k k
tp
x=
.m e/
ht
Neglecting negative sign, we get
s
er si
w
//w
s:
2mg ± (2mg ) 2 + 8 mg hk 2k
tp
x=
o
ta rio
ar m
kx – 2 mg x – 2mgh = 0
w
or
A
.ja
1 -mg (h + x) + kx 2 = 0 2 2
s_
an
in f
a.
co
From energy conservation, for the maximum contraction of the spring, the total energy is conserved. Loss in potential energy is gain in the spring energy. Thus
…(i)
(Because at any extreme position the magnitude of acceleration of an oscillatory body equals ω2 a and is restoring in nature.)
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5.44
Simple Harmonic Motion
passing through the middle of bar. Find the period of small oscillations of the sleeve. At what value of ω will there be no oscillations of the sleeve?
If N be the normal force exterted by the floor on the body B, while the body A is at its upper extreme position, then from Newton’s second law for the body B A
B
M
mg N + k a - 1 = m2 g k mg N = m2 g - k a - 1 k
or
From equation (i),
)
om
) Solution
ita rio s x
.m From Fx = maximum for the sleeve in the frame of rod – kx + mω2x = m x k x = - - ω2 x or …(i) m 2π = 0.7sec ∴ time period, T = k 2 -ω m It is obvious from equation (i) that the sleeve will not perform small oscillations if
ω≥
k = 10rad/sec. m
15. In the arrangement shown in the figure, the sleeve M of mass m = 0.20 kg is fixed between two identical springs whose combined stiffness is equal to k = 20 N/m. The sleeve can slide without friction over a horizontal bar AB. The arrangement rotates with a constant angular velocity ω = 4.4 rad/s about a vertical axis
B
//t ht tp s:
N′ = m1 (g + ω2a) + m2g N′ = (m1 + m2) g + m1 ω2a
From Newton’s third law, the magnitude of the forces are N′ and N respectively.
M
Hence,
mω x 2
kx
rs
t.m
://
tp s
ht
From equation (i),
A
ve
…(iii)
In this, case if N′ be the normal force exerted by the floor on the body B, then from Newton’s second law for body B, we get
mg N′ = k a + 1 + m2 g k
fo
ita
e/ un iv
mg k a + 1 = m1 ( g + ω 2 a ) k
ht tp s:
or
//w
w
er s
w
mg m1 g - k a + 1 = m1 (-ω 2 a ) k
e/ un i
.ja
Fy = may for the body A at this state
Obviously, the sleeve performs small oscillations in the frame of rotating rod. In the rod’s frame, let us depict the force acting on the sleeve along the length of the rod while the sleeve is at a small distance x towards right from its equilibrium position. The free body diagram of block does not contain Coriolis force because it is perpendicular to the rod.
rio s_ in
m
ar an a. c
When the body A is at its lower extreme position, the spring is compressed by the distance m1 g . a + From Newton’s second law in k projection from, i.e.,
N = m2g – m1 (ω2a – g) = (m1 + m2)g – m1 ω2a
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Simple Harmonic Motion
5.45
)
iv
) Explanation The motion will SHM only if
un
t.m e/
k ( x - a)2 , 2
dU = k ( x - a) dx
F = –k (x – 1)
This is the condition for SHM about point x=+a Other function do not satisfy this condition.
5. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downwards push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is
2. Which of the following functions represent S.H.M.? (a) t + sin ωt (b) sin2 ωt (c) sin ωt + 2cos ωt (d) sin ωt + cos 2 ωt d2y acceleration or 2 ∝ - y . dt This is true for (c)
⇒
s: //
F1 = mat = mg sin θ
ht tp
mv l
2
dU dx
er si ta rio s
)
un iv e/ .m
//t
ht tp s:
T – mg cos θ =
ita er s
w //w s:
)
For U( x) =
mg cosθ
) Explanation Considering the F.B.D. of the
rio s_ in fo
na .c om
ar a m .ja
(d) U(x) = constant
) Explanation F = -
tp
ht
mg sin θ
k (x - a2 ) 2 (b) U(x) = k1x + k2x2 + k3x3 (a) U( x) =
(c) U(x) = Ae–bx
w
T
and
4. The potential energy U(x) of a particle executing SHM is given by
θ
bob
3. The time-period of a simple pendulum of small amplitude (a) is directly proportional to the square root of its length (b) is inversely proportional to the square root of acceleration due to gravity (c) is independent of the amplitude (d) all of these Ans. (c)
1. A simple pendulum of length l and mass (bob) m is oscillating in a plane about a vertical line between angular limits –φ and + φ. For an angular displacement θ (where | θ | < φ), the tension in the string and the velocity of the bob are T and v respectively. Then (a) T cos θ – mg mv 2 (b) T – mg cos θ = l (c) magnitude of the tangential acceleration of the bob is |aT| = g sin θ (d) T = mg cos θ
t
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a height 3 Re from the earth’s surface, where Re is earth’s radius. The period of the pendulum will be
1/2
1 k - Aρg 2π M
1/2
1 k + Aρg (b) 2π M
8. A simple harmonic motion has an amplitude A and time period T. The time required by it to travel from x = A to x = A/2 is T T (a) (b) 4 6 T T (d) (c) 4 3
1/2
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π 2π sin = sin t 6 T T t= ⇒ 12 Hence, time required to travel from x = A to A x= will be 2 T T T - = 4 12 6
⇒
9. A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period
Phase difference
7. A second’s pendulum is placed in a space laboratory orbiting around the earth at
A 2π = A sin t 2 T
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π 2π y1 = a sin + Q and y2 = b sin + Q 2 3 2π π π = + Q - + Q = 3 2 6
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π 2π 2 t + Q and y2 = b sin 3 t + Q . The phase difference between these after 1 sec is π (a) π (b) 2 π π (c) (d) 4 6
∴
T and time required to travel A to x = 0 is 4 A from x = to x = 0 is equal to the time 2 A required to travel from x = 0 to x = which 2 is equal to 2π y = A sin t T
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6. Two SHM are given by y1 = a sin
) Explanation t = 1 sec
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) Explanation Time required to travel from x =
m
L Fr = Mg - k ( x + y ) + A + y ρg 2 = -(k + Aρg ) y
)
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released, then
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L Mg = kx + A ρg 2 Let it be given a small downward push y and
)
) Explanation In equilibrium,
k + Aρg M
)
and its time period will be infinite.
1 k + Aρg (d) 2π Aρg
ω2 =
) Explanation The pendulum will not oscillate
1/2
1 k + AρgL2 (c) 2π M
∴
(b) 2 3s (d) infinite
(a) zero (c) 4 s
(a)
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Simple Harmonic Motion
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(c) 2π
L g
L g sin α
(d) 2π
L g tan α
)
) Explanation mg = kx
(b) 2π
L g cos α
(a) 2π
1 (c) T = π mg / x(M + m) 2 (M + m) (d) T = 2π mgx
x(M + m) mg
10. The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined pane of inclination α, is given by
(b) T = 2π
5.47
mg x(M + m)
(a) T = 2π
Simple Harmonic Motion
mg M +m and T = 2π x k
) Explanation Effective gravity inside the vehicle )
k=
is g cos α.
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⇒
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(d)
1 s 4
(b)
1 s 3 1 s 6
6. A block of mass m is tied with a spring of length l0 fixed at A and placed on a smooth horizontal surface as shown. The block is compressed and a smooth vertical wall is placed at l1. Now if the block is further compressed by length a, the period of oscillation of block will be (assume the collision between the block and the wall to be elastic)
1 s 2
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3. A small bob is tied to a light string and is suspended in a lift. The bob is set into oscillations. The time period of oscillations is maximum, when the lift is (a) at rest (b) moving upward with acceleration (c) moving downward with acceleration (d) time period will be same for all conditions
5. A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?
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(d) none of these
4. For a particle undergoing SHM, the velocity is plotted against displacement. The curve will be (a) a straight line (b) a parabola (c) a circle (d) an ellipse
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2. As the acceleration of a vibrating particle executing simple harmonic motion increases, its time period (a) increases (b) decreases (c) remains unchanged (d) first increases, then decreases
(c) π
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m
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UNSOLVED OBJECTIVE TYPE QUESTIONS
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fo
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come to rest (c) undergo vertical SHM with a time
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m2 g and k
m1 k (d) undergo vertical SHM with a time period of 2π
period of 2π
(m1 + m2 ) k
A 3 k 2 M
(d) A
k M
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(a) remain undisturbed
M k
8. The displacement y of a particle executing certain periodic motion is given by y = 4 1 cos2 t sin (1000t). This expression 2 may be considered to be the superposition
(c) A 3
(b)
k M
(a) A
10. Two blocks of masses m1 and m2 are attached to the lower end of a light vertical spring of force constant k. The upper end of the spring is fixed. When the system is in equilibrium, the lower block (m2) drops off. The other block (m1) will (b) move up through a distance
s: tp ht 0–60°
(d) all of these
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an a. co m
7. An insect of negligible mass is siting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude A in front of a plane mirror placed as shown. The maximum speed of insect relative to its image will be
in
l -l k sin -1 0 1 m l0 - l1 + a
si
(d) 2π
(a) speed of the particle is maximum at x = 0 π (b) amplitude of oscillations is 2a (c) time period of small oscillation is, m T = 2π 2 a U0
l -l m k sin -1 o 1 k m l0 - l1 + a
(c) 2π
Then (for the small oscillations)
m k l -l sin -1 o 1 k m a
(b) 2π
9. A particle of mass m is moving in a potential well, for which the potential energy is given by U(x) = U0(1 – cos ax), where U0 and a are constants.
(lo - l1 ) m k sin -1 k m (lo - l1 + a )
(a) 2π
A
of n independent harmonic motions. Then n is equal to (a) 2 (b) 3 (c) 4 (d) 5
l1
Simple Harmonic Motion
M
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o-w
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11. A coin is placed on a horizontal platform, which undergoes horizontal simple hormonic motion about a mean position O. The coin does not slip on the platform. If the force of friction acting on the coin is F, then (a) F is always directed towards O (b) F is directed towards O when the coin is moving away from O, and away from O when the coin moves towards O (c) F = 0 when the coin and platform come to rest momentarily at the extreme position of the harmonic motion (d) F is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion
m
k2 k1
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17. A simple pendulum with a solid metal bob has period T. The metal bob is now immersed in a liquid of density one-tenth that of the bob. The liquid is non-viscous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be 9 (a) T (b) T 10
(d)
10 T 9
9 T 10
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(d)
16. A simple pendulum with angular frequency ω oscillates simple harmonically. The tension in the string at lowest point is T. The total acceleration of the bob at its lowest position is T (a) -g (b) zero m g T (c) g (d) T + m m
(c)
18. A block of mass m compresses a spring of stiffness k through a distance as shown 2 in the figure. If the block is not connected with the spring and the impact of the block with the vertical wall is elastic, the period of motion of the block is
(b) 2 (d) 2
14. Two S.H.M.’s are represented by the π equations, y1 = 10 sin 3 πt + and 4 y2 = 5 (sin 3πt + 3 cos 3πt). Their amplitudes are in the ratio of
k2 k1
(c)
a. co
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13. A simple pendulum has a time period T1 when on the earth’s surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is (a) 1 (c) 4
15. Two bodies M and N of equal masses are suspended from two separate massless springs of spring-constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, then ratio of the amplitude of vibration of M to that of N is k k1 (b) (a) 1 k2 k2
12. In the previous question, the angular frequency of the simple harmonic motion is W. The coefficient of friction between the coin and the platform is µ. The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time (a) at the extreme positions of oscillations µg (b) for an amplitude of W2 (c) at the mean position g (d) for an amplitude of µ W2
5.49
(b) 1 : 2 (d) 1 : 4
(a) 2 : 1 (c) 1 : 1
Simple Harmonic Motion
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5.50
(b) the time period of oscillation will decrease (c) the time period of oscillation will remain unaffected (d) the mean position of simple hormonic motion will shift to the right
Simple Harmonic Motion
k
(d) none of these
7Q
m k
19. A spring-block system undergoes simple harmonic motion on a smooth horizontal surface. The block is now given some positive charge and uniform horizontal electric field to the right is switched on. As a result (a) the time period of oscillation will increase
E
s ita rio
7. (c) 15. (d)
8. (b) 16. (a)
ve
rs
6. (c) 14. (c)
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RACTICE EXERCISE 1
1. A particle executes harmonic oscillations along the x axis about the equilibrium position x = 0 with frequency ω = 4/s. At a certain instant, the particle has a position coordinate x0 = 25 cm and velocity v0 = 100 cm/s. Find the position x and the velocity v of the particle after a time t = 2.4 s. [IIT, 1978] π Ans. x = 35.36sin ωt + 4 ; v = 81.9cm/s
of elasticity for the wire, obtain the frequency of oscillation for simple harmonic motion along the vertical line. [MNR, 1990] 1 YA Ans. 2 π mL
2. A point mass m is suspended at the end of a massless wire of length L and crosssection A. If Y be the Young’s modulus
3. A certain spring, when stretched by 0.25 m, stores 5 joule of energy. It is kept vertical by fastening its lower end to a rigid support. Find the mass of a block which, if attached to the free end of this spring, will execute 5 oscillations per second. [Ans. 0.162 kg]
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5. (d) 13. (d)
.m
4. (d) 12. (a, c) 20. (b)
//t
3. (c) 11. (a, d) 19. (c, d)
ita rio
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2. (c) 10. (c) 18. (c)
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1. (c) 9. (d) 17. (c)
a.
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ANSWERS
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20. The time period of a spring-mass system is T in air. When the mass is partially immersed in water, the time period of oscillation is (a) T (b) < T (c) > T (d) ≤ T
co
(c) (1 + π)
m k
(b) (π + 4)
m k
l/2
l/2
l
(a) 2π
m
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o
nf
s_ i
rio
s
si ta
rio
α l cos.-1 - Ans. 2 g β
12. Consider the situation shown in the figure. The block on the left moves with a velocity v towards block on the right which is placed in equilibrium. Assume all the collisions to be elastic and all the surfaces to be smooth. Neglecting the dimensions of the blocks, prove that the motion of the blocks are periodic and find their periods.
ht tp
s: //
t.m
e/ un
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er
11. Two linear simple harmonic motions of equal amplitudes and frequencies ω and 2ω are impressed on a particle along the axes of X and Y respectively. If the initial π phase difference between them is , 2 find the resultant path followed by the particle. [Roorkee, 1985] [Ans. Parabola]
s: //t .m
ht tp
Mg 2 γP + A 1 A Ans. n = 2π MV0
si ta
er
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s:
ht tp
10. A ball is hung by a thread of length l from a point P of the wall that makes an angle α with the vertical. The thread with the ball is then deviated through a small angle β (β > α) and set free. Assuming the collision of the ball with the wall to be perfectly elastic, find the period of such a pendulum.
a. co m
ar
m
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//w
w w
an
5. A mass M attached to a spring oscillates with a period of 2 sec. If M is increased by 2 kg, the period increases by 1 sec. Find the initial mass M assuming that Hooke’s law is obeyed. [IIT, 1979] [Ans. 1.6 kg]
7. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of the mass M. The piston and cylinder have equal cross sectional area A. Atmospheric pressure is P and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from the equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT, 1981]
Determine the maximum velocity of the point. [Ans. 2.73 aω]
6. The potential energy of a particle of mass M situated in a unidimensional potential field varies as U(x) = U0 (1–cos ax) where U0 and a are constants. Find the time period of small oscillations of the particle about the mean position. m Ans. T = 2π 2 a U0
5.51
8. Find the time period of small oscillations of a ball suspended by a thread of length l = 0.20 m if it is placed in a liquid whose density σ is n = 3.0 times less than the density ρ of the ball. The resistance of the liquid is to be neglected. [Ans. T = 1.1 sec.] 9. A block is resting on a piston which is moving vertically with simple harmonic motion of period 1.0 sec. At what amplitude of motion will the block and piston separate? What is the maximum velocity of piston at this amplitude? [Roorkee, 1985] [Ans. 0.25 m; 1.57 m/s]
4. A point executes two harmonic oscillations simultaneously along the same direction: x1 = a cos ω t and x2 = a cos2 ω t.
Simple Harmonic Motion
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5.52
Simple Harmonic Motion
er si
ht
tp
s:
s
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M A
B
15. Two particles execute simple harmonic motion of same amplitude and frequency along the same straight-line. They pass one another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them? [MNR, 1986] [Ans. 120º]
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s:
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18. A ball of mass m and radius r rolls on a rough concave surface of large radius R. It performs small oscillations about the mean position. Assuming the slipping to be absent, find its time period. [MNR, 1990] 7(R - r ) Ans. T = 2π 5g 19. In the given figure, a sleeve of mass m is fixed between two identical springs, each having a force constant equal to k. The sleeve is free to slide without friction over a horizontal bar AB, rotating with a constant angular velocity ω, about a vertical axis passing through the middle of the bar. Find the time period of small oscillations of the sleeve. For what value(s) of ω, the sleeve will not oscillate at all?
ita
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14. A uniform plate of mass M is placed horizontally and symmetrically on two wheels, separated by a distance L and rotating in opposite directions as shown in the figure. The coefficient of friction between each wheel and the plate is µ. If the plate is slightly displaced along its length and released, show that it executes simple harmonic motion. Find its time period.
tp s:
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(kl = 1.8 N/m, k2 = 3.2 N/m, m = 200 gm) [IIT, 1985] [Ans. 2.83 sec]
L Ans. T = 2π 2µg
ta rio
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s_
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in
a.
fo
co
m
16. A horizontal spring-block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass k m is put on it and the two move together. ν m m Find the new amplitude and frequency of L vibration. [Roorkee, 1988] Ans. Both the amplitude and frequency change by a factor 2L m +π Ans. v k M Ans. Both the amplitude and frequency change by a factor 13. Two light springs of force constants k1 M+m and k2 and a block of mass m are in one 17. A simple pendulum is made by attaching line AB on a smooth horizontal table a 1 kg bob to a 5 m copper wire of diamsuch that one end of each spring is fixed eter 0.08 cm and it has certain period of on rigid supports and the other end is oscillation. Next, a 10 kg bob is substifree as shown in the figure. The distance tuted for 1 kg bob. Calculate the change CD between the free ends of the springs in period if any. is 60 cm. If the block moves along AB (Young’s modulus of copper = 12.4 × 1010 with a velocity 120 cm/sec in between the N/m2) [Roorkee, 1988] springs, calculate the period of oscillation [Ans. 0.004 s] of the block.
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rio s_ in fo
iv
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s: //
22. One end of each of the two identical springs, each of force constant 0.5 N/m, are attached to the opposite sides of a wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched and are collinear in a horizontal plane. To the wooden block is fixed a pointer which touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s, find the equation of path traced by the pointer on the paper
t.m e/
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24. A uniform cylinder of length L and mass M having cross-sectional area A is suspended with its length vertical from a fixed point by a massless spring such that it is half submerged in a liquid of density σ at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with small amplitudes. If the force constant of the spring is k, calculate the frequency of oscillation of the cylinder. [IIT, 1990] 1 k + Aσg Ans. n = 2 π M
er s
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s:
tp
ht
M Mm . x0 ,(b)T = 2π Ans. (a) m k M + (M + m)
ita
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m
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21. A circular ring hanging from a nail in a wall undergoes oscillations of amplitude θ = 2º and period T = 2 s. Find (a) radius of the ring, (b) speed of the particle opposite to the point of suspension as it passes through the position of rest, (c) acceleration of this particle when it crosses the mean position, and (d) acceleration of this particle when it is at one of the extreme positions. Given, g = π2m/s2. [Ans. (a) 50 cm, (b) 10.996 cm/s, (c) 1.19 cm/s2, (d) 34.4 cm/s2]
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2 2 Ans. T1 = 3 s; T2 = 3 s
23. In the given figure, the mass of the trolley is M, that of the block is m, and the force constant of the spring is k. Initially, the entire system is at rest. The spring is now stretched through x and the system is released. Find: (a) amplitude of the resulting simple harmonic motion of the block and the trolley, as seen by an observer from the road, and (b) periods of the two simple harmonic motions.
25. Find the frequency of small oscillations of a thin vertical rod of mass m and length l hinged at the point O, as shown in the figure. The equivalent force constant of the springs is equal to k. Assume the springs to be massless and the rod to be uniform.
Find out the ratio of their amplitudes. What are the time periods of the two motions? [IIT, 1986]
20. Two simple harmonic motions are represented by the following equations: π y1 = 10 sin (12t + 1) 4 y2 = 5 (sin 3πt + 3 cos 3πt)
5.53
and distance between two consecutive maxima of this path. [Roorkee, 1990] [Ans. x = 0.02 cos 10 (t – 10 z), ∆z = 6.28 cm]
1 k ;ω ≥ Ans. T = 2π k m - ω2 m
Simple Harmonic Motion
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5.54
Simple Harmonic Motion
in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 metre. Each spring has a natural length 0.06π metre and spring constant 0.1 N/m. 26. A point particle of mass 0.1 kg is executInitially, both the balls are displaced by ing S.H.M. of amplitude 0.1 m. When the an angle θ = π/6 radian with respect to particle passes through the mean posithe diameter PQ of the circle (as shown tion, its kinetic energy is 8 × 10–3 joule. in figure) and released from rest. Obtain the equation of motion of this (a) Calculate the frequency of oscillation particle if the initial phase of oscillation of ball B. is 45º. [Roorkee, 1999] (b) Find the speed of ball A when A and B are at the two ends of the diameter PQ. A B (c) What is the total energy of the system? π/6 [IIT, 1993] P π/6 Q 1 Ans. (a) n = π cycle/sec,(b) v = 0.0628m/s, (c), (c)3.947 × 1 1 π Ans. (a) n = cycle/sec,(b) v = 0.0628m/s, (c), (c)3.947 × 10-4 J Ans. 0.1sin 4 t + π 4 29. Find the time period of the system shown in the figure. The radius of the pulley is 27. A uniform rod AB of mass m = 1.5 kg is R, its moment of inertia about the axis of suspended by two identical threads of rotation is I, mass of the body is m and equal length l = 0.9 m, as shown in the figthe spring constant is k. ure. When turned through a small angle about a vertical axis passing through its middle point C, the threads are deviated by an angle θ = 5º The rod is then released to allow it to perform small oscillak∆l TO tions. Find the time period of oscillation of the rod. m O [Ans. 1.1 s]
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rio
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28. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed
//t
ht
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tp
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s:
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//t
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tp
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s:
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w
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in
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fo
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m
1 3k 3 g + Ans. n = 2 π m l
x
mg
1 m+ 2 R Ans. T = 2π k
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Simple Harmonic Motion
5.55
QUESTION BANK
1. A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod’s other end is supported by a spring of force constant k. The rod is rotated (in vertical plane) by a small angle θ from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is
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2. Suppose that a grandfather clock (a simple pendulum) is running slowly. That is, the time it takes to complete each cycle is longer than it should be. Should one shorten or lengthen the pendulum to make the clock keep the correct time? Why?
3. The amplitude of a simple pendulum decreases with time. But there is no decrease in the amplitude of a pendulum clock. Why?
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m
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2. A particle of mass m is attached to three identical springs A, B and C each of force constant k as shown in figure. If the particle of mass m is pushed slightly against the spring A and released, then the time period of oscillation is:
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7. A particle is oscillating in simple harmonic motion. The time required for the particle to travel through one complete cycle is equal to the period of the motion, no matter what the amplitude is. But how can this be, since larger amplitude mean that the particles travel farther? Explain.
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2 P
8. A pendulum clock is taken to moon. Will it gain or lose time?
9. Time period of a particle in SHM depends on the force constant k and mass m of the
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m 2k
(d) 2π
m 3k
m k
(b) 2π
(c) 2π
2m k
(a) 2π
m . A simple pendulum 2k executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? particle: T = 2π
$
(d)
k 3m
iv
3k 3 g + m 2L
(b)
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(c)
3k m
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6. Distant explosion can break glass windows. How?
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5. What is the period of oscillation of a simple pendulum if its bob is made of ice?
(a)
4. A car travels over a road that contains a series of equally spaced bumps. Explain why a particularly jarring ride can result if the horizontal velocity of the car, the bump spacing, and the oscillation frequency of the car’s suspension system are properly “matched”.
1. Can pendulum clocks be used in artificial satellite? Explain.
ONLY ONE OPTION IS CORRECT
CONCEPTUAL QUESTIONS
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5.56
3. A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude? (a) s/2 (b) s/6 (c) s/4 (d) s/3
2πt π + (c) x(t ) = a sin 4 4
πt π (d) x(t ) = a cos + 3 2
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4. A system is shown in the figure. The time period for small oscillations of the two blocks will be
Simple Harmonic Motion
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9. On a smooth inclined surface a body of mass M is attached between two spring. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is
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(c)
(a)
13. Three simple harmonic motions in the same direction having the same amplitude A and same period are superposed. If each differs in phase from the next by 45º then (a) The resultant amplitude is (1 + 3) A (b) The phase of the resultant motion relative to the first is 90º (c) The energy associated with the resulting motion is (3 + 2 2) times the energy associated with any single motion. (d) The resulting motion is not simple harmonic.
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18. If the displacement (x) and velocity v of a particle executing simple harmonic motion are related through the expression 4v2 = 25 – x2 then its time period is (a) π (b) 2π (c) 4π (d) 6π
19. A large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizontal circular path. If ℓ is the length of the string from m to the top end of the tube and θ is angle between
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23. A condenser of capacity 20µF is first charged and then discharged through a 10 mH inductance. Neglecting the resistance of the coil, the frequency of the resulting vibrations will be (a) 356 cycles/sec (b) 133 cycles/sec (c) 233 cycles/sec (d) 1333 cycles/sec
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22. A particle of mass m slides back and forth between two inclined frictionless planes as shown. If h is initial height, then time period of motion is
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29. A 1 kg body when suspended from the lower end of a light spring produces a vertical extension of 9.8 cm in it. The time period of the oscillations of the spring will be
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28. Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then (a) Time period of the simple pendulum will be more than that of the springmass system. (b) Time period of the simple pendulum will be equal is that is of the springmass system. (c) Time period of the simple pendulum will be less than of the spring-mass system. (d) Nothing can be said definitely without observation.
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30. Two SHMs are respectively represented by y = a sin(ωt – kx) and y = b cos(ωt – kx). The phase difference between the two is (a) π/2 (b) π/4 (c) π/6 (d) 3π/4
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26. A particle is subjected to two simple harmonic motions given by: y1 = 10 sin ωt, and y2 = 5 sin(ωt + π) The maximum speed of the particle is
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41. A body of mass 0.1 kg is attached to two springs of force constants 6 N/m and 4 N/m and supported by two rigid supports. If the body is displaced along the length of the springs, the frequency of vibrations will be (a) 5 vibrations/sec (b) 10 vibrations/sec (c) 5/π vibrations/sec (d) π/5 vibrations/sec
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37. A particle is executing simple harmonic motion along a straight line 8 cm long. While passing through mean position its velocity is 16 cm/s. Its time period will be
36. A 5 kg. weight is suspended from a spring. The spring stretches by 2 cm/kg. If the spring is stretched and released, its time period will be (g = 10 m/s2) (a) 0.628 sec (b) 6.28 sec (c) 62.8 sec (d) 0.0628 sec
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constant k is 2π
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48. A uniform rod of mass m and length ℓ is hinged at its mid point in such a way that it can rotate in the vertical plane about a horizontal axis passing through the hinge. One of its ends is attached to a spring of spring constant k which is unstretched when the rod is horizontal. If this end is now given a small displacement and released angular frequency of the resulting motion is
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47. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest, its time period is T. With what acceleration should lift be accelerated upwards in order to reduce its time period to T/2? (a) g (b) 2g (c) 3g (d) –3g
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49. The displacement of a particle is given at time t, by: x = A sin(–2ωt) + B sin2 ωt Then,
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45. A man of mass 60 kg is standing on a platform which is oscillating up and down with frequency 2 oscillations/sec and amplitude 50 cm. A machine on the platform indicates weight of the man with respect to time, then
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43. A block of mass ‘m’ is connected to two springs of spring constants ‘2k’ and ‘3k’ respectively, as shown in the diagram, with both the springs being vertical and fixed at their other ends. Both springs are initially unstretched and the mass m is given an initial velocity ‘u’ in the downward direction. The block oscillates. The amplitude of oscillation of the block is given by
(2 + 2) k π m
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46. The bob in a simple pendulum of length ℓ is released at t = 0 from the position of small angular displacement θ. Linear displacement of the bob at any time t from the mean position is given by
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(c)
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3 mω2 A sin ωt (d) µ ( M + m) g + 2
55. A particle of mass m and charge –q moves along a diameter of uniformly charged sphere of radius R and carrying a total charge +Q. The frequency of the SHM of the particle, if the amplitude does not exceed, is
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54. A particle executes S.H.M. starting from its mean position at t = 0. If its velocity is 3bω, when it is at a distance b from the mean position, when ω = 2π/T, the time taken by the particle to move from b to the extreme position on the same side is (a) 5π/6ω (b) π/3ω (c) π/2ω (d) π/4ω
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50. A block of mass m, attached to a fixed position O on a smooth inclined wedge of mass M, oscillates with amplitude A and linear frequency f. The wedge is located on a rough horizontal surface. If the angle of the wedge is 60º, then the force of friction acting on the wedge is given by (coefficient of static friction = µ).
53. The displacements y of a particle executing a certain periodic motion is given by
B 4 (b) the motion of the particle is not SHM, but oscillatory with a time period of T = π/ω (c) the motion of the particle is oscillatory with a time period of T = π/2ω (d) the motion of the particle is aperiodic. A2 +
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(c) the energy associated with the resulting motion is (3 + 2 2) times the energy associated with any single motion. (d) the resulting motion is not simple harmonic
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3. Three simple harmonic motions in the same direction having the same amplitude ‘a’ and same period are superposed. If each differs in phase from the next by 45º, then (a) the resultant amplitude is (1 + 2)a (b) the phase of the resultant motion relative to the first is 90º
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ONE OR MORE THAN ONE O MAY BE CORRECT
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(c) the block weights 1.5 times its weight on the plank halfway down from the mean position. (d) the block weights its true weight on the plank, when velocity of the plank is maximum.
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9. Two blocks of masses 3 kg and 6 kg rest on a horizontal smooth surface. The 3 kg blocks is attached to a spring with a force constant k = 900 N/m which is compressed 2 m from beyond the equilibrium position. The 6 kg mass is at rest at 1 m from mean position. 3 kg mass strikes the 6 kg mass and the two stick together.
11. The amplitude of a particle executing SHM about O is 10 cm. Then (a) When the K.E. is 0.64 of its max. K.E. its displacement is 6 cm from O. (b) When the displacement is 5 cm from O its K.E. is 0.75 of its max. P.E. (c) Its total energy at any point is equal to its maximum K.E. (d) Its velocity is half the maximum velocity when its displacement is half the maximum displacement. 12. A horizontal plank has a rectangular block placed on it. The plank starts oscillating
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Simple Harmonic Motion
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.
(b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1.
(c) Statement 1 is True, Statement 2 is False.
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4. Statement 1: The graph of total energy of a particle in S.H.M. w.r.t. position is a straight line with zero slope. Statement 2: Total energy of particle in S.H.M. remains constant throughout its motion. 5. Statement 1: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Statement 2: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.
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ASSERTION AND REASON QUESTIONS Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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will be 2 3 m (b) Maximum speed of the particle will be 20 7 m/s (c) Magnitude of maximum acceleration of the particle will be 200 7m/s 2 (d) Energy of the resultant motion will be 28 J
2. Statement 1: In simple harmonic motion the total energy is proportional to the square of the amplitude. Statement 2: Total energy is given by 1 2 kA 2
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1. Statement 1: The time period of a simple pendulum is independent of its amplitude of vibration. Statement 2: If the amplitude of a simple pendulum is doubled; its vmax is also doubled.
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(d) Statement 1 is False, Statement 2 is True.
vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then (a) the period of oscillation is (2π/5) (b) the block weighs double its weight, when the plank is at one of the positions of momentary rest. (c) the block weighs 0.5 times its weight on the plank halfway up (d) the block weighs 1.5 times its weight on the plank halfway down
6. Statement 1: The percentage change in time period is 2%, if the length of simple pendulum increases by 4%. Statement 2: Time period is directly proportional to length of pendulum.
Simple Harmonic Motion
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Column I The block’s mass in kg The block’s displacement at t = 0 in metres The block’s acceleration at t = 0.10 s in m/s2 The block’s maximum kinetic energy in joule
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1. A simple harmonic oscillator consists of a block attached to spring with k = 200 N/m. The block slides on a frictionless horizontal surface, with equilibrium point x = 0. A graph of the block’s velocity v as a function of time t is shown. Correctly match the required information in the column I with the values given in the column II. (use π2 = 10)
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10. Statement 1: In oscillatory motion, displacement of a body from equilibrium can be represented by sine or cosine function. Statement 2: The body oscillates to and for about its mean position.
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8. Statement 1: Two simple harmonic π motions are given by y1 = 10sin 3πt + 4 y = 5(sin 3 π t + 3 cos3 π t ). and 2 These have amplitudes in the ratio 1 : 1 Statement 2: y2 represents two waves each of amplitude 5 and so total amplitude is 10, same as that of y1. 9. Statement 1: The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. Statement 2: The potential energy and kinetic energy of a particle in SHM, don’t vary linearly with position.
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7. Statement 1: When a body executing SHM passes through equilibrium it has zero acceleration. Statement 2: When a body executing SHM passes through equilibrium it has maximum energy.
Simple Harmonic Motion
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5.68
3. For a particle under going linear SHM about x = 0, choose the correct possible combination. Symbols have their usual meaning.
PASSAGE BASED QUESTIONS
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(d) Square plate
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6. First time (t) when velocity of block is maximum is
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(a) Ear ring
(d) (e)
PASSAGE–2 You know very well that a block attached to an elastic spring performs SHM. Consider such a spring block system lying on a smooth horizontal table with a block of mass m = 2 kg attached at one end of a spring (k = 200 N/m) whose other end is fixed. The block is pulled so that the spring is extended by 0.05 m. If at this moment (t = 0), the block is projected with a speed of 1 m/s in the direction of increasing extension of the spring.
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2. In this projection φ is (a) π/2 (b) π (c) 3π/2 (d) 0
1. In this projection ω is (a) v/R (b) m2/R (c) R/v (d) None of these
4. A particle of mass 2 kg is moving on a straight line under the action of force F = (8 – 2x)N. The particle is released at rest from x = 6 m. For the subsequent motion match the following (All the values in the column II are in their S.I. units)
(b) velocity is (q) mean position negative (c) acceleration is (r) 0 < x < A negative (s) –A < x < 0 (d) v × a = 0
Column II (p) extreme position
Column I (a) v .a > 0
PASSAGE–1 An object of mass m is moving in uniform circular motion in xy-plane. The circle has radius Rand object is moving clockwise around the circle with speed v. The motion is projected onto the x-axis where it appears as simple harmonic motion according to x (t) = Rcos (cot ωt + φ). The motion starts from point of coordinates (0, R).
Simple Harmonic Motion
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1. A simple pendulum has a time period T = 2 sec when it swings freely. The pendulum is hung as shown in figure, so that only one-fourth of its total length is free to swing to the left of obstacle. It is displaced to position A and released. How long does it take to swing to extreme displacement B and return to A? Assume that displacement angle is always small.
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12. After what time when they were released from rest, shall the blocks collide for the first time m π m (b) πr (a) k 2 k
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(d) None of these
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8. Equation of motion of particle B can be written as (a) y = (2 cm) cos 61 cm (b) Y = (4 cm) – (2 cm) cos 2πt (c) y = (2 cm), sin 2πt (d) y = (4 cm)–(2 cm) sin 2πt
(b)
(c)
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7. Equation of motion of particle A can be written as (a) x = (2 cm) cos 2πt (b) x = (3 cm) – (2 cm) cos 2πt (c) x = (2 cm) sin 2πt (d) x = (3 cm) – (2 cm) sin 2πt
10. The speed of block P just before P and Q are about to collide for the first time
PASSAGE–3 Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequency of 2 cm and 1 Hz respectively. Equilibrium positions of the particles A and B are at the co-ordinates (3 cm, 0) and (0, 4 cm) respectively. At t = 0, B is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin.
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(b) 0.36 s (d) 0.58 s
(a) 0.268 s (c) 0.47 s
Simple Harmonic Motion
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Simple Harmonic Motion $
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7. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track. of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period? 8. Find the frequency of small oscillations of a thin uniform vertical rod of mass m and length ℓ hinged at the point O. The combined stiffness of the springs is k.
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5. A block of mass m is tied to one end of a string which passes over a smooth fixed pulley A and under a light smooth movable pulley B. The other end of the string is attached to the lower end of a spring of spring constant k2. Find the period of small oscillations of mass m about its equilibrium position.
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4. A heavy ball of mass m is tied to a weightless thread of length ℓ. The friction of the ball against air is proportional to friction of the ball against air is proportional to its velocity relative air : Ffr = µv. A strong horizontal wind is blowing at a constant velocity v. Determine the period T of small oscillations, assuming that the oscillations of the ball attenuate in a time much longer than the period of oscillations.
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6. An A-shaped wire is made of bent wire ABC and a straight wire DE. Rings are provided which can slide over the bent wire as well as on straight wire. The arrangement is dipped into a soap solution and is then taken out and is placed in a vertical plane. The wire DE hangs in equilibrium at a distance L = 4.9 cm from the bent B as shown. When DE is displaced vertically parallel to itself it starts oscillating. Find its period of oscillation in milli-seconds. Surface tension of soap solution is 0.07 N/m. (g = 10 m/s2)
3. Determine the period of oscillations of mercury of mass m = 200 g poured into a bent tube whose right arm forms an angle θ = 30º with the vertical. The cross-sectional area of the tube S = 0.50 cm2. The viscosity of mercury is to be neglected. [
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Simple Harmonic Motion
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9. A spring of force constant k = 300 N/m connects two blocks having masses 2 kg and 3 kg, lying on a smooth horizontal plane. If the spring block system is released from a stretched position, find the number of complete oscillations in 1 minute. Take
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1. A thin rod of length L and uniform crosssection is pivoted at its lowest point P inside a stationary homogeneous and nonviscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density d1 of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.
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PREVIOUS YEARS’ IIT-JEE QUESTIONS
12. A smooth horizontal disc rotates about the vertical axis O with a constant angular velocity ω0. A thin uniform rod AB of length l performs small oscillations about the vertical axis A fixed to the disc at a distance a from the axis of the disc. Find the period of these oscillations.
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Find the period of its oscillations if it rolls without slipping.
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11. The pulley of radius r shown in figure has a moment of inertia I about its axis and mass m. Find the time period of vertical oscillations of its centre of mass. The spring has spring constant ‘k and the string does not slip over pulley.
14. A uniform semicircular cylinder of radius r and weight W is displaced through a small angle θ from its equilibrium position as shown in the figure.
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10. (a) What will be the time period of a simple pendulum if its length is equal to radius of earth (=6400 km)? (b) What is the maximum time period which an oscillating simple pendulum can have?
The free end of a thread tightly wound on the pulley carries a deadweight A. At a certain angle α it counterbalances a point mass m fixed at the rim of the pulley. Find the frequency of small oscillations of the arrangement.
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Simple Harmonic Motion
same period are superposed. If each differ in phase from the next by 45º, then [1999] G
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2. A particle of mass m is executing oscillation about the origin on the x-axis. Its potential energy is U(x) = k |x| 3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is [1998]
6. The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination α, is given by [2000]
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7. A particle executes simple harmonic motion between x = –A and x = +A. The time for it to go from O to A/2 is T1 and to go from A/2 to A is T2, then [2000] (a) T1 < T2 (b) T1 > T2 (c) T1 = T2 (d) T1 = 2T2
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8. For a particle executing SHM the displacement x is given by x = Acosωt. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x [2003]
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4. A particle free to move along the x-axis has potential energy given by U(x) = k [1 – exp(–x2)] for –∞ ≤ x ≤ + ∞, where k is a positive constant of appropriate dimensions. Then [1999]
(c) 2π
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9. A solid sphere of radius R is floating in a liquid of density ρ with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing simple harmonic motion. Find the frequency of these oscillations. [2004] 10. A mass m is undergoing SHM in the vertical direction about the mean position y0 with amplitude A and angular frequency ω. At a distance y from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m. Find the distance y, (measured from the mean position) such that the height h attained by the block is maximum (Aω2 > g). [2005]
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∆g g, i.e., × 100 for students I, II and III, g respectively, (a) EI = 0 (b) EI is minimum (c) EI = EII (d) EII is maximum
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PASSAGE A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity V0 = V0iˆ. The coefficient of friction is µ. [2008]
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12. y = A sin2ωt + B cos2ωt + C sinωt cosωt. For what value of A, B and C it will represent SHM? [2006] (a) for all values of A, B and C (C ≠ 0) (b) A = B, C = 2B (c) A = –B, C = 2B (d) A = B, C = 0
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13. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or
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Time period (s)
If EI, EII and EIII are the percentage errors in
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(b) II, IV (d) I, IV
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Simple Harmonic Motion
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15. The centre of mass of the disk undergoes simple harmonic motion with angular frequency ω equal to
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ASSERTION AND REASON QUESTIONS 1. (b) 9. (b)
5. (a, b, c, d)
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(b) (c) (a) (c) (b) (b) (b)
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(c) (c) (c) (a) (a) (b) (c)
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14. The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is (a) –kx (b) –2kx (c) –2kx/3 (d) –4kx/3
Simple Harmonic Motion
MATCH THE COLUMN TYPE QUESTIONS 1. (a) → r (b) → p (c) → q (d) → s
2. (a) → r, s (b) → r, s (c) → s (d) → q
3. (a) → r, s (b) → r, q, s (c) → p, r (d) → p, q, r, s 4. (a) → r (b) → t (c) → q (d) → r (e) → q 5. (a) → r (b) → s (c) → p (d) → q
PASSAGE BASED QUESTIONS 1. (a) 9. (b)
2. (c) 10. (a)
3. (b) 11. (b)
4. (c) 12. (a)
5. (a)
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6. (a)
7. (b)
8. (d)
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Simple Harmonic Motion
5.75
HINTS AND SOLUTIONS
1 Maximum energy = mVm2 = 4 J 5 1 2 or kA = Emax 2 1 1 = × 200 × 0.04 = 4 J = Vm2 = 4 J 2 2
. Inside the satellite, the g effect of g is zero. So pendulum will not oscillate.
3. In the case of simple pendulum, amplitude decreases because energy is lost in overcoming air resistance, friction etc. In the case of pendulum clock, this energy lost is balanced by the energy supplied by the wound spring.
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At t = 0.1 sec, acc. is maximum = –ω2A = –200 m/s2
2. (a) T = 2π /3 g = 1.1 sec; (b) E =
3. T = 0.8 sec 4. It can easily be seen that the ball attains the equilibrium position at an angle α of deflection of the thread from the vertical, which is determined from the condition µv tan α = . mg
m T 2k ⇒ m = 2 = 0.2 kg k 4π
1. 3/2 sec (mgℓα2)/2 = 0.05 J
T = 2π
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SUBJECTIVE QUESTIONS
MATCH THE COLUMN TYPE QUESTIONS V 2π 1. Vm = Aω ⇒ A = m = × (0.2) = 0.20 m; ω 2π
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1 1 = kA2 = × 2 × 4 N-m = 4 j 2 2 As particle has started its motion from positive extreme ∴ Phase constant = π/2
8. The period of oscillation of a simple pendu-
. The value of g is less in g moon than that on the surface of earth. So T increases. The pendulum takes more time to complete one oscillation. Hence it will lose time.
Time to go from x = 2 m to x = 4 m (i.e., from extreme position to mean position) T π = = 4 2 Energy of S.H.M.
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1. No. T = 2π
CONCEPTUAL QUESTIONS
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5.76
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13. ω0 =
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14. T = π
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force F = (mg ) 2 + (µv) 2 and a small drag force (Fig. 148). Consequently, the motion of the ball will be equivalent to a weakly attenuating motion of a simple pendulum with a free-fall acceleration g' given by
1 2 + m r 11. 2π 4k
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Simple Harmonic Motion
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12. (a, b, c)
∆T ∆g ∆ = +2 13. (b) g = 4π2 2 T T g ∆ ∆T ⇒E= +2 , greater the value of t, T lesser the error. Hence, fractional error in the Ist observation is minimum.
14. (d) 2kx – f = ma ⇒ f.R = Iα a = Rα ⇒ ma = 4kx/3 (directed opposite to displacement) 15. (d) –(2kx) R = IPα
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General Physics rio s_ in fo er si ta rio s
General Physics Particle Kinematics Dynamics of Particle Circular Motion Energy and Momentum
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(Volume 2)
Chandigarh • Delhi • Chennai https://t.me/universitarios_infohttps://www.jamarana.com
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Production Editor: Nitkiran Bedi Composition: Tantla Composition Services, Chandigarh Printer: Sanat Printers
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The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertant omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book.
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Copyright © 2010 Dorling Kindersley (India) Pvt. Ltd
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This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated, without the publisher’s prior written consent, in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise) without the prior written permission of both the copyright owner and the abovementioned publisher of this book. ISBN 978-81-317-3444-5 10 9 8 7 6 5 4 3 2 1 Published by Dorling Kindersley (India) Pvt. Ltd, licensees of Pearson Education in South Asia.
Head Office: 7th Floor, Knowledge Boulevard, A-8-(A) Noida-201309, India. Registered Office: 11 Community Centre, Panchsheel Park, New Delhi 110 017, India.
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Chapter 2
Gravitations
Chapter 3
Elasticity
Chapter 4
Hydrostatics and Hydrodynamics
Chapter 5
Simple Harmonic Motion
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Rotational Dynamics
Chapter 1
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Preface
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Contents
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For a science student, physics is the most important subject as it requires logical reasoning and high imagination. Without improving the level of physics, it is difficult to achieve a goal with the kind of competition that exists today. This five part volume covers all parts of general physics—Mechanics, Heat, Wave, Light, Electromagnetism and Modern Physics—which is written in accordance with the latest syllabus of the IIT-JEE and AIEEE. There is no single book that is available in the market that contains a large amount of solved examples.
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■ Entire syllabus is covered in five volumes. ■ Content of each chapter is well defined and builds new concepts from the scratch. ■ Each chapter describes the theory in a simple and lucid style. ■ Covers a wide spectrum of questions to enable the student to develop enough expertise to tackle any problem. ■ Helps students in building analytical and quantitative skills, which, in turn, develop confidence in problem solving. ■ Practice exercises are given at the end of each chapter. ■ Numerous diagrams in every chapter. After studying the entire chapter, students will be able to learn different tricks and techniques of problem solving with suitable level of analytical ability. Suggestions for improving the book are always welcome.
All the best!
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S. C. PANDEY
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1.1 KINEMATICS OF A RIGID BODY
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Rotational Dynamics
→
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A body of finite shape is said to be rigid, if it does not suffer any deformation under the increase of the external force or torque. In the given figure, if 1 and 2 are points on the body, then for the body to be rigid.
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Example A sheet of plywood is in motion on a horizontal floor such that the velocity of two points 1 and 2 are v1 and v2 respectively, as shown in the figure, then (a) v1 cos θ1 > v2 cos θ2 (b) v1 cos θ1 = v2 cos θ2 (c) v2 cos θ2 > v1 cos θ1 Ans. v1 cos θ1 = v2 cos θ2 Translational motion of a rigid body Consider a rigid body in motion. At a certain instant of time, the position vectors of any → → two points say 1 and 2 are r1 and r2 respectively with respect to ground from (x, y, z). →
→
→
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Rotational Dynamics
Differentiating with respect to time, we get →
→
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v1 = v2 + ω × r12
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→ → → → → = a2 + α × r12 + ω × ω × r12
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Again differentiating with respect to time, we get → → dω → → → a1 = a2 + × r12 + ω × v12 dt
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ω is in the z direction. → → → → →→ ω k × ω k × j r12 = ω k × (− i ) ω r12
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1.2 TRANSLATIONAL MOTION →
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Rotational Dynamics
So, for the linear motion of the body
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(iii) In (ii), the first term of the right hand side arrange due to pure translational with point 2 and second term arises due to pure rotation of 1 about point 2. (iv) If point 2 becomes the centre of mass (C.M.) of the rigid body, we can understand the general motion or plane motion of rigid body as translational with centre of mass and rotation about centre of mass.
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Rotational Dynamics →
1.5
In the Fig. (a), the direction of ω changes with time or the line joining 1 and 2 forms a cone during its rotation. Such type of rotation is fixed point rotation (but not the fixed axis rotation). In Fig. (b), sense of ω vector is directed along a fixed line. Hence, the fixed point rotation about the point 2 is also the fixed axis rotation, which passes from the point → 2 and perpendicular the plane of disc. In the sense of ω in Fig. (b), the point 2 is also the centre of circle which the position 1 inscribed. The Fig. (c) also represents the fixed point rotation as well as the fixed axis rotation but the point 2 is not the centre of the circle which the particle inscribed.
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Note: (i) The sense of ω is also the direction of axis of rotation.
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Rotational Dynamics
(ii) In all the three figures, centre of the circle inscribed by all the particles lie on a line know as axis of rotation. In Fig. (b) all the circle are concentric. (iii) Perpendicular from the point on the axis of rotation is its radius. (iv) ω is same for all the points of rigid body in pure rotation. (v) ω is independent of axis of rotation. (vi) As ω (or dθ) and α is same for all the points of a rigid body in pure rotation, then it could define the angular kinematical term at any point of a rigid body. Therefore, all the formula of angular kinematics is equally hold for the rigid body. →
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It is similar to the angular motion in circular motion chapter.
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Example A flywheel of radius 30 cm starts from rest and accelerates with constant acceleration of 0.5 rad/sec2. Compute the tangential radial and resultant acceleration of a point on its circumference,
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Given, α = 0.5 rad/sec2, R = 0.30 m ∴ Radial acceleration = ω2 R = 0
(i) Initially, θ = 0, so ωi = 0
1.6
Tangential acceleration = at = R α = (0.3)(0.5) = 0.15 m/sec2 Net acceleration, anet. = (at ) 2 + (an) 2 = 02 + (.15) 2 = 0.15 m/sec2
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Rotational Dynamics
1.7
2π (ii) After, θ = 120º 3 2π 2π ωf2 = ωi2 + 2α θ = 0 + 2 (0.5) = 3 3 2π rad/sec. 3
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It is a kind of general motion in which the motion of rigid body is confined in a plane motion say X-Y plane and Z co-ordinate of all the points is constant. The plane motion can be regarded as sum of the translational and rotational motion for simplicity.
v = vc + w × r12
→ → a1 = ac + ω2 − r12
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Rotational Dynamics
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1.10
Rotational Dynamics
1.4 INSTANTANEOUS CENTRE OF ROTATION OF ZERO VELOCITY The velocity of a point of a rigid body in a plane motion can be visualized or imagine in pure rotation. The point about which the plane motion can be imagined pure rotation is known as instantaneous centre of rotation. The centre of rotation may lie outside the rigid body. Let us denote the centre of rotation by 2. Velocity of any point of the rigid body in plane motion can be written as → → → vi = ω × ri 2 i ≠ 2
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→ → vc = ω − k × vc2 j →
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(i) With a constant velocity v, while the disc started rotating counterclockwise with a constant angular acceleration α (the initial angular velocity is equal to zero). (ii) With a constant acceleration a (and the zero initial velocity) while the disc rotates counter-clockwise with a constant angular velocity ω.
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(ii) Centre moves with constant acceleration a, initial velocity is zero. 1 x = at 2 2 ∴
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1.13
Rotational Dynamics
1.5 ROLLING MOTION Rolling motion can be considered as combination of rational and translational motion. For the analysis of rolling motion, we deal translation separately and rotation separately and then we combine the result to analyse the overall motion. $ 2 Z
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aB = a 2 + α 2 R2
t.m e/
Similarly, if disc rolls with angular acceleration a and if its center of mass has an acceleration ‘a’ different points will have accelerations given by: aA = a + αR aC = a – αR To write equations of motion for rolling motion, we can apply Fext = Macm for translation motion and τ = lα about axis passing through centre of mass of body. Rolling motion is possible in two ways—rolling without slipping, while in case of rolling with slipping, relative motion takes place between contact points. In the example, if rolling is without slipping we will have VA = 0 ⇒ V = ωR and, ac = 0 ⇒ a = αR
One more important distinction between these two kinds of rolling motion is in case of rolling with slipping the frictional force is a known force of magnitude µN, while in case of rolling without slipping frictional force will be unknown force may take any value between zero and µN. https://t.me/universitarios_infohttps://www.jamarana.com
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Rotational Dynamics
Example A ball of radius R = 10 cm rolls without slipping down an inclined plane, so that its centre moves with constant acceleration a = 2.50 cm/s2, t = 2.00 sec. After the beginning of motion its position corresponding to that is shown in the figure. Find: (i) The velocity of the points A, B and O. (ii) The acceleration of these points. $ &
%
) Solution
rio s_ in fo
na .c om
2
5
e/
tp
s:
[
un
→
v0 = vc + ω × rOC = 0
t.m e/
→
Z5
9& 9%
iv
.m //t
→
ht tp s:
→
Z
er si ta rio s
er s un iv
//w
w
w
\
]
9&Z5 9$
ita
.ja
m
ar a
Let the fixed co-ordinate axis x, y and z as shown in the figure. As the ball is rolling without slipping along the rigid surface, therefore
ht
→ → → → Thus, vc = ωR and ω ↑↑ − k as vC ↑↑ i →
→
and
ac = αR
s: //
→
ωc + α × rOC = 0
...(i)
ht tp
1.14
At the position corresponding to that of figure in accordance with the problem, ac = a, So vc = at vc at = , R k a and, α = R ω=
(i) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the figure. As point O is the instantaneous centre of rotation of the ball at the moment shown in the figure, therefore vO = 0 https://t.me/universitarios_infohttps://www.jamarana.com
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Rotational Dynamics →
→
→
→
Now, vA = vc + ω × rAC → → → → = vc i + ω − k × R j = (vc + ωR) i →
→
→
vA = 2vc i = 2 at i
∴
Similarly, → → → → → → vB = vc + ω × rBC = vC i + ω(−k ) × R i
na .c om
→ → → → = vC i + ωR − j = vC i + vC − j
m
ar a
→ → → → → (ii) aO = aC + ω2 − r0C + α × rOC
.ja
er s
w
ita
2
→ → v ^ aO = ω2 − rOC = 0 − uOC R
rio s_ in fo
So, vB = 2 vC = 2 at
→
∴
a0 =
%
Z5
Z5
2
ht tp
s: //
t.m e/
un
//t ht tp s: D5
er si ta rio s
.m
tp ht
Z5 &
DD5
iv
$
e/
s:
un iv
//w
w
where u^0C is the unit vector along r0C .
D D5 D
vC2 a 2t 2 = and is directed towards the centre of ball R R
→ → → → → Now, aA = aC + ω2 − rAC + α × rAC → → → = a i + ω2 R − j + α − k × R j
( )
→
= (a + αR ) i +
vc2 (− j ) R
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1.15
1.16
Rotational Dynamics
a 2t 2 = 2a i + (− j ) R ∴
aA = 4 a 2 +
at 2 a 4t 4 = 2 a 1 + R2 2R
2
→ → → → → Similarly, aB = aC + ω2 − rBC + α × rBC → → → → = a i + ω2R − i + α − k × R i
na .c om
v2 → = a − C i + αR − j R 2
ar a
m
1.6 SUMMATION OF ANGULAR VELOCITY
er s
w
ita
.ja
2
rio s_ in fo
at 2 a 2t 2 2 aB = a − + a = a 1 + 1 − R R
So,
er si ta rio s
iv Z
ht tp
s: //
t.m e/
un
Z
%
ht tp s:
//t
ht
.m
tp
e/
s:
un iv
//w
w
Relative angular velocity in rigid body Let us consider a rigid body rotating about an axis OA which itself rotates about fixed axis OB as shown in the figure.
Z
$
2
Let the angular velocity of OA with respect to the OB is ω′ and the angular velocity at axis OB is ω0. Then, the angular velocity of rigid body is ω = ω′ = ω0 Again consider a special case in which →
→
angular velocity = ω0 and ω′ = Constant. →
→ d ω′ In this case, angular acceleration α = d ω = = ω′ ω0 dt dt
→
Note: α = ω′ × ω0 https://t.me/universitarios_infohttps://www.jamarana.com
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Rotational Dynamics
1.17
Example A round cone with half-angle α = 30º and the radius of the base R = 5.0 cm rolls uniformly and without slipping over a horizontal plane as shown in the figure. The cone apex is hinged at the point O which is on the same level with the point C, the cone base centre. The velocity of point C is v = 10.0 cm/s. Find the modulii of (i) The vector of the angular velocity of the cone and the angle it forms with the vertical. (ii) The vector of the angular acceleration of the cone.
ar a m
ita
.ja
er si ta rio s
w
er s
w
) Solution
&
D
rio s_ in fo
na .c om
2
iv
→
e/
→
.m
→
ω = ω′ + ω0
Z
un
2
t.m e/
Z
Z
s: //
ht tp s:
//t
ht
tp
s:
un iv
//w
(i) Let the axis of the cone (OC) rotates in anticlockwise sense with constant → angular velocity ω′ and the cone→itself about it’s own axis (OC) in clockwise sense with angular velocity ω0 . Then, the angular velocity of the cone,
ht tp
T T
2
D 5
X
$ →
→
As the rolling is pure, magnitude of the vector ω′ and ω0 can be easily found from the figure. ω′ =
v v , ω0 = R cot α R
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...(ii) https://t.me/universitarios
1.18
Rotational Dynamics →
→
As ω′ ⊥ ω0 , from equations (i) and (ii), we have 2
v v ω = ω′2 + ω02 = + R cot α R =
2
v = 2.3 rad/sec. R cos α
(ii) Vector of angular acceleration →
→
d ω d (ω′ + ω0 ) α= = dt dt →
→
rio s_ in fo
na .c om
→ as ω′ = constant
→
ar a
The vector ω0 which rotates about the OO′ axis with the angular velocity ω′ retains its magnitude. This increment in the time interval dt is equal to
ita
.ja
m
|dω0| = ω0 . ω′ dt
→
er si ta rio s
→
//w
w
er s
→
Thus, α = ω′ × ω0
un iv
w
or in vector form dω0 = (ω′ × ω0) dt
e/
.m
iv
v v v2 = tan α = 2.3 rad/sec 2 R cot α R R2
s: //
v v v2 = tan α = 2.3 rad/sec 2 R cot α R R2
t.m e/
un
//t
∴α=
Example Find the velocity of centre of the body which rolls without slipping on a horizontal floor in terms of v, r and R.
ht tp
α=
α = ω′ ω0 (as ω′ ⊥ ω0)
ht tp s:
ht
tp
s:
The magnitude of the vector α is equal to
5 U
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Rotational Dynamics
1.19
) Solution →
→
→
v0 = ω × r02 → → → v = ω − k × (R − r ) j
3 5GT
Z
v ω= R −r
or
5
T
3
→ v Ri R −r
na .c om
vC =
Note: PP′ = R dθ = ds
Y
P GV
er s
w
ita
.ja
Rω = v Rα = a
ar a
d θ ds = dt dt
m
R
rio s_ in fo
→ → → → vC = ω × rC2 = ω − k × R j
er si ta rio s
iv
0
5
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
un iv
//w
w
Example In the figure shown, a block of mass m and given the velocity v. Calculate the velocity of centre of mass of inclined sphere.
U
1
ht tp
s: //
5 '
U
5 7
U
T
3
P Y
) Solution Let, the angular velocity of sphere 3 is ω. ∴
v = (R3 – r3) ω3
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Rotational Dynamics
v R 3 − r3
or
ω3 =
∴
vT = ω3 2R 3 =
2vR 3 (R 3 − r3 )
vN = ω2 r2 ∴
ω2 =
vN 2vR 3 = r2 (R 3 − r3) r2
vM = ω2 × R 2 =
2vR 2 R 3 = vD (R 3 − r3 )r2
2vR 2 R 3 (R 3 − r3) r2 (R1 + r1)
2v R1 R2 R3 m/sec. (R3 − r3) r2 (R 1 + r1 )
er s
w
Torque About a Point
ita
.ja
m
ar a
vC1 = ω1 R1 =
rio s_ in fo
ω1 =
na .c om
vD = ω1 × (R1 + r1)
er si ta rio s
iv
)
T
T
s: //
t.m e/
ht tp s:
un
//t
2
U
.m
tp
e/
s:
un iv
//w
w
Let a point influence by a force F and the position of some point of application of the force (of the particle). Then the moment of force F about the point O,
ht
0
→
→
ht tp
1.20
τ0 = r × F |τ0| = F (r sin θ) = F × OM directed towards points and line of action of force. Let there are a number of forces acting on a system and we want the net torque about a certain point O. Then, from the principle of super-position (or vector addition) →
→
→
τ = τ0 = ∑ ri × Fi = ∑ τ0 where, τ0 = Torque about point O.
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Rotational Dynamics
1.21
Torque About an Axis Let there is û axis (x, y, z) passing through the point O. Then, the moment of force or torque about the axis (û axis). → → → → ^ ^ τu = u . τ 0 = u . r × F
ux u y uz τu = x y z δx δy δz →
^
→
→
na .c om
if u is collinear to r or F, then δu = 0
er si ta rio s
→
.m
un iv
→
e/
→
τ z = k . τ0
rio s_ in fo
er s
→
= mgr j
→
iv
= k . mgr j
un
//t
ht
tp
s:
//w
w
w
ita
.ja
m
ar a
Example A circular disc rotates in horizontal plane about a fixed axis passing through its centre and perpendicular to the axis. A particle of mass m which is at rest with respect to disc and which is at a distance r from the centre of disc. Find the torque of the weight of the particle about the point O, and also about the axis of rotation of the disc. → → → → ) Solution τ0 = r × F = rc × F − k
t.m e/
ht tp s:
=0
ht tp
s: //
Example In the position shown in the figure, find the net torque about the axis passing through the point O in the sense of angular velocity (or axis of rotation) and also about the axis OA. 2
T O
$
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PJ U
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Rotational Dynamics →
) Solution τ0 = → → = l sin θ i × mg − j →
= −lmg sin θ k →
→
τ z = k .(−lmg sin θ k ) = – lmg sin θ.
5
ar a
Z
un iv e/
P
iv
.m t.m e/
un
//t ht tp s:
7
er si ta rio s
er s
w w //w s: tp
→
5
ita
.ja
m
7
rio s_ in fo
na .c om
Example Find the net torque of the tension forces about the axis of rotation of the disc.
ht
→
s: //
) Solution τT = r × F
0
→
→
= R1 (− i ) × T(− j )
ht tp
1.22
→
= T R1 k →
→
τT 0 = R 2 i × T0 (− j ) →
= − R 2 T0 k ∴
τ0 = (τT + τT0) →
= (TR1 − T0 R 2 ) k →
→
τ z = k . (TR1 − T0 R 2) k = (TR1 – R2 T0)
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Rotational Dynamics
1.23
Angular momentum of a particle about a point Let a particle of mass m follows a curve path. At a certain instant of time, its position →
→
with respect to a reference point O is r and its velocity with respect to point O is v . →
→
→
→
LO = r × p
→
→
LO = r × (m v )
Pĺ Y
rio s_ in fo
na .c om
U
ar a
The angular momentum for a system of particle with respect to the point O, →
m
LO = Σ LiO
→
→
un iv
w
//w
→
→
ht
e/
tp
Lz = O
.m
s:
If k is collinear with r and p . Then
er si ta rio s
→
ita
w
→
Lz = k . L O = k . ( r × p )
er s
.ja
Angular momentum about an axis passing through to the point O,
ht tp s:
un
//t
iv
From the principle of the super-position, the total angular momentum of a system of particle about the z-axis,
t.m e/
Lz = ∑ Li
→
→
→
L0 = r × p
ht tp
s: //
Relationship between torque and angular momentum about the same point. (corresponding to a particle)
→
→
→ d p d L0 d r = × p + r× dt dt dt
→ → d L0 → = v× m v + r × F dt
d L0 = τ0 dt If,
τ0 = 0, then L0 = constant Στ x =
dl dlx dl , Στ y = y , Στ z = z dt dt dt
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Rotational Dynamics
Example The angular momentum of a particle relative to a certain point O varies with time as L0 = a + bt2, where a and b are constant vectors, with a ⊥ b. Find the force moment τ0 relative to the point O acting on the particle when the angle between the vector τ0 and L0 equals 45º.
) Solution τ0 =
d L0 = 2 bt dt
...(i)
Given that the angle between τ0 and L0 is 45º
na .c om
1 2b 2t 3 = 2 2bt . a 2 + b 2 t 4
1 b 2t 4 = 2 2 4 2 a +b t
a2 = b2 t4
w
e/
un
//t
iv
.m
a b
un iv
w
//w
s: tp
τ0 = 2b .
er s
a b Hence form equation (i), t=
or
ita
.ja
m
or
ar a
Squaring we get,
er si ta rio s
or
rio s_ in fo
τ0 .L0 2bt.[a + bt 2 ] = |τ0||L0| 2bt . a 2 + b 2 t4
cos 45° =
∴
ht
Angular momentum about a point and about an axis in polar co-ordinate system →
t.m e/
ht tp s: →
u^ u^ θ k = rρ rθ rz
s: //
L0 = r × p
=
ht tp
1.24
U
XT XU
Pρ Pθ Pz
Lz = rρ Pθ = ρ mvθ = ρm ρ ωz = mρ2 ωz Similarly, for the torque
N
2
τz = ρFθ. where, ρ = perpendicular distance from the point mass to the axis. Note: Projection of Lz and τz are indeed independent choice of the point O on the z-axis relative to which are τ0 and L0.
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Rotational Dynamics
1.25
Angular momentum of a rigid body rotating about a stationary axis Lz = ΣLiz
=
= Σ mi ρi2 ωz = ωz Σmi ρi2
Y ± U P ±
= Iωz
na .c om
where, Σmi ρi2 = I - is known as M.I. of solid body or rigid body about z-axis. In general, continuous mass of a rigid body about an axis can be defined
rio s_ in fo
I = ∫ r 2 dm
ar a
Moment of inertia of rigid body depends on the distribution of mass relative to the axis as well as on the axis of rotation.
iv
er si ta rio s
&
.m
e/
;
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Example Three particles, each of mass m are situated at the vertices of an equilateral triangle ABC of side L as shown in the figure. Find the moment of inertia of the system about the line AX perpendicular to AB in the plane of ABC.
%
/
ht tp
s: //
$
) Solution Perpendicular distance of A from AX = 0
∴
Perpendicular distance of B from AX = L L Perpendicular distance of C from AX = 2 Moment of inertia of the particle at A = 0
Moment of inertia of the particle at B = mL2, and
L Moment of inertia of the particle at C = m 2
2
The moment of inertia of the three-particle system about AX is 2
2
L 5mL 0 + mL2 + m = 4 2
Note that the particles on the axis do not contribute to the moment of inertia. https://t.me/universitarios_infohttps://www.jamarana.com
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1.26
Rotational Dynamics
Radius of gyration If M is the mass and I is the moment of inertia of a rigid body, then the radius of gyration (K) of a body is given by I M ⇒ I = MK2 K=
un iv e/
iv
.m
Annular cylinder (or ring) about cylinder axis, I = 1 M(R12 + R 2 2 ) 2
t.m e/
ht tp s:
un
//t
ht
er si ta rio s
er s
w w //w s: tp
Hoop about cylinder axis. I = MR2
$[LV
s: //
$[LV
ht tp
5
5 5
(b)
ita
.ja
5
$[LV
rio s_ in fo
(a)
m
ar a
na .c om
1.7 MOMENT OF INERTIA OF SOME IMPORTANT BODIES
(c)
(d) /
/ 5
Solid cylinder (or disc) about cylinder axis. 1 I = MR 2 2
Solid cylinder (or disc) about centre diameter 1 1 I = MR 2 + ML2 4 12
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Rotational Dynamics
1.27
$[LV
(e)
/
Thin rod about axis through centre perpendicular to length, I =
1 ML2 12
$[LV /
rio s_ in fo
na .c om
(f)
1 3
(h)
5
ht tp
s: //
t.m e/
ht tp s:
un
iv
5
//t
ht
.m
(g)
e/
tp
s:
un iv
//w
w
er s
w
$[LV
er si ta rio s
ita
.ja
m
ar a
Thin rod about axis through one end perpendicular to the length, I = ML2
Solid sphere about any diameter 2 I = MR 2 5
(i)
Thin spherical shell about any diameter 2 I = MR2 3
Hoop about any diameter. I = 1 MR 2 2
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Rotational Dynamics $[LV
(j)
E
D
1 M(a 2 + b 2 ) 12
rio s_ in fo
I=
ar a
na .c om
Rectangular plate about perpendicular axis through centre.
ita
er s
w
) Solution
.ja
m
Example Find the moment of inertia of a uniform hollow sphere about a diameter.
M (2πR sin θ) (R dθ) 4πR2
t.m e/
M sin θ d θ 2
ht tp
s: //
=
un
//t
ht tp s:
Mass of the ring =
M 4πR2
iv
.m
tp
Mass per unit area of the sphere =
er si ta rio s
un iv
= (2 πR sin θ) (R dθ)
e/
s:
//w
w
Let, mass be m and radius be R, Then area of the ring
ht
1.28
5 GT T
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Rotational Dynamics
1.29
Moment of inertia of this elements ring about OX, M M dI = sin θ d θ (R sin θ)2 = R2 sin3 θ d θ 2 2
0
π
=∫ 0
MR2 sin 3 θ . d θ 2 MR2 (1 − cos2 θ) sin θ d θ 2 π
MR2 2
COS 3θ −COS θ − 3 0
=
MR 2 2
1 2 2 1 + 3 = 3 MR .
na .c om
=
rio s_ in fo
π
I=∫
m
ar a
Two Important Theorems on Moment of Inertia
er si ta rio s
un iv
.m
e/
I = ICM + md2
iv
(ii) Perpendicular axis theorem Consider a lamina (cross-section of rigid body). From the figure
s: //
ht tp
ri2 = xi2 + yi2
t.m e/
ht tp s:
un
//t
ht
tp
s:
//w
w
er s
w
ita
.ja
(i) Theorem of parallel axis The rotational inertia of any body about an arbitrary axis equals the rotational inertia about a parallel axis through the centre of mass plus the total mass times the squared distance between the two axis. Mathematically, the parallel axis theorem has the following form:
]
\
U
[\ [
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Rotational Dynamics
Σmi ri2 = Σmi xi2 + Σmi yi2
Multiplying both sides by Σmi, we get
Iz = Ix + Iy
where, Iz = Moment of inertia of the rigid body about an axis passing through centre of mass.
a
s
a
r
Fundamental Equation of Rotational Dynamics of Rigid Body ( otation bout a tationary xis) In this case, if axis of rotation is denoted by z-axis, then we have Lz = Iω dω = Iα dt
m
rio
ar
s_ i
where, α = angular acceleration of the rigid body.
nf o
τz = I
an
a.
co m
Differentiating with respect to time, we get
w
si
ta
.ja
Kinematic Energy of Rigid Body Rotating About a Fixed Axis
s io ta r
iv
si
un
er
e/
//t .m
1 = Σ mi ρi2 ω2 2
ht
tp s
:// w
w
er
1 T = Σ mi vi2 2
.m
tp
ht tp
s:
1 dW = dT = d Iω2 2
//t
ht
s:
e/ un
iv
1 = Iω2 , 2 where, I = Σ mi ρi2 The work performed by external forces from work-energy theorem,
= I ωz . dωz dW = I ωz . αz dt = I . αz ωz dt = τz dθ where, τz = Iαz and dθ = ωz dt ∆t
W = ∫ τz . d θ
0
Kinetic energy of a rigid body in plane motion,
1 T = T + mvc2 2
1.30
1 1 = IC ω2 + mvC2 2 2
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Rotational Dynamics
1.31
Impulse momentum theorem →
∆ L = ∫ τ dt This equation holds equal in the all the frame (inertial or non-inertial frame) In the non-inertial frame, →
→
τ = Σ τext + Σ τ (system)
Total Momentum of External Force →
→
t.m e/
→
→
τ0 = (r0 + r1 ′) × Fi
→
→
ht tp
→
s: //
τ0 = Σ ri × Fi
→
2
un
//t ht tp s: →
er si ta rio s
un iv
e/ .m
tp
U
U
iv
er s
w //w s:
2
ht →
ita
m w
.ja
U
5LJLG ERG\
rio s_ in fo
)U
ar a
na .c om
Let τ0 be the total moment of force relative to the point O and τ01 relative to the → point O′, whose radius vector is equal to r 0 .
τ0 = (r1 × Fi ) + r0 × Fi →
→
→
τ0 = τ01 + (r0 × F) If F = 0, τ0 = τ01 In other words, when F is equal to zero, the total momentum of external forces do not depend on its choice of the point relative to which it is to be determines. In particular, such is the case when a couple acts on the system with respect to the C frame, which possesses an important characteristic.
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Rotational Dynamics
This reference frame is rigidly fixed to centre of inertia of the system of particles. Since in the general case, the C frame is non-inertial, resultant of the external force must include not only the external forces of introduction but also the inertial forces, i.e., →
...(i)
F = F net + Fin
→
→
Σ τic = 0 →
→
→
ar a
Σ ri × (− mi ) ac = Σmi ri × (− ac )
m
Because in the C frame itself,
ita
.ja
→
rc = 0
rio s_ in fo
na .c om
On the other hand, the system of particles as a whole rest in the C frame. Therefore, → in this frame F = 0 Taking into account the equation (i) we conclude that in the C frame, the total moment of forces (including inertial forces) does not depends on the choice of the point O, and another important conclusion that in C frame the total moment of inertial forces relative to the centre of mass is always equal to zero.
→
t.m e/
Σ τic ( z ) = Iα z
er si ta rio s
un
ht tp s:
//t
iv
→
Σ Fic = m ac ,
.m
ht
tp
e/
s:
un iv
//w
w
er s
w
∴ ∑τic = 0 Note that the angular acceleration αz as well as ωz are equal in both reference frames. Since, the C frame translate relative to the inertial frame (in the frame of ground earth). Basic equation of dynamics of the solid or a rigid body in plane motion is
s: //
How to solve a problem:
ht tp
(i) Draw the free body diagram of each body. (ii) On the basis of force diagram, analyse direction of kinematical term and then write the chemical dependence of the body. (iii) For the translation (linear motion) Σ Fix = m ac x , ΣFiy = m acy
If necessary put τic = 0 For the curve linear motion Σ Fit = m atc , Σ Fin = manc
And if necessary put τic = 0 (iv) For the pure rotation
1.32
Σ τic = I αzc
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Rotational Dynamics
1.33
If necessary put ΣFix = 0, ΣFiy = 0 or ΣFit = 0, ΣFin = 0 (v) In the plane motion use →
Σ Fi = m ac Example A sphere of mass m rolls without slipping on an inclined plane of inclination θ. Find the linear acceleration of the sphere and the force of friction acting on it. What should be the minimum coefficient of friction to support pure rolling?
) Solution Applying the law of motion equation
I
ita
.ja
m
ar a
1
un iv
//w s:
e/
iv
.m s: //
2 a fr = mr 2 5 r
a=
f =
ht tp
2 ma 5 From equations (i) and (iii), we get or
...(iii)
5 2 g sin θ and f = mg sin θ 7 7
For pure roller, µN > f or
µ mg cos θ >
or
µ>
...(ii)
t.m e/
ht tp s:
un
//t
tp ht
a r fr = I α
er si ta rio s
w
er s
w
PJ
T
As the sphere rolls without slipping, thus α=
...(i)
rio s_ in fo
na .c om
mg sin θ – f = ma
2 mg sin θ 7
2 tan θ 7
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1.34
Rotational Dynamics
3. Two identical solid spheres of mass M and radius R are joined together and the combination is rotated about an axis tangent to one sphere and perpendicular to the line connecting them. What is the rotational inertia of the combination?
1. A wheel having moment of inertia 2 kg-m2 about its axis, rotates at 50 rpm about this axis. Find the torque that can stop the wheel in one minute.
)
) Solution Initial angular velocity = 50 rpm 5π rad/sec. 3 Using, ω = ω0 + αt =
ω − ω0 −5π /3 −π = rad/sec 2 = rad/sec 2 t 60 36 The torque that can produces this deceleration is −π τ = I (α) = (2 kg − m 2) rad / sec 2 36 π = N−m 18
co m
α=
5
a.
an
rio
ar
m
) Solution Like masses, rotational inertias of )
si ta
.ja
rio s
ta
si
er
iv
m
2 MR2 + MR2 = 1.4 MR2 5 And for the second sphere, I2 = ICM + Md2
un
//t .
=
tp s
NJ P
:// t
.m
ht tp
e/
s:
iv
un
e/
tp s
ht
NJ P
solid objects add like scalars, therefore total for the two sphere, I = I1 + I2. For the first sphere (the one closer to the axis of rotation) we have, from the parallel axis theorem. I1 = ICM + Md2
er
w
w
:// w
=
)
ht
) Solution Taking moment about pivoted
I = I1 + I 2
1 (m2 g − m1 g ) × = I . α 2
= 1.4 MR2 + 9.4 MR2 = 10.8 MR2
2
α=
7/4
1 2 = 8.4 rad/sec 2.
∴
(5 − 2) × 9.8 ×
1 1 7 = m1 + m2 = kg.m 2 4 2 2
4. A uniform sphere of weight W and radius r is being held by a rope attached to a frictionless wall a distance L above the centre of the sphere as shown in the figure. Find: (i) Tension in the rope. (ii) The force exerted on the sphere by the wall.
2
Moment of inertia about pivoted
2 MR2 + M(3R)2 = 9.4 MR2 5
2. A light rod of length 1 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends as shown in the figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.
5
s_ in fo
WORKED OUT EXAMPLES
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Rotational Dynamics
1.35
with L1 = 20.0 cm and L2 = 80.0 cm. The rod is held in the horizontal position shown in the figure and then released. Calculate the linear acceleration of the two blocks as they start to move.
/
/
/
U P
) Solution )
P
) Solution Taking torque about point of pivot,
nf o
we have (mgL2 – mgL1) = I . α ...(i) Moment of inertia about pivot mL12 + mL22 = m(L12 + L22) Putting the value I in equation (i), we get
s
rio
9.8 × .60 = 7.13 rad/sec 2 (.20) 2 + (.80) 2
er s
iv
e/ un
6. In the given figure, calculate the linear acceleration of the blocks. Mass of the block A = 10 kg Mass of the block B = 8 kg Mass of disc shaped pulley = 2 kg
.m //t ht
1/2
r2 T = W(1 + tan 2 θ)1/2 = W 1 + 2 L
tp
s:
r L
α=
iv un e/ .m
ht tp :
From
∴
Linear accelerations are a1 = α L1 = 0.20 × 7.13 = 1.426 m/sec2 a2 = α L2 = 0.80 × 7.13 = 5.70 m/sec2
//t
1
But from the figure, tan θ =
mg(L2 – L1) = m (L12 + L22) . α
ita
w w 7
s:
ht
tp s:
//w
T
er si
.ja
m
ta rio
...(i)
s_ i
W cos θ
a.
T=
ar an
T = W sec θ
or
or
T cos θ = W
)
co
m
Analysis of the forces From ΣFiy = 0
ΣFix = 0 T sin θ = N
r2 r N = W 1 + 2 . 2 L ( r + L2 )1/2 r L
$
) Solution According to the law of motion,
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m2g – T2 = m2a T1 – m1g = m1a
5. Figure given below shows two blocks each of mass m suspended from the ends of a rigid weightless rod of length L1 + L2,
=W
%
)
1/2
∴
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...(i) ...(ii)
1.36
Rotational Dynamics
Now,
Σ τic = I α
1 (T2 R − T1R) = MR2 . α 2 7
7
PJ
∴
a = αR
in f
an a
rio
ar //t
ht
s:
ht
tp
s
First Method: Let the particle after time t reached the point P. Thus, the velocity and position of point P relative to origin O are. →
→
→
v = [v0 cos α i + (v0 sin α − gt ) j ]
1 r = v0 cos αti + v0 sin α t − g t 2 2
→
)
.m
tp
) Solution
(i) What is the angular acceleration of the rod when it is at an angle θ to the vertical? (ii) What is the tangential linear acceleration of the free end when the rod is horizontal? The moment of inertia 1 of a rod about one end is ML2. 3
ita
rs
ve
un i
e/
s:
(10 − 8) × 10 20 = m/sec 2. 10 + 8 + 1 19
7. A uniform rod of length L and mass M is pivoted freely at one end.
rio
iv un e/
.m //t
ht
=
(m2 − m1) g M m2 + m1 + 2
si
er
w //w
a=
tp
s:
M m2 g − m1 g = m2 + m1 + a 2 or
8. A ball of mass m is thrown at an angle α to the horizontal with the initial velocity v0. Find the time dependance of the magnitude of the ball’s angular momentum vector relative to the point from which the ball is thrown. Find the angular momentum L at the highest point of the trajectory if m = 130 g, α = 45º, and v0 = 25 m/s. The air drag is to be neglected.
ta
m .ja
7
w
7
α=
o
.c
3 g sin θ rad/sec 2 2L When the rod is horizontal, π 3g θ = and α = 2 2L 3g at = α L = m/sec 2. ∴ 2
or
s_
1 Ma ...(iii) 2 From equation (i), (ii) and (iii), we get T2 – T1 =
MgL 1 sin θ = ML2 α 2 3
om
or
Iα =
PJ
PJ
<
) Solution Take torque about pivoted end of rod
3
)
MgL sin θ Torque due to weight = 2 and, torque due to rotational motion = Iα.
D 2
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j
Rotational Dynamics
Angular momentum, →
→
L(t )= Thus, L(t) =
→
L0 = r × p
mv0 gt 2 cos α 2
Thus, angular momentum at maximum height
→
1.37
→
= r×mv
t v0 sin α = → g 2 1 2 = m v0 cos αt i + v0 sin αt − gt j × v0 cos α i + (v0 sin α − gt ) j 2 2 t mv0 g v0 sin α L = cos α → → → 2 g 1 2 → s αt i + v0 sin αt − gt 2 j × v0 cos α i + (v0 sin α − gt ) j 2 mv03 2 = → sin α .cos α 1 2g = m v0 .v0 sin α ⋅ cos αt − v0 gt 2 cos − v02 sin α cos αt + gt 2 v0 cos α k 0 2 = 37 kg-m2/sec. i.e., at m t =
→
→
in fo
rio
s
er si
ta
un t.m ) Solution
→ → → → → 1→ → L(t ) = r × P = v 0 t + g t 2 × m( v 0 + g t ) 2
Second Method:
(i) mg – T = ma ...(i) Torque about the centre T.R=I.α Since tangential acceleration, a = αR, where α is angular acceleration
;
P )
D
ht
PJ
tp s: //
\
5
0
e/
tp s ht
:// t
iv
.m
ht tp
mv 3 = 0 sin 2 α cos α 2g = 37 kg-m2/sec Y
ta rio s_
e/ un
.ja
w w
w
s: //
2
v sin α 1 mg . v0 . 0 cos α 2 g
9. A light thread with a body of mass m tied at its end and is wound on a uniform solid cylinder of mass M and radius R as shown in the figure. At a moment t = 0, the system is set in motion. Assuming the friction in the axle of the cylinder to be negligible, find the time dependence of (i) the angular velocity of the cylinder. (ii) the kinetic energy of the whole system.
iv er si
v0 sin α g
At the highest point, t = L0 =
ar an m
1 = mg t 2 v0 cos α 2
a.
1 → in α ⋅ cos αt − v0 gt 2 cos − v02 sin α cos αt + gt 2 v0 cos α k 0 2
co m
→
π → 1 π → = mv0 gt 2 sin + α − k + mv0 gt 2 sin + α −∴k T.R= 1 MR 2 . a 2 2 2 2 R
1 → = mv0 gt 2 cos α − k 2 https://t.me/universitarios_infohttps://www.jamarana.com
1 I = MR 2. 2 1 T = Ma 2
π → 1 π → mv0 gt 2 sin + α − k + mv0 gt 2 sin + α − k 2 2 2
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...(i)
1.38
Rotational Dynamics
P
0
)
0
7
)
) Solution
...(iii)
1 mR2 . α 2
I
I
rio s_ in fo
)
For the cylinder, linear acceleration + angular acceleration = block acceleration. i.e., ac + α . R = a ...(iv) From equations (ii) and (iii), we get
er si ta rio s
ita er s
1 1 K = Iω2 + mv 2 2 2
w
(ii) Total kinetic energy
w
2mgt ω= αt = R(M + 2m)
ar a
a 2mg = R R(M + 2m)
m
∴ α=
na .c om
D&
2mg M + 2m
f.R=I.α =
Ma = ma 2
or a =
mg −
.ja
From equations (i) and (ii), we get
...(i) ...(ii)
F – f = Ma f = mac
PJ
un
iv
t.m e/
ht tp
un iv
e/
.m
s: //
ht tp s:
//t
ht
tp
s:
//w
f 2f 3f + = m m m 2 2 1 1 2mgt 1 2mgt 2 2 3 f = . MR . or R a= + m. 2 2 R(M + 2m) 2 R(M + 2m) m ma 2 2 ∴ f = ...(v) 2mgt 1 2mgt 2 1 1 3 = . MR2 . + m . R 2 2 From equations (i) and (v), we get R(M + 2m) 2 R(M + 2m) ma F− = Ma mg 2 t 2 3 = . M F 21 + ∴ a= 2m m M+ 3 10. A block of mass M rests on a horizontal 11. A rod of length l is constrained to move frictionless surface. A uniform solid in vertical plane containing its length cylinder with a mass m rests on the upper between two smooth perpendicular surface of the block as shown in the figure. planes. When it is released from the verA horizontal force F is now applied to the tical position by pulling its bottom gently, block. As a consequence, the cylinder find its angular acceleration and angurolls without slipping toward the rear of lar velocity as a function of the angle θ the block as the block accelerates. Derive made by it with horizontal as shown in an expression for the acceleration of the the figure. block in terms of F, M and m. ac + α.R =
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Rotational Dynamics
1.39
From equations (ii) and (v)
l N B − mg = m [α cos θ − ω2 sin θ] 2
N B = mg +
or
ml [α cos θ − ω2 sin θ] 2 ...(vii)
Putting this value in equation (iii), we get
T
ml ml [(−α sin θ − ω2 cos θ)sin θ] − mg + 2 2
)
) Solution
ΣFx = max NA = max ...(i) ΣFy = may or NB – mg = may ...(ii) Taking torque about centre of the gravity, we have τG = (NA sin θ – NB cos θ)
[α cos θ − ω2 sin θ]cos θ =
m
fo
co
a.
s
rio
ta
αl l α − = g cos θ 2 6
rs i
−
ve
or
un i
e/
−4αl = g cos θ 6
or
or
3 3g g cos θ = − cos θ 2l 2l
α = −ω
ω=
dω dθ
d ω 3g = cos θ d θ 2l
l l Now, x = cos θ and y = sin θ 2 2
α=−
Here, α is negative, so our reference is reverse.
tp s
ht
or
:// t.m
ht tp
1 2 ml ⋅ α 12
ml α (NA sin θ – NB cos θ) = 6
αl 2 l lα l ω2 sin θ − ω2 cos 2 θ.sin θ − g cos θ − cos 2 θ + s 2 2 2 2
−
er si
iv
e/ un
or
.m
%
s: //t
ht
T
PJ
or
ta
.ja w
:// tp s
D[
1%
I ⋅ α=
rio
m
s_
ar an
in
l lα l ω2 lα − α cos 2 θ − cos 2 θ + sin θ ⋅ cos θ = 2 2 2 6
αl 2 l lα l ω2 lα sin θ − ω2 cos 2 θ.sin θ − g cos θ − cos 2 θ + sin θ ⋅ cos θ = 2 2 2 2 6
w
−
l [−α sin 2 θ − ω2 cos θ − sin θ] − g cos θ 2
or
w
$ 1$
mlα 6
...(v)
ml [−α sin θ − ω2 cos θ] 2
...(vi)
∴
3g 2l
θ
∫ cos θ d θ
90°
ω2 3g θ = − [sin θ]90 ° 2 2l 3g 3g ° = [sin θ]90 [1 − sin θ] θ = 2l 2l
or
From equations, (i) and (iv) NA =
0
d2y l = [α cos θ − ω2 sin θ] dt 2 2
...(iv)
d 2x l = [− α sin θ − ω2 cos θ] dt 2 2
∴
and
ω
Integrating, ∫ ω d ω = −
ω=
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1.40
at rest. The rod is then given an angular g velocity 2 . Assuming that the rings L are free to slide on the rod, find the distance between the points where the rings strike the floor in the subsequent motion.
12. A uniform thin bar of 4 kg and length 2 metres is bent to make a square. Calculate its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the square.
Rotational Dynamics
)
) Solution Total mass of the bar is m kg and length l metres.
)
) Solution
'
&
Y
Z
Y
na .c om
ita
...(i)
er si ta rio s
2
iv
ML2 3ML2 = 2 2
un
I = ML2 + 2ML2 = 3ML2
g 3 ML2 .2 I0 ω0 2 L = g ω= = 2 I 3ML L
g . L = gL L From energy conservation,
v2 = ω L =
1 1 1 I0 ω02 = Iω2 + .2M(v12 +v 22 ) 2 2 2
∴
3 g g ML2 . 4 . = ML2 . + 2 MgL + 2Mv12 2 L L
or
or 6gL – 3gL = 2v12 ⇒
13. A uniform rod of length 2 L and mass 3 M is free to rotate in a horizontal plane about a vertical axis passing through its mid point O which is fixed at a height L above the horizontal floor. Two small rings each of the mass M are located on the rod on opposite sides of O and at disL tance from O. The system is initially 2
t.m e/ s: //
∴
ht tp
ml 2 1 I = 4 l1 = × kg.m 2 16 3 Putting the value of m and l, we get 4× 4 1 I= × kg.m 2 = 0.333 kg.m2 16 3
e/
.m
ml 2 1 = × kg.m 2 64 3 Similarly, total moment of inertia
1 L .3M.(2L) 2 + 2M. 12 2
= ML2 +
and,
//t
1 1 12 + 4
s:
tp
ml 2 64
I0 ω0 = I ω
where, I0 =
un iv
2
m 1 l m l . + . [Parallel axis theo4 12 4 4 8 rem]
ht
=
//w
2
I1 =
ht tp s:
w
Moment of inertia of bar of side AB
er s
w
.ja
∴
Y
/
Let the angular velocity just after the ring slip-off is ω. From conservation of angular momentum
m
m Then, mass of one side of square is 4 l and length is metres. 4
/
rio s_ in fo
%
ar a
$
Y
v1 =
3 gL 2
Time of fall of each ring is given by 1 2 gt = L 2
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1.41
Rotational Dynamics
2L g
∴
=
Required distance between the points
2µg (1 + µ) R (1 + µ 2 )
t=
⇒
I
= 2 (L + v1t ) 2 + (v2t ) 2
1 1 5 1
= 2L (1 + 3) 2 + 2 I
= 2L 6 + 2 3
PJ
14. A uniform cylinder of radius R is spinned about its axis to the angular velocity ω0 and then placed into a corner shown in the figure. The coefficient of friction between the corner walls and the cylinder is equal to µ. How many turns will the cylinder accomplish before it stops?
Kinematics equation of the rigid body is ω02 – ω2 = 2 α θ From the problem ω = 0
m
in f
s_
rio s
er si ta
iv
e/ un iv
) Solution The given problem is shown in the )
According to the rigid body, torque about the centre
µmg µ 2 mg and, f 2 = 2 1+ µ 1 + µ2
f1 =
15. A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end ‘A’ of the rod with a velocity v0 in a direction perpendicular to AB. The collision is completely elastic. After the collision the particle comes to rest m (i) Find the ratio . M (ii) A point P on the rod is at rest immediately after the collision. Find the distance AP. (iii) Find the linear speed of the point P πL after the collision. at a time (3v0 )
∴
ω02 R(1 + µ 2 ) 8πµg (1 + µ)
.m
ht tp s: //t
f1 + µf1 µ
un e/
t.m
tp s
ht
://
...(i) ...(ii)
mg =
n=
er
w
s: //w ht tp
)
∑Fy = 0 mg = N1 + f2 f1 = N 2 f1 = µN1 And, f2 = µN2 From equation (i)
si
.ja w
∴
) Solution From the Law of motion
ω02 ω02 ω2 R(1 + µ 2 ) = = 0 2α 2. 2µg (1 + µ) 4µg (1 + µ) R 1 + µ2
But θ = 2 π n
5
θ=
ta rio
∴
o
ar an a. co m
= 6.15 L
1 ( f1 + f 2 )R = Iα = mR2α 2
or
µmg µ 2 mg 2 + 1 + µ 2 1 + µ 2 2( f1 + f 2 ) α= = mR mR
figure. %
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Y&
3 [ Z[
$ Y2 P
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1.42
Let the given point P on the rod be situated as shown in the figure. It is given that the point P remains at rest immediately after the collision of the object with the rod. This is possible only when the effect of collision is to cause a linear motion of the rod with velocity vc as shown in the figure and also a counterclockwise rotation of the rod about the axis passing through its centre of mass. Let x be the distance of the point from its centre of mass, then the above condition will be satisfied when at the point P. ∴ vc = ωx ...(i) where ω the angular velocity of the rod. From the conservation of momentum, mv0 = Mvc ...(ii) From the conservation of angular momentum about the centre of mass,
Using equations (ii) and (vi) in equation (i), we get
Distance of P from A, PA = x +
in time
co m
s
si ta iv er un e/
//t .m
Y2 Y2
3
Y2
The position of the rod is as shown in the figure.
πL , the point P is under the in3 v0 fluence of two velocities. Velocities due to the linear motion of the rod (acting in the upward direction) is At time
1 m vc = v0 = v0 4 M
2
m 3m + M M
m 1 = M 4
&
s:
ht tp
1 1 mv 1 1 6mv0 mv0 2 = M 0 + ML2 2 2 M 2 12 ML
or
rio
un
e/ t.m
://
...(vi) Using equations (ii) and (vi) in equation (v), we get
1=
fo
ta r
si
iv er
w w
tp s
ht
mv0 (L/2) mv0 (L /2) 6 mv0 = = Ic (1/12)ML2 ML
m 1 = , we get M 4
6v πL π θ= 0 = 4L 3v0 2
...(v)
From equations (iii) and (iv), we get
2
io s_ in
ar
m
.ja
//w
s:
ht tp
1 1 1 mv02 ) = Mvc2 + Iω2 ( 2 2 2
or
6mv0 πL = ML 3v0
From equation (vii), since
...(iv)
Also, it is given that the collision of the object with the rod is elastic. Therefore, from the principle of conservation of energy, we have
ω=
πL is 3v0
θ = ωt
a.
an
1 ML2 12
Ic =
L 2
L L 2 + = L ...(viii) 6 2 3 The angle through which the rod has rotated =
L mv0 = Ic ω ...(iii) 2 where, Ic is the moment of inertia of the rod about the axis passing through its centre of mass. It is given by the expression.
vc mv0 /M L = = ω 6mv0 /ML 6
x=
Rotational Dynamics
...(vii)
Velocity due to the rotation motion of the rod (acting in the horizontal direction) is
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1.43
Rotational Dynamics
From equations (i) and (ii), we have
6mv0 L vr = ωx = mL 6
mg sin α −
1 m = v0 = v0 4 M Resultant of these two velocities is 2
1 9.8 × g sin α 2 = = 1.6 m/s 2 a= 0.45 I 1 + 2 1 + 0.2 × (.03) 2 mr
2
v v v v = vc2 + vr2 = 0 + 0 = 0 2 2 4 4
17. A man pushes a cylinder of mass m1 with the help of plank of mass m2 as shown in the figure. There is no slipping at any contact. The horizontal component of the force applied by the man is F. Find:
o
rio s si ta
er
un
)
e/ .m //t
1 )
...(i)
ht tp s:
cylinder is shown in the figure.
D
I PJ
P
iv
.m
//t s: tp ht
ΣFi = ma or mg sin α – T = ma Torque about the centre is given by a T.r=I.α=I. r a r2
I
...(ii)
D I
7
Applying the kinematic law f1 + f2 = m1 a1
...(i)
T=I
PJ
or
1 a1 = αR and I = m1R2 2
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Also (f1 – f2) R = I α D
or
P
) Solution Free body diagram of plank and
D
D
)
ita
e/
s: ht tp
P
) Solution
)
un i
//w
ve
rs
w w
.ja
m
ar
rio s_
(i) The acceleration of the plank and the centre of the mass of the cylinder. (ii) The magnitudes and directions of frictional forces at contact point.
an
in f
a.
co
m
16. A spool with a thread wound on it is placed on an inclined smooth plane set at an angle α = 30º to the horizontal. The free end of the thread is attached to the wall as shown in the figure. The mass of the spool is m = 200 g, its moment of inertia relative to its own axis I = 0.45 g m2, the radius of the wound thread layer r = 3.0 cm. Find the acceleration of the spool axis.
The velocity vector will be directed at an angle of 45º with respect to the horizontal.
I mg sin α = 2 + m a r
or ∴
Ia = ma r2
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...(ii)
1.44
Rotational Dynamics
From equations (i) and (ii), we get
x=
R sin θ
1 f1 − f 2 = m1a1 ...(iii) 2 Then, from the equations (i) and (iii), we get
1 f 2 = m1a1 4 Equation of the plank F – f2 = m2a+2 and a2 = 2 α R = 2 a1
...(v)
When x = 2.25
and
om .c
si
Y$ [
rs
ve
ni
= (0.438) 2 + (0.216) 2 = 0.52 m/sec.
19. A spool with thread wound on it of mass m, rests on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to I = KmR2, where K is the numerical factor, and R is the outside radius of the spool. The radius of the wound thread layer is equal to r. The pool is pulled without sliding by the thread with a constant force F directed at an angle α to the horizontal. Find: (i) The projection of the acceleration vector of the spool axis on the x-axis. (ii) The work performed by the force F during the first t seconds after the beginning of motion.
T
ita rio s
)
dx → = vA dt
From the figure
v = (vBx ) 2 + (vBy ) 2
e/ u
$
→
) Solution
vBx = [0.6 + 1.35 × 0.6 (–0.2)] i vBy = 1.35 × 0.8 × 0.2
.m
//t
s:
tp ht Y$
[
dyB R cos θ . d θ = dt dt
er ∴
//t
s:
ht tp P
P
dxB dx R sin θ . d θ = + and dt dt dt
iv
un
ht
18. The end A of the slender rod AB against a fixed semi-cylindrical surface as shown is pulled with vA = 0.6 m/sec (→). If x = 2.25 m, determine ωAB of AB and vB of the end B.
∴ and
.m e/
tp s:
m1F . 3m1 + 8m2
%
fo
ta rio s_ in
ar
//w
w
w
.ja
m
Putting the value of a1 in equations (iv) and (v), we get 3m1F 3 4F f1 = m1. = 4 3m1 + 8m2 3m1 + 8m2
f2 =
dθ 0.6 × 0.6 × 0.6 =ω=− dt 1.35 × 0.8
= – 0.2 rad/sec (anti-clockwise) For point B xB = x – R cos θ and yB = R sin θ
an a
4F a1 = 3m1 + 8m2
⇒
∴
3 F = m1a1 + 2m2 a1 4
F – f1 = 2m2 a1 ⇒
R 1.35 = = 0.6 x 2.25 cos θ = 0.8 sin θ =
and,
sin 2 θvA dθ =− dt R cos θ
⇒
...(iv)
3 f1 = m1a1 4
dx R cos θ d θ =− dt sin 2 θ dt
∴
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Rotational Dynamics
1 1 v = m(ax t ) 2 + I. C 2 2 R
1 1 = max 2t 2 + KmR 2 ax2t 2 2 2
[
=
)
2
(i) ΣFx = max
r F2t 2 cos α − R . Work done = 2m (1 + K)
rio s_
in
fo
rio s
P
5
s: //
e/ un i
ve
t.m
tp
%
//t
.m
ht tp
[
s:
0
) Solution When the cylinder reached at
1 1 = mvC 2 + Iω2 2 2
or or ∴
bottom point, the block M traveled through a distance x. Since, the cylinder is slip of the track, thus, does not possess angular velocity. mxm + MxM = 0
(ii) The friction is no required any work because friction is static. ∴ Work done = Change in kinetic energy
)
r F cos α − R ax = m(1 + K)
∴
ht
tp
From equations (i) and (ii), we get r Fcos α − F. = K max + max R
rs ita
e/ u
$
) D
ht U
si ta
.ja m
s:
5
//w w
r − K max R
ax R
w
or F ⋅ r − f ⋅ R = K mR 2 ⋅ or f = F ⋅
om
.c
a R
20. A block of mass M with a semi-circular track of radius R, rests on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips on the semi-circular frictionless track. How far has the block moved when the cylinder reaches the bottom (point B) of the track? How fast is the block moving when the cylinder reaches the bottom of the track?
ni ve r
Fr − f ⋅ R = I ⋅α = I ⋅
ar an a
where, f = Frictional force, because no sliding occurs. Torque about the centre of spool
...(i)
or F cos α – f = max
m 2 2 t ax [1 + K] 2
Putting the value of ax, we get
) Solution
2
) D
U
5
1.45
m[xmM + xM] + MxM = 0 m[(R – r) – x] – Mx = 0
x=
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m(R − r ) m+M
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1.46
Rotational Dynamics
92
5
D
)
) Solution The cylinder does not jump, it means that normal reaction is always present.
s_ ita rio ∴
ita rio
rs
ni ve
.m e/ u s:
//t
where ω0 =
ht tp
2 g (R − r )m 2 M(M + m)
s
//t
ht tp s:
v2 2 =
PJ
v0 mR 2 and I = R 2
mv0 2 +
mv02 mv 2 = mv12 + 1 − 2mgR (1 − cos α) 2 2
v2 =
2 g (R − r ) .m M(M + m) 2 mv02 mv 2 mv0 + = mv12 + 1 − 2mgR (1 − cos α) 2 2 21. A uniform solid cylinder of radius R = 3 2 3 2 15 cm rolls over a horizontal plane passv0 = v1 − 2 gR (1 − cos α) or 2 2 ing into an inclined plane forming an angle α = 30º with the horizontal shown 4 v0 2 = v12 − gR (1 − cos α) ...(ii) or in the figure. Find the maximum value 3 of the velocity v0 which still permits the From equations (i) and (ii), we get cylinder to roll on to the inclined plane NR 4 section without a jump. The sliding is v0 2 = gR cos α − − gR (1 − cos α) m 3 assumed to be absent.
⇒
in fo
.c
e/ u
1Mv 1 + Mv2 2 2 m 2
Now 2 g (R − r )m 2 = (M 2 + Mm) v2 2 ∴
D
1 1 mv0 2 + I ω0 2 2 2 1 1 2 = mv1 + Iω2 − mg R(1 − cos α) 2 2
.m
ht
=
2
Here, O is the instantaneous centre of rotation and its velocity is v1. From the energy conservation
er s
2
1 M 1 mg (R − r ) = m v2 + Mv2 2 2 m 2
ni v
tp s: //w
w
w
Substituting the value v1 in equation (i), we get
2 2 2
...(i) 1
an a
.ja m
M v1 = v2 m
ar
1 1 mg (R − r ) = mv12 + Mv2 2 2 2 From the conservation of momentum mv1 – Mv2 = 0
mv12 R
om
Hence, the block moved in negative x-direction. From the energy conservation
mg cos α − N =
∴
[
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Rotational Dynamics $
1 NR = gR (7 cos α − 4) − 3 m
1 = gR (7 cos α − 4) 3
→
fo
in
s_ ita
ve rs
un i
e/
.m
//t
s:
ht tp
ita rio s
(ii) Since, the wedge does not undergo any rotation, the magnitude of torque due to F should be completely balanced by the torque due to N. Hence,
where, h is the perpendicular distance between the centre of mass of the wedge and the line of action of the force F.
2mv 3 ∆t
2mv → In the vector form N = mg + k 3 ∆t
23. In the arrangement shown in the figure, a block A of mass m is suspended by a string. The other end of the string is wound over a pulley B having a coaxial axel C. The axel-pulley system is suspended symmetrically by two light strings, one end of each is fixed to the ceiling and the other end would over the axel. The mass of the pulley-axel system
)
During the collision, the total downward force acting on the wedge,
4mv τ = Fh = h 3 ∆t
) Solution The collision is elastic and the sphere
is fixed so velocity of wedge is same and return to the original path. Hence, change in momentum causes the impulse force. Now, change in momentum in x-direction = 2 mv ∴ F cos 30º . t = 2 mv 2mv 4mv = F= or 3 3t ×t 2
4mv 3 → 4mv 1 → . i− × k 3t 2 3t 2
rio
ve rs e/ un i
tp
ht
s:
//t
(i) Find the force F and also the normal force N exerted by the table on the wedge during the time ∆t. (ii) Let h denotes the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge, during time interval ∆t.
→
N = mg + Fsin 30° = mg +
&
.m
%
→
→ 2mv → k = 2mv i − 3 → 1 → k = 2mv i − 3
a. c
ar an
m
w
//w ht tp s:
[
&
)
om
=
.ja w Y
)[
F = (Fcos α i − Fsin α k )
$
5
)
In the vector form
%
22. A wedge of mass m and triangular crosssection (AB=BC=CA=2R) is moving with a → constant velocity – v i towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time ∆t, during which the sphere exerts a constant force F on the wedge.
\
5
Hence, maximum value at v0
]
1.47
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Rotational Dynamics
% 7
&
f = ma1 f = µN = µmg ∴ a1 = µg ∴ v1 = µgt Torque about sphere surface I . α = f . R = µmg . R
) Solution )
ar a
m
un iv
er s
ita
.ja
w
w
//w
e/
t.m e/
5
I
I
s: //
5
P
un
ht tp s:
5
iv
.m
//t
ht
tp
s:
%
0
ht tp
&
P
$
∴
2 I = mR 2 5 α=
µmg .R 5µg = 2 mR 2 2R 5
Angular velocity, ω = αt =
Iac = 3ma + Mac R
Velocity of board, v2 = v − µgt .
m M
For the pure rolling, v2 = v1 + ωR
Mg + 3mg −
...(i)
5µg 2R
Since, there is no slipping, therefore, a = ac + 2 αR = ac + 2ac = 3ac From equations (i) and (iii), multiplying equation (iii) by 3, we get Mg + 3mg – (2TA + 2T) = 3ma + Mac ...(v) From the equations (ii) and (v), we have
where,
→
where, ac = acceleration of centre of mass of the cylinder Torque about common centre a TA 2R + 2TR = I. α = I c …(ii) R Motion of block mg – TA = ma ...(iii) where, a = Acceleration of block
...(i)
Mg + TA – 2T = Mac
)
) Solution
$
P
na .c om
7$
PJ
er si ta rio s
5
3(Mg + 3mg ) I M + 9m + 2 R
24. A board of mass M, whose upper surface is rough and under surface is smooth, rests on a smooth horizontal plane. A sphere of mass m is placed on board and the board is suddenly given a velocity v in the direction of its length. Find the time after which the sphere begins pure rolling, if the coefficient of friction between the board and the sphere is µ.
7 5
a=
rio s_ in fo
or
or
a a Mg + 3mg = 3ma + M + I 2 3 3R
is M and radius of the axel and the pulley are R and 2R respectively. The strings do not step over the pulley and axel. Find the acceleration of block A when the system is left free. The moment of inertia of the pulley-axel system about the common axis is I.
1.48
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Rotational Dynamics
(ii) Work done = change in kinetic energy Since, initial kinetic energy = 0 Therefore, final kinetic energy = work done =Fx
m 5µgt = µgt + .R M 2R
Thus, v − µgt .
µmg 5µgt t + µgt + M 2 v t= 7 m + µg 2 M
1 = F at 2 2 1 F(3m1 + 2m2 ) 2 =F. t 2 m1 (m1 + m2 )
25. A uniform solid cylinder A of mass m1 can freely rotate a horizontal axis fixed to a mount B of mass m2 shown in the figure. A constant horizontal force F is applied to the end K of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent. Find: (i) The acceleration of the point K. (ii) The kinetic energy of this system t seconds after the beginning of motion.
F 2 t 2 (3m1 + 2m2 ) m1 (m1 + m2 )
26. A small body A is fixed to the inside of a thin rigid hoop of radius R and mass equal to that of the body A. The hoop rolls without slipping over a horizontal plane. At the moments when the body A gets into the lower position the centre of the hoop moves with velocity v0 as shown in the figure. At what value of v0, will the hoop move without bouncing?
m
=
fo
in
s iv
un
ht
tp
s: //
(i) The acceleration of cylinder and the mount is same. F a= ...(i) m1 + m2
$
Torque about at the centre of cylinder F. R = I . α F.R 2F ∴ α= = 1 m1R 2 m1R 2 Acceleration of point K = a + αR
) Solution Clearly, the tendency of bouncing of )
)
5
Y2
e/ t.m
tp
s: //
)
ht
) Solution
si ta er
e/ t.m
tp ht %
si ta
un
s: //
iv
w
$
rio
w
er
w
.ja
m
rio
s_
ar an
the loop will be maximum when the small body A, will be at the highest point of the hoop during its rolling motion. Let the velocity of centre of mass of the hoop equals v at this position. The static friction does not work on the hoop, so from conservation of energy. 1
F 2F = + m1 + m2 m1
1
PJ PJ 1
Y2
F(3m1 + 2m2 ) m1 (m1 + m2 )
or
N = mg – N′
=
a.
co
∴
v=
or
1.49
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1.50
= mg + mg − ⇒
2
$ 5
2mgR − NR v = m 2
= 2 gR −
mv mv 2 = 2 mg − R R
NR m
& 3
...(iii)
Rotational Dynamics
4 &
From equations (i) and (iii), we get NR 3 2 gR − = v0 2 − 2 g R m
$
om
o
in f
rio s
ta
rs i
ve
ni
e/ u
Moment of inertia of disc about PQ 2
R I 2 = IC + m (using parallel 4 axis theorem)
//t .m
Initially, the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position
2
25mR 2 5R I1 = m = 16 4
ht
4
Total potential energy decreases 5mgR mgR = + 2 2 = 3mgR Moment of inertial of m about PQ,
tp s:
&
R mgR = mg 2 × = 4 2
ve
un i
.m e/
s: //t ht tp 5
3
rs ita rio s_
ar
.ja m
w
w
w
://
tp s
ht
$
Decrease in potential energy of disc from centre C to C′
3N m
27. A uniform circular disc has radius R and mass m. A particle also of mass m is fixed at a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed horizontal chord PQ R from the centre that is at a distance 4 C of the disc. The line AC is perpendicular to PQ.
v0 = 8gR −
5 5mgR = mg 2 × R = 4 2
)
) Solution The final position of rotation when particle reaches the lowest position is shown in the figure.
=
mR 2 mR 2 5mR 2 + = 4 16 16
Hence, total moment of inertia about PQ = I1 + I 2 25 5mR 2 15 mR 2 + = mR 2 16 16 8 1 Thus, kinetic energy = Iω 2 2 1 15 = . mR 2 ω 2 2 8 =
∴
Decrease in potential energy of the particle
3NR m
an a. c
v0 2 = 8 gR −
or
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1.51
Rotational Dynamics
Now, Decrease in potential energy = Increases in kinetic energy
∴
Ml ω = m(v − v1 ) 6
The linear speed of the particle, at which it reaches the lowest position is,
5R = 5 gR 4
in
s_
rio ta
m(v − v1 ) 3m(v − v1 ) + M M
s
(4m − M) .v (4m + M)
er si
v1 =
ta
or
iv
t.m
e/
un
mv = mv1 + Mv2
Substituting the value v1, we get
t.m
or
s: // tp ht
(4m − M) 2mv .M ∴ Mv2 = mv 1 − = (4m − M) (4m + M) 2m v2 = v 4m + M
From equation (iii) Ml ω 4m − M = m(v − v1 ) = m v − v 6 4m + M
(i) Let after impact, the velocity of puck be v1 and the centre of mass be v2. From the conservation of linear momentum mv = mv1 + Mv2 ...(i) From conservation of angular momentum, we have
(4m − M) . v + Mv2 (4m − M)
ht
Y
)
2M = m v 4m + M
∴ ω=
...(iv)
or Mv = –Mv1 + mv – mv1 + 3mv – 3mv1
mv = m
) Solution
m (v − v1 ) M
(ii) From equations (i),
s: // tp P
v = −v1 +
er si
iv
un e/
s: tp ht FHQWUH
...(ii)
From equations (i) and (iv), we get
ar
.ja
//w
w
w
0
ωl − v1 2 v
fo
co m
a.
an
v = −v1 + v2 + 3
m
(i) What is the velocity of the puck after impact? (ii) What is the velocity of the centre of mass and the angular velocity of the stick after impact?
v2 +
Since, e = 1, therefore, ωl ...(iii) v = −v1 + v2 + 2 From equations (ii) and (iii), we get
28. A metre stick lies on a frictionless horizontal table. It has a mass M and is free to move in any way on the table. A hockey puck m, moving as shown with speed v collide elastically with the stick.
v =ω
vn − vn e= 2 1 = v−0
rio
16 g 5R
ω=
or
or
15 mR 2 ω 2 16
3mgR =
or
2 mv1l I = Ml mvl = Iω + 12 2 2 Ml ω mv = + mv1 6
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1.52
29. A billiard ball initially at rest is given a sharp impulse by a cue. The cue is held horizontally at a distance h above the centre line as shown in the figure. The ball leaves the cue with a speed v0 and because of its forward “English” (rolling and sliding) eventually acquires a final 9 v0 4R speed of . Prove that h = , where 7 5 R is the radius of the ball.
µmgR =α Ic
or
µ mgR ω = ω0 − α t = ω 0 − ...(v) t Ic Since, after time t, the ball is in pure rolling, therefore, ∴
5
a. c
&
rio
ita
rio s
ita
er s
iv
un or
e/ ∴
5
mv 9 2R v0 = 0 h − R 7 I0 7
mv0 9 2R v0 = h− R 2 7 7 2 mR 5
9 5 2R = h− 7 2R 7
or 1
P1
2R 18R = 35 h − 7
f . R = Icα
or
35(7 h − 2R) 7 18 R = 35 h – 10R 18R =
∴
or
In final position, only friction torque is required to decrease the angular velocity. ∴ f = ma or µmg = ma or a = µg ∴ vt = v0 + µgt ...(iv)
or
PJ
v = ωR
2 But, Ic = mR 2 , because ball in solid sphere 5
.m //t
s: tp
ht K
.m
//t tp s:
ht
...(iii)
mv0 h Ic
mv0 h 2mv0 R − Ic Ic
ve
e/ un i
...(ii)
2mv0 R 7Ic
mv0 h mv0 2R ω = = h− Ic Ic 7
Now,
From equations (i) and (ii), we get
ω0 =
=
rs
w
//w w
⇒
ht
tp
s:
...(i)
µ mgR 2v0 . Ic 7µg
s_ in
an ar .ja m
)
or mv0 – 0 = F ∆t From angular momentum theorem Ic ω0 = τz . ∆t Ic ω0 = Fh . ∆t Fh . ∆t ω0 = Ic
2v0 7µg
fo
ω = ω0 − = ω0 −
) Solution From impulse momentum theorem Px = F ∆t
t=
From equation (v)
om
K
⇒
9v0 = v0 + µgt 7
v = ωR From equation (iv)
Rotational Dynamics
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4 R 5 https://t.me/universitarios
30. The axis of the cylinder given in the figure is fixed. The cylinder is initially at rest. The block of mass M is initially moving to the right without friction and with speed v1. If passes over the cylinder to the dashed position. When it first makes contact with the cylinder, it slips on the cylinder, but the friction is large enough, so that slipping ceases before M loses contact with the cylinder. The cylinder has a radius R and a rotational inertia I. Find the final speed v2 in term of v1, M, I and R. This can be done easily by using the relation between impulse and change in momentum.
)
the rod initially rest on its centre of mass, so there is no horizontal displacement. The velocity of centre of mass
v2 =
rio
ta
si
rio
s
er
2
l θ 1 4v 1 . 2sin 2 = ml 2 . 2 c2 + mvc 2 2 2 24 l sin θ 2
er
si ta
iv un e/
un
iv
or gl sin 2
//t .m
θ (1 + 3sin 2 θ)vc2 = 2 6sin 2 θ
e/
s:
t.m :// tp s
mv1 I m + 2 R
θ (sin 2 .sin θ) ∴ vc = 6 gl 2 1 + 3sin θ
32. Two cylinders having radii R1 and R2 and rotational inertias I1 and I2 respectively, are supported by axis perpendicular to the plane shown in the figure. The large cylinder is initially rotating with angular velocity ω0. The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases, and the two cylinders rotate at constant rates in opposite directions. Find the angular velocity ω2 of the small cylinder in terms of I1, I2, R1 and R2 and ω0.
Iv2 R2
31. A rod of mass m and length l is held vertically on a smooth horizontal floor. Now it is released from this position, find the speed of its centre of mass when it makes an angle θ with the vertical.
∴
mv1 = mv2 +
ht
or
mg
Iω R According to the problem v2 = ωR mv1 = mv2 +
or
s_
ar m w
Iω R
ht
m(v1 − v2 ) =
2
l 1 ml 2 2vc 1 2 mg (1 − cos θ) = . + mvc 2 2 12 l sin θ 2 or
...(i) ...(ii)
tp
ht
tp s
From the impulse-momentum m(v1 – v2) = F ∆t Iω = F ∆tR From equations (i) and (ii), we get
...(ii)
Substituting the value of ω in equation, we get
w
)
:// w
) Solution
in
l 1 1 mg (1 − cos θ) = I c ω2 + mvc 2 2 2 2
.ja
5
From conservation of energy
fo
co
m
T
an
Y
...(i)
l vc = sin θ ω 2
a.
Y
1.53
) Solution Since, there is no horizontal force and
Rotational Dynamics
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1.54
on a rough horizontal floor so that it starts-off with a purely sliding motion at t = 0. After t0 seconds, it acquires a purely rolling motion as shown in the figure.
Z
,
(i) Calculate the velocity of the centre of mass of the disc at t0. (ii) Assuming the coefficient of friction to be µ, calculate t0. Also calculate the work done by the frictional force as a function of time and the total work done by it over a time much longer than t0.
)
) Solution
5
From angular impulse equation Lz = I1 (ω1 – ω0)
m
Z
co
5
fo
Y
s
ta r
si
er
un
s:
R1R2 I1 ω0 – R ω2 I1 = R1 I2 ω2
or
ω2 (R12 I2 + R22 I1) = R1 R2 I1 ω0
or
RR Iω ω2 = 2 1 2 1 20 (R1 I 2 + R 2 I1 )
2
33. A uniform disc of mass m and radius R is projected horizontally with velocity v0
2 fr mR
...(iii)
So, angular velocity after time t0
1 mR 2 . α 2
α=
ω = α t0 =
2
∴
or
ht
R1I 2 ω2 I1 (ω0 R1 − ω2 R 2 ) 2 2
Torque about the centre of the disc f .R = I. α =
=
I 2 ω2 ωR I1 ω0 − 2 2 R1
R2 = R1
tp
s:
//t
ht
.m
tp
e/
According to the problem, after impact since they rotate at the constant rate, therefore ω1R1 = ω2 R2 ...(iv) From equations (iii) and (iv), we get
io
//t .m e/
tp
(i) By law of motion fr = ma ...(i) Acceleration of cylinder in the negative direction vx = v0 – at0 ...(ii)
un
...(ii)
...(iii)
ht
)
iv
//w
s:
R2 L 2 ω2 = R1 I1 (ω0 − ω1 )
) Solution
er si
w
For small body f . ∆t R2 = I2 ω2 From equations (i) and (ii), we get
ta
m ...(i)
w
.ja
For large body – f . ∆t R1 = I1 (ω1 – ω0)
rio
ar
s_
an
5
in
a.
Z
5
,
iv
Rotational Dynamics
2 f t0 mR
According to the problem, when the disc is in pure rolling motion, then vx = ωR
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Rotational Dynamics
or
or v0 −
2 f t .R f t0 = r 0 m mR
&0
3f t v0 = r 0 m
Y&
)
v0 2v0 = 3 3
(ii) From equation (iv)
ta
s
ta rio
un e/ .m
O
:// t
,
tp s
Z
Y&
1
ht
ft = −µmg v0t0 − m 2
er si
iv
s: //
ht tp t
1 f 0 = − f r v0t − . r t 2 2 m 0 2 r 0
fo
iv un e/ t
0 f Total work done = − f r ∫ v0 − r t dt m 0
l ml 2 cos θ ∆t = ω−0 2 12 where, m is the mass of the rod N
t.m
ht tp
= – fr vx dt Integrating, we have
s_
an ar m
s: //
w
Work done = fr dx
er si
v0 3µg
=
mv0 3µmg
w .ja
t0 =
w
∴
where, fr = μmg
in
a. co
m
mv0 mf r
t0 =
impulsive force acting on the rod is the normal reaction N acting at the end A of the rod. Since, torque of the impulsive normal reaction N about the centre of mass of the rod gives the clockwise rotation to the rod, the end A of the rod starts to move horizontally towards left just after impact. Further as the impact force N at A is vertical, the centre of mass will continue to move vertically after impact but with some new velo vc′ (say). For the rod from angular impulse momentum theorem about the axis passing through its centre of mass and perpendicular to the plane of figure (say z-axis) in the frame of the centre of mass of the rod: τz ∆t = ∆Lz
rio
vx = v0 −
) Solution During the course of impact, the only
..(iv)
From equation (ii)
T
$
v f ∴ r t0 = 0 m 3
1.55
N
l ml 2 cos θ ∆t = ω 2 12
or
...(i)
∴
From the linear impulse-momentum theorem, the translational motion of the rod is in the vertically downward direction.
or
Fy ∆t = ∆Py
– N ∆t = m v′c – mv
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34. A slender rod of length l forming an angle θ with the horizontal strikes a frictionless floor at A with its centre of mass velocity vc and no angular velocity. Assuming that the impact at A is perfectly elastic, find the angular velocity of the rod immediately after the impact.
$
1 = −µmg v0 t0 + µ 2 mg 2 t0 2 2
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...(ii)
Rotational Dynamics
Since, the point A moves towards left and the centre of mass moves vertically downward, the intersection point I of the normal drawn on the velocity of point A and centre of mass is the instantaneous centre of rotation of the rod just after impact. Hence, kinetic relation between vc′ and ω becomes
m v1 ...(i) 2 where, v′ and v1 are the velocities of the ball and sphere 1 respectively after collision (Remember that collision is head on). As the collision is perfectly elastic, from the definition of coefficient of restitution
co m
o s_ in f
a.
rio si ta
s rio rs ita
P
iv e
From equations (i) and (ii), we get 4v0 , directed towards right. 3 In the centre of mass frame of spheres 1 and 2.
e/
un
v1 =
→
→
P1 = − P2
s: //t .m
tp ht
O
iv er
un
e/
s: //t .m
tp
ht
P
Y
P
an ar
m
.ja
w
w
s: //w
tp
ht
35. A ball of mass m moving with velocity v0 experience a head-on elastic collision with one of the spheres of a stationary rigid dumb-bell as shown in the figure. The mass of each sphere equals m and the distance between them is l. 2 Disregarding the size of the spheres, find the proper angular momentum L of the dumb-bell after the collision, i.e., the angular momentum in the reference frame moving translationally and fixed to the dumb-bell’s centre of inertia. P
...(ii)
...(iv)
6vc cos θ l 1 + 3cos 2 θ
Y
v1 – v′ = v0
or
Solving equations (i) and (iv), we get ω=
v1 − v′ v0 − 0
1=
...(iii)
ωl N ∆t = m vc − cos θ 2
ωl cos θ 2 From equations (ii) and (iii), we get v′c =
mv0 = mv′ +
and
→
→
| P1 | = | P2 | = m | v1 − v2 |
1.56
→
→
→
→
Also, r1c = − r2 c
)
) Solution From conservation of linear
momentum along the direction of incident ball for the system consisting of colliding ball and sphere,
→
→ 1 m / 2 4v0 nˆ therefore, L = 2 2 2 3 where, nˆ is the unit vector in the sense of →
→
r1c × P1 Hence,
P
→
r1c ⊥ P1
As,
→
L = 2[r1c × P1 ]
thus
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→
L=
mv0l 3
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1.57
36. A small body of mass m tied to a nonstretchable thread moves over a smooth horizontal plane. The other end of the thread is being drawn into a hole O shown in the figure with a constant velocity. Find the thread tension as a function of the distance r between the body and the hole if r = r0. The angular velocity of the thread is equal to ω0.
Now, from second law of motion for m
T = F = mω2r Hence, the tension 2
37. One fourth length of a uniform rod of length 2l and mass m is placed on a horizontal table and the rod is held horizontal. The rod is released from rest. Find the normal reaction on the rod as soon as rod is released.
mω04 r04 ω r F = T = m 02 0 r = r3 r
co m
fo rio s_ ta
)
iv er
3 edge τ = mg 4
un
s: //
3 I0 α = mg 4
//t
s:
...(i)
3 I0 = I g + m 4
ht
tp
where,
ht
.m
tp
e/
⇒
=
2
4ml 2 9 2 43ml 2 + ml = 12 16 48 O
t.m
ht
si
) Solution Torque applied by gravity about the
→ As N = mg , the net torque of these two forces about any fixed point must be equal to zero. Tension T, acting on the mass m is a central force, which is always directed towards the centre O. Hence, the moment of force T is also zero about the point O and therefore, the angular momentum of the particle m is conserved about O. Let the angular velocity of the particle be ω. When the separation between hole and particle m is r, then from conservation of momentum about the point O
rio
ve ni
e/ u
s: tp
7
PJ
)
s
rs ita
w .ja w
//w
1
2
in
a. an ar
)
m
) Solution Forces acting on the mass m are shown in the figure.
O
O
Rotational Dynamics
*
O
m(ω0r0) r0 = m(ωr) r
ω=
From equations (i) and (ii), we get
2 0 0 2
ωr r
α=
or
O
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36 g 43l https://t.me/universitarios
...(ii)
1.58
Rotational Dynamics
rod about COM as shown below.
3l 27 g ×α = 4 43
D D[ %\
PJ
Putting this value in equation (iii), we get
Then, ∑Fx = 0 ∴ ax = 0
27 g 16mg N = mg − m = 43 43
ΣFy
ay =
rio
ita
\
Let ax and ay be the linear accelerations of COM and α the angular acceleration of the
s
rio
si ta
iv er
un e/ .m //t
s:
tp
$
39. A sphere of mass m and radius a rolls on other sphere of mass m′ and radius b which rolls on horizontal plane; friction is sufficient to prevent sliding and the motion is all parallel to the some vertical plane. Set up the equation to determine the motion of each sphere.
[ PJ
&
ht
%
D\
&
$
6 3g 4 mg and α = 13 L 13
/D
%
/
7
...(iii)
rs
iv e
un e/ t.m
)
/
τ TL cos30° 3 3T = = m(2L) 2 I 2mL 12
Now just after the string breaks, acceleration of point A in vertical direction should be zero. i.e., ay = Lα cos 30º Solving equations (i), (ii), (iii) and (iv) we get T=
s: // tp ht
) Solution Drawing free body diagram of rod immediately after the string breaks.
...(ii)
s_
an
ar
m
.ja
w
w
//w
s: ht tp
mg − T m
in f
a.
a=
m
=
o
m
and
co
38. A thin uniform bar of mass m and length 2L is held at an angle 30º with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after spring breaks.
...(i)
1
a=
where,
From newton’s second law of motion mg – N = ma ....(iii)
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1.59
Rotational Dynamics
&
Y
mg sinθ – fs = m (acc′(t) + ac′t) 1
IV
mg cosθ – N = m (acc′(n) + ac(n))
IV
2 d θ dv = m (a + b) − sin θ dt dt For rotational motion of C, τcz = Icαz
s_ i
ta rio
ht
Now for sphere C ac = acc′ + ac′ = acc′ (t) + acc′ (n) + ac′
where, acc1 (t ) = (a + b) gent) 2
d 2θ ut ′ (along the tandt 2
dθ acc ′ (n) = (a + b) un (along the normal) dt
....(vi)
un
e/
2 d ω′ f x b + f s ′ b = m′ b 2 ...(vii) 5 dt These seven equation are sufficient to determine the motion of each sphere or in other words the obtained equation are sufficient to calculate all the kinematical terms. We might alternately have written down the equation of energy for the whole system instead of equations of dynamics because neither friction forces nor normal reactions do work on system. Therefore
.m //t
tp
s:
...(ii)
...(v)
For the rotational motion of sphere C′ τc′z = Ic′ αz
.m
dθ dt
ta
e/
iv er si
N′ – m′g – Ncosθ – fs sinθ = 0
//t
s:
tp
ht
bw’ = – aw + (a + b)
dv dt
From the equation Fy = macy
un
s:
ht
Hence,
− f s ′ − Í sinθ + f s cos θ = m′
iv er si
m
.ja
w
//w
w
tp
vpc′ (t) = vpC′(t) + vcc′ (t)
a.
an
ar
and on the other hand vp = vpc′ + vc′ Also the point of contact P of two spheres have no relative tangential velocity i.e., the tangential velocity, vp(t) must be same for both the spheres. Hence, vpc′(t) + vc′(t) = vpc (t) + vcc′ (t) + vc(t) and
o
2 dω f s a = ma 2 ...(iv) 5 dt For sphere C′ corresponding of its linear motion Fx = macx
their velocities at an arbitrary time t. Let v denote the velocity of the sphere m′, and let C′ C makes an angle θ with the vertical at time t. As the lower sphere is pure rolling, v = bω′ ...(i) Also, vp = vpc + vc = vpc + v’cc’ + vc′
nf
)
) Solution Let C and C′ be the centres and ω, ω′
co m
1
s
P J
I V
From equation Fn = macn
PJ
T 1
d 2 θ dv = m (a + b) 2 + cos θ dt dt
rio
T
&
the centre for each sphere, we have for sphere C, for wave-linear motion: Ft = mact
3
Z
1 2 12 2 2 1 12 mvc + ma ω + m′vc2 + m′b 2 ω2 + mg (a + b) cos 2 2 5 2 2 5
dv i along1x – axis 12 1 12 mvc2 + ma 2 ω2 + m′vc2 + m′b 2 ω2 + mg (a + b) cos θ = constant dt 2 2 5 2 2 5 From the equations of dynamics for each But vc = vcc′ + vc′ ...(viii) sphere; resolving and taking moment about
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and ac =
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1.60
where,
Rotational Dynamics
vcc ′ = (a + b)
dθ and vc ′ = v dt
3VHXGRIRUFH PD
D D
and angle between vcc′ and vc′ equals θ 2
dθ ∴ vc 2 = (a + b) 2 + v 2 dt dθ + 2(a + b) v cos θ dt
I IULFWLRQ
...(ix)
in
io
ta r
ta
si
er
vr = at
4 2 = 2 3 = m/s (leftwards) 3 3
In the same time, truck has acquired a forward velocity. vt = a0t
ht
tp s
://
ht
t.m
tp
e/
s:
un
//t
iv
.m
ht tp
s:
e/ un
iv
rio
s
er
si
w .ja
//w
w
m
ar
s_
an
a.
fo
co
m
Drawing free body diagram of cylinder with respect to truck. ma0 – f = ma f a = a0 − ...(i) or Thus from equations (viii) and (ix), we have m 2 1 2 dθ 2 2f 2 τ 2 2 fR 2 dθ 21 2 m v + (a + b) + 2(a + b) v cos θ + a 2 ωAlso, m′ αv= + =b ω + =mg (a + b) cos θ ...(ii) 1 2 dt dt 5 2 5 I mR 2 mR 2 2 dθ dθ 2 2 2 1 2 2 2 2 Since cylinder rolls without slipping on the + 2(a + b) v cos θ + a ω m′ v + b ω + mg (a + b) cos θ truck, we have dt dt 5 2 5 a = Rα ...(iii) 1 2 2 2 2 Solving equations (i), (ii) and (iii), we get m′ v + b ω + mg (a + b) cos θ = costant ...(x) 2 5 2 2 2 a = a0 = (1) = m/s 2 40. A cylindrical pipe of diameter 1m is kept on 3 3 3 the truck as shown in the figure. If the truck With above acceleration, cylinder covers a now starts moving with constant acceldistance s = 4 m on the truck in time t given 2 eration of 2m/s , the pipe rolls backward by without slipping on the floor of the truck 1 12 and finally falls on the road. If the pipe or s = at 2 = t 2 2 2 3 moves a total length of 4m on the floor of the truck, find how much distance the pipe t2 For s = 4m, 4 = moves on the road before it finally stops. 3 The coefficient of friction between the pipe The linear velocity of the cylinder relative 2 and the road is 0.4 (g = 10m/s ) to truck is
= (1) 2 3 m/s (rightwards)
a0 = 1 m/s2 (towards right) Let ‘a’ be the linear acceleration of centre of mass of cylinder (towards left) with respect to the truck and α be its angular acceleration (anticlockwise) about its centre of mass.
or
)
) Solution Acceleration of truck
Hence, at the moment cylinder leaves the truck, it is moving with a linear velocity v = vr – vt 4 2 = −2 3 = − m/s 3 3
v=
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2 m/s (rightwards) 3 https://t.me/universitarios
Rotational Dynamics
1.61
nf o
$
s
[COM of semicircular ring lies at a dis-
2r from centre] π
ta rio
tance of
1 2 Since, P and M are the centres of mass of the two parts A and B. 1 2 2r 2 MO = PO = = = 0.32 m ∴ π π
) Solution Here, r = m = 0.5m.
e/ un
s: //t .m
Let C be the centre of mass of the whole ring. MP = 2(0.32) = 0.64 m
ht tp
ht
tp s: //
τ µ mgR 2µg 2 × 4 α' = = = = = 16 rad/s 2 I 1 mR 2 R 0.5 2 Linear velocity becomes zero in time t1 given by 2 1 v t1 = = 3 = s a′ 4 2 3
)
t.m
ht tp
and angular retardation,
%
iv er si
un i e/
s: //
w
f = µg = (0.40(10)) = 4 m/s 2 m
w
ve r
w
Friction acts backward. Linear retardation,
si ta
.ja m
I IPD[ PPJ
a' =
ar a
Z
P V
rio s_ i
Y
(c) friction force
na .c
om
Distance moved by cylinder during this time v 4 8 and its angular velocity, ω = r = = rad/s (anticlockwise) 2 R 3 × 0.5 3 v 4 8 2 ω= r = = rad/s(anticlockwise) (anticlockwise) rad/s R 3 × 0.5 3 v 2 3 1 = = m d= ′ 2 a 2(4) 6 Once the cylinder leaves the truck, there is no external torque on it about centre of mass 41. A ring shown in figure is made up of two axis of rotation. Hence, its angular velocity semicircular rings A and B of markes and linear horizontal velocity do not change. 2 kg and 4 kg respectively. The ring has Therefore, when it touches the ground, it does the diameter of 1m. The ring rolls withnot perform pure rolling. out slipping. Find 8 (a) angular acceleration ω= rad/s 3 (b) normal reactions and
and angular velocity becomes zero in time t2 given by 0
8 1 ω t2 = = 3 = s α’ 16 2 3
3
D
4
1 Then, MC = (0.64) = 0.213 m 3
Since, t1 = t2 i.e., linear velocity and angular velocity become zero simultaneously, hence, 1 s. the cylinder stops in time, t = 2 3
&2
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1.62
Rotational Dynamics
Now, (mA + mB) ay = ΣFy
∴
QC = (OQ) 2 + (CO) 2
or
(mA + mB) a sin θ = 60 cos 30º – N
∴ Normal reaction, N = 60 cos 30º – (mA + mB)a sin θ
CO = 0.32 – 0.213 = 0.107 m
∴
2
2
= (60) (0.86) – (2 + 4) (3.5) sin 11.9º
= (0.5) + (0.107) = 0.511 m
Force diagram is shown below:
∴ Frictional force, f = 60 sin 30º – (mA + mB) a cos θ
4
:HLJKW 1
co m
= (60) (0.5) – (2 + 4) (3.5) cos 11.9º n
fo
in
rio s_
m
ita
ita
rs ve un i
P
$
)
Let α be the angular acceleration of ) Solution 2 = 6.84 rad/s the rod. Acceleration of B with respect to A is 1.2 α at an angle of 60º with horizontal as shown in figure: Using sine law, we get aBA aB aA = = sin 45° sin 60° sin 75°
ht
tp s:
1 (2 × 4 + 2 × 2) 2 (0.5) + (60) (0.86) (0.107) 2 = 6.84 rad/s 2 1 2 × 4 + 2 × 2) 2 (b) Acceleration of centre of mass, C: a = (QC) (α) in the direction shown in figure i.e., perpendicular to QC a = (0.511) (6.84) m/s2 = 3.5 m/s2
//t .
2
e/
s:
tp
ht
(60) (0.5) (0.5) + (60) (0.86) (0.107)
rio
s
rs ve un i
e/
(60sin 30°) (QO) + (60cos30°) (CO) (2mA + 2mB )r 2
=
(a) the angular acceleration of rod (b) the reactions at A and B (g = 9.8 m/s2) %
//t .m
ht tp
s:
//w
w
F1 = 60 cos 30° (a) The ring will rotate about point Q. τ angular acceleration, α = Q IQ
=
42. The extremities of a 1.2m rod weighting 25 kg can move freely and with no friction along two straight tracks as shown. If the rod is released with no initial velocity from the position shown, determine
m
.ja w
θ = 11.9. W = 60 newton F1 = 60 sin 30°
a. ar an
∴
∴ = 9.45 ewton
QO 0.5 cos θ = = CQ 0.511
D
:
or (mA + mB)a cos θ = 60 sin 30º – f
1
)
T T
)
(c) (mA + mB)ax = ΣFx
2
&
n
= 47.27 ewton
D%
[
D
2
T T
T
LQT
DV
D$
$
4
or aA = 1.64 α and aB = 1.47 α
VT
R DF
&
*
%
\
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1.63
Rotational Dynamics
aG = aA = aGA
% *
1%
Therefore, acceleration of G can now be obtained by
$
Z 1
1$
D$
I
D%
I
%
D%$ D
NB = 7.37 α
...(iv)
or
\
NB sin 45º = (25) (1.34 α)
or
rio s_ in fo
o D*$ D
\
Now ΣFx = max
[
na .c om
o D$
$
D\
%
D[
D
aG into x and y components, we
Resolving have
[
*
iv uestions
un type
t.m e/
L = mv h
Putting, h=
(d) m 2 gh3
(∵ r sin θ = h)
⇒
where, θ = angle between v and r
s: //
Angular momentum the particle about O L = mvr sin θ
v02 sin 2 θ v02 sin 2 45° v02 = = , we get 2g 2g 4g L=
Putting,
v = 4 gh , we get L = m 2 gh3
Speed of the particle at the top = horizontal component of the speed of projection. v v = v0 cos θ0 = v0 cos 45° = 0 ⇒ 2
mv02 4 2
)
) Explanation
Q
mv 4 2g
ht tp
(c)
er si ta rio s
er s
un iv
e/
.m
ht tp s:
//t
...(iii)
1. Particle of mass m is projected with a velocity v0 making an angle of 45º with horizontal. The magnitude of angular momentum of the projectile about the point of projection at its maximum height is mv 3 (a) Zero (b) 2g 2 0
ita
.ja
w
w
//w
s:
tp
ht
= 0.173 NA – 0.19 NB
SOLVED OBJECTIVE
ar a
m
Also ΣFy = may ax = 1.64 α – 0.6 α cosº = 1.34 α ...(i) or 245 – NA – NB cos 45º = (25) (0.52 α) N and ay = 0.6 α sin 60º = 0.52 α N A...(ii) (0.6cos30°) − B (0.6cos30° + 0.6sin 30°) or ...(v) τ 2NA + 0.707 NB = 245 – 13 α Free body diagram of rod is α = G = 2 (25)equations (1.2) IG Solving (iii), (iv) and (v), we get 12 N α = 2.33 rad/s2 N A (0.6cos30°) − B (0.6cos30° + 0.6sin 30°) τ 2 and NA = 137 N, NB = 110 N α= G = (25) (1.2) 2 IG 12
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1.64
2. A homogenous rod AB of length L and mass M is pivoted at the centre O in such a way that it can rotate freely in a vertical plane. The rod is initially in a horizontal position. An insect S of the same mass falls vertically with speed V at the point C, midway between the points O and B and stick to the rod. Immediately after the insect fall, the angular velocity of the system is
7 D
5
PJ
) Explanation Let a be the acceleration of the )
6
cylinder. Then mg – T = ma If α be the angular acceleration, then
...(i)
mR 2 T. R. = I α = α 2
...(ii)
m
ita
%
/
mg (b) g and 2 g mg (d) and 2 3
3 a up the plane from 2 a from the centre 2 3
(d) at a distance down the plane 1 '
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P1
& 2
%
$
PJVLQD
2g mg and (a) 3 3 g mg (c) and 3 2
3. A string is wrapped several times round a solid cylinder and then the end of the string is held stationary while the cylinder is released from rest with an initial motion. The acceleration of the cylinder and tension in the string will be
3 a up the plane from 2
ht tp
&
(b) at a distance the centre (c) at a distance the centre
s: //
t.m e/
12 V 7 L
2
$
4. A cube in limiting equilibrium on an inclined plane forming an angle of 30º with the horizontal. The line of action of the normal reaction of the plane on the cube is (a) at the centre of the cube
un
//t
ht tp s:
ω=
2 mg g and T = 3 3
iv
.m
ht
∴
e/
s:
tp
)
ML ML MV = + ω 4 3
⇒
un iv
L2
a=
er si ta rio s
er s
w w //w
ML2
L
+ M ω ) Explanation (MV) = 4 12 16
As the string unwinds without slipping a=Rα ...(iii) Solving these equations, we get
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Rotational Dynamics
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Rotational Dynamics
)
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equilibrium mg sin 30º = µN and mg cos 30º = N 1 or µ = tan 30º = 3
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6. A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is
5. A uniform rod of length l and mass M is suspended on two vertical inextensible string as shown in the figure. Calculate tension T in left string at the instant, when right string snaps.
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7. A horizontal disc rotes freely about a vertical axis through its centre. A ring, having the same mass and radius as the
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zero.
and the acceleration of the centre of mass of the rod downwards be a.
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∴
Ratio of heat produced to initial kinetic E − E2 energy = 1 E1
disc, is gently placed on the disc. After some time, the two rotate with a common angular velocity. (a) Some friction exists between the disc and the ring. (b) The angular momentum of the ‘disc plus ring’ is conserved. 2 rd of (c) The final angular velocity of 3 the initial angular velocity of the disc. 2 rd of the initial kinetic energy (d) 3 changes to heat.
Rotational Dynamics
=
8. A long horizontal rod has bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ and gravity is neglected, then the time after which the bead starts slipping is
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1 Initial kinetic energy, E1 = I1ω12 2 1 Final kinetic energy, E 2 = (I1 + I 2 )ω2 2 2 Heat produced = less in kinetic energy = E1 – E2
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Rotational Dynamics
(d)
e1 + 1 2 gh e2 − 1
e1 + 1 2 gh e2 + 1
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W Z W
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the axis of rotation on the system of platform and tortoise, angular momentum will remain unchanged. As the tortoise moves moment of inertia of system will first decrease and then increase. Hence, angular velocity will first increase and then decrease. Also variation will not be linear.
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10. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform ω (t) will vary with time t as
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1. A semicircular lamina of mass m has radius r and centre C. Its centre of mass is at a distance x from C. Its moment of inertia about on axis through its centre of mass and perpendicular to its plane is 1 1 (a) mr 2 (b) mr 2 2 4 1 1 (c) mr 2 + mx 2 (d) mr 2 − mx 2 2 2
exercise
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2. A uniform rod of length 2 ℓ is placed with one end in contact with the horizontal table and is then inclined at an angle α to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be
unsolved
OBJECTIVE TYPE QUESTIONS
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mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is 1 1 (b) MR 2 (a) MR 2 4 2
3 g sin α (a) ω = 2
(d)
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3. A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear velocity v0 and angular velocity ω0 as shown. The disc comes to rest after moving some distance to the right. It follows that (a) 3 v0 = 2ω0 r (b) 2 v0 = ω0 r (c) v0 = ω0 r (d) 2 v0 = 3 ω0 r
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4. A disc of radius R rolls without slipping with a velocity v0. Find the speed of the point A on the rim of disc at the moment shown.
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6. Two particles A and B are moving as shown in the figure. At this moment of time the angular speed of A with respect to B is
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Rotational Dynamics
5. One quarter sector is cut from a uniform circular disc of radius R. This sector has
9
7. A sphere kept on a rough inclined plane is in equilibrium by a string wrapped over it. If the angle of inclination is θ, the tension in the string will be equal to 2mg (a) mg sin θ (b) 5 mg sin θ (d) none of these (c) 2
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Rotational Dynamics ;
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;
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(d)
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13. A stick of length ℓ and mass m lies on a frictionless horizontal surface on which it is free to move in any direction. A ball of same mass m moving with speed v0 perpendicularly to this length and collide at the end of stick. Find the angular velocity when the collision is perfectly inelastic.
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(c)
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10. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX’ is ρL3 8π 2
(d) 3
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12. A string is wrapped around a cylinder of radius R. If the cylinder is realased from rest, the velocity of the cylinder, after it has moved h distance will be
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9. As shown in figure, a small block of mass m is at rest with respect to a smooth hemispherical bowl of radius R rotating with angular velocity ω about its axis. Find the radius of the circle (r) over which block is moving as seen from ground.
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11. A particle P is placed inside a hemispherical bowl which rotates about its vertical axis with a constant velocity ω. It is just prevented from sliding down when OP is inclined at 45º with the axis, O being the center of the bowl. Radius of bowl = 10 2 and µ, the coefficient of friction between the particle and bowl is 0.5, g = 10 m/s2. The value of ω is 1 1 (b) (a) 3 3
8. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending.
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1.70
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18. A solid sphere is rolling on a rough surface, whose centre of mass is at C at a certain instant. Find at that instant it has angular velocity ω. Its radius is R. Find the angular acceleration at that instant mass of sphere is m. (b)
5 gd 7 R2
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15. A spherical body of mass m, radius r and moment of inertia I about its centre moves along the x-axis. Its centre of mass moves with velocity = v0, and it rotates about its centre of mass with angular velocity = ω0.
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17. The surface density of a thin circular disc of radius ‘a’ depends on the distance from the centre as f(r) = A + Br, where A and B are constants. Its moment of inertia about the line perpendicular to the plane of the disc through its center is
14. A hollow sphere of mass 2 kg is kept on a rough horizontal surface. A force of 10 3 N is applied as shown in the figure. Find the minimum value of ‘µ’ so that the sphere starts pure rolling. (Take g = 10m/s2) (a) 0.3 (b) 0.2 (c) 0.1 (d) for no value of µ it will start pure rolling
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16. A sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotav tional velocity about the centre 0 . The 2r rotational velocity when the sphere starts pure rolling is v v (b) 0 (a) 0 3 3 v0 2v0 (c) (d) 5 7
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Rotational Dynamics
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(d) none of these
ω 2
ω 3
22. A uniform circular disc of radius r placed on a rough horizontal surface has initially a velocity v0 and angular velocity ω0 as shown in the figure. The disc comes to rest (neither translates nor rotates) after v moving some distance. Then 0 is rω0 1 (a) 1 (b) 2 3 (d) 2 (c) 2
(c)
(b)
(a) ω
1.71
19. A disc is freely rotating with an angular speed ω on a smooth horizontal plane. If it is hooked at a rigid pag P and rotates without bouncing. Its angular speed after the impact will be equal to
Rotational Dynamics
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21. An this figure a body is moving with constant velocity v = 3m/s along x-axis and also rotating about is centre with angular velocity ω = 4 rod/sec. Find the co-ordinate of the axis of pure rotation when centre of the body is 4 m from the origin.
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23. A cylinder of radius 10 cm rides between two horizontal bars moving in opposite direction as shown in the figure. The location of the instantaneous axis of rotation and the angular velocity of the roller are respectively (There is no slipping at P or Q) (a) 8 cm from Q, 1.25 rad/s2 (b) 8 cm from Q, 2.50 rad/s2 (c) 15 cm from Q, 1.25 rad/s2 (d) 15 cm from Q, 2.50 rad/s2
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24. A solid sphere of radius R is rolled by a force F acting at the top of the sphere as shown in the figure. There is no slipping and initially sphere is in the rest position, then
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20. A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear velocity v0 and angular velocity ω0 as shown. The disc comes to rest after moving some distance to the right. It follows that (a) 3 v0 = 2ω0 r (b) 2v0 = ω0 r (c) v0 = ω0 r (d) 2v0 = 3ω0 r
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(a) H1 = H (b) H1 > H (c) H1 < H (d) can not be decided
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28. Three mass particle A, B and C having masses m, 2m and 3m respectively are rigidly attached to a ring of mass m and radius R which rolls on a horizontal surface without slipping. At a certain instant the velocity of the centre of the ring is v0 as shown in the figure. The kinetic energy of the system is (a) 7mv02 (b) 9mv02 7 (c) mv02 (d) mv0 2 2
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25. A disc of mass M and radius R moves in the x–y plane as shown in the figure. The angular momentum of the disc at the instant shown is
27. A rod of mass M is dropped on a horizontal smooth surface from the position shown in the figure. It collides elastically the surface. After collision with the surface, the center of mass of rod rises to a height H1. Then
(a) work done by force F when the centre of mass moves a distance S is 2FS. (b) speed of the c.m. when c.m. moves a 20 RS distance S is 7 M (c) Work done by the force F when c.m. moves a distance S is FS (d) speed of the c.m. when c.m. moves a 4RS distance S is M
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Rotational Dynamics
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26. A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. A horizontal impulse P is applied to the end B perpendicular to the rod. The time taken by the rod to turn through a right angle is
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29. A triangular block ABC of mass m and sides 2a lies on a smooth horizontal plane as shown. Three point masses of mass m each strikes the block at A, B and C with
2
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Rotational Dynamics
speeds v as shown. After the collision the particle come to rest. Then
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30. A uniform rod is released from its vertical position with its bottom on a smooth surface. If v is the speed of centre of mass of the rod and ω is its angular velocity when it makes an angle θ with the horizontal, then
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2. 10. 18. 26.
(a) (d) (d) (a)
3. 11. 19. 27.
(b) (b) (b) (c)
4. 12. 20. 28.
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ANSWERS
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(a) the centre of mass of ABC remains stationary after collision (b) the centre of mass of ABC moves with a velocity v along x-axis after collision (c) the triangular block rotates with an 2 3mva angular velocity ω = about its I centre of mass (here I is the moment of inertia of triangular block about its centroid axis perpendicular to its plane)
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5. 13. 21. 29.
(a) (b) (c) (b, c)
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6. 14. 22. 30.
(d) (b) (d) (c)
7. (c) 15. (d) 23. (a)
8. (b) 16. (b) 24. (a, b)
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Rotational Dynamics
EXERCISE 2 cylinder is less than 1/3 tan θ, where, θ is the angle of inclination of the plane with the horizontal.
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7. A weightless thread with a body of mass m tied to its end is wrapped around a uniform solid cylinder of mass M and radius R shown in the figure At t = 0, the system is set in motion. Find: (a) angular velocity of the cylinder, and (b) kinetic energy of the whole system, as a function of time. There is no friction anywhere.
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4. Three forces are applied at the three corners of a square plate, as shown in figure. Find the resultant force (both in magnitude and direction) and its point of application, if this point is taken on the side BC. )
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3. A uniform rod AB of mass m = 1.0 kg moves linearly with an acceleration a = 2.0 ms2 due to two anti-parallel forces F1 and F2 = 5.0 N. The distance between the points of application of these two forces is equal to l = 0.2 m. Find the length L of the rod. [Ans. 1.0 m]
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2. A solid sphere starting from rest rolls for 5.3 second without slipping along a smooth inclined plane 1 m in length, the angle of inclination being θ = 0.573º. What is the acceleration due to gravity at that place? [Ans. 9.8 m/s2]
6. A disc of radius R/2 is cut from a uniform disc of radius R = 0.2 m, as shown in the figure. The mass of the remaining shaded portion of the disc is m = 7.3 kg. Find the moment of inertia of the shaded portion of the disc about an axis passing through its centre of mass and perpendicular to its plane. [Ans. 0.15 kg-m2]
1. Four particles each of mass m are placed at the four corners of a square of edge a. Find the moment of inertia of the system about an axis perpendicular to the plane of the square and passing through its centre. [Ans. 2ma2]
unsolved
1.74
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5. Show that a cylinder will slip on an inclined plane if the coefficient of static friction between the plane and the
R
5 7 7
[Ans. Resultant will be 2F acting at the middle point of AB and parallel to the diagonal AC]
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Rotational Dynamics
cal sections of the threads, assuming that the thread does not slip over the pulley and the friction is absent.
7
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3 v2 , what is the M.I. of the body? 4 g What shape(s) it may have?
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12. A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16 kg and length is 3 metre. Two particles, each of mass 0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity of 10 m/s and the other with 6 m/s, as shown in the figure. The first particle strikes the bar at the point A and the other at point B. Points A and B are at a distance of 0.5 m from the centre of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. Calculate the loss of kinetic energy of the system in the above collision process. [IIT, 1989]
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11. A square plate of mass M = 120 g and of side a = 5.0 cm is rotated about one of its sides as axis. Find the torque acting on the plate, if its angular acceleration is α = 0.2 rad/s2. [Ans. 2.0 × 10–5 N–m]
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m1 ( 4m2 + M ) Ans. m2 ( 4m1 + M )
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9. A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has an elevated section and a horizontal part. The horizontal part is 1.0 m above the ground level and the top of the track is 2.4 m above the ground. Find the distance on the ground with respect to the point B (which is vertically below the end of the track as shown in the figure) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centre of mass? [IIT, 1987]
7
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If h =
MR 2 Ans. I = 2
5
8. A body of mass M and radius R is rolling on a smooth horizontal plane without slipping with a velocity v. In the course of its rolling, it meets a rising slope and rolls up to a height h before stopping.
gt 1 mg 2t 2 Ans. (a) , (b) M 2 M + 1 R 1 + 2m 2m
[Ans. 1.99 m] 10. As shown in the figure, the mass of the pulley of radius R is M, and the masses of the two blocks are m1 and m2. Find the angular acceleration of the pulley and T the ratio of the tensions, 1 , in the vertiT2
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[Ans. v = 4 m/s ; ∆E = 2.76 joules]
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1.76
13. A uniform cylinder of radius R, rotating about its axis with an angular velocity ω, is placed in a corner, as shown in the figure. The friction coefficient between the corner walls and the cylinder is µ. Find the number of turns the cylinder will complete before it stops.
without slipping on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of cylindrical part of the carpet when its radius reduces to R/2. [IIT, 1990]
Ans.
17. A uniform solid cylinder of mass m = 8.0 kg and of radius r = 1.3 cm as shown in the figure starts falling at time t = 0 due to gravity. Find: (a) tensions in the threads and the angular acceleration of the falling cylinder, and (b) instantaneous power developed by the force of gravity as a function of time. Take mass of the thread negligible.
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Rotational Dynamics
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18. A uniform disc of mass M and radius R is initially at rest with its axis fixed through the point O as shown in the figure. A block of mass m initially moving with speed v1 on a frictionless surface passes over the disc to the dotted position. When it first makes contact with the disc, it slips on it. But the friction between the block and the disc is large enough so that slipping ceases before the block loses contact with the disc. Show that the final velocity of the block is given by v1 v2 = M 1+ 2m
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19. A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v, respectively, strike the bar as shown in
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15. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Another disc M of same dimension but of mass is 4 placed gently on the first disc. Find the angular velocity of the combined system. [IIT, 1986]
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Rotational Dynamics
force constant k of the spring is 3.0 N/m. If the system is released from rest at a position in which the spring is stretched 0.25m, find the translational and rotational kinetic energy of the cylinder as it passes through the equilibrium position.
the figure and stick to the bar after collision. Denoting the angular velocity (about the centre of mass), total energy and centre of mass velocity by ω, E and vc respectively, find the total energy of the system after the collision. P D
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27. A man of mass M stands on the edge of a horizontal uniform disc of mass m and of radius R, which can rotate freely about a vertical axis passing through its centre. At time t, he starts moving along the edge of the disc, and after moving through an angle θ relative to the disc, he stops. In the process of motion, his velocity with time varies as v(t). Assuming m << M, find: (a) angle through which the disc had turned by the time the man stopped, and (b) moment of the force about the rotation axis with which the man acted on the disc in the process of motion.
(a) Determine the angular velocity ω in terms of v and l. (b) If the insect reaches the end B, when the rod has rotated through an angle of 90º, determine v. [IIT, 1992]
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26. A thin uniform rod of mass m and of length l rotates about a vertical axis with uniform angular velocity ω. The upper end of the rod is hinged. Find the angle θ that the rod makes with the vertical.
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(a) the force of friction on each disc. (a) the angle θc through which the cylinder rotates before it leaves contact with the (b) the magnitude and the direction of the edge. frictional torque acting on each disc about the centre of mass O of the object. (b) the speed of the centre of mass of the Express the torque in the vector form in cylinder before leaving contact with the edge, and terms of unit vectors iˆ, ˆj and kˆ in the (c) the ratio of the translational to rotax, y and z directions. [IIT, 1997] tional kinetic energies of the cylinder when its centre of mass is in horizontal Ans. (a ) 6Niˆ, (b) 0.6 2 N − m line with the edge. [IIT, 1995] 31. A wedge of mass m and triangular cross6 mgr = BC = CA = 2R) is moving E 4 4 section (AB Ans. (a) θc = cos −1 , (b) v = gR , (c) T = 7 = 6 velocity −viˆ towards 7 E R 1with mgra constant 7 radius R fixed on a smooth a7 sphere of horizontal table as shown in the figure. 6 mgr ET 4 1 4 7 The wedge makes an elastic collision gR , (c) = = 6 , (b) v = 1 7 7 E with the fixed sphere and returns along mgr R 7 the same path without any rotation. Neglect all friction and suppose that the 30. Two thin circular discs of mass 2 kg and wedge remains in contact with the sphere radius 10 cm each are joined by a massfor a very short time ∆t, during which less rod of length 20 cm. The axis of the the sphere exerts a constant force F on rod is along the perpendicular to the the wedge. planes of the discs through their centres as shown in the figure. This object is kept (a) Find the force F and also the normal on a truck in such a way that the axis of force N exerted by the table on the the object is horizontal and perpendicular wedge during the time ∆t. to the direction of the motion of the truck. (b) Let h denote the perpendicular distance Its friction with the floor of the truck is between the centre of mass of the large enough so that the object can roll on wedge and the line of action of F . Find the truck without slipping. Take x-axis as the magnitude of the torque due to the the direction of motion of the truck and normal force N about the centre of the z-axis as the vertically upward direction. wedge, during the internal ∆t. If the truck has an acceleration of 9 m/s2, [IIT, 1998] calculate
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9. Why there are two propellers in a helicopter? 10. You are given two circular discs of equal masses and thickness but made of different materials. Which one will have a larger moment of inertia about its central axis and why?
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12. The cap of the pen can easily be opened with the help of two fingers than with one finger. Why?
(a) 0 m/s2 (c) 10 m/s2
11. How can you distinguish between a hard boiled egg and a raw egg by spinning them on a table top?
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8. A man standing on the turn table raises his hands suddenly. What happens?
7. How a swimmer jumping from a height, is able to increase the number of loops made in the air?
3. In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1 m respectively and the acceleration of block A is 5 m/s2 in the downward direction, then the acceleration of block B will be
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6. A person sits near the edge of a circular platform revolving with a uniform angular speed. What will be the change in the motion of the platform? When the person starts moving from the edge towards the centre of the platform?
2. Two points of a rod move with velocities 3 v and v perpendicular to the rod and in the same direction, separated by a distance r. Then the angular velocity of the rod is (a) 3 v/r (b) 4 v/r (c) 5 v/r (d) 2 v/r
5. A thin wheel can stay upright on its rim for a considerable length of time when rolled with a considerable velocity, while it falls from its upright position at the slightest disturbance, when stationary. Explain.
4. Why it is more difficult to revolve a stone by tieing it to a longer string than by tieing it to a shorter string?
1. The moment of inertia of a door of mass m, length 2ℓ and width ℓ about its longer side is 11m 2 5m 2 (a) (b) 24 24 2 m (c) (d) None of these 3
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13. Why the speed of whirlwind in a tornado is alarmingly high? ONLY ONE OPTION IS CORRECT
2. Why spokes are fitted in the cycle wheel?
1. A body is rotating. Is it necessarily being acted upon by an external torque?
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5. Consider the following two statements (A) Linear momentum of the system remains constant. (B) Centre of mass of the system remains at rest. (a) A implies B and B implies A (b) A does not imply B and B does not imply A (c) A implies B but B does not imply A (d) B implies A but A does not imply B
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6. A sphere is released on a smooth inclined plane from the top. When it moves down its angular momentum is (a) conserved about every point (b) conserved about the point of contact only (c) conserved about the centre of the sphere only (d) conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball. 7. A ball hits a fixed surface with a velocity u at an angle θ with the normal to the surface at the point of impact and rebounds from it at an angle β with the surface. Then (A) β ≤ θ (B) β ≥ θ tan θ (C) Coefficient of restitution = tan β tan β (D) Coefficient of restitution = tan θ (a) B, C (b) A, D (c) C, D (d) A, B
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One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is (a) 3 v cos θ (b) 2 v cos θ (c) 3/2 v cos θ (d) ( 3/2 v cos θ)
The coordinates of point of application of forces is given by
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17. A rotating star has a period of 30 days about an axis passing through its centre. The star undergoes an internal explosion and converts to a neutron star. Initial radius of the core was 1.0 × 104 km, whereas, final radius is 3.0 km. Determine the period of rotation of the neutron star. (a) 2.7 × 10–6 days (b) 1.7 × 10–6 days –6 (c) 4.1 × 10 days (d) 5.2 × 10–6 days
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16. In the above question, the distance the sphere will have moved relative to the ground at time t1 is
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12. A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc
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22. In above question find the condition when bobbin moves to right (a) R sin α = r (b) R sin α > r (c) R sin α < r (d) None of these
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21. The free end of a thread wound on a bobbin is passed round a nail A hammered into the wall. The thread is pulled at a constant velocity. Assuming pure rolling of bobbin, find the velocity v0 of the centre of the bobbin at the instant when the thread forms an angle α with the vertical. (a)
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Find minimum friction coefficient between floor and ladder for this equilibrium (a) 2/3 (b) 1/2 (c) 3/4 (d) 1/3
24. Two points of a rod move with velocities 3 v and v perpendicular to the rod and in the same direction, separated by a distance ‘r’. Then the angular velocity of the rod is 3v 2v (b) (a) r r 5v 4v (d) (c) r r 25. A uniform disc of radius R lies in the x-y plane, with its centre at origin. Its moment of inertia about z-axis is equal to its moment inertia about the line y = x + c. The value of c will be R (a) –R/2 (b) ± 2 (c) +R/4 (d) –R
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35. A dancer on ice spins faster when she folds her hand. This is due to (a) Increase in energy and increase in angular momentum (b) decrease in friction at the skates (c) constant angular momentum and increase in kinetic energy (d) increase in energy and decrease in angular momentum
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37. Two discs of same mass m and same thickness t are made from two different materials of densities d1 and d2 respectively. The ratio of the moment of inertia of the two about an axis passing through the centre and perpendicular to the plane of disc is (a) d1 : d2 (b) d2 : d1 (c) (d1/d2) : 1 (d) 1 : d1 d2
41. For the same total mass which of the following will have the largest moment of inertia about an axis passing through the C.G. and perpendicular to the plane of the body (a) a disc of radius a (b) a ring of radius a (c) a square lamina of side 2a (d) four rods forming a square of side 2a
36. A uniform rod of mass 2 kg is 1 metre long. Its radius of gyration about an axis through its one end and perpendicular to its length in SI units is (b) 1/2 3 (a) 1/ 3 (c) 2/3 (d) 2/ 3
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52. The moment of inertia of a body about a given axis of rotation depends upon (a) the distribution of mass (b) distance of the body from the axis of rotation (c) shape of the body (d) all of the above
49. When an explosive shell travelling in a parabolic path under the gravity explodes, the centre of mass of the fragments will continue to move (a) along a hyperbolic path (b) vertically upwards and then vertically downwards (c) along the original parabolic path (d) horizontally and then follows the parabolic path
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47. A solid iron sphere A rolls down an inclined plane, while an identical sphere B slides down the plane in a frictionless manner. At the bottom of the inclined plane, the total kinetic energy of sphere A is (a) less than that of B (b) equal to that of B (c) more than that of B (d) sometimes more and sometimes less
46. When a sphere of moment of inertia I about its centre of gravity and mass m rolls from rest down an inclined plane without slipping, its kinetic energy is 1 1 (b) mv 2 (a) Iω2 2 2 1 2 1 2 (c) I ω + mv (d) I ω + mv 2 2
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45. A train of mass M is moving on a circular track of radius R with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be (a) 0 (b) 2 MV/π (c) MVR (d) MV
Rotational Dynamics
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64. If the force applied on a particle is zero then the quantities which are conserved are (a) only momentum (b) only angular momentum (c) momentum and angular momentum (d) only potential energy
65. A string is wrapped several times round a solid cylinder and then the end of the string is held stationary while the cylinder is released from rest with no initial motion.
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59. When a sphere of moment of inertia I rolls down an inclined plane, then the percentage of its rotational kinetic energy is (a) 100% (b) 50% (c) 28% (d) 72%
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62. A solid sphere, a solid cylinder, a disc and a ring are rolling down an inclined plane. Which of these bodies will reach the bottom simultaneously (a) solid sphere and solid cylinder (b) solid cylinder and disc (c) disc and ring (d) solid sphere and ring 63. The angular velocity of a body is ω = 2iˆ + 3 ˆj + 4kˆ ω = 2iˆ + 3 ˆj + 4kˆ and a torque τ = iˆ + 2 ˆj + 3kˆ acts on it. The rotational power will be (a) 20 watt (b) 15 watt (c) 17 watt (d) 14 watt
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58. A fly wheel is so constructed that almost whole of its mass is concentrated at its rim, because (a) it increases the moment of inertia of the fly-wheel (b) it decreases the moment of inertia of the fly-wheel (c) it increases the speed of the fly-wheel (d) it increases the power of the fly-wheel
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61. A ring, a disc, a sphere and a spherical shell are simultaneously released to roll down from the top of an inclined plane of height h. The four bodies will reach the bottom in the following order (a) sphere, disc shell and ring (b) ring, spherical shell, disc and sphere (c) sphere, spherical shell, disc and ring (d) sphere, spherical shell, ring and disc
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57. The ratio of kinetic energies of two spheres rolling with equal center of mass velocities is 2 : 1. If their radii are in the ratio 2 : 1, then the ratio of their masses will be (a) 2 : 1 (b) 1 : 8 (c) 1 : 7 (d) 2 2 :1
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56. The moment of inertia of a circular disc of mass 200 gm and radius 5 cm about a tangential axis normal to the plane of disc will be (a) 750 g cm2 (b) 7500 g cm2 2 (c) 75 g cm (d) zero
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60. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The velocity of its center of mass at the bottom of the plane will be:
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55. A long thread is wrapped round a reel. If one end of thread is held in hand and the reel is allowed to fall under gravity, then the acceleration of the reel will be (a) g (b) (2/3)g (c) (3/2) g (d) zero
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54. The moment of inertia of a hollow sphere of mass 1 kg and inner and outer diameters 0.2 and 0.4 meter respectively about its diametric axis will be (a) zero (b) 0.177 kg m2 (c) 0.0177 kg m2 (d) 177 kg m2
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Rotational Dynamics
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Rotational Dynamics
71. A small block of mass ‘m’ is rigidly attached at ‘P’ to a ring of mass ‘3 m’ and radius ‘r’. The system is released from rest at θ = 90º and rolls without sliding.
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67. A uniform rod of length 2 ℓ is placed with one end in contact with the horizontal table and is then inclined at an angle α to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be 3 g sin α (a) ω = 2
70. A metre stick is held vertically with one end on the floor and is then allowed to fall. The speed of the other end when it hits the floor assuming that the end at floor does not slip, is (g = 9.8 m/sec2) (a) 3.2 m/sec (b) 5.4 m/sec (c) 7.6 m/sec (d) 9.2 m/sec
66. A string is wrapped around a cylinder of mass m and radius R. The string is pulled vertically upward to prevent the centre of mass from falling as the cylinder unwinds the string. The length of the string unwound when the cylinder has reached a speed ω will be R 2 ω2 Rω (b) (a) 4g 4g R 2 ω2 Rω (d) (c) 8g 8g
The acceleration of the cylinder and tension in the string will be 2 mg mg (a) g and (b) g and 3 3 2 g mg g mg and and (d) (c) 2 3 3 2
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68. A mass M is supported by a massless string wound round a uniform cylinder of mass M and radius R. On releasing the system from rest, the acceleration of mass M is (a) g (b) g/2 (c) (2/3) g (d) depends on R
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73. A disc of mass M and radius R can rotate freely about a horizontal shaft at S distant r
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69. Three thin uniform rods each of mass M and length L are placed along the three axis of a cartesian coordinate system with one end of each rod at the origin. The M.I. of the system about z-axis is (a) ML2/3 (b) (2/3)ML2 2 (c) ML /6 (d) ML2
The value of ω for which the force exerted by the hinge on rod is O in this position is
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from the centre of mass of the disc. The disc is released from the position shown in the figure, its angular acceleration will be
76. A force of (2iˆ − 4 ˆj + 2kˆ) Newton acts at a point (3iˆ + 2 ˆj − 4kˆ) metre from the origin.
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77. A thin wire of length ℓ and mass m is bent in the form of a semicircle.
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Rotational Dynamics
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Its moment of inertia about an axis joining its free ends will be (a) zero (b) mℓ2 m m 2 (c) 2 (d) π 2π 2
78. Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be
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On releasing from rest the block falls through 5 m height in 2 s. The moment of inertia of the wheel will be (a) 1 kg m2 (b) 3.2 kg m2 2 (c) 2.5 kg m (d) 1.5 kg m2
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75. If the rotational kinetic energy of a body is increased by 300% then the percentage increase in its angular momentum will be (a) 600% (b) 150% (c) 100% (d) 1500%
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79. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will take same time.
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4. A rod AC of length ℓ and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes rod at point B making angle 37º with the rod. The collision is elastic. After collision
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3. A ring rolls without slipping on the ground. Its centre C moves with a constant speed u. P is any point on the ring. The speed of P with respect to the ground is v (a) 0 ≤ v ≤ 2u (b) v = u, if CP is horizontal (c) v = u, if CP makes an angle of 60º with the horizontal and P is below the horizontal level of C (d) v = 2u , if CP is horizontal
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80. In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will achieve same kinetic energy. ONE OR MORE THAN ONE OPTION MAY BE CORRECT
2. In the figure, the disc D does not slip on the surface S. The pulley P has mass, and the string does not slip on it. The string is wound around the disc.
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Rotational Dynamics
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1.92
Rotational Dynamics
of the same mass m. The wedge is free to move on a smooth horizontal surface. The ball rolls without sliding on the wedge. The required height of the wedge are mentioned in the figure. Which of the following statement about the above system is/are true
5. A disc of circumference s is at rest at a point A on a horizontal surface when a constant horizontal force begins to act on its centre.
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9. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with
7. A small ball (uniform solid sphere) of mass m is released from the top of a wedge
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8. P is the centre of mass of four point masses A, B, C and D which are coplanar but not collinear. (a) P may or may not coincide with one of the point masses. (b) P must lie within the quadrilateral ABCD (c) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C and D three at a time. (d) P must lie on a line joining two of the point A, B, C and D
6. A particle of mass m is travelling with a constant velocity v = v0iˆ along the line
of the wedge is gh (b) The total kinetic energy of the ball just before it falls on the ground is (31/18) mgh (c) The horizontal separation between the ball and the edge ‘PQ’ of wedge just before the ball falls on the ground is
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Between A and B there is sufficient friction to prevent slipping, and the surface is smooth to the right of B. AB = s. The disc moves from A to B in time T. To the right of B (a) the angular acceleration of the disc will disappear, linear acceleration will remain uncharged (b) linear acceleration of the disc will increase (c) the disc will make one rotation in time T/2 (d) the disc will cover a distance greater than s in further time T
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Rotational Dynamics
12. A solid sphere of mass M and radius R is pulled horizontally on a sufficiently rough surface as shown in the figure.
10. A solid cone and a solid sphere is arranged as shown in the figure. The centre of mass is
Choose the correct alternative. (a) The acceleration of the centre of mass is F/M. (b) The acceleration of the centre of mass is 2 F . 3M (c) The friction force on the sphere acts forward. (d) The magnitude of the friction force is F/3.
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(a) at 3R if m1 = m2 = m and R1 = R2 (b) at 2R from the centre of mass of solid cone if ρ1 = ρ2 and R1 = R2 (c) If ρ1 = 2ρ2 then distance from the centre of solid cone is 11R/3 (d) (a) and (b) 11. The diagram shows a uniform solid cylinder of mass M and length 2ℓ.
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Note that B is the common mid-point of the semicircular rings, while O is the centre of the rings. If m is the total mass of the system and R is the radius of the rings. Which of the following statement is true. (a) Moment of inertia of the system about the axis OB is given by mR2. (b) Moment of inertia of the system about the axis OB is given by mR2/2. (c) Radius of gyration of the system about the axis OB is given by R/ 2. (d) Radius of gyration of the system about the axis OB is given by R. 14. A disc of radius R rolls on a horizontal surface with linear velocity V and angular velocity ω.
XY and PQ are axes passing through the centre of the cylinder and parallel to its plane face. (a) If the moment of inertia of the cylinder about the axis XY is I, the moment of inertia about PQ is also I. (b) The moment of inertia of the cylinder about its axis X' Y' is 2I (c) The moment of inertia about a diameter of one end is I + M ℓ2. (d) All of the above
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velocity v and gets caught in the bag. Then for the combined system (bag + bullet) (a) Momentum is mv/(M + m) (b) KE is (1/2) Mv2 (c) Momentum is mv (d) KE is m2v2/2(M + m)
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(c) and axis parallel to the z-axis and pass10 2 ing through a corner is m 3 5 (d) one side is m 2 2 17. A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a common angular velocity. (a) Some friction exists between the disc and the ring. (b) The angular momentum of the ‘disc plus ring’ is conserved. (c) The final common angular velocity is (2/3) rd of the initial angular velocity of the disc. (d) (2/3) rd of the initial kinetic energy changes to heat.
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1. Statement 1: When you lean behind over the hind legs of the chair, the chair falls back after a certain angle. Statement 2: Centre of mass lying outside the system makes the system unstable. 2. Statement 1: A rigid disc rolls without slipping on a fixed rough horizontal surface with uniform angular velocity. Then the acceleration of lowest point on the disc is zero. Statement 2: For a rigid disc rolling without slipping on a fixed rough horizontal surface, the velocity of the lowest point on the disc is always zero.
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16. Four identical rods, each of mass m and length ℓ, are joined to form a rigid square frame. The frame lies in the xy plane, with its centre at the origin and the sides parallel to the x and y axis. Its moment of inertia about 2 (a) the x-axis is m 2 3 4 2 (b) the z-axis is m 3
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(a) Friction force decelerates the disc M1 and accelerates the disc M2. (b) The final angular speeds of the discs are ω R in the ratio 1 = 1 . ω2 R2 (c) The angular momentum of the system (M1 + M2) is conserved about the point of contact. (d) The final rotational energy is less than the initial rotational energy.
Direction: Each question contains statement 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct.
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Rotational Dynamics
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Direction: Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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2. A uniform disc rolls without slipping on a rough horizontal surface with uniform angular velocity. Point O is the centre of disc and P is a point on disc as shown.
6. Statement 1: A wheel moving down a frictionless inclined plane will slip and not roll on the plane. Statement 2: It is the frictional force which provides a torque necessary for a body to roll on a surface.
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Column I Column II (a) If the coefficient (p) Downward of friction is tan θ, along the then linear accelerinclined ation of the disc is plane (b) If coefficient of (q) Zero friction is less than tan θ, then linear acceleration of the disc is (c) Direction of fric- (r) Perpendicution force is lar to plane of disc (d) Angular accelera- (s) Upward tion of disc is along the inclined plane
Under the given condition, the horizontal surface exerts normal reaction nonuniformly on lower surface of the block. Statement 2: For the cubical block given in statement-1, the horizontal force F has tendency to rotate the cube about its centre in clockwise sense. Hence, the lower right edge of cube presses the horizontal surface harder in comparison to the force exerted by lower left edge of cube on horizontal surface.
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7. Statement 1: When a sphere rolls on a horizontal table it slows down and eventually stops. Statement 2: When the sphere rolls on the table, both the sphere and the surface deform near the contact. As a result, the normal force does not pass through the centre and provide an angular deceleration. MATCH THE COLUMN TYPE QUESTIONS
4. Statement 1: If two different axes are at same distance from centre of mass of a rigid body, then moment of inertia of the given rigid body about both axis will always be same. Statement 2: From parallel axis is theorem I = Icm + md2, where all terms have usual meaning. 5. Statement 1: A uniform cubical block (of side a) undergoes translational motion on a smooth horizontal surface under action of horizontal force F as shown.
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3. Statement 1: A uniform thin rod of length L is hinged about one of its end and is free to rotate about the hinge without friction. Neglect the effect of gravity. A force F is applied at a distance x from the hinge on the rod such that force always is perpendicular to the rod. As the value of x is increased from zero to L, the component of reaction by hinge on the rod perpendicular to length of rod increases. Statement 2: Under the conditions given in statement 1 as x is increased from zero to L, the angular acceleration of rod increases.
Rotational Dynamics
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(c) Acceleration of point C (d) Acceleration of point D
In each situation of column I a statement is given and the corresponding results are given in column II. Match the statements in column I with the results in column II. 3
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4. In each situation of column I, a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface as shown. At t = 0 (initially) the angular velocity of disc is ω0 and velocity of centre of mass of disc is v0 (in horizontal direction). The relation between v0 and ω0 for each situation and also initial sense of rotation is given for each situation in column I. Match the statements in column I with the corresponding results in column II.
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Rotational Dynamics
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Rotational Dynamics
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The mass of the prism is M and it is uniformly distributed. The length of each side of the cross-
2. The kinetic energy of the prism just before and after impact is similarly Ki and Kf. We may write Kf = rKi and value of the coefficient r (a) 121/17 (b) 121/289 (c) 12/17 (d) 17/12
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sectional hexagon is a. The moment of inertia I of the hexagonal prism about its central axis is I = (5/12) Ma2 The moment of inertia I' about an edge of the 17 prism is I ′ = Ma 2 . 12 The prism is initially at rest with its axis horizontal on an inclined plane which makes a small angle with the horizontal (figure). Assume that the surfaces of the prism are slightly concave so that the prism only touches the plane at its edges. The effect of this concavity on the moment of inertia can be ignored. The prism is now displaced from rest and starts an uneven rolling down the plane. Assume that friction prevents any sliding and that the prism does not lose contact with the plane. The angular velocity just before a given edge hits the plane is ωi while ωf is the angular velocity immediately after the impact. We may write ωf = s ωi.
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Rotational Dynamics
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Rotational Dynamics
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10. Where will be the centre of mass after the removal of square 1 (a) In the first quadrant (b) In the second quadrant (c) In the third quadrant (d) In the fourth quadrant
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6. The velocity of centre of mass immediately after the collision is (a) v/3 (b) 2v/3 (c) 3v/2 (d) 2v/5
12. In the above question, where will be the centre of mass after the removal of squares 1, 2 and 3 (a) In the third quadrant (b) In the fourth quadrant (c) In the first quadrant (d) At the origin
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PASSAGE-2 A uniform sphere with mass 2m and radius 2r is at rest in a place where no external forces are applied. P U
9. The kinetic energy of the body after the collision is mv 2 11mv 2 (b) (a) 108 54 2 17 mv 17 mv 2 (d) (c) 54 108
4. If the condition of previous Question is satisfied, the kinetic energy Ki, will approach a fixed value Ki, 0 as the prism rolls down the incline Ki, 0 = K Mg a, K is equal to tan θ cos θ (b) (a) 1− r 1− r sin θ 1− r (c) (d) 1− r sin θ
8. The angular momentum about the center of mass shortly before and after the collision (a) increases (b) decreases (c) remains same (d) none of these
1 (c) δ = (1 − tan(30° − θ)) r 1 (d) δ = (1 − sec(30° − θ)) r
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PASSAGE-4 The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90º.
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Rotational Dynamics
PASSAGE-5 Rod AB of length 20 m is moving on an smooth horizontal plane with v0 = 10 m/s and rotating with ω = 2 rad/sec in anti-clock wise direction. $
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(b) 2v0 (d) v0/2
(a) 2 v0 (c) v0
22. Velocity of point P at time t = t0 is
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21. Distance travelled by C.O.M. during time t = 0 to t = t0 is (a) 7.5 π (b) 5 π (c) 2.5 π (d) None of these
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18. Velocity of ends A and B when velocity of CM is perpendicular to rod are (a) 30 m/s, 10 m/s (b) 10 m/s, 30 m/s (c) 15 m/s, 25 m/s (d) 25 m/s, 15 m/s
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Rotational Dynamics
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8. A uniform solid body consists of a cylinder of radius 2a and height 2h with another cylinder of radius a and height h with one plane face placed centrally on one plane face of the first cylinder. Find the position of the centre of gravity of the solid.
1. Three particles A, B, C, of mass 2, 3, 4 kg are at points (1, 4), (3, 6), (2, 1) in the xy plane. Find the coordinate of their centre of gravity. 2. Three particles of mass 5, 3, 7 kg are at the points A, B, C whose position vectors are iˆ − 2 ˆj , 7iˆ + ˆj , −3iˆ + 5 ˆj. Find the position vector of their centre of mass. Also find the position vector of the centroid of the points A, B, C.
9. A rod of length R and mass M is free to rotate about a horizontal axis passing through hinge P as in figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the rod hits the block B of mass m and stops. Find the ratio of masses such that the block B completes the circle. Neglect any friction.
SUBJECTIVE QUESTIONS
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6. Show that the centre of gravity of a uniform lamina in the form of a parallelogram is at the point of intersection of the diagonals.
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5. The vertices of a triangle are at the points iˆ + ˆj , 3iˆ − ˆj , 2iˆ + ˆj. Find the position vector of the centre of gravity of the triangle, assuming it to be a uniform lamina.
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4. A uniform lamina is in the form of a trapezium ABCD where, AB and DC are the parallel sides. Show that the centre of gravity of the trapezium lies on the line joining the mid-points of AB and DC.
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3. A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at a distance x from A. Find the normal reactions at A and B.
23. Choose the incorrect statement.
PASSAGE-6 A ring of mass M and radius R sliding with a velocity v0 suddenly enters into rough surface where the coefficient of friction is µ, as shown in figure
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Rotational Dynamics
16. An automobile engine develops 100 kilowatt when rotating at a speed of 1800 rev/ min. What torque does it deliver?
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17. A 1 m long road has a mass of 0.12 kg. What is its moment of inertia about an axis passing. (i) the centre and perpendicular to the length of rod (ii) one end and perpendicular to rod
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18. The diameter flywheel increases by 1% What will be percentage increase in moment of inertia about axis of symmetry?
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19. A disc of mass 2 kg and radius 20 cm is free to rotate about an axis through its centre and perpendicular to the disc. If a force of 50 N is applied tagentially, calculate the angular acceleration.
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20. A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 rad/s. If this mass is brought to rest in 10 sec by a brake, what is the torque that is applied?
22. A solid sphere and a solid cylinder having the same mass and radius, roll down the same incline. What is the ratio of their accelerations? 23. The moment of inertia of sphere is 20 kg-m2 about the diameter. What is the moment of inertia about any tangent?
21. If a flywheel of mass 20 kg and diameter 1 m is rotating 300 revolutions per minute, what is its kinetic energy?
24. The moment of inertia of a wheel is 1000 kg m2. At a given instant, its angular
13. A flywheel starting from rest acquires in 10 sec an angular velocity of 240 revolutions per minute. What is its angular acceleration assumed constant?
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12. A vertical post of mass M is to be driven into the ground. A pile-driver of mass m strikes the post vertically with a velocity v. Assuming that the pile-driver does not bounce off the post, find the velocity with which the post enters the ground. If the combined mass comes to rest when the post has beep driven into the ground to a depth h find the constant force with which the ground resists penetration.
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The angle between the two legs is 60º. The fat person sitting on the ladder has a mass of 80 kg. Find the contact force exerted by the floor on each leg and the tension in the crossbar.
15. A wheel starting from rest, is rotating with a constant angular acceleration of 3.0 rad/ sec2. An observer notes that it traces an angle of 120 radians in a 4.0 sec interval. For how long the wheel has been rotated when the observer started his observation?
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11. The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle.
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14. The angular velocity of a wheel increases from 1200 to 4500 rev/min in 10 sec. Compute its angular acceleration and the number of revolutions made during this time.
10. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal, to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. Find the minimum value of F for which the cube begins to tip about an edge. (assume that cube does not slide).
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Rotational Dynamics
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2. An isolated particle of mass m is moving in horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses m/4 and 3 m/4. An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at [1997]
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(b) y = +20 cm (d) y = –20 cm
(a) y = –5 cm (c) y = +5 cm
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This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the co-ordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position. [1996]
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1. A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The masses of large and small spheres are 4 M and M respectively.
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29. A thin hoop of mass M and radius r is placed on a horizontal plane. At the initial instant, the hoop is at rest. A small washer of mass m with zero initial velocity slides from the upper point of the hoop along a smooth groove in inner surface of the hoop. Determine the velocity u of the centre of the hoop at the moment when the washer is at a certain point A of the hoop, whose radius vector forms an angle ϕ with the vertical (Fig.). The friction between the hoop and the plane should be neglected.
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PREVIOUS YEARS' IIT-JEE QUESTIONS
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28. A hoop of radius R rests on a horizontal surface. A similar hoop moves past it at a velocity v. Determine the velocity vA of the upper point of “intersection” of the hoops as a function of the distance d between their centres, assuming that the hoops are thin, and the second hoop is in contact with the first hoop as it moves past the latter.
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27. It was found that the winding radius of a tape on a cassette was reduced by half in a time t1 = 20 min of operation. In what time t2 will the winding radius be reduced by half again?
26. A solid sphere of mass 0.5 kg and diameter 1 meter rolls without slipping with a constant velocity of 5 m/sec along a smooth straight line. Calculate its total energy.
30. A horizontal weightless rod of length 3ℓ is suspended on two vertical strings. Two loads of mass m1 and m2 are in equilibrium at equal distances from each other and from the ends of the string (Fig.). Determine the tension T of the left string at the instant when the right string snaps.
25. A sphere is rolling down without slipping the incline plane from a vertical height h. Find the angular velocity and linear velocity as it reaches the ground if its mass is m and radius is r.
velocity is 10 rad/s. After the wheel rotates through an angle of 100 radians, the wheel’s angular velocity is 100 rad/s. Calculate (i) the torque applied on the wheel (ii) the increase in rotational kinetic energy
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3. A wedge of mass m and triangular crosssection (AB = BC = CA = 2R) is moving with a constant velocity- viˆ towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that wedge remains in contact with the sphere for a very short time ∆t during which the sphere exerts a constant force F on the wedge. [1998]
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7. A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is
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8. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown.
(b) mg/4 (d) mg (1 – µ)
(a) Infinitesimal (c) mg/2
5. A disc of mass M and radius R is rolling with angular speed ω on a horizontal as shown. The magnitude of angular momentum of the disc about the origin O is [1999]
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4. A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB respectively. Then [1999]
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6. A cubical block of side a moving with velocity v on a horizontal smooth plane as shown. It hits a ridge at point O. The angular speed of the block after it hits O is [1999]
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(a) Find the force F and also the normal force N exerted by the table on the wedge during the time ∆t (b) Let h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge during the interval ∆t.
(a) ωA < ωB (c) ωA = ω
1 MR 2 ω 2 3 (c) MR 2 ω 2 (a)
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Rotational Dynamics
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Rotational Dynamics
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The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its centre of mass is 1.2 kg-m2. Two robber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5 m from the line AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacle is 6 N-s. [2001]
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(a) Angular velocity and total energy (kinetic and potential) (b) Total angular momentum and total energy (c) Angular velocity and moment of inertia about the axis of rotation. (d) Total angular momentum and moment of inertia about the axis of rotation.
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9. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down; One along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as beads slides down are [2000]
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11. Two heavy metallic plates are joined together at 90º to each other. A laminar sheet of mass 30 kg is hinged at the line AB joining the two heavy metallic plates. The hinges are frictionless.
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10. A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in the direction perpendicular to AB. The collision is completely elastic. After the collision the particle comes to rest. (a) Find the ratio m/M (b) A point P on the rod is at rest immediately after the collision. Find the distance AP. (c) Find the linear speed of the point P at a time πL/(3v0) after the collision. [2000]
The moment of inertia of the loop about the axis XX' is [2000]
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12. One quarter section is cut from a uniform circular disc of radius R. This section has a mass M.
The body is set into rotational motion on the table about A with a constant angular velocity ω. (a) Find the magnitude of the horizontal force exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T. [2002]
Rotational Dynamics
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14. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity ω0. When the tortoise move along a chord of the platform with a constant velocity (with respect to the platform). The angular velocity of the platform ω(t) will vary with time t as: [2002]
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Rotational Dynamics
19. A block of mass m is held fixed against a wall by a applying a horizontal force F. Which of the following option is incorrect
Centre of the circle Outside the circle Inside the circle Point on circumference
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16. Two particles each of mass M are connected by a massless rod of length ℓ. The rod is lying on the smooth surface. If one of the particle is given an impulse MV as shown in the figure then angular velocity of the rod would be [2003]
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20. A disc has mass 9 m. A hole of radius R/3 is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre ‘O’ of the disc and perpendicular to the plane of the disc is [2005]
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18. A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is (a) 2K (b) K/2 (c) K/4 (d) 4K
(a) (b) (c) (d)
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21. A particle moves in circular path with decreasing speed. Which of the following is correct [2005]
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(a) vQ > vC > vP
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23. A rod of length L and mass M is hinged at point O. A small bullet of mass m hits the rod as shown in the figure.
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.m
ht tp
e/
un
si ta
rio
s
ta si er
iv e/ un
t.m s: //
27. The ball rolls down without slipping (which is at rest at a) along ab having friction. It rolls to a maximum height hc where bc has no friction. Ka, Kb and Kc and kinetic energies at a, b and c. [2006]
s: //t
24. If n number of balls each of mass m collide elastically with the plate of mass M per second with a velocity v elastically and reflect back.
D
E
Q
Find the velocity of balls such that plate is at equilibrium. [2006] Given: n = 100, m = 0.01 kg, M = 3 kg
(a) Ka = Kc, ha = hc (c) Kb > Kc, ha < hc D
KF
E
(b) Kb > Kc, ha = hc (d) Kb > Kc, ha > hc
ht
KD
F
tp
(a) The friction force is dissipative. (b) The friction force is necessarily changing. (c) The friction force will aid rotation but hinder translation. (d) The friction force is reduced if θ is reduced.
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s_
an
ar
tp
/0
m
26. A solid cylinder is rolling down the inclined plane without slipping. Which of the following is/are correct? [2006]
co
a.
The bullet gets embedded in the rod. Find angular velocity of the system just after impact. [2005] 4
22. A cylinder of mass m and radius R rolls down an inclined plane of inclination θ. Calculate the linear acceleration of the axis of cylinder. [2005]
(b) only direction of L is constant (c) acceleration a is towards the centre (d) it will move in a spiral and finally reach the centre
25. A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis about the edge and perpendicular to the plane is I. The relation between radius R and r. [2006] 2 2 (a) r = R (b) r = R 15 15 2 2 R (d) r = (c) r = R 15 15
(a) L is constant
Rotational Dynamics
PASSAGE BASED QUESTIONS Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an
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1.109
Rotational Dynamics
s
rs
ita rio
ve e/ un i (a) (d) (a) (c) (a) (a) (d) (c) (c) (b)
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4. 12. 20. 28. 36. 44. 52. 60. 68. 76.
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(d) (b) (a) (a) (c) (d) (d) (c) (a) (c)
s:
3. 11. 19. 27. 35. 43. 51. 59. 67. 75.
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(d) (a) (c) (c) (d) (b) (c) (a) (b) (d)
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2. 10. 18. 26. 34. 42. 50. 58. 66. 74.
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(c) (b) (a) (b) (c) (d) (c) (a) (a) (d)
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tp
ONLY ONE OPTION IS CORRECT 1. 9. 17. 25. 33. 41. 49. 57. 65. 73.
ita rio
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nswers
A
an ar m //w
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3I ω 2t
w
(d)
w
9I ω 4t
(c)
31. Statement 1: If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. because Statement 2: The linear momentum of an isolated system remains constant. (a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement-1 (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True [2007]
a. co m
9I ω 2t
29. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is 2I ω (a) 3t
(b)
Iω2 3 Iω2 (d) 6
(b)
Iω2 2 Iω2 (c) 4 (a)
30. The loss of kinetic energy during the above process is
angular velocity ω by a spring having the same spring constant and compressed by a distance x2. Both the discs rotates in the clockwise direction. [2007] 28. The ratio x1/x2 is (a) 2 (b) 1/2 (c) 2 (d) 1/ 2
5. 13. 21. 29. 37. 45. 53. 61. 69. 77.
(d) (a) (a) (b) (b) (b) (b) (a) (b) (d)
6. 14. 22. 30. 38. 46. 54. 62. 70. 78.
(d) (a) (b) (b) (a) (d) (c) (b) (b) (a)
7. 15. 23. 31. 39. 47. 55. 63. 71. 79.
(a) (c) (c) (d) (b) (b) (b) (a) (b) (d)
8. 16. 24. 32. 40. 48. 56. 64. 72. 80.
(a) (a) (a) (c) (c) (c) (b) (c) (b) (b)
ONE OR MORE THAN ONE OPTION MAY BE CORRECT 1. (a, b, d) 2. (a, b, d) 3. (a, c, d) 4. (a, b, c) 5. (b, c, d) 6. (a, d) 9. (c, d) 10. (a, b, c) 11. (a, c) 12. (b, c) 13. (b, c) 14. (c, d)
7. (b, c, d) 8. (a, c) 15. (b, d) 16. (a,b,c,d)
17. (a, b, d)
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1.110
Rotational Dynamics
ASSERTION AND REASON QUESTIONS 1. (c)
2. (d)
3. (d)
4. (d)
5. (c)
6. (a)
7. (a)
MATCH THE COLUMN TYPE QUESTIONS 2. (a) → p, (b) → q, s (c) → p, (d) → q, s
3. (a) → q, (b) → s, (c) → r, (d) → p
4. (a) → p, q, r (b) → p, q, r (c) → p, q
5. (a) → s, (b) → p, (c) → q, (d) → r
6. (a) → p, q, (b) → q, r, (c) → p, q, (d) → p, q
1. (a) → q, (b) → p, (c) → s, (d) → r
(d) → p, q, r 7. (a) → r, (b) → s, (c) → r, (d) → q
8. (a) → q, (b) → s, (c) → p, r, (d) → p
3. (b) 11. (a) 19. (a)
4. (c) 12. (a) 20. (a)
5. (a) 13. (a) 21. (b)
8. (c) 16. (c) 24. (b)
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ita
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3. A fly wheel is a large heavy wheel with a long cylindrical axle passing through its centre and its centre of mass lying on its axis of rotation. It is constructed in such a way that most of its mass is concentrated at the rim of fly-wheel. This increases the moment of inertia about the axis of rotation. The function of moment of inertia is to oppose any change in uniform rotatory motion. Therefore, when a fly-wheel of large moment of inertia is coupled to an engine, it assures the running of engine smoother and steadier.
5. Equilibrium of the wheel when it is still and standing up-right on its rim is unstable. Moreover, when the wheel is rolling, it possesses and angular momentum in the horizontal direction which is conserved quantity of the motion. The angular momentum does not changes unless friction slows down the angular velocity. Since, falling results in a change of angular momentum direction, the wheel does not fall, till is loses its angular velocity. 6. The system tends to keep its angular momentum constant. When the person sits near the edge of the platform, the moment of inertia
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t.m e/
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ht tp s:
4. The torque τ required to cause an angular acceleration α is given by τ = Iα, where, I (= mr2) is the moment of inertia. More is the distance of the stone from the axis of rotation, greater is its moment of inertia and hence, more torque is required to produce same angular acceleration.
un iv
w w //w s: tp
2. The cycle wheel is constructed in such a way so as to increase its moment of inertia with minimum possible mass. This is done to ensure uniform speed.
7. (a) 15. (a) 23. (c)
HINTS AND SOLUTIONS
1. No, torque is required only for angular acceleration.
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CONCEPTUAL QUESTIONS
6. (a) 14. (c) 22. (a)
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(a) (a) (c) (a)
m
1. 9. 17. 25.
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PASSAGE BASED QUESTIONS
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Rotational Dynamics
13. In a tornado, the air from the surrounding whirls inwards as the air at the centre rises. Since, the moment of inertia decreases, the angular velocity of the inward whirling wind increases according to the law of conservation of momentum.
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7. (a) Initial velocity of centre of mass of given system is zero and net external force is in vertical direction. Since, there is shift of mass downward, the centre of mass has only downward shift. (b) Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts a force on string which has a horizontal component towards right. Hence, centre of mass of system has a rightward shift. (c) Both block and monkey moves up, hence, centre of mass of given system shifts vertically upwards.
12. When two fingers are applied, two equal and opposite forces act on the cap, thereby constituting a couple.
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11. When the raw and the hard boiled eggs are spun together the raw egg will have a greater moment of inertia as the fluid in it will concentrate along its surface. Assuming that the same rotational torque is applied to both the eggs, raw egg will come to rest first as it will have lesser angular velocity.
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10. We know that I = Σmr2 i.e., the moment of inertia of a body not only depends upon the mass but also depends upon the distribution of mass from the axis of rotation. The disc made of less dense material will have larger moment of inertia because it will have greater surface area, i.e., the particles are at the greater distance from the axis of rotation.
5. (a) Since, all forces on disc pass through point of contact with horizontal surface, the angular momentum of disc about point on ground in contact with disc is conserved. Also the angular momentum of disc in all cases is conserved about any point on the line passing through point of contact and parallel to velocity of centre of mass. The K.E. of disc is decreased in all cases due to work done by friction. From calculation of velocity of lowest point on disc, the direction of friction in case a, b and d is towards left and in case c is towards right. The direction of frictional force cannot change in any given case.
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9. If there were only one propeller in the helicopter, the helicopter itself would have turned in opposite direction due to conservation of angular momentum.
3. (a) Speed of point P changes with time. (b) Acceleration of point P is equal to ω2x (ω = angular speed of disc and x = OP). The acceleration is directed from P towards O. (c) The angle between acceleration of P (constant in magnitude) and velocity of P changes with time. Therefore, tangential acceleration of P changes with time. (d) The acceleration of lowest point is directed towards centre of disc and remains constant with time.
8. By raising the hand, the moment of inertia of the system increases and thus angular velocity ω decreases. Now the turn table will slow down as I ω remains constant in the absence of any external torque.
MATCH THE COLUMN TYPE QUESTIONS
7. The swimmer can increase the number of loops by pulling his legs and arms inwards i.e., by decreasing the moment of inertia. By doing so the angular velocity ω increases because I ω remains constant.
of platform increases and hence, its angular velocity decreases. When the person starts moving towards the centre of platform, the moment of inertia decreases and hence, the angular velocity increases.
1.111
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1.112
Rotational Dynamics
(d) Net external force on given system is zero. Hence, centre of mass of given system remains at rest. 8. Taking into the plane –ve (a) L = MvR + Iω > 0 → v f forward
1ˆ j 3
7h from the base of the larger cylinder 6
8.
5. 2iˆ +
9. Minimum velocity required by block m to complete the motion in 5gR
2v (b) L = MvR – Iω = MvR −1 R MvR – 2 MvR < 0 for ring ⇒ v f forward 4 MvR − > 0 MvR for solid sphere ⇒ v f 5 left (c) L = MvR + Iω > 0 ⇒ v f left
Conserving mech. energy
1 2 R I ω = Mg × ⇒ ω = 2 2
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ML3 MR 2 (since L = R) = 3 3
M = 15 m
10. For troppling about edge xx'
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Fmin
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3a a = mg 4 2
Fmin =
2mg 3
Z
N1 = mg – N2 = mg –
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mgx d −x = mg d d
NA = w (1 – x/d), NB = wx/d
D
11. The forces acting on different parts are shown in figure. Consider the vertical equi
mgx N2 = d
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er
si
ta
w .ja s:
Putting I =
1
[
Μ gR = mR 5 gR I
MgRI = m2R25gR
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I ω = mR 5 gR
I
3. As shown in FBD → equation in vertical direction N1 + N2 = mg Taking moments about A mgx = dN2 1
J5
Conservation of angular momentum with respect to P before and after collision.
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s: tp
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SUBJECTIVE QUESTIONS
1 1 ˆ (5i + 28 ˆj ), (5iˆ + 4 ˆj ) 3 15
P
rio
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v MvR − MR 2 > 0 ⇒ v f left 2R
2.
fo
50
a.
2 v MvR − MR 2 > 0 ⇒ v f left 5 2R
19 10 1. , 10 3
3 Z
m
(d) L = MvR – Iω
MgR I
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Rotational Dynamics
o
rio
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19. 250 rad/sec2
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20. 0.9 Nm
π − α with the horizontal and 2 v v = is given by v A = 2cos ϕ 2sin α angle ϕ =
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21. 2465 Joules
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(ii) 0.04 kg-m2
18. 2 %
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17. (i) 0.01 kg-m2
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1800 × 2π ; 60
10000 = 531 N -m 6π
O1 is at rest, the velocity vA of the upper point A of “intersection” of the hoops must be directed along the tangent to the circle with centre O1 at any instant of time (Figure) At any instant of time the segment AB divides the distance d = OO1 between the centres of the hoops into two equal parts, and hence, the horizontal projection v of the velocity vA is always equal to . 2 Consequently, the velocity vA forms an
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15. 8.0 sec.
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22. 15 : 14
23. 70 kg-m2 2
(ii) 4.95 × 106 Joules
24. (i) 4.95 × 104 N-m
5 D
M G
2
%
2 gh 2gh 25. ω = 2 and v = 2 2 2 1 + K /R r +K 27. Let the initial winding radius be 4r. Then the decrease in the winding area as a result of the reduction of the radius by half (to 2r) will be S = π (16r2 – 4r2) = 12 π r2.
2
d Since, sin α = 1 − cos 2 α = 1 − 2R the velocity of the upper point of “intersecv tion of the hoops is v A = 2 d 2 1− 2R
26. 8.75 Joule
m
14. 475
τ=
28. Since, the hoop with the centre at point
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13. 2.51 rad/sec2
100 × 103 = τ
a. co
m 2 v 2 + 2 gh( M + m) 2 mv , 2 h ( M + m) M +m
an
12.
2 2 = (392 N ) × = 450 N 3 3
or T = N
or N = (40 kg) (9.8 m/s2) = 392 N Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, N (2 m) tan 30º = T (1 m)
Which is equal to the product of the length ℓ1 of the wound tape and its thickness d. The velocity v of the tap is constant during the operation of the tape-recorder, hence ℓ1 = vt1, and we can write 12 π r2 = vt1 d ...(i) When the winding radius of the tape on the cassette is reduced by half again (from 2r to r), the winding area is reduced by π (4r2 – r2) = 3 π r2, i.e., 3 π r2 = vt2 d …(ii) Where t2 is the time during which the winding radius will be reduced in the latter case. Dividing equations (1) and (2) term wise, t we obtain t2 = 1 = 5 4
librium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg)g and the contact force N + N = 2N due to the floor. Thus 2N = (80 kg)g
1.113
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1.114
2
\
29. The forces acting on the hoop-washer system are the force of gravity and the normal reaction of the plane. These forces are directed along the vertical. Consequently, the centre of mass of the system does not move in the horizontal direction. Since, there is no friction between the hoop and the plane, the momentum conservation law, at any instant of time we have Mu + mvx = 0 .........(1) where u and vx are the horizontal components of the velocities of the centre of the hoop and the washer. Since, vx periodically changes its sign, u also changes sing “synchronously”. The general nature of motion of the hoop is as follows: the centre of the hoop moves to the right when the washer is on segments BC and BE, and to the left when the washer is on segments CD and DE (Figure).
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Y
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u = m cos ϕ
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2 gr (1 + cos ϕ) ( M + m)( M + m sin ϕ)
30. At the moment of snapping of the right string, the rod is acted upon by the tension T of the left string and the forces N1 and N2 of normal pressure of the loads of mass m1 and m2 (Figure)
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(
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........(3)
Solving equations (1) and (3) together, we determine the velocity of the centre of the hoop at the instant when the radius vector of the point of location of the washer forms an angle ϕ with the vertical
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The velocity v of the washer and u of the hoop are connected through the energy conservation law: mv 2 Mu 2 mgr (1 + cos ϕ) = + ......... (2) 2 2 The motion of the washer relative to a stationary observer can be represented at any instant as the superposition of two motions; the motion relative to the centre of the hoop at a velocity vt directed along the tangent to the hoop, and the motion together with the hoop at its velocity u having the horizontal direction (Figure). The figure shows that
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Since, the rod is weightless (its mass is zero), the equations of its translatory and rotary motions will have the form – T + N1 – N2 = 0, N1ℓ = 2N2ℓ. The second equation (the condition of equality to zero of the sum of all moments of force about point O) implies that
Rotational Dynamics
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Rotational Dynamics
(c)
2L 3
v0 2 2
11. (a) I = 0.1 m, (b) w' = 1 rad/s (c) laminar sheet will never come to rest
12. (a)
3mω2
(c)
(b) Fx = F/4
3mω2
13. (a)
x=
14. (c)
m 1 = (b) M 4
10. (a)
9. (b)
N1 = 2N2 .....(1) Combining these conditions, we get T = N1 – N2 = N2 .....(2) At the moment of snapping of the right string, the accelerations of the loads of mass m1 and m2 will be vertical (point O is stationary, and the rod is inextensible) and connected through the relation a2 = 2a1. Let us write the equation of motion for the loads at this instant: m1g – N '1 = m1a1, m2g + N '2 = m2a2, Where N'1 and N '2 are the normal reaction of the rod on the loads of mass m1 and m2. Since, N1' = N1 and N2' = N2, we have m1g – 2N2 = m1a1 m2g – N2 = 2 m2a1
16. (a)
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24. 10 m/s
26. (c, d)
27. (d)
28. (c)
29. (a) 30. (b)
25. (b)
31. (c)
8. (d)
20. (b)
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18. (b)
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PREVIOUS YEAR ’ IIT-JEE QUESTIONS 1. (L + 2R, 0)
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15. (a)
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Hence, we can find N2 and consequently the m1m2 tension of the string T = N 2 = g m1 + 4m2
1.115
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ita
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c h a p t e r
2
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Gravitation
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2.1 NEWTON’S UNIVERSAL LAW OF GRAVITATION
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“Each body in the universe attracts to the another body”, this is inverse square law, this force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Let, the two particles are of mass m1 and m2 and distance between them is r. Therefore m m F∝ 12 2 r G m1 m2 F= r2 where, G = Universal constant = 6.67 × 1011 Nm2/kg2
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2.2 GRAVITATIONAL POTENTIAL ENERGY Let, the two particles are separated by a distance r, mass of the particles are m and M. Hence, the gravitational force between them F = G m2 M r P
0
P
U
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2.2
Gravitation
Work done = change in the gravitational potential energy. ∞ ∞ W = ∫ F⋅dr = −GmM ∫ dr2 r r r Negative sign indicates that work done is against the gravitational force ∞
1 W = −G m M − r = − G mr M r
M Hence, gravitational potential energy = − G m r
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Potential energy decreases, when the separation between them increases. Gravitational potential energy is zero at the infinity.
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2.3 ACCELERATION DUE TO GRAVITY
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If the mass of the earth is M, radius of the earth is R and a particle of mass m is at the surface of the earth, thus the attraction force between them is
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m
F = G m2M R
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This force imports the acceleration of the particle m, acceleration due to gravity. According to the Newton’s second law, F a=m
Here,
⇒
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∴
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F = ma
a = g = G m2M = GM R2 R m
On the surface of the earth = GM R2
2.4 VARIATION IN ACCELERATION DUE TO GRAVITY (i) Above the surface: Let a particle is above the earth surface at a height h. Hence the gravitional force F′ = G m M2 (R + h) https://t.me/universitarios_infohttps://www.jamarana.com
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Gravitation
2.3
P K 5 2
m
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From previous, g = GM R2 From equations (i) and (ii), we get 2
...(i)
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or g ′ = G M 2 (R + h)
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g′ = g R + R h where, g = accelevation due to gravity at the surface.
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(ii) Below the surface of the earth: For this part, radius of shadded part is (R – h).
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Volume of the shadded part = 4 π (R − h)3 3 Density of the earth =
M 4/3 π R3
Hence, mass of the shadded part M m′ = 4 π(R − h)3 ⋅ 3 4 πR3 3 =
(R − h)3 M R3
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...(ii)
Gravitation
So g ′ = G m
(R − h)3 M R3 (R − h) 2
=G m M
(R − h) R3
∴ g ′ = g 1 − h R
where, g is the acceleration due to gravity. Note: It should be noted that, the value of g decreases if we move above the surface or below of the surface of the earth.
2
U 5
$
PZU
er si ta rio s
e/
PJ
iv
ș
un
//t
t.m e/
ht tp s:
s: //
Here, r is the radius of the circular path followed by m. ⇒ r = R cos θ The resultant force of masses m is given by
ht tp
%
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
rio s_ in fo
na .c om
(iii) With latitudes: Consider a point A on the earth surface where the line joining centre O to A makes an angle θ with the equatorial plane as shown. The angle θ is said to be the latitude of the point A´. Let us place a small particle of mass m at A. Analysing the forces of on m, from the reference frame (or relative to) the earth, we find two forcesgravitational force (mg) and pseudo force (mω2r).
ht
2.4
F = m 2 g 2 + m 2 ω4 r 2 + 2(mg )(mω2 r ) cos(π − θ) The effective gravity at point A, F = g 2 + ω4 r 2 − 2ω2 g r cos θ g′ = m 2
2 2 g ′ = g 1 + ωgR cos 2 θ − 2 ωgR cos 2 θ where, g is the acceleration due to gravity at the surface of the earth neglecting the effect of the rotation.
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2.5
(a) At poles:
θ = 90º g’ = g
⇒
Hence, the rotation of the earth has no effect on the gravity at poles. (b) At equator: θ = 0º 2 g = g 1 − ωgR
2.5 ESCAPE VELOCITY
rio s_ in fo
un iv e/ .m
YH
er si ta rio s
w
er s
w
(DUWK UDGLXV U PDVV 0
//w s: tp ht
P
ita
.ja
m
ar a
na .c om
It is the minimum velocity with which a body must be projected from the surface of the earth, so that it permanently overcomes and escapes the gravitational field of the earth. We can also say that a body projected with escape velocity will be able to go to point which is at infinite distance from the earth.
or
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
Let us imagine what happens to a body of mass m, if it is thrown from the earth with a velocity ve (escape velocity). As the body moves away from the earth, it slows down (due to gravitational pull of the earth) and, hence the kinetic energy is converted into gravitational potential energy of the mass earth system. Let us imagine that it is just able to reach up to infinity (where gravitational potential energy is zero). Kinetic energy lost by mass m = Gain in gravitational potential energy of the mass earth system 1 mv 2 = (G.P.E.) − (G.P.E.) f i 2 e 1 mv 2 = 0 − − GMm 2 e R ve =
2GM R
or
ve = 2 gR
Subsitituting the value of g = 9.81 m/sec2 and R = 6400 km, we get ve = 11.3 km/sec
Hence, any object thrown with a velocity of 11.3 km/sec or more will escape the gravitational field and will never come back to the earth.
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2.6
Gravitation
2.6 MOTION OF A SATELLITE OF MASS M AROUND THE EARTH Consider a satellite of mass m revolving in a circle around the earth which is located at the centre of its orbit. If the satellite is at the height h above the earth’s surface, the radius of its orbit is r=R+h where R is the radius of the earth. The gravitational force between m and M provides the centripetal force necessary for circular motion. K
S 5
rio s_ in fo
m
P
ita
.ja
P
un iv
er si ta rio s
er s
w w //w
Orbital Velocity
s:
P
(DUWK
ar a
P
na .c om
P
iv
//t
ht
.m
tp
e/
The velocity of a satellite in its orbit is called the orbital velocity. Let v be the orbital velocity of the satellite, then
un t.m e/
v = GM r
Also v =
s: //
or
ht tp s:
GMm = mv 2 r r2
GM R+h
ht tp
Hence, the orbital velocity of a satellite is decided by the radius of its orbit or its height above the earth’s surface. GM v = GM r ≈ R = gR
Time Period The time taken to complete one revolution is called the time period. It is given by r 2π ⋅ r r 2πr T = v = 2πr = GM GM ⇒
4π2 3 T2 = r Gm
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Gravitation
2.7
Note: For a satellite whose time period is 24 h (same as the time period of the earth’s rotation). Radius of orbit, r = 3
GMT 3 (T = 24 hrs) 4π 2
This satellite seems stationary if observed from the surface of the earth. This is also known as Geostationary satellite.
2.7 TOTAL ENERGY OF THE SATELLITE
=
1 GM Gm M m − r 2 r
=
GMm GmM − 2r r
s_ in f si ta
GMm 2r
s
GmM 2r
rio
ni v
e/ u
tp s:
Binding energy = – (Total energy) =
ita
//w w
w
=−
er
.ja m
rio
ar a
na
o
.c o
1 2 Gm M mv + − r 2
=
m
Total energy = kinetic energy + potential energy
er s
iv m e/ un
ht
:// t.m
The energy that must be given to the orbiting satellite to make it escape to infinity is known as binding energy.
tp s
2.8 TRAJECTORY OF A SATELLITE FOR DIFFERENT SPEEDS
//t .
ht tp s:
ht
Let v be the velocity given to satellite. Let νc respresent the orbital velocity and νe be the escape velocity.
vc =
GM 2GM and ve = r r
where, r is the distance of the satellite from centre at the earth.
(i) ν < νc The satellite follows an elliptical path with the centre of the earth as the farther. In this case, if statellite is projected from near the surface of the earth, it will hit the earth’s surface without completing the orbit. (ii) ν = νc The satellite follows a circular orbit with the centre of the earth as the centre of orbit. (iii) νc < ν < νe The satellite follows an elliptical orbit with the centre of the earth as the focus nearer to the point of projection. (iv) ν = νe The satellite escapes the field of the earth along a parabolic trajectory. (v) ν > νe The satellite escapes the field of the earth along a hyperbolic trajectory.
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2.8
Gravitation
2.9 TIME PERIOD OF A SATELLITE AS OBSERVED FROM THE EARTH Consider a satellite in the circular orbit with the time period Ts. The earth also rotates with the time period TE = 24 h. If an observer on the earth sees this satellite, the angular velocity of the satellite relative to the earth will be
7(
7V
→
ωSE = ωS −ωE
rio s_ in fo
→
→
na .c om
(DUWK
m
ar a
Hence, the time for one revolution will appear to be different from Ts if observed from the earth. This time period can be calculated as TSE from the above equation.
er s un iv
w //w
er si ta rio s
⇒ TSE = TSTE TE − TS
w
2π 2π 2π = − TSE TS TE
ita
.ja
(i) If satellite and the earth are rotating in the same direction.
e/
TSTE TS + TE
iv
.m
TSE =
ht tp s:
un
//t
ht
tp
s:
(ii) If satellite and the earth are rotating in opposite directons.
t.m e/
2.10 KEPLER’S LAWS
ht tp
s: //
(i) The Law of the Orbits: Each planet revolves around the sun in an elliptical orbit with the sun at one focus of the ellipse. (ii) The Law of Areas: This law state that the radius vector from the sun to the planet sweeps out equal areas in equal time intervals. Both shaded areas are equal if the time from A to B is equal to the time from P to Q. 3ODQHW
4
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Gravitation
2.9
(iii) The Law of Periods: It states that, the square of the time taken by the planet about the sun is proportional to the planet’s mean distance from the sun. If T be the time period of the planet and r be the mean distance of planet from the sun (average of maximum and minimum distances from the sun), then
⇒
r=−
rmin + rmax 2
T2 is same for all planets. r3
na .c om
rio s_ in fo
x = Separation between two stars. d = Distance between two stars (centre to centre) = 90 × 103 m
ita iv
un
or
v
)
) Solution
First method: m1 = m2 = m = 0.74 × 1029 kg r1 = r2 = r = 11.25 × 103 m
dv 2Gm =− 2 dx x
V
d /2
0
d
∫ vdv = −2Gm ∫
dx x2
or
2. Two neutron stars, each of mass 0.74 × 1029 kg and radius 11.25 km are separated by a centre to centre distance of 90 km. They are initially at rest with respect to each other. (i) How fast are they moving when their saparation has decreased to one-half of its initial value? (ii) How fast are they moving just before they collide? (ignore elastivistic effects)
ht tp
= 1.117 × 104 ms–1
GM x2 Since, the stars approach to each other, therefore 2GM a = 2a1 = 2 x dv 2GM or =− 2 dt x [negative sign indicates that x decreases as velocity increases] a1 = a2 =
t.m e/
= 2 × 9.8 × 6.37 ×106
s: //
v = 2 gR
//t
2GM = 2gR R
ht tp s:
n2 =
∴
.m
tp
ht
∴
D
G
e/
s:
potential energy 1 2 GmM 1 1 mv = −GM − m = or 2 R ∞ R
D
er si ta rio s
un iv
)
//w
w
) Solution Gain in kinetic energy = loss in
er s
w
.ja
m
ar a
1. A body is released at a distance far away from the surface of the earth. Calculate its velocity, when it is near the surface of the earth. g = 9.8 ms–2 and R = 6.37 × 106 m. Assume no resistance.
WORKED OUT EXAMPLES
d /2
v2 1 1 1 − = −2Gm − = 2Gm 2 /2 x d d d 2 v 1 = 2Gm. 2 d
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2.10
∴
∴
3. A hypothetical planet of mass M has three moons each of the equal mass m each revolving in the same circular orbit of radius R. The mass are equally spaced and thus form an equilateral triangle. Find (i) The total potential energy of the system. (ii) The orbital speed of each moon such that they maintain this configuration.
Gm v=2 d v v1 = v2 = 2
Gravitation
Gm 6.67 × 10−11 × 0.74 × 1029 = d 90 × 103 = 7.41 × 106 m/sec
=
2r
v2 1 1 1 = −2Gm − = 2Gm − 2 2r d x d
om
v2 = (6.67 × 10–11)
( 0.74 ×10 ) 22.51×10
v = 1.28 × 107 m/sec
s_ ta si
er iv un
e/
GmM R WB = Work done to place mass m at B =
GM Gm = + m AB R WC = Work done to place mass m at C GM Gm Gm = + + m BC CA R From ∆ ABC, cos 30° =
29
3
−
1 90 × 103
&
W0 = Work done to place mass M at O from infinity = 0 WA = Work done to place mass m at A
or
or
1 1 Gm − = v2 r1 + r2 d
⇒
rio
si ta //t
s:
ht tp
Gm = 7.41 × 106 m/sec. d Gm1m2 Gm1m2 1 1 − = m1v12 + m2 v 22 (ii) r1 + r2 d 2 2 v=
or
in fo
a. c .m
tp s
ht
∴
2
%
:// t.m
ht
GM = v2 d
)
∴
BC/2 R
BC = 3 R = BA = AC
And AO = BO = CO = R
tp s
e/ un
iv
w
://
Second method: (i) Using the energy conservation Loss in potential energy = gain in kinetic energy Gm1 m2 Gm1 m2 1 1 or − = m1v 12 + m2 v 22 d /2 d 2 2 But m1 = m2 and v1 = v2
)
s
w
3 × 6.67 × 10 × 0.74 × 10 90 × 103 = 1.28 × 107 m/sec
=
29
rio
−11
$ )
er
w
.ja
or
P
) Solution )
v 3 Gm v1 = v 2 = = 2 90 × 103
0 5
P
an
3 Gm 90 × 103
v=2
ar
∴
P
m
1 1 = 2Gm − 3 3 22.5 × 10 90 × 10 2 Gm × 3 = 90 × 103
(ii)
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Gravitation
Let the distance x from the larger block the force is zero. Then GmM 16 MGm = (10a − x) 2 x2
3Gm m M+ R 3
or x2 = 16 (10a – x) or x = 4 (10a – x) or x = 8a From the energy conservation at point P
GMm along AO R2 Force on A due to mass m at Gmm C, F2 = AC 2
Gm 2 along AC 3R2
or
∴ Net force on A = F1 + F2 sin 60° + F2 sin 60°
D
in
s_
s
rio
ta
3 5GM 2 a
si er
e/
un
iv
5. Taking the earth to be a uniform sphere of radius 6400 km and the value of g at the surface to be 10 m/sec2, calculate the total energy needed to raised a satellite of mass 2000 kg to a height 800 m above the earths’s surface and to set it into circular orbit at that altitude.
.m
) Solution )
//t
s:
tp
ht
D
[
GmM 45 a 8 Gm 45 v2 = ⋅ a 4 =
v=
e/ .m
ht
)
) Solution
D±[
∴
//t
s:
tp
4. Distance between the centre of two stars is 10a. The mass of these stars M and 16 M and their radii a and 2a respectively. A body of mass m is fired staright from the surface of the larger star towards the smaller star. What should be its minimum intial speed to reach the surface of the smaller star? Obtain the expression in terms of G, M and a.
D
rio
si er un
iv
//w
s:
tp
ht ∴
w
or
w
Gm Gm v= + R 3R
ta
.ja
or
For circular motion at m1, mv 2 Gm m = 2 M + R R 3
1 2 GmM 64 − 16 + 1 − 4 mv = 2 a 8
an
ar
or
m
Gm2 3 GMm = + 2 2 2 R 3R 2 Gm m m+ 2 R 3
1 2 GmM 16 16 1 1 mv = − + − 2 a 2 8 8 2
a.
1 2 GmM Gm(16M) mv − − 2 8a 2a GmM Gm(16M) =− − 2a 8a
co m
=
=
fo
(ii) Force on A due to mass M at O, F1
(i) Potential energy of the system = – [W0 + WA + WB + WC] =−
2.11
GMm R where, m = mass of the satellite and M = mass of the earth GmM Final energy = (R + h)
Initial energy =
Change in energy = Final energy – Initial energy GmM GmM GmM GmM = −− − = R+h R R R+ h
GM Rh h = GmM = 2 .m R(R + h ) R (R + h)
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2.12
s
rio
si ta
er
un
e/ )
t.m :// tp s
ht
rmin + rmax r + R = 2 2
ht
iv
//t
tp
s:
r=
Now, T 2 =
∴
3
T=π
( r + R )3 2GM
8. A planet A moves along elliptical orbit around the sun. At the moment when it
4π 2 r 3 GM
r+R 4π 2 2 3 2 = π (r + R) = GM 2GM
9.8m −2 × ( 6.4 × 106 m )2 = 7.8 × 106 m ) ( = 7174 ms–1 (iii) Let t = time taken between two successive times at which the satellite is overhead to an observer fixed at a position on the equator. θ1 = Angle turned through by the rotating earth during t seconds.
s_
rio
ita rs
iv e un
e/
.m
tp
ht
1/2
= 7417 sec
7. A planet of mass M moves around the sun along the ellipse, so that its minium distance from the sun is equal to r and the maximum distance is equal to R. Using Kepler’s laws, find its period of revolution around the sun. ) Solution According to the Kepler’s Laws,
1/2
in fo
ar an
m
w .ja
//w
s:
1/2
4π 2 r 3 (i) Period of the satellite, T = GM 1/2 2 4π = 2 r3 gR
1/2
2π Here, TS = ω = Period of the satellite S = 6831 sec. 2π and TE = ω = Period of rotation of E
the earth = 24 × 3600 sec. T ⋅T 24 × 3600 × 6861 t= S E = sec TE − TS 24 × 3600 − 6861 ∴ T ⋅T 24 × 3600 × 6861 t= S E = sec TE − TS 24 × 3600 − 6861
w
)
Radius of the satellite orbit, r = R + h = (640 + 1400) km = 7.8 × 106 m
4π2 × (7.8 × 106 m)3 = −2 6 9.8ms × (6.4 × 10 m) = 6831 sec (ii) Circular orbital speed,
a. co m
= 1.6 × 106 J 6. Assuming the radius of the earth to be 6400 km, calcluate the period of revolution of a satellite which is describing an equatorial orbit at 1400 km above the surface and its speed in its orbit. If the satellite is travelling in the same direction as the rotation of earth, i.e., west to east, what is the interval between two successive times at which it will appear vertical overhead to an observer at a fixed point on the equator?
GM R2 vc = = g r r
6400 × 103 × 800 = 10 × 2000 3 6400 × 10 + 800
) Solution
θ2 = Angle turned through by the revoling satellite during t seconds. θ2 2π θ1 = + Then, θ2 = 2π + θ1 and t t t θ2 = Angular speed of the But, t satellite = ωS θ1 = Angular speed of the earth = ωE t 2π = ωS − ωE ∴ t 1 ω ω 1 1 ∴ = S − E = − t 2π 2π TS TE
Rh Energy needed = g . m R+h GM where, g= 2 R ∴ Energy needed
Gravitation
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2.13
Gravitation
was at the distance r0 from the sun its velocity was equal to v0 and the angle between the radius vector r0 and the velocity vector v0 was equal to α. Find the maximum and minimum distance that will separate this planet from the sun during its orbital motion.
angular momentum L of this planet to the centre of the sun. )
) Solution From the conservation of momentum
1 2 Gs ms m 1 2 Gs ms m mv1 − = mv2 − 2 r1 2 r2
)
) Solution From the conservation of angular
momentum about the sun mv0r0 sin α = mv1r1 = mv2r2 or v1r1 = v2r2 = v0r0 sin α
in
ta r
(r2 + r1 ) 1 = 2Gs ⋅ ms r22 r1 ⋅ r2
er si ta rio
s
v12
iv
v12 = 2Gs ⋅ ms
.m
or
e/ un
//t .
s:
ht tp
2Gs ms v12 r12 2Gs ms = 2 − r1 r2 r2
r2 1 1 v12 1 − 12 = 2Gs ms − r2 r1 r2
si
m
or
e/ un
iv
or
er
w
w
:// w
tp s
2G m −2Gs ms ± 4G s2 ms2 + 4 ( v02 r02 sin 2 α ) v02 − s s r0 r1 = 2G m 2 v02 − s s r0
ht
...(iii)
io s_
m
.ja
.c om
an a
v12 −
or
2 2G m 2 2 2 or v0 − s s r1 + 2G s ms − v0 r0 sin α = 0 r 0 or
mv1 r1 v1 r1 = m r2 r2
From equations (ii) and (iii), we get
ar
v02 Gs ms v02 r02 sin 2 α Gms − = − 2 r0 2r12 r1
From equation (i), v2 =
...(ii)
fo
From equation (i)
2Gs ms 2G m = v22 − s s r1 r2
1 2 G s ms m 1 2 G s ms m mv0 − = mv1 − 2 r0 2 r1
v12 −
or
...(i)
For the conservation of energy
v1 = 2Gs ms
r2 r1 ( r1 + r2 )
r2 r1 (r1 + r2 )
s: //t
From equation (i) Angular momentum = mv1 r1
ht tp
= m 2Gs ms
r1r2 (r1 + r2)
where,
η=
v02 r0 Gs ms
) Solution )
Case I When r < R Let us divide the sphere in two parts by imagining a concentric spherical surface passing through P.
9. A planet of mass m moves along an ellipse around the sun, so that its maximum and minimum distance from the sun are equal to r1 and r2 respectively. Find the
m = mass of sun; and Gs = gravitational constant of sun.
10. There is uniform sphere of mass M and radius R. Find the strength E and potenial V of the gravitational field of this sphere as a function or the distance r from its centre (with r < R and r > R). Draw the approximate plates of the function E (r) and V (r).
1 ± 1 − (2 − η)η sin 2 α = r0 (2 − η)
2 v2 1 ± 1 − v02 r02 sin α − 0 r0 rms = 2 v0 − r0 Gs ms
...(i)
mv1 r1 = mv2 r2 From conservation of energy,
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Gravitation
The inner part of mass 5
GM − R3 r for r ≤ R g= GM − r for r ≥ R R3
Hence, 3 U
G[
4 Mr 3 × πr 3 = 3 4 3 3 R πR 3
M
The potential V due to this inner part GM′ V1 = − r M 3M dm = 4πx 2 dx = 3 x 2dx 4/3πR3 R Gdm 3M d V2 = − = − 3 xdx x R
m
rio si ta
er
Let v1 and v2 be the linear speeds with respect to the planets S1 and S2 and R1, R2 their radii respectively. From the Kepler’s law the square of the time period is proportional to the cube of the radius.
iv
un e/ or
R23 = η2 R13 R2 = η2/3 R1
or
2
R 2 ηT = R1 T
As the time period of S1 is T, therefore
...(i)
2πR1 T 2πR 2 Similarly, v2 = T v1 =
tp
s:
.m
3
∴
2πη2/3 R1 ...(ii) T Clearly, at the closest separation, they are moving in the opposite directions. Hence, the speed of S2 with respect to S1 becomes =
When r > R G dm G dV = − = − ∫ dm r R GM =− r GM gr = 3 r r
Case II
ht
GM = − 3 r R
//t
ht
GM for r ≤ R R
dV 3 2r GM = − 0 − 2 dr 2 3R R
6
s
er
//t
s:
tp
3 r = − 1 − 2 2 3R gr =
) Solution
.m
ht
GMr 2 3GM 2 − (R − r 2 ) R3 2 R3 GM = − 3 ( 3R2 − r 2 ) 2R =−
6
5
iv
un
e/
V = V1 + V2
2
5
si ta
.ja
w
w
...(ii)
tp s:
//w
3 GM 2 2 R − r =− 2 R2
o
s_
R
GM 3GM x2 V2 = ∫ − 3 xdx = − 3 R R 2 r r
)
R
rio
m
ar an
in f
a.
co
...(i)
11. Two satellite S1 and S2 revolve around a planet in coplaner circular orbits in the opposite sense. The period of revolutions are T and ηT respectively. Find the angular speed of S2 as observed by an astronaut in S1 when they are closest to each other.
M′ =
3 r 2 Gm − − 1 for r ≤ R 2 3R3 R V= GM − R for r ≥ R
[
2.14
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2.15
Gravitation
Hence, total field at point P2 g = g1 + g 2
2πR1 ...(iii) (1 + η2/3 ) T At the closest separation, the distance between S1 and S2 is
=
T ( η2/ 3 − 1)
si
ta
rio
( 6a )
2
(ii)
GM 36 a 2
=
At point P2, due to the thin sphere GM g2 = 2 ( 5a )
si
GM e T 2 = 2 4π
1/3
gr 2 T = e2 4π
1/3
10 × ( 6400 × 103 )2 ( 8.64 × 104 )2 = 2 4 3 . 14 ( ) 1/3 GM 10 × ( 6400 × 103 )2 (8.64 × 104 )2 = ≅ 4.28 × 107 m = 6.7 r 25a 2 = 2 e 4 ( 3.14 ) https://t.me/universitarios_infohttps://www.jamarana.com https://t.me/universitarios
Mm M = −Gm e + R − r r r1 = Altitude of geosynchronous orbit 1/3
GM
g1 =
At point P2, due to inner sphere.
er
iv
GM 16 a 2
Here, re = radius of the earth R = distance from the earth to the moon = 60 re M = 11,000 kg, rm = radius of the moon Potential energy of the body = Potential energy of the body due to the earth + Potential energy of the body due to the moon GM e m GM m m =− − r R −r
un
e/
ht tp
Since thin sphere inside the field is zero, so at point P1, gravitational field g=
ta
)
GM GM = 2 (4a ) 16a 2
rio
s
er
iv
ht
g1 =
s: //t .m
)
tp s
) Solution Field at point P1 due to the inner sphere,
(i)
un
) Solution
:// t
ht
.m
2
3
e/
D
3
s_ i
an
ar
m
.ja
w w
:// w
D
tp s
D
nf o
a. co m
12. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a shown in the figure. The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P1 and P2 shown in the figure.
2π ( η2/ 3 + 1)
=
v 2 +v1 R 2 − R1
ω=
Hence, the relative angular speed
GM GM 61 GM + = 36 a 2 25a 2 900 a 2
13. (i) Find an expression for the gravitational potential energy of mass m located at a distance r from the earth’s centre along a line joining the earth and the moon. (ii) Find the work done in boosting the body of mass 11,000 kg from geosynchronous orbit to the moon surface. (iii) Hence or otherwise, find an expression for the net gravitational force and find a point where the force vanishes. Mass of earth = 81 × mass of moon Radius of earth = 3.68 × radius of moon = 6400 km Acceleration due to gravity near the surface of the earth = 10 ms–2 Distance of the moon from the earth = 60 × radius of the earth
...(iv)
R2 – R1 = R1(η2/3 – 1)
v1 + v2 =
≅ 4.28 × 107 m
2.16
U′geo = Potential energy of system at geosynchronous orbit
M Mm = −Gm e + r R − r1 1 GM m m GM e m − R − rm rm
Umoon = −
1 = Work needed to lift the body from geosynchronous orbit to the moon, W = Umoon – Ugeo M Mm M Me = Gm e + − e− r R − r r R − rm 1 m 1
14. A spherical hollow is made in a lead sphere of radius r, such that its surface touches the outside surface of the lead sphere and passed through its centre. The mass of the sphere before hollowing was M. With what force according to the Law of Universal Gravitation will the hollow lead sphere attract small sphere of mass m, which lies at a distance d from the centre of lead sphere on the straight line connecting the centres of the spheres and of the hollow?
Gravitation
a. co
m
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∴
Mm M = e 2 r2 (R − r )
r M m = (R − r ) me r=
R Me Me + Mm
(60re ) 81 = = 54 rr = 3.46 × 108 m 51 + 1
) Solution Let the mass A is at a distance r from
Force is zero, when
15. A tunnel is dug through the earth along one of its diameters. Show that the body dropped in this tunnel will execute simple harmonic motion about the centre of the earth. Assume that the earth is a sphere of uniform density. )
GM e m GM m m + r2 (R − r ) 2
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2 M P M M M 5 e e e e = Gm + − − r r1 81( 60re − r1 ) 81 re 60r − e e 3.68 3.68 ) Solution Mass of the hollow sphere 3 4 R π Me Me Me − − M 3 2 = ×M = 81( 60re − r1 ) re 60r − re 4 81 8 e × πR 3 3.68 3.68 3 Force of attract = Due to the bigger sphere – 1 1 1 1 = M g re + − − due to the smaller sphere 6.7 81(53.3) 22 59.7 GMm GmM = − ≈ 6.14 × 1010 J. 2 2 d R 8 d − (iii) Gravitational force lies along the earth2 moon line dU ∴ F=− 1 1 dr = GmM 2 − 2 d d Me Mm R = Gm + 8 d − dr r R −r 2
the centre of sphere
M 4 3 Gm ⋅ πr 3 GmM 4/3πR 3 = r F= 2 R3 r
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Gravitation
Hence it is S.H.M. F = Kr GmM K= R3
$ P
⇒
5
Time period, T = 2π
m K
mR3 R3 = 2π GMm GM R 6400 × 103 = 2π = 2 × 3.14 × g 9.8
The magnitude of this force is proportional to r and its direction is towards centre. Hence, this acts as a restroing force towards the equilibrium position at centre O.
2.17
= 2π
= 84.6 min
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3. P is a point at a distance r from the centre of a solid sphere of radius a. The gravitational potential at P is V. If V is plotted as a function of r, which is the correct curve?
2. A simple pendulum of length one metre is taken to a height R, which is the radius of earth. If the acceleration due to gravity on the surface of earth is π2 m/s2, what is the period of the pendulum at this height?
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2
(a)
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l . g where g is acceleration due to gravity on the surface of earth and is given as π2 m/s2. At height R, the acceleration due to gravity, T = 2π
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we get ω =
(b) 2 s (d) 4 s
) Explanation Time period of simple pendulum,
) Explanation Putting ge = 0, in the expression ge = g0 – rω2
q
(a) 1 s (c) 3 s
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1. The period of rotation of the earth so as to make any object weightless on its equator is (a) 84 minutes (b) 74 minutes (c) 64 minutes (d) 54 minutes where M are r are the mass and radius of earth.
m
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2.18
5. Three particles each of mass m are placed at the corners of an equilateral triangle of side d. The potential energy of the systems is
U D
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3Gm 2 d
Gm 2 d
(d) none of these
) Explanation For the system of two particles, )
(c)
9
(b)
U
2
3Gm2 d
(a)
(b) 9
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2
Gravitation
gravitational energy is given as Gm1m2 r UA = U2 + U23 + U31
U
U=−
U A = −3
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Hence (c) is the correct. –ve sign resembles that the particles are bounded by their mutual gravitational field.
6. Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west and B, from west to east. Which train will exert greater force on the tracks? (a) A (b) B (c) They will exert equal force (d) The mass and the speed of each train must be known to reach a conclusion.
Given g = g′
G 4 3 G 4 πR ρ = 2 × πR′3 (2ρ) R2 3 R′ 3 R R′ = or 2
∴
4 3 3 πR (2ρ)
) Explanation Let, v = speed of each train relative
G (R′) 2
to the earth’s surface, vE = speed of earth’s surface relative to the earth’s axis, vA, vE = speeds of A and B relative to the earth’s axis.
For the planet, g ′ =
)
where ρ is the density of the earth.
GM G 4 3 = πR ρ R 2 R 2 3
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P
P
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4. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of earth R will be R R (b) (a) 2 4 R R (d) (c) 3 8
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−GM 2 2 ) Explanation V = 3 (3a − r ) inside, and 2a −GM = outside the sphere. r Hence (c) is the correct.
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Gravitation
Then, vA = vE – v, vB = vE + v
)
) Explanation At infinity, the total energy of the
body is zero. Therefore the total energy of the body just before hitting the planet of P will be zero according to the conservation of energy ⇒ Fp = E∞ = 0
v NA = mg − m R
2 A
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8. A body of mass m is approaching towards the centre of a hypothetical hollow planet of mass M and radius R. The speed of the body when it passes the centre of the planet through its diametrical chute is
3GMm 8r 2
(d)
(b)
(c)
GMm 8r 2
(a)
GMm 4r 2
GMm 8r 3
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2GM R (d) none of these (b)
GM R (c) Zero (a)
mgr h 2 1 + r
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9. A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is at a distance 3 r from the centre of the sphere as shown in the figure. The gravitational force exerted by the sphere on the ring will be
ve r
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v=
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=−
na
)
) Explanation Energy required to raise it to a heigh h E1 = change in potential energy
−
Since the force imparted on a particle inside spherical shell is zero, therefore the veloctiy of the particle inside the spherical shell remain constant. Therefore, the body passes the centre 2GM of the planet with same speed v = . R
⇒
Up + Kp = 0
7. The ratio of the energy required to raise a satellite up to a height h above the earth to that required to put it into orbit there is (a) h : R (b) R : h (c) 2h : R (d) R : 2h
⇒
v2 N B = mg − m B R NA > N B
∴
2.19
at a distance
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2.20
Gravitation
GMm 3r r 2 +
=
2 3/2
( 3r )
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E=
3 GMm 8r 2
10. A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of co-ordinates. Two spheres of equal radii 1 unit with their centers at A (–2, 0, 0) and B (2, 0, 0) respectively are taken out of the solid leaving behind spherical cavities as shown in the figure. (a) The gravitational field due to this object at the origin is zero (b) The gravitational potential is the same at all points of the circle y2 + z2 = 36 (c) The gravitational potential is the same at all points on the circle y2 + z2 = 4 (d) All of these
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) Explanation Here cavities and mass of the
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,
(b)
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OLVED OBJECTIVE TYPE UESTIONS
1. In a gravitational force field a particle is taken from A to B along different paths as shown in the figure. Then (a) work done along path I is more (b) work done along path III is less (c) work done along path IV is more (d) work done along all paths is the same
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∴ Force = M × E =
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Gravitation
F1 r22 = if r1 > R and r2 > R F2 r12
(c)
F1 r1 = if r1 > R and r2 > R F2 r2
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11. The ratio of the radius of the earth to that of the moon is 10. The ratio of the acceleration due to gravity on the earth to that on the moon is 6. The ration of the escape velocity from the earth’s surface to that from the moon is (a) 4 (b) 6 (c) 8 (d) 12
8. A spaceship moves in the absence of external force with constant velocity. To change the direction of motion, a jet engine is switched on. If it starts ejecting gas relative to the spaceship and directed at right angle to the spaceship’s motion, then (a) angular velocity will be constant (b) the magnitude of velocity will be change (c) it will follow circular path (d) none of these
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10. A point p lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from and reaches C under gravitational attraction only, if speed at C will be
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7. If E be the energy require to raise a satellite to a height h above the surface of earth and E′ be the energy require to put the same satellite into orbit at that E height, then is equal to E′ 2R 3h (a) (b) h 2R R 2h (c) (d) h R
m
(d)
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6. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be (a) 1/2 h (b) 1 h (c) 2 h (d) 4 h
(b)
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5. Let the acceleration due to gravity be g1 at a height h above the earth’s surface, and g2 at a depth d below the earth’s surface. If g1 = g2, h < < R and d < < R, then (a) h = d (b) h = 2d (c) 2h = d (d) It is not possible for g1 to be equal to g2
4H 3g
(d)
9. The magnitude of the gravitational field at distance r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then F r (a) 1 = 1 if r1 < R and r2 < R F2 r2
H g
2H 3g
(b)
(c)
2H g
(a)
height H < < R above the earth’s surface, and reaches the earth’s surface in time t. Then t is equal to
2.21
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2.22
Gravitation
tion due to gravity on the surface of the earth will (a) decrease by 2% (b) decrease by 0.5% (c) increase by 2% (d) decrease by 0.5% 13. A particle of mass m is placed inside a spherical shell, away from its centre. The mass of the shell is M. The particle will (a) move towards the centre. (b) move away from the centre, towards the nearest wall. (c) move towards the centre if m < M, and away from the centre if m > M. (d) remain stationary
d will increase, v will increase d will increase, v will decrease d will decrease, v will decrease d will decrease, v will increase
17. A sphere of mass M and radius b has a concentric cavity of radius a as shown in figure. The graph which shown variation of gravitational potential V with distance r from the center of sphere is 9
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16. A satellite going round the earth in a circular orbit loses some energy due to a collision. Its speed is V and distance from the earth is d. Then
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(d)
E
U
D
18. A body of mass m is raised to a height h (h << R) from the surface of the earth
E
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(d) none of these
15. A planet is moving around a sun in circular orbit of radius R. If we increase the velocity of the planet 2 times. Find the path of the planet. (a) ellipse (b) circular (c) parabola (d) hyperbola
V1r12 r2
U
14. A planet is revolving in an elliptical orbit around the sun as shown in figure. What will be the areal velocity? rV Vr (a) 1 1 (b) 2 1 3 q (c)
E
(a)
a. c
(a) (b) (c) (d)
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2.23
Gravitation
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7. (c) 15. (c)
EXERCISE 1
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6. (c) 14. (a)
ita er s
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5. (c) 13. (d)
1. Four identical particles, each of mass m, move along the circumference of a circle of radius r under the action of their mutual gravitational attraction. Determine the speed of each particle. Gm 2 2 +1 Ans. 4r
25 9 5 (d) 27 (b)
20. A satellite is revolving around a planet in an elliptical orbit under its gravitational field. It is seen that the linear momentum of the satellite varies with the radius vector as R–1/2. Then the angular momentum of the satellite is proportional to (a) R1/2 (b) R 3/2 (c) R (d) R0
na .c om ar a
4. (a) 12. (c) 20. (d)
m
3. (b) 11. (c) 19. (d)
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w
1. (d) 9. (b) 17. (c)
19. Two satellites A and B of masses m1 and m2 respectively are moving in the same circular orbit of radius R, in the opposite m 5 directions, where, 1 = . Just after a m2 3 perfectly elastic collision between them, the ratio of their kinetic energies will be
ANSWERS
2. If a double star (a system of two stars moving around the centre of inertia of the system due to gravitation) has a mass M and its period of revolution is T, then find the distance between the components of the double star.
&0
3 5 9 (c) 25 (a)
where the acceleration due to gravity is g. The loss in weight is mgh mgh (a) (b) R 2R 2mgh mgh (c) (d) R 4R
1/3 GmT 2 Ans. 2 4π
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P
2.24
8. Find the gravitational attraction due to a semicircular wire of length L and of mass M on a particle of mass m placed at the centre of the semicircular wire.
3. Find the gravitational potential energy of a particle of mass m and a thin uniform rod of mass M and of length L, if they are situated at a distance l apart. Find the force of interaction between them.
2πGMm Ans. L2
GMm GMm L Ans. U= L log e 1 + l ;F = l (L + l )
rio s_
10. A planet P moves in an elliptical orbit round the sun. At the instant when its distance from the sun was r0, its velocity was v0 and the angle between the radius vector r0 and the velocity vector v0 was equal to θ. Find the maximum and the minimum distances of the planet from the sun during its motion.
e/
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ta
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rs
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w
w
//w
s:
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3
6
9
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s:
tp
6. Using the conservation laws, show that the total mechanical energy of a planet of mass m moving in an elliptical orbit round the sun of mass M depends only on its semi-major axis a. Find this energy as a function of a.
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2GMr1r2 r1 + r2
Ans.
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5. A planet of mass m moves in an elliptical orbit round the sun of mass M so that its farthest and the nearest distances from the sun are r1 and r2 respectively. Find the angular momentum of the planet about the centre of the sun.
9. A satellite is put in an orbit just above the earth’s atmosphere with a velocity 1.5 times the velocity for a circular orbit at that height. The initial velocity is parallel to the surface. What should be the maximum distance of the satellite from the earth, when it is in the orbit? [Roorkee, 1980] [Ans. 2R + 3h]
4. Consider an earth satellited position so that it appears stationary to an observer on earth and serves the purpose of a fixed relay station for intercontinental transmission of television and other communication. What should be the height at which the satellite should be positioned and what would be its direction of motion? (Radius of earth R = 6400 km) [IIT, 1973] 6 [Ans. 35.94 × 10 m]
Gravitation
tp s:
9
ht
r0 r0 2 rmax = 2 − η 1 + 1 − η(2 − η)sin θ ; rmin = 2 − η 1 − 1 − η( GMm Ans. E = − 2a , where, 2a = r1 + r2 = major axis of the ellipse. r0 r0 2 2 rmax = 2 − η 1 + 1 − η(2 − η)sin θ ; rmin = 2 − η 1 − 1 − η(2 − η)sin θ m , where, 2a = r1 + r2 = major axis ofthe ellipse. 11. What will be the acceleration due to 7. If the radius of the earth were to shrink gravity on the surface of the moon if by one pernocent, its mass remaining the 1 its radius were th the radius of the same, the acceleration due to gravity on 4 1 the earth’s surface would decrease by earth and its mass is th of the mass what amount? [IIT, 1981] 80 of the earth? [Ans. 2%]
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R Ans. 2 gR 2 − 1 v0
17. The astronauts in the satellite orbiting the earth feel weightless. Does the weightlessness depend upon the distance of the satellite from the earth? If so, how? Explain your answer. [Roorkee, 1985] [Ans. The satellite does not exert any force on the astronaut and hence the astronauts in a satellite are in a state of weightlessness. This state of weightlessness does not depend upon the distance of the satellite from the earth.]
12. A planet of mass m moves towards the sun with a velocity v0 (when far away from the sun) and directed along a line whose perpendicular distance from the sun is d. Assuming the mass of the sun to be M, find the minimum distance of the planet from the sun.
2.25
R and acceleration due to gravity on its surface is g.
What will be the escape velocity on the surface of the moon if it is 11.2 km/s on the surface of the earth? [Roorkee, 1983] [Ans. 2.5 km/s]
Gravitation
19. A satellite of the earth has to move in an equatorial plane close to the surface of the earth either in the direction of rotation of the earth or against it. Calculate by how many times the kinetic energy of the satellite in the latter case exceeds that in the former. [Ans. 1.27] 20. Two satellites S1 and S2 revolve round a planet in coplanar circular orbit in the same sense. Their periods of revolution are 1 hour and 8 hour respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1, find: [IIT, 1986] (a) the speed of S2 relative to S1 (b) the angular speed of S2 actually observed by an astronaut in S1. [Ans. (a) 3.14 × 104 km/hour. (b) – 2.907 × 10–4 rad/sec]
21. A solid sphere has a mass M and radius R. A spherical cavity is made in it in such
16. On the north pole of the earth a body is thrown vertically upward with a velocity v0. Find the maximum height attained by the body if the radius of the earth is
18. Find the rate at which the earth should rotate about its axis so that the apparent value of acceleration due to gravity at the equator becomes zero. What will be the length of the day in this case? [Ans. 1.41 hour]
15. If a planet of mass m be suddenly stopped in its circular orbit of radius r, show that it would fall into the sun in a time ( 2 / 8), times the period of revolution of the planet.
ht
ht
14. Imagine a planet having mass and radius half of the corresponding quantities for earth. The surface temperature of this planet reaches upto 800 K. Estimate whether there is a possibility of presence of oxygen molecule in the planet’s atmosphere. Given escape velocity from earth = 11.2 km/second, Boltzman constant k = 1.38 × 10–23 J/K, mass of a molecule of oxygen = 5.3 × 10–26 kg. [Roorkee, 1984] [Ans. There is possibility of oxygen molecule on the surface of the planet as the molecule cannot escape from its surface.]
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3 Gm 2 Ans. 2 a
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13. Three identical particles, each of mass m, are placed at the three vertices of an equilateral triangle ABC of side a. Find the work required to be done to increase the distance between the particles from a to 2a.
v4d 2 GM Ans. 2 1 + 02 2 − 1 v0 G M
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2.26
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t.m e/
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s:
22. An artificial satellite of moon revolves in a circular orbit whose radius exceeds the radius of the moon n times. Assuming that the satellite experiences a resistive force due to cosmic dust that depends upon the velocity v of the satellite as F = kv2, where k is a constant, Find how long the satellite will stay in the orbit until it falls on to the moon’s surface?
26. Three particles, each of mass m are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle move in a circle while maintaining the original separation a. Find the initial velocity that should be given to each particle and also the time period of circular motion. [IIT, 1988; MNR, 1990]
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1 Ans. F = GMm 1 − 2 r2 R 8 1 − 2r
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25. Find the distance of a point from the centre of the moon at which the strength of the earth’s and the moon’s gravitational fields are equal and opposite and the resultant field is equal to zero. Take the mass of the earth to be η = 81 times greater than that of the moon and the distance between the centres of these planets to be n = 60 times greater than the radius of the earth R. [Ans. 3.84 × 104 km]
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24. A space-ship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the space-ship in the orbit to overcome the gravitational pull? (Radius of the earth = 6400 km, g = 9.8 m/s2) [Roorkee, 1988] [Ans. 3.2804 km/s]
a way that the boundary of the cavity touches the surface of the sphere and passes through its centre, as shown in the figure. Find an expression for the gravitational force on a particle of mass m at a distance r from O along the line joining the centres.
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[Ans. 7.826 × 10–4 rad/sec]
28. The mass M of a planet-earth is uniformly distributed over a spherical volume of radius R. Calculate the energy
much as present. Take the equatorial radius as 6400 km. [Roorkee, 1988]
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27. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape m ( n − 1) moon’s from sufacethe earth; Ans. Time after which the satellite will fall on to thevelocity k height gR of the satellite (a) Determine the above the earth’s surface. m ( n − 1) which the satellite will fall on to the moon’s suface (b) If the satellite is stopped suddenly in k gR its orbit and allowed to fall freely into the earth, find the speed with which it 23. Determine the speed with which the earth hits the surface of the earth. has to rotate on its axes so that a person [IIT, 1990] 3 on the equator would weigh th as [Ans. (a) h = 6400 km, (b) v = 7.919 km/s] 5
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astronaut fall on a straight-line from the earth’s centre. The radius of the earth is 6400 km. [Roorkee, 1998] –2 [Ans. 3 × 10 N] 30. A body is projected vertically upwards from the bottom of a crater of Moon R , where R is the radius of of depth 100 Moon with a velocity equal to the escape velocity on the surface of the Moon. Calculate maximum height attained by the body from the surface of the Moon. [Ans. 99.50 R]
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29. A cord of length 64 m is used to connect a 100 kg astronaut to a space-ship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth’s surface. Also assume that the space-ship and the
needed to deassemble the planet against the gravitational pull amongst its constituent particles. Given, MR = 2.5 × 1031 kg. m and g = 10 m/s2. [IIT, 1992] [Ans. 1.5 × 1032 joules]
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UESTION BANK
1. Can a satellite coast in a stable orbit in a plane not passing through the earth’s centre? Explain your answer.
6. Does the escape speed of a body from the earth depend on (a) the mass of the body (b) the location from where it is projected (c) the direction of projection (d) the height of the location from where the body is launched?
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CONCEPTUAL UESTIONS
Q
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7. An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 ms–2. What is the acceleration of the astronaut, the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.)
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4. You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
9. An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity? 10. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed
5. A satellite in low earth orbit experiences a small drag force from the earth’s atmosphere. What happens to its speed because of this drag force?
3. Explain why the moon has no atmosphere.
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(a) Which of the dashed elliptical orbits shown in the figure will the shuttle then take? (b) Is the orbital period T of the shuttle (the time to return to P) is greater than, less than, or the same as in the circular orbit?
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8. Two planets A and B have the same material density. If the radius of A is twice that of B, then the ratio of the escape velocity VA/VB is (a) 2 (b) 2
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3. A simple pendulum has a time period T1 when on the earth’s surface, and T2 when taken to a height R above the earth’s surface, where R is radius of earth. The value of T2/T1 is
(d)
6. The gravitational potential of two homogeneous spherical shells A and B of same surface density at their respective centres are in the ratio 3 : 4. If the two shells coalesce into single one such that surface charge density remains same, then the ratio of potential at an internal point of the view shell to shell A is equal to (a) 3 : 2 (b) 4 : 3 (c) 5 : 3 (d) 5 : 6
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Gm along + x-axis 2 Gm (b) along + y-axis π 2 2πGm (c) along + x-axis 2 2πGm (d) along + y-axis 2 2. Let ω be the angular velocity of the earth’s rotation about its axis. Assume that the acceleration due to gravity on the earth’s surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth’s surface at a pole (d <
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5. A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass 2 M/3 and a point mass M/3 at the centre. Change in the force of gravity experienced by the man is
1. Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length ℓ is
(c)
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ONLY ONE OPTION IS CORRECT
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4. If the radius of the earth be increased by a factor of 5, by what factor its density be changed to keep the value of the same (a) 1/25 (b) 1/5
(c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
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9. The escape velocity of a planet is Ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be (a) ve (b) ve / 2 (c) ve/2 (d) zero
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3m, 3m and m respectively. The gravitational force on a fourth particle S of mass m is equal to
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11. The kinetic energy of a satellite in an orbit close to the surface of the earth is E. What should be its kinetic energy so that it escapes from the gravitational field of the earth? (b) 2E (a) 2E
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10. A (non-rotating) star collapses onto itself from an initial radius Ri with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration ag on the surface of the star as a function of the radius of the star during the collapse? DJ
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14. An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) = E0. Its potential energy is
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Gravitation
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(b) 1.5 E0 (d) E0
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23. A particle is projected vertically upward from the surface of the earth with a speed 3 gR, R being the radius of the earth of 2 and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance) (a) 3R/4 (b) R (c) 2R (d) 3R
24. A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit of radius r + ∆ r (∆r << r) with time period T + ∆ T then, ∆T 2 ∆r ∆T 3 ∆r = = (b) (a) T 3 r T 2 r ∆T ∆r ∆T ∆r = =− (d) (c) T r T r
19. A satellite can be in a geostationary orbit around earth at a distance r from the centre. If the angular velocity of earth about its axis doubles, a satellite can now be in a
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21. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be (a) 1/2 h (b) 1 h (c) 2 h (d) 4 h
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17. Three identical stars, each of mass M, form an equilateral triangle (stars are positioned at the corners) that rotates around the centre of the triangle. The system is isolated and edge length of the triangle is L. The amount of work done, that is required to dismantle the system is 3GM 2 3GM 2 (a) (b) 2L L 2 3GM GM 2 (c) (d) 4L 2L
20. A planet of mass m is in an elliptical orbit about the sun (m << Msun) with an orbit period T. If A be the area of orbit, then its angular momentum would be (a) 2 mA/T (b) mAT (c) mA/2T (d) 2 mAT
16. A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be (a) zero 3GM (b) R
15. Two point masses of mass 4 m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be (a) 1 : 4 (b) 1 : 5 (c) 1 : 1 (d) 1 : 2
geostationary orbit around earth if its distance from the centre is r r (b) (a) 2 2 2 r r (c) (d) 1/3 (4) (2)1/3
(a) –E0 (c) 2 E0
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25. A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is – (opposite to the centre of gravity) [Here g = GM/R2, where M is the mass of the sphere] mg 3mg (a) (b) 2 8 mg (c) (d) None of these 16
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1. Assuming the earth to be a sphere of uniform density the acceleration due to gravity (a) at a point outside the earth is inversely proportional to the square of its distance from the centre (b) at a point outside the earth is inversely proportional to its distance from the centre (c) at a point inside is zero (d) at a point inside is proportional to its distance from the centre.
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28. A shell is fired vertically from the earth V with speed esc , where N is some number N greater than one and Vesc is escape speed for the earth. Neglecting the rotation of the earth and air resistance, the maximum altitude attained by the shell will be (RE is radius of the earth) N 2 RE NR (a) (b) 2 E N 2 −1 N −1 R RE (d) E2 (c) 2 N N −1
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ONE OR MORE THAN ONE OPTION MAY BE CORRECT
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3. A double star is a system of two stars of masses m and 2 m, rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars then their time
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30. The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is M s r2 (a) × 100 M m r1
(b)
26. A satellite is launched in the equatorial plane in such a way that it can transmit signals upto 60º latitude on the earth. The angular velocity of the satellite is GM GM (a) (b) 8R3 2 R3 (c)
(a) 8 hours (c) 24 hours
Gravitation
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earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall and the acceleration of the particle varies with x (distance of the particle from the centre) according to
4. Mark the correct statement/s (a) Gravitational potential at curvature centre of a thin hemispherical shell of radius R and mass M is equal to GM/R. (b) Gravitational field strength at a point lying on the axis of a thin, uniform circu
period of rotation about their centre of mass will be proportional to (a) r3/2 (b) r (c) m1/2 (d) m–1/2
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9. Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start under their mutual gravitational attraction (a) acceleration of m1 is more than that of m2. (b) acceleration of m2 is more than that of m1. (c) centre of mass of system will remain at rest in all the reference frame (d) total energy of system remains constant
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
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Direction: Each question contains statement 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c), and (d) out of which only one is correct.
1. Statement 1: Space rockets are usually launched in the equatorial line from west to east.
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ASSERTION AND REASON UESTIONS
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14. Inside a uniform spherical shell (a) the gravitational potential is zero (b) the gravitational field is zero (c) the gravitational potential is same everywhere (d) the gravitational field is same everywhere
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13. A planet is revolving round the sun. Its distance from the sun at Apogee is rA and that at Perigee is rp. The mass of planet and sun is m and M respectively, vA and vP is the velocity of planet at Apogee and Perigee respectively and T is the time period of revolution of planet round the sun. π2 (a) T 2 = (rA + rP )3 2Gm
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(c) its angular momentum about the earth decreases (d) its period of revolution around the earth increases
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MATCH THE COLUMN TYPE UESTIONS
Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II. 1. Two identical spacecrafts, each with a mass of 3250 kg, are in the same circular orbit at a height of 270 km above the earth’s surface. Spacecraft A leads spacecraft B by 105 s, that is, A arrives at any fixed point 105 s before B. At a particular point P, the pilot of B fires a short rocket burst in the forward direction, reducing the speed of B by 0.95%. The orbital parameters of B will change after the “burn”.
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7. Statement 1: Two satellites A and B are in the same orbit around the earth, B being behind A. B cannot overtake A by increasing its speed. Statement 2: It will then turn into a different orbit.
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6. Statement 1: Comets move around the Sun in elliptical orbits. The gravitational force on the comet due to Sun is not normal to the comet’s velocity, but the work done by the gravitational force over every complete orbit of the comet is zero. Statement 2: Gravitational force is a conservative force and the work done by a conservative force over a closed path is always zero.
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11. Statement 1: Escape velocity is independent of the angle of projection.
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5. Statement 1: If an object is projected from earth’s surface with escape velocity, path of object will be parabola. Statement 2: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity. Path of object will be ellipse.
9. Statement 1: For the planets orbiting around the sun, angular speed, linear speed and K.E. changes with time, but angular momentum remains constant. Statement 2: No torque is acting on the rotating planet. So its angular momentum is constant.
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4. Statement 1: Consider a planet in an elliptic orbit around sun. Its kinetic energy changes with time and angular momentum remains constant. Statement 2: Gravitational force is a central force. As a result, no torque acts on the planet about the position of sun and so angular momentum remains conserved.
3. Statement 1: If earth suddenly stops rotating about its axis, then the value of acceleration due to gravity will become same at all the places. Statement 2: The value of acceleration due to gravity is independent of rotation of the earth.
8. Statement 1: The escape speed does not depend on the direction in which the projectile is fired. Statement 2: Attaining the escape speed is easier if a projectile is fired in the direction, the launch site is moving as the earth rotates about its axis.
2. Statement 1: In a free fall, weight of a body becomes effectively zero. Statement 2: Acceleration due to gravity acting on a body having free fall is zero.
Statement 2: The acceleration due to gravity is minimum at the equator.
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3. Assume v0 be the velocity of projection of a body from the surface of the earth enables it to become a satellite of the earth close to the earth surface. If R be the radius of GM the earth then v0 = gR where g = 2 ; R G – universal gravitational constant and M is the mass of the earth. If ve be the escape v 1 . If v be velocity on the earth then 0 = ve 2 the velocity of projection of body on the surface of earth then, match the following
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3GM e m (a) Total energy of par- (p) − a ticle (b) Minimum distance (q) − 5 GM e m 8 a of particle from the earth (c) Maximum distance (r) 5a/3 of particle from the earth (s) a
4. A satellite of mass m is orbiting the Earth at a height h from its surface. [M is mass of the earth and its radius is R] Column I Column II −GMm (a) Kinetic energy of the (p) ( R + h) satellite 1 GMm (b) Potential energy of (q) 2 ( R + h) the satellite −GMm (c) Total energy of the (r) 2( R + h) satellite GMm (d) The energy must be (s) ( Rr + h) spent to pull the satellite out of the earth’s gravitational field.
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Column I Column II (p) Path of body is circular (a) v = v0 (q) Path of body is ellipti(b) v < v0 cal and return to earth (c) v0 < v < ve (r) Path of body is elliptical but not return to earth (s) Path of the body is (d) v = ve parabolic
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Column I Time period of B before the burn (in sec) Total energy of B after the burn (in J) Semi major axis of B after the burn (in km) Velocity of B before the burn (in m/s)
Gravitation
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Gravitation
(d) A satellite of mass m is revolving around planet of mass M its total energy is given GMm as − then it 2r0 may follow
The ratio of the gravitational force on A and B is the ratio of their gravitational masses; that is, FAC m′A = FBC mB′
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Now, suppose that the third body C is the earth. Then FAC and FBC are what we have called W m′ the weights of bodies A and B. Hence, A = A WB mB′
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In other words, the weights of bodies at the same place on the earth are exactly proportional to their inertial masses as well. Hence, inertial mass and gravitational mass are at least proportional to one another. In fact, they appear to be identical.
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The word mass has been used in two quite different experimental situations. For example, if we try to push a block that is at rest on a horizontal frictionless surface, we notice that it requires some effort to move it. The block seems to be inert and tends to stay at rest, or if it is moving it tends to keep moving. Gravity does not enter here at all. It would take the same effort to accelerate the block in gravity free space. It is the mass of the block which makes it necessary to exert a force to change its motion. This is the mass occurring in F = ma in our original experiments in dynamics. We call this mass m the inertial mass. Now, there is a different situation which involves the mass of the block. For example, it requires effort just to hold the block up in the air at rest above the earth. If we do not support it, the block will fall to the earth with accelerated motion. The force required to hold up the block is equal in magnitude to the force of gravitational attraction between it and the earth. Here inertia plays no role whatever; the property of material bodies, that they are attracted to other objects such as the earth, does play a role. m′M e The force is given by: F = G , Re2 When m’ is the gravitational mass of the block. Consider two particles A and B of gravitational masses m'A and m'B acted on by a third particle C of gravitational mass m'C. Let the third particle be at an equal distance r from the other two. Then, the gravitational force exerted on A by C
Therefore, the law of universal gravitation contains within it the result that the weights of various bodies, at the same place on the earth, are exactly proportional to their gravitational masses. Now, suppose we measure the inertial masses mA and mB of the particles A and B by dynamical experiments. Having done this, we then let these particles fall to the earth from a given place and measure their accelerations. We find experimentally that objects of different inertial masses all fall with the same acceleration g arising from the earth’s gravitational pull. But the earth’s gravitational pulls on these bodies are their weights, so that using the second law of motion we obtain. W m WA = mA g , WB = mB g or A = A WB mB
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3. If inertial mass m and gravitational mass m' of body are not same then equation of motion for freely falling bodies will have to be modified. Suppose a body falls with initial velocity zero by a height h then speed of it at any instant will be given by V = k 2gh here k is a constant which will be equal to
4. For a spring mass system if time period of oscillation on a horizontal frictionless surface be T then the same for the oscillation in vertical plane of the aforementioned system will be (a) ( m′/m′ )T (b) T ( m /m′ ) (c) T (d) T (m'/m)
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While we’re on the subject of space travel, let’s talk about free fall and something that is misleadingly called “weightlessness”. As you are reading this, you can probably feel your chair pushing upwards on you with a force of several hundred newtons. If your feet are not touching the ground this is an upwards force equal in magnitude to your weight (a downwards force). My weight is 680 N downwards, so I know that the force from the chair is about this much, upwards, You can also feel your abdominal muscles holding your abdominal organs in place. �����������������������������������������
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Two satellites A and B are revolving around the earth in circular orbits of radius r1 and r2 respectively with r1 < r2. Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. Focal lengths of the elliptical path are r1 and r2 respectively. At position 2, A is given another impulse so that velocities of A and B at 2 become equal and the two move together. For any elliptical path of the satellite time period of revolution is given by Kepler’s planetary law as T2 ∝ r3 where, ‘a’ is semi major axis of the r +r ellipse which is 1 2 in this case. 2
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Also angular momentum of any satellite revolving around the Earth will remain a constant about Earth’s centre as force of gravity on the satellite which keeps it in elliptical path is along its position vector relative to the earth centre.
Gravitation
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Gravitation
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11. This passage refers to weight that is felt by an observe if he himself moves under different circumstances. In this context of
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9. From the above passage we can say that weight of a body is (a) Gravitational pull by earth while it is stationary relative to the earth. (b) Gravitational pull by all the celestial bodies including earth on the body while body is stationary relative to earth. (c) Net gravitational pull by all the celestial bodies including earth on the body irrespective of whether body is accelerating or not. (d) None of these.
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In an orbiting spacecraft, the spacecraft and the cosmonaut are both accelerating towards the centre of the earth at the same rate (their centripetal acceleration is ac = v2/r, where v is the orbital speed and r the radius). Their weight is what keeps them in orbit: W = mac. Because they are both accelerating towards the centre of the earth at the same rate, there is on average no force between the cosmonaut and the spacecraft. This absence of forces from seat, floor, abdominal wall etc. is what is commonly but misleadingly called ‘weightlessness’: the cosmonauts in the space station are not without weight, in fact they have (almost) their usual weight. It’s just that they donot feel the force of chairs on their bums and they donot feel their abdomens holding in their organs. If the cable of a lift (elevator) has broken, both the lift and its passenger are in free fall, accelerating downwards at g. The passenger no longer feels the force on his abdominal muscles, and so, like the cosmonaut, he might say that he feels ‘weightless’. He is not, of course, without weights. His weight is still W = mg. A NASA airplane (nicknamed the ‘vomit comet’) cuts the power in its engines and, for about 25 seconds, travels in a trajectory that is nearly parabolic. Both the plane and its occupants accelerate towards the Earth at g: all are in free fall. Astronauts are thus exposed to free fall and ob-
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tain brief periods of experience in working in this conditions. Some physicists tend to use the word ‘weightless’ in scare quotes to make it clear that they are not talking about a situation in which there is no weight. Many physicists prefer to avoid the word altogether and talk instead about free fall. There are some similarities between the passenger (mass m) in the lift (let’s put it at the equator) and a cosmonaut (mass m) in low Earth orbit. The weight of each is about mg. Both accelerate towards the centre of the Earth at approximately g. The difference is that the spacecraft makes a circle around the Earth in about 90 minutes, whereas, the lift makes a circle around the Earth in about 24 hours. The acceleration g is just enough to keep an object in low Earth orbit with a period of 90 minutes. It is far too great for the ‘orbit’ of the helpless passenger in the lift. If a satellite loses speed, it gradually spirals in towards the Earth. The horizontal speed of the passenger in the lift is so low that his ‘spiral’ towards the centre of the Earth is almost a straight line.
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These force and some others give you the sensation of having weight. You do not readily sense your weight directly very much, because it is applied homogeneously over your whole body. When the forces from the chair or on your abdominal wall are reduced or zero, you may feel ‘weightless’ the feeling you get when a lift starts to accelerated rapidly downwards, or when you go quickly over a peak on a roller coaster.
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12. In case II in which man is in free fall with the lift, suddenly makes a jump upward with velocity 5 m/s relative to the lift which is of height 10 m and then collide with the ceiling elastically and again hits the bottom of the ceiling and sticks to it. For masses of the lift and man being same, how much time he spent in the air (a) 4 seconds (b) 2 seconds (c) 8 seconds (d) None of these
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14. Two commercial satellites, Seeker I and Seeker II, have circular orbits of R and 2 R, respectively, about the same planet. What is the orbital velocity of Seeker II if the orbital velocity of Seeker I is v? (a) v/2 (b) v/(2)1/2 1/2 (c) v(2) (d) 2v 15. Two boxes of negligible mass are placed 2 metres apart on the surface of Planet X. Initially 15 identical steel ball bearings are placed in each box. Which statement is most accurate if 10 of the ball bearings are transferred from box 1 to box 2? (a) The force of gravity between the two boxes remains the same
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(a) Newton’s second law, F = ma, applies (b) Newton’s law of universal gravitation states that every body attracts every other body with a force of magnitude Fgravity = GMm/r2, where M and m are the masses of two interacting bodies, r is the distance between
An enterprizing physics student has developed a software package that calculates gravity-based parameters for actual and theoretical planets and satellites. The program is based on the following simple principles and approximations
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Since, the more massive planet is stationary, EK depends only on the motion of the satellite. The potential gravitational energy increases from a large negative value to zero as the distance increases to infinity. The speed needed for a projectile to completely escape the pull of the planet is called the escape velocity. A projectile is completely beyond the gravity of a planet has a total mechanical energy of zero with respect to that planet. Conservation of energy requires that the total mechanical energy of the projectile when it was at the planet’s surface must also equal to zero.
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13. In the case of vomit comet, the airplane remains in the parabolic path for 25 seconds and again comes to the same altitude. At the new position it will be (a) Moving with speed with its nose down as against initial situation. (b) Moving with same speed with its nose up as in the initial case irrespective of the movement of the passengers inside the plane. (c) Orientation of the plane at t = 25 s will depend upon movement of the passengers inside the plane. (d) None of these
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their centers of mass, and G is the gravitational constants. For a pair of interacting bodies if one is significantly more massive than the other, the heavier body is considered stationary with the smaller body orbiting around it. The orbit of the smaller body is a circle. The smaller body experiences a centripetal force. Fcentripetal = mv2/r, directed towards the centre of the more massive body. For bodies on or near the surface of a planet, r is simply the planet’s radius, R. Planets are spheres with volume, V = (4/3) πR3. The total mechanical energy of a satellite is the sum of its kinetic and gravitational potential energy E = EK + EP = mv2/2 – GMm/r = constant
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weight. If he is moving in a circle whose radius is 10 m with a constant speed of 10 m/s and plane of the circle is horizontal, how much weight will he feel if his mass is 50 kg [Take g = 10 m/s2] (a) 500 N (b) 1000 N 500 (c) 500 2 N (d) N 2
Gravitation
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= 5.4 × 109 kg distributed uniformly over its circumference. A highly dense particle of mass m2 = 6 × 108 kg is placed on the axis of the ring at a distance x0 = 3 m from the centre. Neglecting all other forces, except mutual gravitational interaction of the two, calculate (a) displacement of the ring when particle is closest to it (b) speed of the particle at that instant.
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6. A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is M and radius of the earth is R.
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5. A body of mass m is suspended on a spring balance in a ship sailing along the equator with a speed v from east to west. If ω be the angular velocity of’ the Earth, derive an expression for the scale reading.
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7. A small body is dropped into a narrow channel drilled along the Earth axis. Considering the Earth to be a homogeneous sphere and disregarding air resistance. Find (a) the magnitude of the body’s acceleration a(r) as a function of r, (b) the magnitude of the body’s speed v(r) as a function of r, (c) the body’s speed v(0) at the instant when it reaches the centre of the Earth. Compare v (0) with the orbital velocity and (d) the time T after which the body returns to the initial point.
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1. Given a thin homogeneous disk of radius a and mass m1. A particle of mass m2 is placed at a distance ℓ from the disk on its axis of symmetry. Initially both are motionless in free space but they ultimately collide because of gravitational attraction. Find the relative velocity at the time of collision. Assume a << ℓ.
4. A pendulum beats seconds on the surface of the Earth. Calculate as to how much it loses or gains per day if it is taken to: (a) a mine 8 km below, (b) a point 8 km above, the surface. (Radius of the Earth = 6.4 × 106 m)
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19. What is the minimum velocity required to launch a rocket from the surface of Planet Z ? The planet has a mass of M and a radius of R? (a) (2GM/R2)1/2 (b) (2GM/R)1/2 (c) (GM/R2)1/2 (d) (GM/R)1/2
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18. Which expression is an accurate description of the total mechanical energy for a satellite of mass m in a circular orbit of radius r around a planet of mass M? (a) GMm/r (b) –GMm/r (c) GMm/2r (d) –GMm/2r2
SUBJECTIVE UESTIONS
3. A particle is fired vertically upwards from the surface of Earth and reaches a height 6400 km. Find the initial velocity of the particle if radius of the Earth R = 6400 km and g at the surface is 10 m/s2.
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17. Which expression is an accurate description of the kinetic energy for a satellite of mass m in a circular orbit of radius r around a planet of mass M? (a) 2GMm/r (b) GMm/r (c) GMm/2 r (d) GM/r2
16. Planet Y has density D, and surface gravitational acceleration, g. The radius of Planet Y is suddenly doubled while its density remains the same. Compared to the original g, the new value of the surface gravitational acceleration would be, which of the following? (a) 4 g (b) 2 g (c) g (d) g/2
(b) The force of gravity between the two boxes remains the same (c) The force of gravity between the two boxes increases (d) Any change in Fgravity between the boxes cannot be determined without knowing the mass of the steel ball bearings
2.41
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Gravitation
8. A body on the equator of a planet weights half its weight at the pole. The density of matter of the planet is 3 g/cm2, Determine the period of rotation of the planet about its axis. 9. Find the potential energy of the gravitational interaction of a mass point of mass m and a thin uniform rod of mass M and length ℓ, if they are located along a straight line at a distance a from each other, also find the force of their interaction.
16. The masses of two stars are m1 and m2, and their separation is ℓ. Determine the period T of their revolution in circular orbits about a common centre.
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18. The radius of Neptune’s orbit is 30 times the radius of the Earth’s orbit. Determine the period TN of revolution of Neptune around the sun.
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12. A satellite of mass Ms is orbiting the earth in a circular orbit of radius r. It starts losing energy at a constant rate C due to friction. Show that the satellite falls to the earth in GM e M s 1 1 a time t given by t = − 2C Re r where Re and Me are the radius and mass of earth.
17. A meteorite approaching a planet of mass M (in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in the circular trajectory of radius R. The mass of the station is ten times as large as the mass of the meteorite. As a result of collision, the meteorite sticks in the station which goes over to a new orbit with the minimum distance R/2 from the planet. Determine the velocity u of the meteorite before the collision.
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11. A synchronous satellite is one whose velocity in its circular orbit allows it to remain above the same position on the rotating earth’s surface. Calculate the required distance H above the earth’s surface to the satellite. Locate the position of the orbital plane of the satellite and calculate the angular range β of longitude on the surface of the earth for which there is a direct line of sight to the satellite.
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10. What is the minimum work that has to be performed to bring a spaceship of mass m = 2 × 103 kg from the surface of the Earth to the Moon?
15. A body is projected vertically upwards from the surface of earth with a velocity, sufficient to carry it to infinity. Find n if the time taken by it to reach a height of thrice the radius of earth (Take radius of earth = R and acceleration due to gravity as g) is (n/3) ( 2 R /g ).
the satellite as F = αv2, where, α is a constant, find how long the satellite will stay in orbit until it falls onto the Moon’s surface.
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(b) 129 (d) 730
(a) 64.5 (c) 182.5
2. Distance between the centres of two stars is 10a. The masses of these stars are M and 16 M and their radii a and 2a respectively. A body of mass m is fired at night from the surface of the larger star towards the smaller star. What should be its minimum initial speed to reach the surface of the
14. An artificial satellite of the Moon revolves in a circular orbit whose radius exceeds the radius of the Moon η times. In the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming the resistance force to depend on the velocity of
1. If the distance between the earth and the sun were half its present value, the number of days in a year would have been [1996]
13. A planet of mass M moves around the Sun along an ellipse so that its minimum distance from the Sun is equal to r and the maximum distance to R. Making use of Kepler’s laws, find its period of revolution and angular momentum around the Sun.
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PREVIOUS YEARS’ IIT-JEE UESTIONS
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3. An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) E0. Its potential energy is [1997] (a) –E0 (b) 1.5 E0 (c) 2E0 (d) E0
(b) TA > TB (if mA > mB) 2
T r (c) A = A TB rB (d) TA = TB
8. A spherically symmetric gravitational system of particles has a mass density ρ for r ≤ R ρ= 0 where ρ0 is a con 0 for r > R stant. A test mass can undergo circular motion under the influence o the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by [2008]
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6. A body is projected vertically upwards from the bottom of a crater of moon of depth R/100 where, R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body from the surface of the moon. [2003]
(b)
(a) W1 = W2 = W3 (c) W1 = W2 > W3
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7. A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then [2006] (a) TA > TB (if rA > rB)
4. A cord of length 64 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. [1998]
smaller star? Obtain the expression in terms of G, M and a. [1996]
Gravitation
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9. Statement 1: An astronaut in an orbiting space station above the earth experiences weightlessness. because Statement 2: An object moving the earth under the influence of Earth’s gravitational force is in a state of “free-fall”. [2008] (a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. [2008]
Gravitation
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5. (a, b, c, d) 6. (a, c) 12. (a, c) 13. (b, c, d)
4 (b, c) 11 (b, c, d)
3. (a, d) 10. (a, b, c)
2. (a, b, c, d) 9. (a, d)
1. (a, d) 8. (a, c)
Q
ASSERTION AND REASON UESTIONS 1. (b) 9. (a)
2. (c) 10. (a)
3. (c) 11. (a)
4. (a)
5. (b)
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6. (a)
7. (a)
8. (b)
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Gravitation
2.45
Q
MATCH THE COLUMN TYPE UESTIONS 1. (a) → s, (b) → r, (c) → r, (d) → q
2. (a) → p, (b) → s, (c) → r
3. (a) → p, (b) → q, (c) → r, (d) → s
4. (a) → q, (b) → p, (c) → r, (d) → q
5. (a) → p, q (b) → p, q, r, (c) → r, s (d) → r, s
Q
PASSAGE BASED UESTIONS 2. (c) 10. (c) 18. (d)
3. (b) 11. (c) 19. (d)
4. (c) 12. (a)
5. (b) 13. (c)
na .c om
1. (b) 9. (c) 17. (c)
6. (b) 14. (b)
7. (b, c) 15. (c)
8. (c) 16. (b)
ar a un iv
er s
ita
.ja
w
w
//w
iv
.m
e/
s:
tp
un
s: //
t.m e/
ht tp s:
//t
ht
5. As due to air drag some mechanical energy of the satellite will be converted into heat energy, there will be loss of ME of the satellite, i.e., E = – (GMm/2 r) will decrease.
ht tp
3. The escape velocity for the moon is smaller than the root mean square speed of molecules of the gas (e.g., oxygen and nitrogen). Therefore, the gas molecules cannot remain attracted due to moon’s gravitation; but escape from the surface of moon.
2
This in turn implies that the radius of the orbit will decrease and so the satellite will follow a spiral path towards the earth as shown in figure. Now as in case of satellite motion, GMm GMm PE = and KE = r 2r So with decrease in r, PE will decrease but KE and hence, speed will increase. Increase in speed (or KE) inspite of decrease in mechanical energy is accounted by the fact that the rate at which PE decreases (due to decrease in r) is more than the rate at which mechanical energy decreases.
7. Since, there are no nearby stars to exert gravitational force on him and the small
er si ta rio s
Q
1. In case of satellite motion centripetal force is provided by gravitation between the earth and the satellite. If earth is not at the centre of the orbit, i.e., if the plane of the orbit is not passing through the centre of earth, the gravitational force F will not be directed towards the centre of the orbit O’ and hence, the orbit will not be stable. The component of gravitational force perpendicular to the plane of orbit will pull the satellite till its plane passes through the centre of the earth as shown in figure.
6DWHOOLWH
m
CONCEPTUAL UESTIONS
rio s_ in fo
HINTS AND SOLUTIONS
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2.46
Gravitation
fo
VA M A RB = BB M B RA
rio
ta
.ja
rio
ta
si
er
iv
e/ un
then,
V M RA = VA R RB
=
2
σ4π( RA2 + RB2 ) RA = 2 2 1/2 ( RA + RB ) σ4πRA2
ωR d 2. (a) g 1 − = g − ω2 R; d = g R
3. (d) The acceleration due to gravity at earth’s surface is g and at a distance R from earth’s surface it is g/4. Hence,
=
7. (a) V = ωR g = g0 – ω2R [g = at equator, g0 = at poles] g0 = g 0 − ω2 R; 2 g g R ω2 R = 0 ; V 2 = 0 2 2 Ve = 2 g 0 R = 4V 2 = 2V
4 Dd πR 3 GM 3 4. (b) g = 2 = [d = density] R R2 4 d = Gd πR 3
T1 = 2 [ T = 2π 1/g ] T2
RA2 + RB2 5 = RA 3
s: //t .m
%
[
σ4πR 2 = σ4π( RA2 + RB2 )
ht tp
2 U
s
si iv
un
e/
s: tp ht GT
T
2
VA RA 3 4 = = then RB = RA VB RB 4 3 for new shell of mass M and radius R M = M A + M B = σ4πRA2 + σ4πRB2
GP
(
σ4πRA2 RB RA = σ4πRB2 RA RB
Given
.m //t
ht
\
=
er
w
w
2Gλ ˆ 2GM ˆ 2Gmπ ˆ j= j= j (along y-axis) r r 2
tp
=
π π Gλ ˆ ˆj sin θ i cos θ + ∫ ∫ r 0 0
s: //w
E=
B
s_
A
m
Gλrd 0 ∫0 dE = ∫0 r 2 (iˆ cos θ + ˆj sin θ)
an
π
ar
π
in
a.
co m
Since, g is constant and R increased by factor 5, d decreases by factor 1/5. M G m GMn 5. (a) Change in force of gravity = 2 − 32 R R M 8. (a) path 1 decreased E (more negative) GMn G m 3 (only due to mass M/3 due to gives decreased semi major axis “a”; = 2 − R R2 (b) less decreased “a” gives decreased T. shell gravitational field is zero (inside the shell)) ONLY ONE OPTION IS CORRECT 2GMm m = 1. (d) Let mass per unit light of wire, λ = R2 6. (c) M A = σ4πRA2 , M B = σ4πRB2 , where, σ and πr = , r = . mass of element, dm = π is surface density Gdm −GMA GMB λrdθ then dE = 2 VA = , VB = r R R
spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.
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2.47
fo
tp s:
ht
P
2
G 3m Gm + cos30° 3d 2 4d
G &0
P 2
1 d 1 4mω2 4× KEm 2 5 = 25 = 1 = 2 16 4 KEm 1 4d mω2 25 2 5
16. (d) Gravitational potential at P,
G 3m 2 cos 60° 12d 2
GMm = 2 E0 . r
15. (a) Due to mutual force of attraction they will revolve about centre of mass with same ω.
In vertical direction,
+
io s
ta r
si
er
iv
un e/
.m
Therefore, P.E. = −
Gm 2 Gm 2 = − =0 8d 2 8d 2
2
GMm GMm GMm − − 2r r 2r
G
Gm 2 cos 60° 4d 2
Net force =
...(ii)
ta si er
iv un
e/
E0 =
//t
−
G 2mm cos30° 12d 2
1 2 GMm mv = 2 2r
or
1 2 mv in equation 2 (i) from equation (ii), we get
//t
s:
tp
ht
Net force =
mv 2 GMm = 2 r r
Substituting the value of
.m
tp
ht
12. (c) In horizontal direction
in
s_
rio
ar
w .ja
//w w
s:
Further;
If the body is to escape, the velocity at surface of earth is 2 gR . If E’ is the kinetic energy corresponding to this velocity then 1 E ′ = m( 2 gR ) 2 ⇒ E ′ = 2 E 2
14. (c) Total energy = kinetic energy + Potential energy 1 GMm E0 = mv 2 − ...(i) 2 r
m
11. (b) Orbital velocity close to surface 1 of earth is gR . So, E = m( gR ) 2 2 1 ⇒ E = mgR 2
m
1 R2 R decreasing g increase curve b represent variation.
10. (b) g ∝
3Gm 2 1 + 8 + 3 3Gm 2 along SQ. d 2 24 2d 2
13. (a) In figure A force due to 5M, 2M cancel and similarly in figure B force due to 2M, 2M cancel, hence, FA = FB. In fig. C also force due to 5M cancel but net force will be more in comparison to figure A or figure B as man at the centre is double similarly think for figure D.
a. co
v R 2π = gR = e 2 π R /g 2
an
=
=
GMm 1 2 GMm = 2 gR mve = ; ve = R 2 R
In tunnel body will perform SHM at centre Vmax = Aω (see chapter on SHM)
3Gm 2 3Gm 2 3Gm 2 + + 2 2 24d 3d 8d 2
9. (b)
4 = R 2Gd π as Ve ∝ R for same density, 3 VA =2 VB
=
4 2Gd πR 3 2GM 3 8. (a) Ve = 2 gR = R= R R2 R2
Gravitation
vp =
−GM 5R
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2.48
05
5
P
2
18. (d) TE in the orbit at a height equal to radius of earth 1 GMm −1 GMm = mv 2 − = 2 2R 4 R
−GM R
Gravitational potential at O, vO =
Gravitation
mv 2 GMm = 2 2 R (2 R)
3
Change in energy = −
s
rio
rs ita
ve
e/ un i t.m
3/2
ht tp
GM 2 mv 2 = 2 L ( L / 3)
22. (a) Remember Keplers law are applicable for satellite also. dA J 1 = = vr Areal velocity = dt 2m 2
tp s:
T2 6400 = T1 36000
or
3/2
0
3
A =J T
21. (c) We know that
Integrating, 2m
s: // Therefore,
[J = angular momentum]
6400 or T2 = × 24 ≈ 2 h 36000
q
0
dA J = dt 2m
T2 ∝ R3 or (T2/T1) = (R2/R1)3/2
ht &20
un iv
.m e/
//t
ht
0
20. (a) Using Kepler’s 2nd law
w
w
s: //
tp
GMM GM cos30° = 3 2 L2 L This acts as centripetal force. ⇒ 2
2
T2 ∝ r'3 r r ′ = 1/3 4
er
2 L × L cos30° = 3 3 Force on each star M due to the other two
si ta rio s_ in fo
ar m
w
.ja
17. (b) Stars move around COM. Distance of each star from
=
GM to dismantle the system, L energy equal to total energy of the system must be provided.
⇒ v=
1 GM mv 2 GMm r = 2 2 r r r
23. (d)
GM 2 1 3 GM 2 i.e., 3 × MV 2 + 3 × =− 2 L 2 L
Areal velocity ∝ velocity = 1 : 2
2GM 1 (1 − ) R 5
COM is
19. (c) Angular speed of earth = angular speed of geostationary satellite. If ω is double, time period become half using
an a. co m
GM GM 1 2 m − = mv or 5R 2 R
3 GMm 3 = mgR 4 R 4
=
work KE theorem W = ∆K ⇒ m[vp – vO] 1 = mv 2 2
1 GMm GMm −− 4 R R
r . Hence ratio of areal
GM 1 2 GMm GMm mv0 − =− and g = 2 R 2 R R+h
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Gravitation
1 3 R2 g × × gR − Rg = − 2 2 R+h We get h = 3R
GMm = m(OB )ω2 (OB) 2
Hence,
$
24. (a) Since, T2 = kr3 5
Differentiating the above equation ∆T 3 ∆r ∆T ∆r = =3 ⇒ T 2 r T r
o
s_
ar
rio
ita rs
rio
si ta
un iv
GMm 2rs ...(ii) The angular momentum L is given by
e/
.m //t
1/2
GM L = ms vs rs = ms rs
tp s: ht
28. (c) By conservation of energy
−
GMm 1 m GM GMm + =− RE 2 N 2 2 RE H
⇒ H=
R ⇒ OB = 2 R OB Here gravitational force will provide the required centripetal force.
26. (a) In ΔAOB: cos 60º =
M 1 & ε0 = 3 (4 / 3)πR 4πG
rs = (GMms2rs)1/2
..........(iii) From eqs. (ii) and (iii), we get L = (2Emsrs2)1/2
5
&0
3mg then F = 8
GMms rs
er
P.E. = −
mρR Net force on m → F = mE = 8ε 0 Here, ρ =
s
1 1 GM K.E. = ms vs 2 = ms ; 2 2 rs
e/ un iv e
tp ht 5
...(i)
∴ Total energy E = K.E. + P.E. = −
s: //
ht
t.m
tp
s:
//w
Net gravitational field ρR ρR E = E1 + E2 = − 6ε 0 24ε 0 ρR = 8ε 0
P
GM GMms ms vs2 = ∴ vs = 2 rs rs rs
.ja m w
(−ρ) R −ρR =− 24ε0 R 2 24ε0
w
=−
in f
27. (a) The velocity vs of the satellite is given by
an a.
Gravitational field at mass m due to cavity (–ρ) (−ρ)( R /2)3 r ρa 3 E2 = ........ using E = 3ε0 R 2 3ε 0 r 2 3
%
GM GM GM ⇒ ω= = 3 3 (OB) (2 R) 8R3
⇒ ω=
co m
0
q
(DUWK
0
q
2
25. (b) Gravitational field at mass m due to full solid sphere ρr ρR 1 E1 = = ........ ε 0 = 4πG 3ε 6ε
⇒ 2
Therefore,
2.49
N 2 RE N 2 −1
Altitude = H – R =
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RE N 2 −1
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2.50
3. (a, d).
2π 2π = π 2 2π ω1 + ω2 + T1 T2
29. (c) Given, 8 −
Gravitation
U
T1 = 24 hours for earth. ⇒ T2 = 12 hours (T2 being the time period of satellite, it will remain same as the distance from the centre of the earth remains constant).
m co
fo
in
rio
si ta
s
rio
er
iv
un e/ .m
//t
s:
tp
ht
2GM m r12 ∆a × 100 = × × 100 aav r22 GM s
7. (b, c) Net force towards centre of earth mgx = mg ′ = R
si ta
iv
un
//t
s: tp
Fav GM s = 2 Me r1 Percentage change in acceleration Average force on earth, Fav =
=
Thus pressing force N =
Mg constant and independent of x. 2 Hence, (b). Tangential force F = ma = mg' cos θ
2
1 PJ T 5 [
ONE OR MORE THAN ONE OPTION MAY BE CORRECT
g=
GM r R3
mgx R R 2x
N=
r Mm = 2 1 × 100 r2 M s
1. (a, d) Outside earth g =
32π2 r 3 27GM
Normal force N = mg' sin θ
ht
∆F 2GM m = Me r22
e/
.m
ht
Change in acceleration of earth ∆a =
2GM m M e r22
w
//w
s:
tp
Change in force, ∆F = F1 − F2 =
T22 =
6. (a, c) Due to air resistance satellite will experience tangential force due to which it will not remain in its orbit and approach towards earth hence r decreases and v increases. Thus KE increases. Due to tangential force angular momentum decreases.
er
GM s M e GM m M e − r12 r22
4π2 r23 GM
T2 ∝ r3/2; T2 ∝ m–1/2
a.
an
ar
w
.ja
m
F2 =
T22 =
s_
GM s M e GM m M e + r12 r22 When moon goes on the opposite side of earth. Effective force of attraction, F1 =
P
2mr 2r = m + 2m 3
r2 =
2π 2π = = 23 hours ω2 = ω1 2π − 2π T2 T1
30. (c) During total eclipse Total attraction due to sun and moon,
⇒ T=
&0
U
P
U
GM , Inside earth r2
Q = g ′ cos θ =
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gx R
R2 − x2 4 x
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Gravitation
gx R2 − 4x2 R
GM e m 1 At any distance ‘r’ T.E. = mu 2 − r 2 but angular momentum conservation,
T.E. at any distance ‘r’
1 5 GM e a GM e m = m − but through conr 2 4 r2 servation of total energy
3GM e m 1 5 GM e a GM e m = m = =− r a 2 4 r2
ita rio s_ in
ar
2. Angular momentum of particle = m(v0 + v)
GM e 5 mv0 a.....v0 = 4 a Total energy of particle
a=
GM e m 1 = m(v0 + v) 2 − 2 a GM e m 1 5 = × mv02 − a 2 4
=
3GM e m 5 GM e m GM e m − =− 8 a a a
s
rs
e/ un i
8. (c) In the continuation with same comprehension
(r2 + r2 /3) = 2r2 /3 2 Let time period of A in elliptical path is T0 r=
2
T r ∴ 0 = 2 T r
3
Q
MATCH THE COLUMN TYPE UESTIONS
ht tp s:
//t
.m
//t
tp s:
ht
ve
.m
ht
R R+h Curved area AB on earth = 2πR2(1 – sinθ) Area on earth escaped from satellite = 4πR2 – 2πR2(1 – sin θ) = 2πR2(1 + sin θ) Minimum colatitude, sin θ =
Q
ita
ni e/ u
tp s: %
GMm −GMm for and K.E. = 2r r same m, If r1 > r2, KE1 < KE2 and U1 > U2 Also, angular momentum of A before reaching the position 2 (during elliptical path) was less than that of B. Since, some impulse is needed (at position 2) for A in direction of its motion to equal its speed and angular momentum equal to that of B.
7. (b, c) U = −
∴ T0 =
T (2 2) 3 3
∴ Time taken by A to move from 1 to 2,
K
//w
w
5
PASSAGE BASED UESTIONS
T
w
T
ve rs
.ja
m
$
On solving, 3r2 – 8ar + 5a2 = 0 ⇒ (r – a) (3r – 5a) = 0 ⇒ r = a, r = 5a/3 minimum distance = a, maximum distance = 5a/3
fo
co
an a.
12. (a, c) From figure
T
m
11. (b, c, d) Inside shell gravitational field GM intensity = 0 and potential = − R
5
5GM e 5 GM e a a⇒u = 4a 4 r2
mur = m
9. (a, d) Same force acts on both masses 1 (F = ma) Hence a ∝ m In absence of external force (remember mutual gravitational force is an internal force for the system) total energy remains constant.
R Curve is parabolic and at x = , a = 0. 2 Hence (C).
rio
a=
2.51
t = T0/2 =
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2.52
Gravitation
SUBJECTIVE UESTIONS
Q
or v 2 = ve2 +
3. If a body is projected from the surface of Earth with a velocity v and reaches a heights h, by conservation of energy. Kinetic Energy = Change in potential Energy
i.e., i.e.,
dr 2R =R dt r 1
∫ dt = t=
an a.
R
3/2 1 2 R h 1 + − 1 as h = 3R; 3 g R
s_ i
7 2R ;n=7 3 g
rio
ita
s
rio
ta
er si
ni e/ u
..........(1)
un iv
.m //t
m e/ //t .
ht tp s:
s:
ht tp
11. H = 35800 km, β = 162.6º 13. T = π (r + R)3 /2 γM
G (m1 + m2 ) and the required period of 1 revolution of these stars is
14. t ≈ ( η − 1)m /α gR 15. If at a distance r from the centre of earth, the body has velocity v, by conservation of mechanical energy,
T = 2π
G (m1 + m2 )
17. Let v1 be the velocity of the station before the collision v2 the velocity of the station and the meteorite immediately after the col
1 2 −GMm 1 2 −GMm mv + = mve + 2 r 2 R
10. W = 1.3 × 105 kJ
γmM ln 1 + , F = − a ( a + ) a
mM 9. U = −γ
r1/2 dr
Here ω1 are the angular velocities of rotation of the stars ℓ1 and ℓ2 are the radii of their orbits, F is the force of interaction between the stars, equal to Gm1 m2/ℓ2 where ℓ is the separation between the stars, and G is the gravitational constant. By the definition of the centre of mass. m1ℓ1 = m2ℓ2, ℓ1 + ℓ2 = ℓ ..........(2) solving equation (1) and (2) together, we obtain G (m1 + m2 ) ω1 = ω2 = = −1 13
8. 3.16 × 10–4 rad/sec
∫
m1ω12 1 = F , m2 ω22 1 = F
(d) T = 2π R /g
R+h
16. Since, the system is closed, the stars will rotate about their common centre of mass in concentric circles. The equations of motion of the stars have the form
ve rs
w
:// w
g 0 R = v1
ht
tp s
(b) v(r ) = g 0 ( R 2 − r 2 )/R (c) v(0)
t=
ar
w .ja m
7. (a) α(a) = g0(r/R)
6. d = 2R
(b) 107.9 sec
5. m(g ± 2ωv)
v2 = gh, i.e., v = 10 × 6400 × 103 = 8 km/s 4. (a) 54 sec
1 R 2g
nf o
Here, h = R = 6400 km and g = 10 m/s so,
i.e.,
co m
0
2
GM R2
2 gR 2 R v2 = 2gR + 2gR − 1 ⇒ v = r r
(b) 18 m/sec
2. (a) 0.3 m
1 2 mgh mv = 2 [1 + (h /R)]
R − 1 r
But as ve = 2 gR and g =
1/2
2 1 1. 2G (m1 + m2 ) − a
2GM R
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Gravitation
velocity v2 of the station immediately after the collision through the relation
VR = v2 x R. 2
Solving equations (1) = (4) together and considering that v1 = GM /R , we determine the velocity of the meteorite before the collision: u =
rio s_ in fo
ar a
Here, mN, mE, ωN, ωE, RN and RE are the masses, angular velocities, and orbital radii of Neptune and the Earth respectively, and M is the mass of the sun. We now take into account the relation between the angular velocity and the period of revolution around 2π 2π the sun ωN = , ωE = TN TE
ita
un iv
e/
iv
.m
un
//t
Here TN and TE are the periods of revolution of Neptune and the Earth. As a result, we find that the period of revolution of Nep-
t.m e/
ht tp s:
5 0
er s
w
.ja
m
w
//w
s:
tp ht
X
RN3 ≅ 165 RE3 years. A similar result is obtained for elliptical orbits from Kepler’s third law. tune around the sun is TN = TE
ht tp
s: //
Y
11mM 11m 2 + (v2 x + v22 y ) R 2
Q
1. (b)
2. Vmin =
3 5GM 2 a
3. (c)
4. T = 3 × 10–2N 5. (a)
Where V is the velocity of the station at the moment of the closest proximity to the planet. Here we have used the formula for the potential energy of gravitational interaction of two bodies (of mass m1 and m2): Wp = –Gm1 m2/r. According to kepler’s second law, the velocity V is connected to the
.........(3)
11mM 11m 2 + v , R /2 2
PREVIOUS YEARS’ IIT-JEE UESTIONS
= −G
−G
6. h = 99R
\
GMmE GMmN , mE ω2E RE = RN2 RE2
er si ta rio s
[
mN ω2N RN =
........(1)
and mu = 11 mv2y .........(2) After the collision, the station goes over to an elliptical orbit. The energy of the station with the meteorite stuck in it remains constant during the motion in the elliptical orbit. Consequently,
58GM R
18. We shall write the equation of motion for Neptune and the Earth around the sun (For the sake of simplicity, we assume that the orbits are circular)
na .c om
Hence, v1 = GM /R . In accordance with the momentum conservation law, the velocities u, v1, and v2 are connected through the following relation mu + 10 mv1 = 11 mv2 We shall write the momentum conservation law in projections on the x– and y-axes
10mv12 10mM =G R R2
10 mv1 = 11 mv2x
.........(4)
lision, m the mass of the meteorite, and 10 m the mass of the station. Before the collision, the station moved around a planet in a circular orbit of radius R. Therefore, the velocity v1 of the station can be found from the equation
2.53
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7. (d)
G (4/3)πρ0 R 3 v 2 1 = r ≤ R. Hence, v ∝ r2 r r
4 v 8. (c) G πρ0 r = , r ≤ R. Hence, v ∝ r 3 r
s_ in
fo
a. c
om
9. (a)
an
si ta rio
ar .ja m
s
er
w w ht
tp
s:
//t
ht
.m
tp
e/ u
s:
ni
//t
ve
.m
rs
e/
ita
un
iv
//w s: tp ht
2
rio
Gravitation
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c h a p t e r
ar a
3.1 DEFORMING FORCE
3
rio s_ in fo
na .c om
Elasticity
iv
un
t.m e/
ht tp s:
3.2 ELASTICITY
er si ta rio s
un iv
.m
//t
ht
e/
s:
The force acting on the spring (F1) is the deforming force. The force exerted by the spring to oppose the deforming action is the restoring force (F2). By Newton’s third law, we can say that F1 and F2 equal and opposite forces turns forming an Action-Reaction pair.
tp
//w
w
er s
w
ita
.ja
m
It is the external force applied to a body which tends to change the size or shape of the body. Under the action of deforming force, a body opposes any change in its shape and size due to net effect of internal (molecular) forces. The resulting force which opposes the deformation is known as restoring force. When one tries to stretch a spring:
ht tp
s: //
The property of a body due to which it opposes the action of the deforming forces is called as elasticity. A material is said to be elastic if it returns back to its original shape or size, when the deforming forces are removed. On the other hand, plastic material remain parmanently distorted when the deforming forces are removed. There are a number of elastic modulii (Young’s modulus, bulk modulus, shear modulus) that measure of response of an elastic solid which is acted upon by the deforming force.
The deforming force is measured (described) in terms of a physical quantity, the stress that is developed in the body. The deformation of the solid is described in terms of physical quantity the strain, that is created in the body as results of deformation force.
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3.2
Elasticity
3.3 STRAIN AND STRESS When an elastic body deforms, under deforming forces, its length dimension, volume dimension and/or its shape may change. Then Strain =
change in dimension original dimension
Tensile strian = Bulk strain =
change inlength ∆l = original length l
change in volume ∆V = original volume V
rio s_ in fo
deforming force area of application
er si ta rio s
un iv
//w
w
er s
w
.ja
Hence at equilibrium, stress =
ita
m
ar a
na .c om
(i) External deforming forces cause deformation (strain) in the body. (ii) Strain cause internal restoring forces, called elastic forces to come into action. (iii) Internal elastic force per unit area of their application is called stress. (iv) As the strain increases, stress also increases. (v) At steady equilibrium state elastic force = deforming force
s:
3.4 LONGITUDINAL STRESS AND LONGITUDENAL STRAIN
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
Consider a rod (or a string) of length l and cross-sectional area A. Suppose that the a force F is applied at the ends parallel to the length so as to stretch the rod by ∆l. The rod is said to be under tension. The tensile stress is defined as the ratio of the force of tension to the cross-section area. F Tensile stress = A The resulting tensile strain is defined as the ratio of the change in length to the original length. ∆l l If the ends of a rod are subjected to equal pushes, then rod is said to be under compression. F Compressive stress = A Tensile strain =
l + ∆l l
Compressive strain =
∆l l
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Elasticity
3.3
Tensile and compressive stresses and strains are also known as longitudinal stress and longitudinal strain. Example A bob of mass m is suspended from a light steel wire of length l and radius r. Assuming mass of wire to be negligible, what is the longitudinal stress and strain?
) Solution Let T be the tension developed in the steel wire. As the bob is stationary T = mg
T
e/
tp
s:
3.5 VOLUME STRESS AND VOLUME STRAIN
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
ar a
Longitudinal strain ∆l = l
l+∆ ∆ll
m
rio s_ in fo
na .c om
Longitudinal stress mg = 2 πr
l
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
Sometimes, the deforming force acting on the body tends to change its volume. Let the deforming force changes the volume of a body from V to V + ∆V (∆V is negative, if volume is decreases). We define volume strain as the ratio of change in V to the original volume.
V + ∆V
∆V V The stress associated with the volume strain is called as volume stress. It equals the deforming forces acting on a unit area. It is usually called as pressure (P). F Volume stress = P = A Volume strain =
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3.4
Elasticity
Example A cube is subjected to pressure of 5 × 105 N/m2. Each edge of the cube is shortened by 1%. Determine volume strain. Given: Volume stress = P = 5 × 105 N/m2
) Solution Let l be the length of cube (undeformed) ∴
99 Distroted volume, V f = l 100 ∆V = Vf – Vi
3
3
na .c om
∴
99 = l − l3 100 3 ∆V 99 Volume strain = = − 1 ≅ – 0.03 Vi 100
rio s_ in fo
3.6 SHEAR STRESS AND SHEAR STRAIN
e/
B
un t.m e/
Q
P
ht tp
s: //
ht tp s:
C
iv
.m
A
//t
ht
tp
s:
un iv
//w
D
er si ta rio s
w
er s
w
ita
.ja
m
ar a
If the deforming force tries to change the shape of a body, a shear stress is developed in the body. In the diagram we see a force F parallel to the surface ABCD of area A, deforms the rectangular area ABPQ into a parallelogram.
Let AA′ = ∆x and AQ = y Hence, due to F, shape of the body is altered. Shear stress =
shearing force F = cross-section of ABCD A ∆x
A
A'
B
B'' B
y θ
Q
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P
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Elasticity
3.5
Shear strain is defined as: ∆x Shear strain = = tan θ (θ is very small) y
rio s_ in fo
un iv e/ .m
tp
er si ta rio s
er s
w //w
s:
3.7 HOOK’S LAW
Motion
ita
µmg
w
.ja
m
ar a
na .c om
Example A block of weight 15 N slides on a horizontal table, the coefficient of sliding friction = 0.4, the area of the block in contact with table is 0.05 m2. What is the shear stress? µ mg ) Solution Shear stress = A 0.4 × 15 = = 120 N/m 2 0.05
iv
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
If the deforming forces are within a limit (known as elastic limit), the stress created in the body is proportional to the resulting strain. i.e., stress ∝ strain stress The ratio is known as modulus of elasticity. strain According to various type of stresses, we have three modulii of elasticity. (i) Young’s modulus of elasticity (Y): Y=
longitudinal stress F/A = longitudinal strain ∆l /l
Fl A∆l The following forms of the result must be carefully noted. Y=
∆l = ∆l =
Fl ∆l and F = AY AY l
Fl ∆l and F = AY AY l
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3.6
Elasticity
(ii) Bulk modulus (K): volume stress p =− volume strain ∆V/V pV ⇒ K=− ∆V ‘–’ sign indicates decrease in volume The reciprocal of K is known as modulus of compressibility. K=
na .c om
(iii) Shear modulus (η): shear stress F/A η= = shear strain θ F η= Aθ n is called as modulus of rigidity
rio s_ in fo
un iv
//w
w
3.8 ELASTIC ENERGY
er si ta rio s
er s
w
ita
.ja
m
3 1 1 = + Y η 3K
ar a
Relation between Y, K and η The three modulus of elasticity are related as
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
When an elastic body is deformed, work is done by applied force. This work is stored as elastic potential energy and is released when the body returns back to its original shape or size. We have already come across an example of elastic potential energy in case of a compressed or stretched spring. The general expression for the elastic potential energy in a unit volume of a deformed body is given by 1 Elastic energy stored per unit volume = (stress) (strain) 2 1 = (modulus of elasticity) (strain)2 2 In case of a longitudinal stress (compressive or tensile)
l
x
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Energy stored 1 ∆l = Y Volume 2 l
Elasticity
3.7
2
1 YA (∆l ) 2 2 l 1 Total energy = F ∆l 2
Total energy =
Work done = ∫ F.dx F/A x /l
Y=
co
m
3.9 THERMAL STRESS
s
rio
si ta
iv er
un
s: //
ht
t.m
tp
s:
e/ un
iv
//w
w
er
w
si
ta
.ja
m
rio
ar
s_
an
in f
a.
o
If the ends of the rod are rigidly fixed as to prevent expansion or contraction and the temperature of the rod is changed, elongation or compressive stress, called thermal stress, will be set up in the rod. This strain may become very large, so as to strain the rod beyond its elastic limit or even beyond its breaking strength. Hence, in the design of any structure which is subject to change in temperature, some provision must be made for expansion. In a long steam pipe this is accomplished by the insertion of expansion joints or a section of pipe in the form of a · V. In bridge, one end may be rigidly fastened to its abutment while the other rests on rollers. It is easy to compute the thermal stress set up in a rod which is not free to expand or contract. Suppose, that a rod at a temperature t has its ends rigidly fartened and that while they are thus held, the temperature is reduced to a lower value t0. The fractional change in length if the rod were free to contract would be
Hence,
.m
ht
tp
s: //t
ht tp
e/
∆L = α (t − t0 ) = α ∆ t L Since, the rod is not free to contract, the tension must increase by a sufficient amount to prevent the same fractional change in length. But from the definition of Young’s modulus. F/A Y= ∆L/L F = AY
Substituting for
∆L , we get L
F = AYα ∆ t
∴
∆L L
Stress in the rod =
F = Y α∆t A
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3.8
Elasticity
3.10 TORSION Consider a cylinder whose upper ends is rigidly fixed and the other end is twisted through an angle θ about the axis of the cylinder. The twisted cylinder exerts a restoring torque given by τ=
1 πηr 4 θ 2 l
Fixed twisted trhough an angle θ
rio s_ in fo
na .c om
l
ar a
where, n is the modulus of rigidity, r is the radius and l is the length of the cylinder. If we express torque as τ = Cθ
er s
w
un iv
//w s:
t.m e/
s: //
ht tp
Stress = Force per unit area mg 2000 × 980 = πr 2 π(0.05) 2
= 2.496 × 108 dyne cm–2 change in length Linear strain = original length =
0.24 × 10−3 = 12 × 10−5 2
iv
//t
ht tp s:
) Solution By the formula
Young’s modulus of elasticity =
un
ht
.m
tp
e/
WORKED OUT EXAMPLES
1. A wire 2 m long, 1 mm in diameter when stretched by a weight of 2 kg increases in length by 0.24 mm. Find the stress, strain and Young’s modulus of the material of the wire. Assume that the change in diameter is negligible.
=
er si ta rio s
ita
.ja
m
πηr 4 θ , is known as the torsional rigidity. 2l
w
then, C =
=
stress strain
2.496 × 108 dyne cm −2 12 × 10−5
= 20.8 × 1011 dyne cm–2 2. If the young’s modulus of the material of wire is 2.2 × 1012 dyne cm–2. What mass must be suspended from the wire fixed at one end to stretch it by 1 mm? Length of the wire is 2.5 m and diameter is 1.0 mm.
) Solution Stress =
weight of body mg = 2 area of wire πr
Strain =
change in length ∆l = original length l
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ta
si
rio
ita
rs
un i e/
//t
s:
tp
= –1.024 × 10–8 m2 = 10.24 mm3
4. What would be greatest length of a steel wire, which when fixed at one end can hung freely without breaking? Density of steel = 7.8 gm–3 and breaking stress for steel is equal to 7.8 × 109 cm–2
) Solution Let the greatest length of steel is l. )
−P ∆V/V
7 3 − PV − ( 2 × 10 ) (0.04) ∆V = = K 1.25 × 1011
ht
∆l 0.035 = = 4.67 ×10−4 l 75 stress 5.9 × 107 Y= = strain 4.67 × 10−4
∆V ∆V = V (0.04)3
.m
ht )
= 5.9 ×107 N/m 2
Strain =
Volume strain =
Bulk modulus, K =
π ( 6.5 × 10
−4 2
=
force F = area A 8 × 9.8
ve
.m
tp s:
//t
ht
(iv) Volume stress, P = 2 × 107 N/m2
)
(i) Stress =
s
er
un iv
e/
s:
tp
face displaced 4.0 mm relative to the bottom surface. What are the shearing stress, shearing strain and shear modulus? (iv) Compute the volume change of a solid copper cube, 44 mm on each edge, when subjected to a pressure of 2 × 107 N/m2.
) Solution
fo
in
s_
rio
.ja
w
//w
w
an a.
m ar
co m
3.9 mg stress πr 2 mgl Young’s modulus of elasticity = = 2 (ii) Young’s modulus, Y = ∆l strain πr ∆ l stress mg l mg = = 2. l stress πr 2 mgl strain πr ∆l Young’s modulus, Y = = = 2 2 ∆ l strain πr ∆ l Yπr ∆l ∴ Required mass, m = l gl mgl ∆l = 2 2.2 × 1012 × 3.14 × 25 × 10−4 × 1× 10−1 πr Y = 980 × 250 80,000 × 9.8 × 4 = = 7049 g π(4.5) 2 × 10−4 × 1.9 × 1011 = 7.049 kg. = 2.6 × 10–3 m = 2.6 mm 3. (i) A metal wire 75 cm long and 0.13 cm tangential force is diameter stretches 0.035 cm, (iii) Shear stress = area of face when a load of 8 kg is hung on its 0.5 ends. Find the stress, strain and = = 333 N/m 2 Young’s modulus. 15 × 10−4 (ii) A solid cylinder steel column is 4 m ∆x (displacement) Shear strain, θ = = long and 9 cm in diameter. What y height will be decrease in its length when 0.4 carrying a load 80,000 kg? Given: = = 0.133 11 2 Y = 1.9 × 10 N/m 3 (iii) A box shaped piece of gelatine shear stress Shear modulus, η = dessert has a top area of 15 cm2 shear strain and height of 3 cm. When a shear333 ing force of 0.5 N is applied to = = 2500 N/m 2 0.133 the upper surface, the upper surElasticity
2
Then, weight of steel = ρAlg
= 1.27 × 10 N/m 11
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3.10
stress 7.8 × 109 = ρg 7.8 × 980
Young’s modulus, Y =
5. A wire of radius r stretched without tension along a straight line is tightly fixed at A and B. A mass m is suspended from the mid point of the wire. Due to the weight of mass, the wire is pulled into the shape ACB. Find the depression ‘d’ in the wire. The length of the wire is 2l and its Young’s moduls is Y.
=
θ
∆l l
2
+ l2 ) − l
o
nf
s_ i
s
rio
er si ta
iv
un
e/ ∴
Strain =
=
T mg = A 2A sin θ
(d
)
mg 2sin θ
Stress =
T=
T l × πr 2 ∆l
T=
Y × πr 2 ∆l l 2
l
0.315 −2 2 × 102 × π× × 10 × 0.1 2 = 100 = 1.559 × 106 dyne
1/2
d2 d2 = 1 + 2 − 1 ≈ 2 l 2l
stress strain
T T l A = = × ∆l A ∆l l
ht
C m
⇒
t.m
T
when the extension is 1 mm.
Young’s modulus of elasticity =
s: //
θ
θ
B
tp
T
6. A sphere of mass 50 g is attached to one end of the steel wire, 0.315 mm diameter and one metre long. In order to form a conical pendulum, the other end is attached to a vertical shaft which is set rotating about its axis. Calculate the number of revolutions necessary to extend the wire by 1 mm. Young’s modulus of elasticity of steel = 2 × 1012 dyne cm–2.
) Solution Let T be the tension in the wire,
s: //
ht
θ
tp
2l
mgl 2l 2 . 2πr 2 d d 2
rio
iv un e/ t.m
)
ht
) Solution Let the T be tension of wire A
stress strain
1/ 3
er
w w
s: tp
m
2 T sin θ = mg
stress strain
si ta
.ja
B
//w
C
(Neglecting d2,
mg d = 2 2πr Y
m
2l
A
ar an
a.
∴
co m
= 1.02 × 106 cm.
d + l2
Young’s modulus, Y =
⇒
l=
stress 7.8 × 109 = ρg 7.8 × 980
2
because d2 << l2) mgl Stress = 2πr 2 d 2
l=
d
sin θ =
∴
As
weight ρAlg = Area A (where ρ = density of the steel) Now, Stress =
Elasticity
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Elasticity
Frequency of revolution =
∴
θ
1 0.3561 = 2.808 revs–1 = 168.5 rev mm–1
T
T cos θ
7. A truncated solid cone of length l radius of its ends equal to r1 and r2 (r1 < r2). The bigger end is fixed. A force F is applied at the other end distributed uniformly over its area of cross-section. Find the total elongation in the cone. Young’s modulus of cone is Y.
θ
T sin θ mg
m
l = lm
m
ita
.ja
rio ita rs
ni
e/ u
where, r is the radius of circle described. r = l sin θ = 100 × 1 = 100 cm
dl =
⇒
∴
From equation (i), rT sin θ m
r −r Yπ r1 + 2 1 x l
2
r
F 2 dx 2 Y ∫r1 (r2 − r1 ) π r1 + x l
Let,
∴
100 × 1.559 × 106 × 1 = 1766cms −1 50 2πr Period of revolution = v 2π× 100.1 = = 0.3561 sec. 1766 =
F dx
∫ dl =
v=
(r2 − r1 ) x l
F stress A = Young’s modulus = strain dl dx
//t
.m
Thus, r = r1 +
s:
2
ve
)
from the lower radius is r.
ht tp
mv r
s
rs iv e t.m
) Solution Let the radius of cone at distance x
s: //
tp
ht
sin θ = 1 − cos 2 θ = 0.9998 ≈ 1 T sin θ =
dx
x
un e/
s:
tp
ht
∴
rio
ar
r1
mg 50 × 980 cos θ = = = 0.0314 T 1.55 × 106
∴
r2
r
s_ i
an a.
nf o
co
l
F
w w //w
r
From the figure T cos θ = mg
1 period
=
T
θ
3.11
(r2 − r1 ) x = z lim of z , r → r2 l (r2 − r1 ) dx = dz l
r1 +
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3.12
Elasticity
wires. One wire is of steel and the other two are of copper. All the wires have the same areas of cross-section. Consider the modulus of elasticity of steel to be double than that of copper.
r
r
=
1 1 Fl − πY(r2 − r1 ) r1 r2
=
Fl π Y r1 r2
8. A solid uniform rod of length l is hanging vertically from its upper end. Find the total elongation of the rod due to its own weight if ρ is the density and Y the young’s modulus of the materials of the rod.
c
s
c
)
m
s
si
ta
rio
s Ts
c Tc
er
m
iv
//t
According to problem YS = 2YC ∴ TS = 2TC From equations (i) and (ii), we get Mg Mg TC = and TS = 4 2
...(ii)
ht
tp
s:
TC TS = YC YS
.m
tp
e/
un
//t s:
or
ht (l–x)
ta si er un
iv
c Tc
.m e/
tp ht x
fo
in
s_
.ja
w w
//w
ρAlg dx lA Y
s:
dl = (l − x)
an m
ar
∴
For the equilibrium, mg = 2TC + TS ...(i) Since elongation of all wires is same, therefore T /A T /A dl = C = S YCl YSl
rio
)
T T dx A Young’s modulus, Y= = dl A dl dx
C
) Solution
a.
W ) Solution Tension,T = (l − x) l where, W = ρ lg A
m B
A
co
2 2 Fl dz Fl 1 l= = − 2 ∫ πY(r2 − r1 ) r1 z πY(r2 − r1 ) z r1
l
9. A homogenous block with a mass m hungs on three vertical wires of equal length arranged symmetrically. Find tensions of
ρg x2 1 ρ g l2 = lx − = Y 2 0 2 Y
10. Two rods made of different materials are placed between massive walls. The cross-section of the rod is A and their respective lengths are l1 and l2. The rods are heated by t degrees. Find the force with which rods act on each other if their coefficient of linear thermal expansion are α1 and α2 and modulii of elasticity of their materials Y1 and Y2 are known.
l
ρg (l − x)dx Y ∫0
Integrating, ∫ dl =
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Elasticity
Y1
Y2
α1
α2
l1
l2
B
) Solution )
C
Y1
A
Y2 F
α2
) Solution Since, the blocks moves with uniform )
l2
velocity (i.e., acceleration = 0)
ar a
Expansion of the wire due to temperature is balanced by the contraction of the compressed wire. Fl dl1 = 1 …(i) Y1A
µN
m
un
mA.g
From block B T – µmB g = 0 There is no friction force acting on A, since the normal reaction exerted by C on A is zero. mA ·g = T mA 2 = = 10 kg µ 0.2
mB =
Stretching force on the spring, T = mAg = 2 × 9.8 N Tension of the spring l= spring constant 2 × 9.8 = = 10−2 m 1960 Potential energy stored in the spring 1 = (2 × 9.8) (10−2 ) = 9.8 × 10–2 J 2
ht tp
s: //
t.m e/
ht tp s:
A
11. Two blocks A and B are connected to each other by a string and a spring, the string passes over a frictionless pulley as shown in the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surface of block is 0.2. Force constant of the spring is 1960 Nm–1. If the mass of block A is 2 kg, calculate the mass of block B and the energy stored in the spring.
T
iv
…(iii)
Fl Fl + 2 = (l1α1 + l2 α 2 ) t Y1A Y2 A (l α + l α ) t A F= 1 1 2 2 l1 l2 + Y1 Y2
ita
un iv e/ .m
//t
ht
∴
tp
Also dl1 + dl2 = l1 α1 t + l2 α2t = (l1α1 + l2 α2) t
er s
w //w
…(ii)
s:
Fl2 Y2 A
T
B
.ja
w
And dl2 =
N
er si ta rio s
l1
rio s_ in fo
F
na .c om
α1
⇒
3.13
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3.14
∴
Putting the value of x in equation (iii), we have YA mg l (mg ) l E= = 2YA 2l YA
If the energy is converted to heat when the bob get snapped, we have E = Heat absorbed by the wire
( mg )
0
( mg )
2
2YA 2ρc
ita
.ja e/
Hence
.m
{(100kg) (10ms )}
t.m e/
2
−3
= 0.00457 K
13. A street shaft of radius 10 cm and of length 2 m is transmitting power of 80 kW at 300 r.p.m. If n = 8 × 1010 Nm–2, calculate the twist between the two ends. )
) Solution Power, P = torque × angular velocity = Tω P ∴ T= ω ⇒
T=
1 πr 2 P θ= 2 l ω
∴
θ=
2Pl πηr 4 ω
mg mg l where Y = A = x Ax l
…(iii)
x
YA YA 2 E = ∫ x dx = x l 2l 0
From equations (i) and (ii)
2
( 7860 kgm )
s: // ht tp
−1
(2) ( 2.1× 1011 Pa ) (1.256 × 10−3 m 2 )
un
//t
iv
un iv
er si ta rio s
Area of cross-section of the wire, A = πr2 = (3.14) (2 × 10–3 m)2 = 1.256 × 10–5 m2
er s
w ht tp s:
…(ii)
= (ALρ) c ∆ T
rio s_ in fo
ar a
∆T =
w //w s: tp ht
YA F= x l
∴
L
2AY
∆T =
m
Fl = Ax
2
where, m is mass of the wire. Rearranging the above expression, we get
m
F stress A = Young’s modulus, Y = strain x l
i.e.,
…(i)
na .c om
Energy stored due to rod = ∫ F.dx
)
x
2
2
) Solution
mg l x= YA
12. A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature at the wire snapped, and change in temperature of the wire ignoring radiation losses. (For the steel wire, Young’s modulus = 2.1 × 1011 Pa, Density = 7860 kg/m3, specific heat capacity = 420 J/kg– K)
Elasticity
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Elasticity
η = 1010
)
) Solution
Torsional torque, T = Force × radius of cylinder
N/m–2, r = 0.1 m 300 × 2π rad/sec. 60 = 10π rad s −1
ω = 300 rev min −1 =
T = Fr =
2
1 π ( 6 × 10 = 2
10
= 4.05 × 10–4 rad.
14. A cylinder of length 2m and radius 8 mm is camped at one end and a force tangential to its cross-sectional boundary and in the plane of the cross-section is applied at the other end to produce one radian twist in the cylinder. If the modulus of rigidity of the material of the cylinder is 6 × 1010 Nm–1, calculate the force applied.
) (8 ×10 )
−3 3
×1
2 F
= 2.41 × 104 N.
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∴
2 × 80 × 103 × 2 4 = × 10−3 π × 8 × 1010 × 10−4 π2
θ=
∴
1 πηr 4 θ 2 l 1 πηr 3θ F= 2 l
⇒
Here, P = 80 × 103 W, l = 2m,
3.15
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Hence
(d)
d1 Y = 1 d2 Y1
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(b)
(c)
d1 Y1 = d 2 Y2
(a)
L is maximum in case of (a). A
3. Two wires of different materials support a massive object suspended from the centre of a uniform horizontal bar with its ends connected to two vertical wires of equal length. What must be the ratio of the diameters of the wires so that the bar may remain horizontal when the object is removed? [Young’s modulus of the wires are Y1 and Y2 respectively].
2. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied to all? (a) L = 50 cm; D = 0.05 mm (b) L = 100 cm; D = 1 mm (c) L = 200 cm; D = 2 mm (d) L = 300 cm; D = 3 mm
L
(∵ F and Y are common)
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∆L1 F1 L1 A2 40 × 1 22 = × = × =1 ∆L2 F2 L2 A1 80 × 2 12
⇒
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) Explanation Young’s modulus, F /A F /A Y= 1 1 = 2 2 ∆L1 / L1 ∆L2 / L2
FL
) Explanation ∆L = AY ∝ A )
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1. Two steel wires of length 1 m and 2 m have diameters 1 mm respectively. If they are stretched by forces of 40 N and 80 N respectively their elongation will be in the ratio of (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 1 : 8
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SOLVED OBJECTIVE TYPE QUESTIONS
d1 Y2 = d 2 Y1 d1 Y = 2 d2 Y1
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3.16
)
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∆l = l
Mg 3KA
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(c)
(d)
⇒
F 1 = Lρα A 2
8. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless pistion of area A floats on the surface of the liquid. When a mass m is placed on the pistion to compress the liquid, the fractional change in the radius δR of the sphere, is R Mg Mg (a) (b) KA 2 KA
1 1 = l ∆l
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Stress =
⇒
⇒
F 2F 2 2 πr = 4πr ∆ L 3L
L Aρ.α 2
F=
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∴
s:
F /A
( ∆l /l )
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the mid point of the rod is
tp
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) Explanation Y =
(d) 4l
(b) l
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7. A uniform rod of length L and density ρ is being pulled along a smooth floor with a horizontal acceleration α. The magnitude of the stress at the transverse cross-section through the mid-point of the rod is Lρα (a) Lρα (b) 2 2 (d) none of these (c) Lρα 3
) Explanation Force at the cross-section throught
5. A wire of length L, radius r, when stretched with a force F, changes in length by l. What will be the change in length in a wire of same material having length 2L, radius 2r, radius 2r and stretched by a forca 2 F?
∆V Change in pressure = V Bulk modulus
3 × 107 − 105 299 = × 10−5 × 100% 8 × 1010 8 = 0.0374% ≈ 0.04%
=
Y1 α1 3 = = Y2 α 2 2
∴
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) Explanation
) Explanation Thermal stress = Yα∆t
4. By what fraction is the volume of an aluminium sphere be reduced as it is lowered from the surface to the depth of 3 km in ocean? Bulk modulus of Al = 8 × 1010 N/m2. At the surface of ocan P = 105 N/m2. At a depth 3 km, P = 300 atmospheres. (a) 20% (b) 10% (c) 1% (d) 0.04%
d1 Y = 2 d2 Y1
or
(∵ A ∝ d)
)
d12Y1 = d22 Y2
or
6. Two rods of different materials having coefficients of thermal expansion α1, α2 and Young’s modulii Y1, Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo then same increase in temperature. There is no bending of the rod. If α1 : α2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9
Fl AY For the bar to remain horizontal, ∆l1 = ∆l2 ∴ A1Y1 = A2Y2 (∵ F and l are common)
) Explanation ∆l =
Elasticity
Mg 4 KA
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Elasticity
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7. For a given wire, the Young’s modulus is 2.4 times that of its rigidity. The Poisson’s ratio is (a) 0.5 (b) 0.2 (c) – 0.5 (d) –1.
8. Two blocks of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is 3.18 × 1010N/m2, then the radius of the wire is
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4. One end of a steel wire of length 1 m and cross-sectional area 1 mm2 is clamped at rigid support. The increase in its length when 1 kg-wt is suspended from it will be (if Y = 20 × 1010 pascal) (a) 5 cm (b) 0.5 mm (c) 0.25 mm (d) 0.05 mm
6. A solid sphere of radius R made of material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the suface of the liquid. When a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is Mg Mg (a) (b) MA 3BA 3Mg Mg (c) (d) 4 BA 4 BAR
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3. A uniform pressure P is exerted on all sides of a solid cube. It is heated through ∆tº in order to bring its volume back to the value it had before the application of pressure. Then P B (a) ∆t = (b) ∆t = Bγ Pγ Bγ (c) ∆t = BγP (d) ∆t = P where γ is the cubical coefficient of expansion.
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2. The length of a wire increased by 1% by a load of 10 kg suspended from it. If the same load is suspended from the wire of same material but of double length and double radius, by what percentage is the length increased? (a) 0.25% (b) 0.5% (c) 2% (d) 4%
is twisted by an angle of 30º. The angle of shear is (a) 12 (b) 1.2 (c) 0.12 (d) 0.012
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1. A force of 400 kg weight can break a wire. The force required to break a wire of double the cross-section will be (a) 800 kg-wt (b) 200 kg-wt (c) 1600 kg-wt (d) 100 kg-wt
OBJECTIVE TYPE QUESTIONS
δR Mg = R 3KA
5. The upper end of a wire 1 m long and 4 mm radius is clamped. The lower end
δV 3δR = V R
S
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or
olved
)
∴
∴
δV = 4πR2δR
∴
Mg / A δV / V δV Mg = V KA 4 2 V = πR 3
) Explanation K =
3.17
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1 kg
4 kg
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3.18
10. The correct graph verifying Hooke’s law is
(d)
K ρP ρK . P
13. Two capillaries of same diameter are dipped into liquids of specific gravity 0.4 and 0.8 respectively. If their surface tensions are in the ratio 6 : 5 then the ratio of heights of liquids in them will be (a) 12 : 25 (b) 5 : 12 (c) 4 : 8 (d) 8 : 4
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14. A stone of 0.5 kg mass is attached to one end of a 0.8 m long aluminium wire of 0.7 mm diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such that the wire makes an angle of 85º with the vertical. Find the increase in the length of the wire (Y for aluminium = 7 × 1010 N/m2, sin 85º = 0.9962 and cos 85º = 0.0872) (a) 0.7 mm (b) 1.7 mm
(d)
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Extension
Load
ρP K
(b)
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Extension
(c)
P ρK
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Extension
Load
(a)
(c)
m
Load
12. A material has normal density ρ and bulk modulus K. The increase in the density of the material, when it is subjected to an external pressure P from all sides is
a. c
Extension
(a)
(b)
(d) 2 YA α2 Lt2
1 YA α 2 Lt 2 4
1 YA α 2 Lt 2 2
(d) 17 cm
(c) 1.7 cm
Load
(c)
(b)
(a) YA α2 Lt2
9. A steel rod of length L, area A, Young’s moduls Y and coefficient of linear expansion α is heated to tºC. The work that can be performed by the rod when heated is
11. A wire 3 m in length and 1 mm in diameter at 30ºC is kept in a low temperature at –170ºC and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is [Y = 2 × 1011 N/m2, g = 10 m/s2 and α = 1.2 × 10–5/ºC] (a) 5.2 mm (b) 2.5 mm (c) 52 mm (d) 25 mm
(b) 2 × 10–5 m (d) 4 × 10–5 m
(a) 1 × 10–5 m (c) 3 × 10–5 m
Elasticity
ANSWERS 1. (a) 9. (b)
2. (a) 10. (c)
3. (a) 11. (a)
4. (d) 12. (c)
5. (c) 13. (a)
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6. (b) 14. (b)
7. (b)
8. (d)
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Elasticity
of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section 10–3 m2 and the other is of brass of cross section 2 × 10–3 m2. Find out the position along the rod at which a weight may be hung to produce (a) equal stresses in both wires, (b) equal strains in both wires.
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l2, α2, γ2
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l1, α1, γ1
A
4. A horizontally oriented rod of length 2L = 1.0 m is being rotated about vertical axis passing through its centre. Find the minimum number of revolutions at which this rod will rupture. The tensile strength of copper is σm = 0.30 GPa and its density is ρ = 8.9 g/cm3. [Ans. 0.8 × 102 r.p.s.]
B
C
Ans. A (l1α1 + l2 α 2 ).t l1 l2 + Y1 Y2
5. A light rod of length 2.00 m is suspended from the ceiling horizontally by means
7. A force of 106 N/m2 is required for breaking a material. If the density of material is 3 × 103 kg/m3, then what should be the
6. Two rods of different metals having the same area of cross-section A are placed between two massive walls as shown in the figure. The first rod has a length l1, coefficient of linear expansion α1 and Young’s modulus Y1. The corresponding quantities for the second rod are l2, α2 and Y2. The temperature of both the rods are now raised by tº C. Find the force with which the rods act on each other (at higher temperature) in terms of given quantities. [IIT, 1975]
Young’s modulus of steel = 20 × 1010 N/m2 Unstretched length of wire = 500 cm Radius of wire = 0.05 cm [IIT, 1972] [Ans. 3.76 m/s]
Young’s modulus of steel = 2 × 1011 N/m2 [IIT, 1974] [Ans. (a) 1.33 m, (b) 1.00 m]
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3. A sphere of radius 10 cm and mass 25 kg is attached to the lower end of a steel wire which is suspended from the ceiling of the room. The point of support is 521 cm above floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate velocity of the ball at its lowest position.
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Given, α for steel = 1.2 × 10–5/º C. [IIT, 1971] [Ans. 1.92 × 103 N]
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2. A steel rod 25 cm long has a crosssectional area of 0.8 cm2. What force would be required to stretch this rod by the same amount as expansion produced by heating it through 10º C?
Given: Young’s modulus of brass = 1011 N/m2
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[IIT, 1970]
practice
EXERCISE 1
1. A steel wire of diameter 0.8 mm and length 1m is clamped firmly at two points A and B which are 1m apart and in the same horizontal plane. A body is hung from the middle point of the wire such that the middle point sags 1 cm lower from the original position. Calculate the mass of the body. Given, Y = 2 × 1011 N/m2 [Ans. 8.2 × 10–2 kg]
3.19
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3.20
Elasticity
culate the displacements of the points B, C and D using the data on Young moduli given below (neglect the weight of the rod). YAB = 2.5 × 1010 N/m2, YBC = 4.0 × 1010 N/ m2, YCD = 1.0 × 1010 N/m2. [Roorkee, 1986] [Ans. Displacement of point B, = 3.92 × 10–6 m] Displacement of point C = 8.82 × 10–6 m Displacement of point D = 23.52 × 10–6m]
13. A simple pendulum is made by attaching a 1 kg bob to 5 m long copper wire of diameter 0.08 cm and it has certain time period of oscillation. Next a 10 kg bob is substituted for the 1 kg bob. Calculate the change in time period if any (Young’s modulus of copper = 12.4 × 1010 N/m2). [Roorkee, 1988] [Ans. 0.0035 s]
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Total extension in the entire rod,
tp
1 ρω2l 2 3 Y
15. A copper wire of negligible mass, 1 m length and cross-sectional area 10–6 m2 is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotating with an angular velocity of 20 rad/sec. If the elongation in the wire is 10–3 m, obtain the Young’s modulus. If on increasing the angular velocity to 100 rad/sec, the wire breaks down, obtain the breaking stress. [Roorkee, 1992] [Y = 4.00 × 1011 N/m2].
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12. A rod AD, consisting of three segments AB, BC and CD joined together, is hanging vertically from a fixed support at A. The lengths of the three segments are respectively 0.1 m, 0.2 m and 0.15 m. The crosssection of the rod is uniformly 10–4 m2. A weight of 10 kg is hung from D. Cal-
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mω2l r 2 Ans. T = 1 − ; 2 l2
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11. A ring made of lead, of radius r = 25 cm is rotated about a vertical axis passing through its centre and perpendicular to its plane. Find the number of revolutions per second at which this ring will rupture. The density of lead ρ= 11.3 g/cm2 and its tensile strength σ = 0.015 GPa. [Ans. 23 rps.]
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[Roorkee, 1982]
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10. A metallic wire is stretched by suspended weight to it. If α is the longitudinal strain and Y, its Young’s modulus of elasticity, show that the elastic potential energy per 1 unit volume is given by Yα2. 2
[Ans. 2.0 × 10–5/º C; 1.1 × 1011 N/m2]
15. A uniform iron rod of length l and of mass m is rotated with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Find an expression for the tension in the rod as a function of the distance r from the axis of rotation. What is the elongation of the rod?
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9. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of the constituents do not change with rise in temperature. Find the Young’s modulus and coefficient of linear expansion of the second rod. Given: αcu = 1.7 × 10–5/º C, Ycu = 1.3 × 1011 N/m2 [IIT, 1979]
8. A composite rod is made by joining a copper rod end to end with a second rod of different material, but of same cross section. At 25º C, the composite rod is 1 m in length, of which the length of the copper rod is 30 cm. At 125º C, length of the composite rod increases by 1.91 mm.
length of the wire made of material so that it breaks by its own weight? [Roorkee, 1979] [Ans. 34.01 m]
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Elasticity
3.21
QUESTION BANK equal to the jumper’s weight? Explain why or why not.
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7. A bungee jumper leaps from a bridge and comes to a stop a few centimeters above the surface of the water below. At that lowest point, is the tension in the bungee cord
1. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W1 is suspended from its lower end. If S is the area of crosssection of the wire, the stress in the wire at a height (3L/4) from its lower end is (a) W1/S (b) [W1 + (W/4)]/S (c) [W1 + (3W/4)]/S (d) (W1 + W)/S
6. A cylindrical steel bar is compressed by the application of forces of magnitude F at each end. What magnitude forces would be required to compress by the same amount (a) a steel bar of the same cross-sectional area but one half the length? (b) a steel bar of the same length ut one half the radius?
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ONLY ONE OPTION IS CORRECT
2. Two rods of different materials having coefficients of thermal expansion α1 and α2 and Young’s moduli Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 and α2 are in the
5. The columns built by the ancient Greeks and Romans to support temples and other structures are tapered, they are thicker at the bottom than at the top. This certainly has an aesthetic purpose, but is there an engineering purpose as well? What might it be?
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4. Does it take more force to break a longer rope or a shorter rope? Assume the ropes are identical except for their lengths and are ideal there are no weak points. Does it take more energy to break the long rope or the short rope? Explain.
9. Read each of the following statements carefully and state, with reason if it is true or false. (a) The modulus of elasticity of rubber is greater than that of steel. (b) The stretching of a coil is determined by its shear modulus. (c) When a material is under tensile stress, the restoring forces are caused by interatomic attraction. While under compressional stress the restoring force is due to interatomic repulsion. (d) Elastic forces are strictly conservative when Hooke’s law is obeyed. (e) A piece of rubber under an ordinary stress can display 100% strain, yet when unloaded it returns to its original length. This shows that the restoring force in the rubber piece are strictly conservative.
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3. A karate student hits downward on a stack of concrete blocks supported at both ends. A block breaks. Explain where it starts to break first, at the bottom or at the top. (The block experiences shear, compressive and tensile stresses. Recall that concrete has much less tensile strength than comprehensive strength. What part of the block is stretched and which is compressed when the block bends in the middle?
2. What are the advantages of using the concepts of stress and strain to describe deformations?
8. Are you more likely to find steel rods in a horizontal concrete beam or in a vertical concrete column? Is concrete more in need of reinforcement under tensile or compressive stress?
1. Young’s modulus for diamond is about 20 times as large as that of glass. Does that tell you which is stronger? If not, what does it tell you?
CONCEPTUAL QUESTIONS
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3.22
6WUHVV
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13. When temperature of a gas is 20ºC and pressure is changed from p1 = 1.01 × 105 Pa to p2 = 1.165 × 105 Pa then the volume changed by 10%. The bulk modulus is (a) 1.55 × 105 Pa (b) 0.115 × 105 Pa (c) 1.4 × 105 Pa (d) 1.01 × 105 Pa
12. A uniform rod of mass m, length L, area of cross-section A and Young’s modulus Y hangs from a rigid support. Its elongation under its own weight will be (a) zero (b) mgL/2YA (c) mgL/YA (d) 2 mgL/YA
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11. A load W produces an extension of 1 m in a thread of radius r. Now if the load is made 4 W and radius is made 2 r all other things remaining same, the extension will become (a) 4 mm (b) 16 mm (c) 1 mm (d) 0.25 mm
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8. The length of elastic string, obeying Hooke’s law is ℓ1 metres when the tension 4 N and ℓ2 metres when the tension is 5 N. The length in metres when the tension is 9 N is (a) 5ℓ1 – 4ℓ2 (b) 5ℓ2 – 4ℓ1 (c) 9ℓ1 – 8ℓ2 (d) 9ℓ2 – 8ℓ1
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7. If the work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of the same material but double the radius and half the length y 1 mm is (a) 16 J (b) 8 J (c) 4 J (d) (1/4) J
(b) YA = YB (d) YA = 3YB
10. A steal wire 1 m log and 1 mm2 cross section area is hung from rigid end. When weight of 1 kg is hung from it then change in length will be (given Y = 2 × 1011 N/m2) (a) 0.5 mm (b) 0.25 mm (c) 0.05 mm (d) 5 mm
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6. Work done in increasing the length of wire 1 m long and of cross section area 1 mm2 by 1 mm by 20 kg-wt will be (a) 0.1 Joule (b) 1 Joule (c) 10 Joule (d) 5 Joule
;
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(a) YB = 2YA (c) YB = 3YA
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5. When load of 5 kg is hung on a wire then extension of 3 m takes place, then work done will be (a) 75 Joule (b) 60 Joule (c) 50 Joule (d) 100 Joule
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4. How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? [Young’s modulus for brass = 0.9 × 1011 N/m2] (a) Nearly 17 N (b) Nearly 34 N (c) Nearly 51 N (d) Nearly 68 N
%
14. Two bodies are suspended from two rods of same length and same area of cross-section.
(d)
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(c)
strain 2 Y× volume stress × strain × volume strain 2 × volume 2Y 1 Y × stress × strain × vol. 2
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3. The elastic energy stored in a wire of Young’s modulus Y is (a)
9. The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and YB are the Young’s moduli of the materials, then
ratio 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9
Elasticity
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Elasticity
(a) (b) (c) (d)
15. The Young’s modulus of a metal is 1.2 × 1011 Nm–2 and its inter atomic force constant is 3.6 × 10–9 N/m, then the average distance between atoms is (a) 15 Å (b) 3 × 10–16 m (c) 3 × 10–20 m (d) None of these
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21. The work required to be done in producing an extension of 1.0 mm in a wire of length 50 cm and area of cross-section 1 mm2 will be (Y = 2 × 1010 N/m2) (a) 6 × 10–2 J (b) 4 × 10–2 J –2 (c) 2 × 10 J (d) 1 × 10–2 J
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22. The work done in increasing the length of a wire of area of cross-section 0.1 mm2 by 1% will be (Y = 9 × 1011 Pa) (a) 2 × 102 J (b) 4.5 × 102 J 2 (c) 3 × 10 J (d) 6 × 102 J
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(d) 9 : 4
(c) 3 3 : 2 2
(b) 3 : 2
3: 2
(a)
19. Find the bulk modulus of elasticity of helium at 1 atmospheric pressure if the variation of pressure and volume is adiabatic in nature.
24. Two springs of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 2 : 3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of
18. A metallic rod of Young’s modulus 2 × 1011 Nm–2 undergoes a strain of 0.05%. The energy stored per unit volume of the rod will be (a) 2.5 × 108 Jm–3 (b) 0.5 × 104 Jm–3 (c) 2.5 × 104 Jm–3 (d) 0.5 × 108 Jm–3
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23. The following four wire (length L and diameter D) are made of the same material. Which of these will have the largest extension when the same tension is applied? (a) L = 50 cm, D = 0.5 mm (b) L = 100 cm, D = 1 mm (c) L = 200 cm, D = 2 mm (d) L = 300 cm, D = 0.5 mm
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20. A steel wire 1.5 m long and of radius 1 mm is attached with a load 3 kg at one end the other end of the wire is fixed. It is whirled in a vertical circle with a frequency 2 Hz. Find the elongation of the wire when the weight is at the lowest position. [Y = 2 × 1011 N/m2, g = 10 ms–2] (a) 1.77 × 10–3 m (b) 7.17 × 10–3 m (c) 3.17 × 10–7 m (d) 1.37 × 10–7 m
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16. A steel wire of uniform cross-sectional area 2 mm2 is heated up to 50ºC and is stretched by clamping its two ends rigidly. The change in tension when the temperature falls from 50ºC to 30ºC is given by – (α = 1.1 × 10–5ºC, Y = 2.0 × 1011 N/m2) (a) 88 N (b) 5 N 10 (c) 1.5 × 2.0 N (d) 2.5 × 2.010 N
7.31 × 105 Pa 1.73 × 105 Pa 5.73 × 107 Pa None of the above
Their Young’s modulii are in the ratio 2 : 3. Then (a) Their lengths will increase in the ratio 3:4 (b) They will be subjected to stresses which are in the ratio 2 : 3 (c) When bodies are displaced vertically and released, they will oscillate with time periods in the ratio 3 : 2 (d) When bodies are displaced and released, time periods will be in the ratio √3 : √2
3.23
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3.24
4. For a substance undergoing extension without any change in its volume (a) σ shows drop in radius with increase in length (b) The poisson’s ratio is σ = 0.5 (c) σ can be –ve (d) σ can be +10
1. A ring made by cutting directly from a metal plate has radius R. The arc ABC makes angle θ at the centre as shown in the figure. If the ring gets heated
5. A wire of radius 1 mm and length 2 m is twisted through 90º. If the modulus of rigidity is 5 × 1010 N/m2, then (a) Angle of shear is 7.85 × 10–4 radians at a point midway. (b) Angle of shear is 7.85 × 10–4 radians on the surface. (c) Couple per unit twist is 3.92 × 10–2 N-m/rad (d) Torsional couple is 6.16 × 10–2 N/m
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
Elasticity
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8. Arrangement of three metallic rods of equal cross sectional area riveted to each other is as shown in figure. α Al is the coefficient of linear expansion for aluminium and αS (where α Al > α S ) for steel. The arrangement is heated, then
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3. A light rod of 2 m length is suspended from the ceiling horizontally using two vertical wires of equal length tied to its ends. One of the wire is made of steel of cross-sectional area 0.1 cm2 and young modulus 20 × 1011 dyne/cm2, while the second is made of brass of C.S. area 0.2 cm2. and Young’s modulus 10 × 1011 dyne/cm2. Then (a) For stress to be same, a mass can hang at 0.6 m from brass wire end. (b) For stress to be same, a mass can hang at 0.6 m from steel end. (c) For equal strain, the mass should be hung at mid-point of the rod. (d) For equal force, the mass should be at 1/4th from steel end.
6. A beam of light of infinity at an angle 60º with the normal falls on a partially reflecting surface of reflectivity 0.8 there will be (a) tangential stress on the surface (b) normal pressure on the surface (c) only normal pressure on the surface (d) normal pressure will be 3 times the tangential stress on the surface
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Elasticity
(a) Tensile stress is developed in the aluminium rod. (b) Compressive stress is developed in the aluminium rod (c) No stress is developed in the aluminium rod (d) Stress developed in aluminium rod is twice that in steel rod
mg where ℓ is the incre2 ment in length of wire.
stretched wire is
Statement 2: In the above statement, loss in gravitational energy is mgℓ while the loss mg . in energy to surrounding is 2
3. Statement 1: Young’s modulus for a perfectly plastic body is zero. Statement 2: For a perfectly plastic body, restoring force is zero.
ASSERTION AND REASON QUESTIONS
4. Statement 1: Identical springs of steel and copper are equally stretched. More work will be done on the steel spring. Statement 2: Steel is more elastic than copper.
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6. Statement 1: Steel is more elastic than rubber. Statement 2: When same deformation is produced in two identical bodies of these material, greater restoring force develops in the steel body.
(d) Statement 1 is False, Statement 2 is True.
1. Statement 1: The stress-strain graphs are shown in the figure for two materials A and B are shown in figure. Young’s modulus of A is greater than that of B.
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Direction: Each question contains state-ment 1 (Assertion) and statement 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct.
3.25
1. Match the following:
(b) Normal stress
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2. Statement 1: If a metal wire is attached to the ceiling of a room and mass m is attached to another end, the energy stored in the
Column II (p) Micrometer with a spirit level (q) Bulk modulus
Column I (a) Hooke’s law
Statement 2: The Young’s modules for stress small strain is, Y = = slope of linear strain portion, of graph; and slope of A is more than slope that of B.
Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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MATCH THE COLUMN TYPE QUESTIONS
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3.26
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5. Match Column I (Elastic properties of an isotropic elastic material) with Column II (nature of strain produced) and select the correct answer using the codes given below the lists.
1. For which range, during unloading the above curve will be retraced? (a) up to OA only (b) up to OB (c) up to C (d) Never retraced its path
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(d) Volume of a substance
Column II (p) Depends on temperature (q) Depends on length (r) Depends on area of crosssection (s) Depends on the nature of material
It is observed that for a substantial range of stresses, the stress-strain relation is linear, and the specimen recovers to its original dimensions when the load is removed.
Column I (a) Young’s modulus of a substance (b) Bulk modulus of a substance (c) Modulus of rigidity of a substance
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(d) Position’s ratio
PASSAGE-1 Figure shows qualitatively the relation between the stress and the strain as the deformation gradually increases.
PASSAGE BASED QUESTIONS
(s) energy
(c) Compressibility
Column II (p) pressure (q) energy density (r) angle
Column I (a) Stress (b) Strain (c) Modulus of elasticity (d) Torque
2. Match the following:
Column I Column II (a) Young’s modulus (p) Shear strain (b) Modulus of rigidity (q) Normal strain (c) Bulk modulus (r) Transverse strain (d) Poisson’s ratio (s) Volumetric strain
(r) Change in volume depends on normal stress (d) Searle’s method (s) Elastic limit
(c) Velocity of sound
Elasticity
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4. Two different types of rubber are found to have the stress-strain curve as shown (i) to absorb vibrations one would prefer rubber A (ii) For manufacturing car tyre one would prefer B
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PASSAGE-2 In Searle’s apparatus for determining the Young’s modulus for a wire, a student is taking measurements of different parameters. Length of the wire, L = 30 cm measured with a millimeter scale Diameter of the wire d = 1 mm measured with a screw gauge Least count of the screw gauge = 0.01 mm Least count of the spherometer attached to the apparatus frame = 0.005 mm Spherometer measured the extension (e) of the wire W = weight of the loads.
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5. Stress vs strain for the elastic tissue of the aorta, the large tube (vessel) carrying blood from the heart, will be: [stress is proportional to square of the strain for the elastic tissue of the aorta] [
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3.28
8. The load vs extension graph should be (if the extension is in elastic limit) (a) straight line (b) curved line (c) broken line (d) any of the above
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Elasticity
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1. A human bone is subjected to a compressive force of 5.0 × 105 N/m2. The bone is 25 cm long and has an approximate cross sectional area 4.0 cm2. If the ultimate compressive strength of the bone is 1.70 × 108 N/m2, will the bone be compressed or will it break under this force?
(a) Which material has greater Young’s modulus? (b) Which material is more ductile? (c) Which is more brittle? (d) Which of the two is stronger material?
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6. A rod AD consisting of three segments AB, BC and CD joined together is hanging vertically from a fixed support at A. The lengths of the segments are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly 10–4 m2. A weight of 10 kg is hung from D. Calculate the displacements of points B, C and D using the data on Young’s moduli given below (neglect the weight of the rod). YAB = 2.5 × 1010 N/m2, YBC = 4.0 × 1010 N/m2 and YCD = 1.0 × 1010
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5. A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m stretches under a load by 0.7 mm. Calculate the load, given that the Young’s modulus for copper is 1.1 × 1011 Pa and that for steel is 2.0 × 1011 Pa (1 Pa = 1 Nm–2)
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Flat faces at the narrow end of the anvil have a diameter of 0.5 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil? 3. The stress versus strain graph for two materials A and B are shown in figure. The graphs are on the same scale.
N/m2.
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7. The radii of two rods of the same length and same material are in the ratio r1 : r2. If these rods are twisted by applying the same torsional torque, then the ratio of the angle of twist produced in the two rods will be 8. A steel wire of length 4.7 m and crosssection 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-section 4.0 × 10–5 m2 under a
4. What is the density of ocean water at a depth, where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3? Compressibility of water = 45.8 × 10–11 N–1 m2]
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2. Anvils made of single crystals of diamond, with the shape as shown in figure, are used to investigate behaviour of materials under very high pressures.
SUBJECTIVE QUESTIONS
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Elasticity
with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance r from the rotation axis. Find the elongation of the rod. Density of rod is ρ.
9. A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section 10–3 m2 and the other is of brass of cross section 2 × 10–3 m2. Find out the position along the rod at which a weight may be hung to produce, (a) equal stresses in both wires (b) equal strains on both wires Young’s modulus for steel is 2 × 1011 N/m2 and for brass is 1011 N/m2.
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The wires A and C are of copper 1 mm2 in cross-section and B of iron 2 mm2 in crosssection. If the Y for copper and iron is 1 × 1011 and 2 × 1011 N/m2 respectively, find the tension in each wire.
(b) 2 × 10–11 N/m2 (d) 2 × 1013 N/m2
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3. In Searle’s experiment, which is used to find Young’s modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer aof least count 0.001 cm). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error. [2004]
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2. The adjacent graph shows the extension (∆ℓ) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10–6 m2, calculate the Young’s modulus of the material of the wire. [2003]
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1. A thin rod of negligible mass and area of cross-section 4 × 10–6 m2, suspended vertically from one end, has a length of 0.5 m at 100ºC. The rod is cooled to 0ºC, but prevented from contracting by attaching a mass at the lower end. Find (i) its mass and (ii) the energy stored in the rod, given for the rod. Young’s modulus = 1011 N/m2, coefficient of linear expansion 10–5 K–1 and g = 10 m/s2. [1997]
PREVIOUS YEARS’ IIT-JEE QUESTIONS
given load. What is the ratio of the Young’s modulus of steel to that of copper?
3.29
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3.30
4. The pressure of a medium is changed from 1.01 × 105 Pa to 1.165 × 105 Pa and change in volume is 10% keeping temperature constant. The bulk modulus of the medium is
lar reading, the student measures the extension in the length in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young’s modulus obtained from the reading is [2007]
Elasticity
[2005] 204.8 × 105 Pa 102.4 × 105 Pa 51.2 × 105 Pa 1.55 × 105 Pa
(a) (b) (c) (d)
ASSERTION AND REASON QUESTIONS 2. (a)
3. (a)
4. (a)
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(a) (b) (c) (d)
8. (b, d)
MATCH THE COLUMN TYPE QUESTIONS 1. (a) → s (b) → q, r (c) → q (d) → p
2. (a) → p, q (b) → r (c) → p, q (d) → s
3. (a) → q (b) → s (c) → p (d) → r
4. (a) → p, s (b) → s (c) → p (d) → p
5. (a) → q (b) → p, (c) → s, (d) → r
PASSAGE BASED QUESTIONS 1. (b)
2. (b)
3. (b)
4. (d)
5. (a)
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6. (b)
7. (c)
8. (d)
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Elasticity
3.31
HINTS AND SOLUTIONS 5 0 9 0 4 0 = = A ( − ) A( 3 − 0 ) A(1 − 0 ) 2 0
Y=
9. (a) False: For a given load, Young’s modulus is inversely proportional to strain which is more for rubber. So elasticity of rubber is less than that of steel. (b) True: Under the application of deforming force, stretching of coil changes its shape without any change in the length of wire used. (c) True: When a material is subjected to tensile stress, the interatomic separation increases and becomes more than r0, due to which interatomic force of attraction comes into play. In case of compressional stress, interatomic separation decreases and becomes less than r0, due to force for repulsion comes into play. (d) False: Elastic forces are conservative as long as loading and unloading curves are identical even if the curves are not linear. (e) False: In case of rubber, elastic hysteresis takes place which signifies nonconservative forces.
These give 4 5 = 1 − 0 2 − 0
CONCEPTUAL QUESTIONS
8. (b) Let ℓ0 be the unstretched length and ℓ3 be the length under a tension of 9 N. Then
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F 5 × 105 = = 12.5 × 108 N/m2 A 4 × 10−4
Since this stress exceeds the ultimate compressive stress of the lone, the bone will break. 2. Here F = 50,000 N, d = 0.5 mm = 5 × 10–4 m
Work necessary in the second case = 8 × 2 = 16 J
F L . A
SUBJECTIVE QUESTIONS
∴
T = 2π
∴ F2 = 8F1 ⇒ W2 = 8W1
YA = tan 60° = 3 YA =3 ⇒ YB = tan 30° = 1/ 3 YB
F YA = 1 L Where ℓ is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by
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Force constant k =
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ONLY ONE OPTION IS CORRECT 7. (a) Work W =
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3.32
7. Torisional torque required for θ twist is ηπr 4 θ 2 Thus for same τ, ℓ and η, τ = Cθ =
4
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9. (a) 1.33 m, (b) 1 m mg 10. ∆ = 2AY
4
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F 50000 × 4 = = 2.5 × 1011 Pa A 3.14 × (5 × 10−4 ) 2 3. (a) For a given strain, stress is more for material A. stress Young’s modulus = strain Hence, Young’s modulus is greater for A than that of B. (b) For material A, there is a larger stress strain variation beyond elastic point, therefore A is more ductile, than B. (c) For material B, there is a small stressstrain variation beyond elastic point, therefore, B is more brittle than A. (d) The strength of a material, is the measure of the amount of stress required to cause fracture, therefore A is stronger than B.
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Elasticity
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Hydrostatics and Hydrodynamics
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Let, us consider two points A and B separated by small vertical height dz. Imagine a horizontal area ∆S1 containing A and identical horizontal area ∆S2 containing B. Then area, ∆S1 = ∆S2 = ∆S ) %
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4.2
Hydrostatics and Hydrodynamics
The Vertical forces Acting on this fluid are (i) F1, vertically upward by the fluid below it. (ii) F2, vertically downward by the fluid above it. (iii) Weight W, vertically downward. Let the pressure at the surface A be p and the pressure at B be p + dp. Then, F1 = p ∆S And F2 = (p + dp) ∆S The volume of the fluid considered is (∆S) (dz). If the density of the fluid at A is ρ, the mass of the fluid considered is ρ (∆S) (dz) and, then its weight
F1 = F2 + W p ∆S = (p + dp) ∆S + ρg ∆S dz dp = – ρg dz
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W = ρ (∆S) (dz) g For vertical equilibrium
…(i)
0
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p2 – p1 = – ρgh
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As we move up through a height dz, the pressure decreases by ρg dz, where ρ is the density of the fluid at that point. Now, consider two points z = 0 and z = h. From equation (i)
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or p1 = p2 + ρgh The pressure of same horizontal line is equal.
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4.2 pASCAL’S LAW If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude. https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
4.3
Simply, if the pressure at one point is increased then all other points pressure is also increased because pressure difference in two points is equal. As an example, suppose a flask fitted with a piston is filled with a liquid as shown in the figure. Let an external force F be applied on the piston. If the cross-section area of the piston is A, the pressure just below the piston is increased by F/A. By Pascal’s law, the pressure at any point B will also increase by the same amount F/A. ) $
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4.3 prINCIpLE Of ArCHIMEDES
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Principle of Archimedes states that when a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to weight of the displaced fluid. According to Archimedes’ principle B = mg = weight of fluid displaced where, B = buoyant force B = Vρg where, V = volume occupied by the solid inside the liquid.
4.4 fLOATING BODY Whenever the buoyant force on a body equals to its weight, the net force acting on the body becomes zero. Hence, the body floats in equilibrium. Let a body of total volume V and density d be floating on a liquid of density ρ. Let v be the volume of the body inside the liquid. Then B = mg (for a floating body) or vρg = Vdg or vρ = Vd https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
na .c om
Note: 1. When a body is in air, the net downward force on it is due to earth’s gravity only. This net force = mg = Vdg. This force is known as weight of the body. 2. When the same body is immersed in a liquid of density ρ, the net downward force on it = mg – B where, B = buoyant force ⇒ Fd = Vdg – vρg This downward force is also known as apparant weight in the liquid. 3. When a body is immersed in a liquid, it feels lighter due to buoyant force. Hence, the buoyant force is also known as loss in weight (= vρg)
)
m
) Solution
rio s_ in fo
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Example 90 g of sulphuric acid of specific gravity 1.8 is mixed with 90 g of water. If the specific gravity of the mixture is 1.5, find the contraction when the liquid are mixed. Density of water = 1 gm–3.
un iv
s:
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tp ht
er si ta rio s
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w
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Density of water = 1 gm–3 ∴ Density of sulphuric acid = 1.8 times density of water = 1.8 g cm–3 Volume of 90 g of water = 90 cm3 90 Volume of 90 g of sulphuric acid = = 50 cm3 1.8 ∴ Total volume before mixing = 140 cm3 After mixing, total mass is 180 g and specific gravity of the mixture is 1.5, i.e., density of the mixture is 1.5 g cm–3. 180 = 120 cm3 1.5 1 Hence the mixture contracts by 20 cm3 i.e., of its original volume. 7 Volume of the mixture =
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Example A vertical U tube of uniform inner cross-section contains mercury in both of its arms. A glycerine (density 1.3 g cm3) column of length 10 cm is introduced into one of the arms. Oil of density 0.8 g cm–3 is poured in the other arm until the upper surface of the oil and glycerine are in the same horizontal level. Find the length of oil column. Density of mercury is 13.6 g cm–3. )
) Solution
Let the height of x in the column is poured. When the horizontal line is same of the oil and glycerine. For our previous concept, pressure in horizontal line is equal to each point. So, the pressure at B and E is equal.
4.4
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Hydrostatics and Hydrodynamics
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er si ta rio s
123 = 9.6 cm 12.3
t.m e/
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ht tp s:
∴
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s:
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w
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Let the atmospheric pressure p0. Then, pressure at B = p0 + ρ1g × 10 where, ρ1 = density of glycerine Pressure at, E = p0 + ρ2 gx + p3 (10 – x) g where, ρ2 = density of liquid and p3 = density of mercury Hence, p0 + ρ1 g (10) = p0 + ρ2gx + ρ3 g (10 – x) or 1.3 × 10 = 0.8 × x + 13.6 × (10 – x) or 13 = 0.8x + 136 – 13.6 x or 12.8x = 123
tp ht
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Example A stone of density 2.5 g cm–3 completely immersed in sea water is allowed to sink from rest. Calculate the depth to which the stone would sink in two seconds. Neglect the effect of friction. Specific gravity of sea water is 1.025 and acceleration due to gravity is 980 cms–2. % 9SJ
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Hydrostatics and Hydrodynamics
) Solution
Let the volume of the stone is V. Then weight of stone = Vdg where, d is density of stone. Buoyant force of sea water, B = Vρg Thus, net downward force,
or
Vdg – Vρg = Vda
or
a=
na .c om
(d − ρ) g d
(2.5 − 1.025) × 980 cm/s 2 2.5 = 578.2 cm/s2 1 s = ut + at 2 2 1 = 0 + × 578.2 × 4 2
ar a
=
er si ta rio s
s:
un iv
//w
= 1156.4 cm = 11.56 m
er s
w
w
ita
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m
∴
rio s_ in fo
F = Vdg – Vρg From Newton’s Law F = ma
e/
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
Example A tank of square cross-section (2m × 2m) is filled with water upto a height of 2.5 m. Find the thrust experienced by the vertical and bottom face of the tank. (g = 10 m/s2).
ht
4.6
h
) Solution Here, the pressure is uniformly distributed over the bottom surface. Thrust at the bottom surface. = Pressure × Area = (2.5 ρg) × A = 2.5 × 1 × 103 × 10 × 2 × 2 = 1 × 105 N
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Hydrostatics and Hydrodynamics
4.7
On the vertical surface in contact with liquid the pressure is not same at all points. Points at greater depth experience large pressures and hence greater thrust. Consider a vertical surface ABCD (l × h) in contact with liquid of height h. Divide the vertical surface ABCD into rectangular strips of height dx and area ldx. h ρg l h 2 F = ∫ pd A = ∫ ( xρg ) l dx = 2 0 Expressing F as F = [Pressure at C.G. of vertical surface] × [area of surface] h = ρg hl 2
ar a
= 6.25 × 104 N
rio s_ in fo
na .c om
2.5 = × ρ× g (2 × 2.5) 2
iv P
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;
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m
Example The figure shows a cross-sectional view of a mercury dam whose length perpendicular to the diagram in 30 m. The depth of water behind the dam is 10 m. The masonary of which the dam is constructed has a density of 3000 kg/m3.
s: //
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2
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(i) Find the dimensions x and 2x if the weight of the dam is to be 10 times the horizontal force extended on it by water. (ii) Check the stability of the dam with respect to its overturning about the edge through point O.
) Solution (i) Weight of the dam = volume × density
1 = ( x + 2 x)12 × 30 3000 kg 2 = 1620000 x kgf https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
Horizontal thrust due to water in kg f (on a vertical surface) = Pressure at G × Area = 5 × 103 × (30 × 10) = 15 × 105 kg f According to problem
1620000 x = 10 (15 × 105) ∴
x=
2x = 18.52 m
and
1500 = 9.26m 162
Anti-clockwise torque exerted by weight of the dam
w
11 2 x g 9 = 169.78 × 106 g Clockwise torque due to liquid thrust
e/
un iv
//w
= 5000 ρg
x3 3
10
iv
0
ht tp
= 30 ρg 5 x 2 −
s: //
0
t.m e/
ht tp s:
//t
= ∫ ( x ρg )(30 dx) (10 − x)
un
10
.m
ht
tp
s:
= 162 × 104 ×
er si ta rio s
er s
w
ita
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11 = W x 9
rio s_ in fo
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3 2 (12 x) x + (6 x) x 2 3 = 11 x x= 12 x + 6 x 9
m
na .c om
(ii) Let G be the centre of gravity of the dam. Taking O as origin and left as + x-axis
4.8
= 5 × 106 g As the torque due to liquid thrust is much less than that due to weight of the dam, hence the dam is stable.
Example A cube of wood supporting a 200 gm mass just float in the water. When the mass is removed, the cube rised by 2 cm. What is the size of cube?
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rio s_ in fo
(a)
FP
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er si ta rio s
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w w //w
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4.9
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Hydrostatics and Hydrodynamics
t.m e/
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) Solution From Fig. (a)
ht tp
s: //
Vertical force = mg = Weight of the cube or 200 g = Weight of the mass B = l3ρg upward where, l = side of the cube At equilibrium (m + 200) g = l3 ρg From figure (b), mg = l2 (l – 2) ρg From equations (i) and (ii), we get 200 = ρ [l3 – (l – 2) l2] Where ρ = 1 g/cc of water Solving, we get l = 10 cm
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…(i) …(ii)
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4.10
Hydrostatics and Hydrodynamics
Example An expansible balloon filled with air floats on the surface of a lake with 2/3 of its volume submerged. How deep must it be sunk in the water so that it is just in equilibrium, neither sinking further nor rising? It is assumed that the temperature of the water is constant and height of the water barometer is 10 metres. )
) Solution
p1V1 = p2V2
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From Boyle’s law,
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Let V1 be the volume of the balloon, when floating Mass of the balloon = Mass of water displaced 2 = × V1 × 1g 3 2 = V1 gf 3 Let V2 be the volume of the balloon when it is sunk in the water. Since, mass of the balloon is constant, therefore 2 V1 = V2 3 2 −10 × V1 = p2 × V1 3 ∴ p2 = 15 m of water At the surface of the lake pressure is 10 m of water Hence the balloon must be sunk at a depth of (15 – 10) = 5 m.
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.m
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Steady low
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When liquid velocity at any point is constant in magnitude and direction at all times, the motion of the liquid is said to be steady. A flow-line in a velocity field is an imaginary line in that space, where the tangent to the line at any point on the line gives the direction of the velocity at that point. A flow line is also called a velocity line or a stream line. The flow of a liquid can be classified as:
4.5 HYD ODYNAMICS
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(i) Stream-line flow In stream-line flow, the stream lines in the flow space remains steady as time progresses. (ii) A Turbulent-flow In turbulent-flow, the stream lines in the flow space shift their positions as the time progresses. In a stream-line flow, a group of stream lines from a tubular volume of the flow space, the surface of which is tangential to the stream lines, forming the lateral boundary of that tubular volume. Such a tubular volume in the flow space is called a tube of flow.
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Hydrostatics and Hydrodynamics
4.11
6
7XEHRIIORZ
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(iii) Steady state flow It is the flow in which the liquid proporties at any point in the velocity field do not change with time.
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4.6 EQUATION Of CONTINUITY
Let us consider two cross-sections of a tube of flow at the point A and B. Let the area of cross-section at A be A1, and that at B be A2. Let the speed of the fluid be v1 at A and v2 at B. dM = ρ Av = constant. dt If the fluid is incompressible, density is constant at all points, hence equation of continuity is Av = constant A1 v1 = A1 v2
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4.12
Hydrostatics and Hydrodynamics
The product of the area of cross-section and the speed remains the same at all points of a tube of flow. The equation of continuity is law of conservation of mass in fluid dynamics.
4.7 BErNOULLI’S EQUATION Suppose, the area of cross-section at A = A1 The area of cross-section at B = A2
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Speed at A = v1 Speed at B = v2 Pressure at A = p1 Pressure at B = p2 and, Density of liquid = ρ PQ = horizontal line. Let, δ l1 = length of the fluid flowing into the tube at A1 per second. δ l2 = length of the fluid flowing out of the tube at A2 per second. Since, the fluid is incompressible, therefore A1 δ l1 = A2 δ l2 ...(i) Total energy of the fluid at any cross-section = pressure potential energy + gravitational potential energy + kinetic energy 1 Total energy at A = p1 (A1 δ l1) + (A1 δ l1 ρ) v12 + (A1 δl ρ) gh1 2 where, ρ is the density of the liquid. Total energy at B = p2 (A2 δ l2) + Then, p1 (A1 δ l1) +
1 (A2 δ l2 ρ) v22 + (A2 δ l2 ρ) gh2 2
1 (A1 δ l1 ρ) v12 + (A1 δ l1 ρ) gh1 2
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Hydrostatics and Hydrodynamics
= p2 (A2 δ l2) +
1 (A2 δ l2 ρ) v2 + (A2 δ l2 ρ) gh2 2
4.13 ...(ii)
Dividing both sides by ρ A1 δl1 = ρ A2 δ l2 = rate of the mass flow, we get p1 v12 p ν2 + + gh1 = 2 + 2 + gh2 ρ 2 ρ 2 Generalizing and removing suffixes, we get p ν2 + + gh = constant ρ 2
4.8 VENTUrIMETEr
rio s_ in fo
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This equation is called Bernoulli’s equation for steady non-viscous and incompressible fluid flow.
un
ν12 ν2 = pB + ρ 2 2 2
s: //
pA + ρ
t.m e/
ht tp s:
for Liquid
iv
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v2 = velocity at B.
er si ta rio s
un iv
e/
v1 = velocity at A
//t
ht
tp
s:
//w
w
er s
w
ita
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m
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This is an instrument for measuring the rate of flow of liquids and gases [fluids]. The instrument is connected horizontally in the tube in which the rate of flow is to be measured. If pA is pressure at A and pB is pressure at B, then pA – pB = h ρg where, h = difference of liquids of density ρ in vertical tubes. Applications of Bernoulli’s equation at A and B in two cases (i.e., liquid, gas).
2 2 ν 2 2 − ν12 = ( pA − pB ) = hρg ρ ρ
or
ν 2 2 − ν12 = 2hg
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4.14
Hydrostatics and Hydrodynamics
or
Q2 Q2 − = 2 gh A 22 A12
∴
Q = A1 A 2
(since, Q = Aν)
2hg A12 − A 22
for Gases Let d be the density of the gas pA – pB = hρg
2hρg d (A12 − A 22
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4.9 pITOT TUBE
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Q = A1 A 2
2hρg d
ar a
v2 2 − v12 =
rio s_ in fo
v12 v2 = pB + d 2 2 2
na .c om
pA + d
or
Pitot tube is used to measure the speed of flow of a fluid. The cross-section of tube at B is perpendicular to the direction of flow and tube is at rest, so that at B, velocity of fluid becomes zero, i.e., vB = 0 Let v be the velocity of fluid at A and pA and pB are pressures at A and B respectively. ⇒ pB – pA = h ρg where, ρ = density of liquid in vertical tubes.
for Liquids If points A and B are in same horizontal line and area of cross-section is same at A and B, gravitational head is same and the pressure difference is due to difference in velocity. https://t.me/universitarios_infohttps://www.jamarana.com
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Hydrostatics and Hydrodynamics
Apply Bernoulli’s theorem at A and B
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∴
v=
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or
rio s_ in fo
1 pA ρv 2 = pB + 0 2
na .c om
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ita
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m
2( pB − pA ) ρ
er si ta rio s
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//w
w
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w
= 2gh
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s:
for Gases
iv
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pB – pA = h ρg
un
//t
ht
tp
Let ρ be the density of the liquid in the vertical tube
= v=
2( pB − pA ) d 2hρg d
ht tp
v2 =
s: //
1 pA + dv 2 = pB 2
t.m e/
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4.15
4.16
Hydrostatics and Hydrodynamics
4.10 VELOCITY Of EffLUX Let us find the velocity with which liquid comes out of a hole at a depth h below the liquid surface. Using Bernoulli’s Theorem
A
B hole
v
rio s_ in fo
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na .c om
h
ita
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1 1 pA + ρ vA 2 + ρghA = pB + ρ vB 2 + ρghB 2 2
er si ta rio s
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//w
w
er s
w
1 1 Patm + ρ vA 2 + ρgh = patm + v 2 + 0s 2 2
or
v=
a2 2 v + 2 gh A2 2 gh a2 1− 2 A
iv
s: //
v2 =
ht tp
∴
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
Note: (pB = patm, because we have opened the liquid to atmosphere.) v2 = vA2 + 2gh By equation of continuity, AvA = av where, A = area of cross-section of vessel. and, a = area of hole
≈ 2 gh
(if the hole is very small)
Example Water is in stream line flow through a horizontal tube. If the velocity at a crosssection of 4 cm2 is 100 cms–1, calculate. (i) The rate of laminar at that section. (ii) The velocity of the flow at another cross-section of 1 cm3. (iii) Pressure difference between the two sections. (ρ = density of water = 1 g cm–3)
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Hydrostatics and Hydrodynamics
4.17
) Solution (i) Rate of flow = (cross-section) (velocity) = (4 cm3) (100 cms–1) = 400 cm–3 s–1 400 cm −3s −1 (ii) Velocity of 1cm 2 = 1 cm 2 = 400 cm s −1 1 1 p1 + ρv12 = p2 + ρv2 2 2
(iii) We know,
na .c om
or
= 7.5 × 104 dynes cm–2
rio s_ in fo
1 ρ(v12 − v2 2 ) = ( p2 − p1 ) 2 1 ( p2 − p1 ) = × 1(42 − 12 ) 2
or
er si ta rio s
w
er s
w
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= 7.5 × 103 Nm–2
ita
m
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1 = (7.5 × 104 ) 10−5 × −4 10
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t.m e/
ht tp s:
un
//t
ht
tp
s:
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Example Water shoots out of a pipe and nozzle as shown in the figure. The cross-sectional area for the tube at point A is four times that of the nozzle. The pressure of water at point A is 40.78 × 103 Nm–2 (guage). Find the height h above the nozzle to which water jet will shoot. Neglect all the losses occurred in the above process. K OO P
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) Solution By Bernoulli’s equation between point A and top of the nozzle, we have 1 1 pA + ρvA 2 + ρghA = pN + ρ vN 2 + ρg hN 2 2 Now, pN = 0, hA =0, hN = l ∴
1 1 pA + ρvA 2 = ρvN 2 + ρgl 2 2
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…(i)
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Hydrostatics and Hydrodynamics
Using continuity equation AA vA = AN vN vN =
∴
AA vA = 4 vA AN
…(ii)
From equation (i) and (ii), we get pA = ρgl + 7.5 v A 2 =
15 2 ρvA 2
(40.78)103 − (9.8)(1.1) 103
vA = 2 ms–1
⇒
…(iii)
na .c om
∴
ar a
1 pA + ρ vA 2 = ρg (l + h) 2 h = 3.97 m.
m
⇒
rio s_ in fo
Applying Bernoulli’s equation between point A and the water jet, we have
er si ta rio s
un iv
//w
w
er s
w
ita
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Example Water stands at a depth H in a tank whose sides are vertical. A hole is made in one of the walls at a depth h, below the water surface, calculate
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s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
s:
(i) Range of emerging steam (ii) Value of h for which this range is maximum
ht
K
ht tp
4.18
KROH %6WUHDP
) Solution (i) Vertical distance travel by steam = (H – h) When path of the stream is parabolic, then, the time taken by the stream to reach the bottom surface is
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Hydrostatics and Hydrodynamics
2(H − h) g
…(i)
t=
In horizontal direction there is no acceleration
Velocity of stream at B = 2gh 2(H − h) g
R = x = v × t = 2 gh 2(H − h) g
R = x = v × t = 2 gh
ar a
R2 = 4 hH –4h2
ita
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er s
w
un iv
H 2
ht tp
iv
s: //
t.m e/
h=
e/
ht tp s:
//t
h=
.m
H 2 Hence, for maximum range ∴
ht
tp
s:
//w
dR 2 = 4 H − 8h = 0 dh
un
For maximum range
w
m
= 4 hH − 4 h 2
er si ta rio s
R = 4h (H − h)
rio s_ in fo
(ii) Horizontal Range,
na .c om
= 4h (H − h)
or
1. A wooden plank of length 1 m and uniform cross-section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. (exclude the case θ = 0º)
P
R
WO KED OUT EXAM LES
4.19
ș
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4.20
Hydrostatics and Hydrodynamics
volume by the pressure and the gravity forces as the fluid flows from point P to point Q.
)
) Solution
Weight of the plank = Aldg where, d = density of the plank. B = (Volume of submerged part) × ρg Let V′ = submerged volume = A (0.5) sec θ where, (0.5) see θ is submerged length. Taking moment at hinged point, we have
4
3 P P
) Solution )
%
a. co
fo
in
s
ta rio
si
er
iv
un
v2 =
=
A1 v1 A2
4 × 10−3 × 1 = 0.5 m/sec. 8 × 10−3
ht tp s
From equation (i) 1 p2 − p1 = ρ (v12 − v2 2 ) + ρg (h1 − h2 ) 2 1 = × 1000(1 − 0.25) + 1000 × 9.8(2 − 5) 2 1 = × 1000 × 0.75 + (−29400) 2 = 29025 Jm–3 Work done by gravity = ρg (h2 – h1) = 1000 × 9.8 (5 – 2) = 2.94 × 104 Jm–3
3. A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous
2. A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points P and Q at height of 2 metres and 5 metres are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work-done per unit
si er
e/
:// t
ht
.m
tp s
iv
un
e/
:// t.m
ht tp
or
io
ta r
m .ja
s:
//w
w
w
l 1 W × sin θ = B ⋅ (0.5sec θ) ⋅ sin θ 2 2 l Al dg sin θ = A × 0.5 sec θ ρg × 0.5 or 2 1 sec θ ⋅ sin θ 2 or d = 0.5 × 0.5 ρ sec2 θ Here, d = 0.5 ρ ∴ 0.5 ρ = 0.5 × 0.5 ρ sec2 θ or sec2 θ = 2 1 or cos 0 θ = 2 1 cos θ = or 2 ∴ θ = 45°
…(i)
Given that h1 = 2 m, h2 = 5 m A1 = 4 × 10–3 m2 A2 = 8 × 10–3 m2 v1 = 1 m/s Applying continuity equation, we have ρ1 A1 v1 = ρ2 A2 v2 But ρ is constant, therefore A1 v1 = A2 v2
ar
ș
1 1 p1 + ρv12 + ρgh1 = p2 + ρv2 2 + ρgh2 2 2
s_
an
9GJ
m
Applying Bernoulli’s theorem
&*
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4.21
Hydrostatics and Hydrodynamics
where,
and incompressible liquids of density d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure p0.
of liquid of density d.
and
$
Hence from equation (i) D =
G
(ii) Total pressure at the bottom P = patm + pressure due to liquid of density d + pressure of density of liquid 2d + pressure of the solid cylinder.
%
na .c om
K
rio s_ in fo
un iv
er si ta rio s
er s
ita
L = dg . 4
3
iv
.m
un
s: //
This pressure is due to the cylinder. Hence, total pressure p = p0 + dg
ht tp
3
(b) (i) Using Bernoullis equation
)
A A 3L A L LDg = . dg + . 2dg 5 5 4 5 4 …(i)
H H L + 2dg + dg 2 2 4
3H L = p0 + + dg 2 4
) Solution
(a) (i) Weight of solid cylinder is balanced by buoyant force of two liquids
G G
t.m e/
ht tp s:
L 3L or p1 − p2 = d ⋅ g − 2d ⋅ g 4 4
e/
s:
//t
ht
tp
//w
w
w
.ja
m
ar a
(a) A homogeneous solid cylinder of length H A L L < , cross-sectional area is 2 5 immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid. Determine: (i) density D of the solid; and (ii) total pressure at the bottom of the container. (b) The cylinder is removed and the original arrangement is restored. A tiny hole of area s (s << A) is punched on the vertical side of the container at a height h (h < H/2). Determine: (i) The initial speed of efflux of the liquid at the hole. (ii) The horizontal distance x travelled by the liquid initially. (iii) The height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm (Neglect the air resistance in these calculations).
H H + 2dg + pressure 2 2 due to cylinder
or p = p0 + dg
;
5d 4
1 pA + ρ1 vA 2 + ρ1 ghA 2
1 = pB + ρ2 vB 2 + ρ2 ghB 2 Here, ρ1 = d and ρ2 = 2d
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G
+
AL 2dg = Buoyant force of 54
liquid of density 2d.
+
A 3L dg = Buoyant force 5 4
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…(i)
4.22
one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ (> ρ). )
) Solution
The centre of mass composite body. (Wooden stick + small piece of metal) can be obtained by using the expression.
pA = pB because we have opened the liquid to atmosphere and also sv Since, vA = B , therefore A vA = 0 According to the problem given that (s << A), hence from equation (i) H H 1 + 2d − h = 2d vB2 2 2 2
xcm =
dg
3 vB 2 = H − 2h g 2
3 vB = H − 2h g 2 (ii) After liquid injected from the hole vertical velocity is zero. So its free fall 1 h = gt 2 2
o nf
a. rio
ar
3H 3H 3 3H − 4 × = H 8 8 4
4. A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its
rio
i.e.,
ita
rs
iv e
un e/ t.m
tp
ht
dx 2 = 3H − 8 h = 0 dh 3H ∴ h= 8
If l is the length of wooden stick dipped in the liquid, the centre of mass of the buoyant force will be obtained at l xb = 2 For the composite body floating vertically in equilibrium, we will have xcm = xb
ML l = 2(M + m) 2
or
M(L/2) + m(0) M+m ML = 2(M + m)
xcm =
s: //
or x2 = h (3H – 4h) dx For the maximum value = 0, dh d 2x so is also zero. dh
or xm =
s
iv er un
.m
//t
x = h (3H − 4h)
(iii)
ht tp s:
= h (3H − 4h)
where, the subscripts A and B represents wooden stick and piece of metal respectively. Taking the origin at the small piece of metal, we get
e/
s:
ht
tp
2h 3 = H − 2h × g 2
si ta
m
.ja
w
//w
w
Horizontal distance, x = v × t
O
s_ i
an
2h g
∴ t=
/
/
co
∴
MA + MB
m
or
M A xA.cm + M B xB.cm
l=
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M L M+m
Hydrostatics and Hydrodynamics
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…(i)
4.23
Hydrostatics and Hydrodynamics
Now, from the mass of displaced liquid we write (M + m) = (πR2) lσ Substituting l from equation (i), we get
)
ita
un
t.m e/
s: //
ht tp
er si ta rio s
Ah ρB g = ma = A (hA + hB + h) d a
iv
.m
//t
ht tp s:
er s
un iv
e/
s:
tp
ht
1.8 − 1.6 = 0.25 cm 0.8 (iii) When the cylinder dipped in the liquid completely, top of the cylinder is top of the liquid. It means the equilibrium condition is destored. If the volume is more than the cylinder volume dipped in liquid, this extra volume is buoyant upward force. Hence or h =
or
m
//w
w
w
.ja
3
a=
0.25 × 1.2 g g = 2.25 × 0.8 6
6. Two identical cylinderical vessels with their bases at the same level contain a liquid of density ρ. The area of either base is A, but in one vessel the liquid height is h1 and the other h2. Find the work done by gravity in equalizing the levels when the two vessels are connected.
5. Uniform solid cylinder at density 0.8 g/cm floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm3 and 1.2 g/cm3 respectively. The height of liquid A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm (i) Find the total force exerted by liquid A on the cylinder. (ii) Find h, the length of the part of the cylinder in air. (iii) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.
= πR 2 L [ ρσ − ρ]
ar a
rio s_ in fo
m = πR 2 L ρσ − M
or (1.2 + 0.8 + h) × 0.8 = 1.9 × 0.7 + 1.2 × 0.8 or (2.0 + h) × 0.8 = 0.84 + 0.96
M + m = (πR 2 L) ρσ
or
na .c om
or (M + m)2 = πR2 MLσ The mass M of the wooden stick is given by M = πR2 Lρ Substituting this on the right side of equation (ii), we get (M + m)2 = (π R2 L)2 ρσ or
(i) Liquid exerts only horizontal force but this force is balance to each other. Buoyancy force is applied in liquid B. (ii) At equilibrium, weight of the cylinder is balanced by the buoyant force of the liquid. Let the area of the cylinder is A. Then, Volume of the cylinder, V = A (hA + hB + h). From equation of equilibrium A (hA + hB + h) dg = A hA ρAg + A hB ρB g
…(ii)
M (M + m) = πR 2 Lσ M+m
) Solution
K DLU $
3
K$ K
%
3
KD
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K
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Hydrostatics and Hydrodynamics
) Solution )
) Solution )
Obviously, the work done by gravity in equalising the levels is equal to the decrement of the potential energy of the liquid mass. Taking the reference level of potential energy at the common level of the base, the initial height of the C.G. of the system.
Z
\
\
Ah1ρgh1 Ah2ρgh2 + 2 2 = Ah1ρgh1 + Ah2ρgh2
2
fo
in
a.
s_
an
ta
si
rio
iv
un
ρgy2
∫
e/
s:
.m
tp
dp =
x =l
∫ ρω x dx
g ( y2 − y1 ) =
ω2l 2 2
∴
y2 − y1 =
ω2l 2 2g
s:
or
8. A cylindrical vessel with a liquid moves linearly with acceleration a. Find the shape of the surface of the liquid in the vessel.
) Solution )
7. The manometer shown below contain a liquid of density ρ. Find the difference in the levels when the manometer rotates at a constant angular velocity ω about one of its vertical limbs.
ht
tp
h 2 + h22 (h + h ) = A ρg (h1 + h2 ) 1 − 1 2 4 g (h1 + h2 )
1 = A ρg ( h1 − h2 )2 4
2
0
ρgy1
//t
ht
W =A ρg (h1 + h2) (H1 – H2)
ta
h1 + h2 4 Hence, the required work done =
s
er
//t
ht
.m
tp
e/
mg (h1 + h2 ) / 4 + mg (h1 + h2 ) / 4 mg + mg
un
s:
iv
//w w
w
.ja
m
rio
ar
Finally, when the levels are equalized, each cylinder holds the same mass (say m) of liquid on account of the quality of the base area, (h + h ) and the level in each is at a height 1 2 2 from the common base level. Therefore, final height of the C.G. of the system
H2 =
%
m …(i)
h +h 2(h1 + h2 )
;
Let A be the cross-section of the lower limb. The element of liquid of length dx has the mass dm = ρ (A dx) and its acceleration an = ω2x, normally towards the axis of rotation. If the pressure at the end of the mass element are p and p + dp at the distance x and x + dx respectively from the point O, we have Fn = man (p + dp) A – pA = ρ (A dx) ω2x or dp = ρω2 dx …(i) But at point O, p0 = ρgy1 and at B, p = ρg y2
co
2 2
=
2 1
G[
si
m1 gh1 m2 gh2 + 2 H1 = 2 m1 + m2
er
4.24
Let us consider a liquid element of crosssectional area A and of length l as shown in
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4.25
Hydrostatics and Hydrodynamics
h1 − h2 a = = constant l g
9. A rod of length 6 m has a mass of 12 kg. It is hinged at one end at a distance of 3 m below the water surface. (i) What weight must be attached to the other end of the rod so that 5 m of the rod are submerged? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material is 0.5.
the figure. Using Newton’s second law of motion for the liquid element F x = m ax or p1 A – p2 A = (A lρ) a ...(i) As the liquid has no vertical acceleration, therefore p1 – p2 = (h1 – h2) ρg ...(ii) From equation (i) and (ii), we get
)
) Solution
in f
m
ita ta si er ni v
AG = 3 m = 300 cm Density of the material of the rod
e/ u
.m
//t
= 0.5 g cm–3 Volume of the submerged part
s:
…(ii)
10 kg 0.5g cm −3
= 2 × 104 cm3 Force of buoyancy FB = Weight of water displaced = Vdg = 2 × 104 × 1.0 gf
Using equations (ii) and (iii), we have dp = – ρa dx – ρg dy But on the free surface dp = 0, thus from equation (iii) dy a =− dx y Thus, the sought shape is a straight line where a slope is − . g
=
mass of 5m of rod Density of rod
ht
tp
V=
But,
And
rio s
rs
.m
s:
tp
ht
∂p = ρa ∂x ∂p − − ρg = 0 ∂x ∂p ∂p dp = dx + dy ∂x ∂y
−
W
θ
AG' = 2.5 m = 250 cm
//t
Let us indicate the positive x-axis in the horizontal direction directed toward right and positive y-axis in upward direction. Then, we have ax = 0 ay = 0 and gy = – g, gx = 0 …(i)
W = Mg
G'
R φ
e/ un i
//w
ht tp s: →
−∇ p + ρg = ρa
FB G
L
ve
w
w
.ja
That’s why the shape of the liquid in the vessel is a straight line as shown in the figure. Another Method From Euler’s equation:
B
Weight of the rod, W = 12 kg f = 12 × 103 gf (i) Let R be the reaction at the hinge. Taking the moments of all forces acting on the rod about A, for equilibrium of the rod, the algebraic sum of moments of the forces must be zero. Therefore
l
So,
o
om
ar
rio s_
2
1
an a
.c
h2 h1
In the figure, G is the centre of gravity of the rod and G' is the central gravity of the submerged part of the rod. Hence, G' is also the centre of buoyancy through which the force of buoyancy FB acts vertically upwards.
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4.26
A ⇒ v2 = 1 (v1 ) = 4 v1 A2
m
ht tp s:
s
rio
si ta
ni
//t .m
e/ u
s: //
ht tp
= 280 × 103 + (– 108) × 103
= 172 ×103 Nm–2 (ii) Let Fx and Fy be the external forces on the pipe bend along x-direction and y-direction respectively. Equation of fluid is
t.m
ht tp
10. Oil enters the bend of a pipe in the horizontal plane with velocity 4 ms–1 and pressure 280 × 103 Nm–2 as shown in the figure. Find (i) The pressure of oil at the point Q. (ii) The force required to hold the bend in the place. (Take specific gravity of oil as 0.9 and sin 37º = 0.6)
of oil = 0.9 × 103 kg m–3]
φ = 0, R is vertical and negative sign just shows that it is directed vertically downwards (θ = 180º).
nf o
s_ i
si ta
ve r
e/ un i
s: //w
w
17 7 = + 12 − 20 kgf = − kgf 3 3
rio
m
w
.ja
∴ R = w + W – FB
∴ φ = 0 or 180º
Also, (R cos φ + FB) = (w + W)
1 1 p1 + ρ v12 = p2 + ρ v2 2 2 2 (Pressure in gravity does not include because plane is horizontal.) 1 p2 = p1 + ρ (v12 − v2 2 ) 2 1 = p1 + × 0.9[16 − 256] × 103 [density 2
co
ar an a.
⇒ R sin φ = 0
Applying Bernoulli’s equation between points P and Q, we have
7 or w = × 103 g = 2.33 kgf 3 Hence, a weight of 2.33 kg should be suspended from the end B.
p2 = A2
+y
V2 Q
+x p1A1 V1 37° P
Q
V2
A2 = A1 4
A1= 0.2 m V1= 4 ms-1
2
37° P
p1 A1 + p2 A2 cos 37° + Fx = [ρQ] (– v2 cos 37° – v1) A 4 ⇒ p1 A1 + p2 1 × + Fx 4 5 4 = ρ A1 v1 −(4 v1 ) − v1 5
)
or w (600) = (2 × 104 × 250) – (12 × 103 × 300)
w · AB = FB AG´ – W × AG
(ii) Resolving R into horizontal and vertical components for translation equilibrium of the rod, we must have
(i) Let A1 and A2 be the cross-sectional area of the pipe at points P and Q respectively. Let v1 and v2 be the velocities of oil at the points P and Q respectively. By conservation of mass. Q = A1v1 = A2v2
∴
(Taking clockwise moments to be negative)
) Solution
Hence, –W · AG sin θ – w · AB sin θ + FB AG´ sin θ = 0
ve r
Hydrostatics and Hydrodynamics
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4.27
Hydrostatics and Hydrodynamics
Fx p 16 = p1 + 2 + (ρ v12 ) 1 + A1 5 5
upward at the back wall of the vessel. To prevent water from flowing out of the hole H, the acceleration of the vessel should have such a 172 21 3value that it occupies a face area DBH and a Fx = −(0.2) 280 + + 0.9 × 16 × × 10 5 5 A width of the vessel given by L 172 21 Fx = −(0.2) 280 + + 0.9 × 16 × × 103 1 5 5 Area of ∆DBH = bc 2 ≈ – 7.5 × 104 N (to the left) D L Along y-direction – p2 A2 sin 37° + Fy ρQ (v2 sin 37°) −
o
ita rs
w
∴
rio
s
1 A Volume of liquid retained = bc × 2 L Let ρ be the density of the water. bc Aρ Mass of the water retained = 2L
er iv
un
//t
Mass of cart and water = M +
bc Aρ 2L
m e/
s: tp
b
ht
tp
s:
//t .
ht c
.m e/
L
ht
tp
s:
un
//w
iv e
w
F = Fx2 + Fy2 = 76 × 103 N.
F
H
rio
ar
m
.ja
= 12.1 × 103 N
s_
B
12 3 × 172 3 ⇒ Fy = (0.2) 0.9 × 16 × + × 10 5 20 ∴
in f
a.
12 3P = ρ v12 + 2 5 20
A1
b
si ta
Fy
⇒
A1 3 3 × + Fy = ρ A1 v1 4 v1 4 4 5
co m
⇒ − p2
an
∴
F
R
b mg
a
11. A vessel with a hole in its bottom is fastened on the cart. The mass of the vessel and the cart is M. With what force F should the cart be pulled so that a maximum amount of water remains in the vessel. The dimensions of the vessel are as shown in the figure.
As the cart is drawn by a force F, the water in the vessel takes up a slant position rising
)
) Solution
Let us consider a small element of the water of mass m. Now, the element is in equilibrium under the action of following forces: (i) Its weight mg acting downwards. (ii) The inertial force ma acting in a direction opposite to the acceleration a of the cart. (iii) The normal reaction, R.
θ
c
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4.28
Hydrostatics and Hydrodynamics
If θ be the inclination of the water surface with the horizontal, therefore
2 1
ma mg
tan θ =
b c Required force, F = mass × acceleration
or
a = g tan θ = g ×
Rate of flow Q = A1 v1
om
in
.m e/
si
ta
un iv
er
ni v
//t
m
e/ u
13. A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional are A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contain a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate: (i) The acceleration of the container. (ii) Its velocity when 75% of the liquid has drained out.
) Solution )
A2 − A2 1 p1 − p2 = ρ v12 1 2 2 2 A2
ht
2
//t .
…(ii)
1 1 vA p1 + ρ v12 = p2 + ρ 1 1 2 2 A2
2 × 8 × 10−2 × 9.8 × 13.6 × 103 103 × π2 (54 − 34 ) × 10−6
= 13991 × 10–5 m3/sec = 139.9 cc/sec
Putting the value of v2 in equation (i), we get
or
s_
rio
ta
Q = π (3 ×10–2)2 π (5 × 10–2)2
tp s:
tp s:
v1A1 A2
ht
v2 =
ht A 2 v2 A1
v1 =
Substituting the numerical values, we get
er si
w .ja
tp s
://
w w
1 1 p1 + ρ v12 = p2 + ρ v2 2 …(i) 2 2 where, v1 and v2 are the velocities at these points respectively A1 v1 A2 v2
or
fo
an a. c
ar
)
In the Venturimeter, if p1 is the pressure at 1 and p2 the pressure at part 2, we have
2( p1 − p2 ) ρ (A12 − A 22 )
where, (p1 – p2) is the pressure difference and A1 and A2 are the area of cross-sections of pipeline and throat.
m
) Solution
= A1 A 2
s
12. The diameter of the throat of a venturimeter is 6 cm. When it is inserted in a horizontal pipe line of diameter 10 cm, the pressure difference between the pipe and throat is 8 cm of mercury. Calculate the rate of flow of the water.
rio
bc A ρ gb = M + 2 L c
h
∴
v1 = A 2
2( p1 − p2 ) ρ (A12 − A 22
Let the height of the liquid column in the container during its motion at an arbitrary instant
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4.29
Hydrostatics and Hydrodynamics
of time t is z (t) and the speed of container at that moment is v(t) as shown in the figure.
∴
t=
200 1/2 1/2 ( z0 − z ) 2g
Initially, m0 = ρA z0 1 and Finally m0 = ρAz 4
ρA −ρ A dz = 2 gzdt 100
s_ in fo
dt =
−100 dz ⋅ 1/2 2g z
s
rio
ita
rs
ve
e/ un i
ht
dV dh =A dt dt
…(ii)
or dV = A dh or A dh = a u dt From equations (i) and (iii), we get A dh dt = a 2 gh
Now at t = 0, v = 0 g v= t ∴ 50 For the liquid in the container
or
u = 2 gh On the time interval dt, volumn of the water ejected through orifice dV = au dt ...(i) Volume of water in the vessel at time t, V = Ah ...(ii) Differentiating equation (ii) with respect to time, we get
or
t
or
dv dm = Fext + vrcl dt dt dv g = = 19.6 × 10−2 m/s 2 dt 50
Let at any time t, height of water in the vessel be h. Thus, speed of flow of the water through the orifice at that moment
∫ dt = 0
∴
m
tp
s:
//t
ht
.m
tp s:
ρA 2 gzdt …(i) 100 [This equation (i) can also be derived by using the well known equation for a variable mass system (in this case container with liquid)], ρ Az dv =
)
e/
.m
//t
ht tp s:
un iv
//w
time dt. So, from the conservation of linear momentum for the container with remaining liquid + ejected mass of liquid system along horizontal direction gives
) Solution
w
w
ρA A 2 gz 2 gz = 2gzdt in mentum ρ 100 100
14. A cylindrical vessel of height H and base area A is filled with water. An orifice of area a is opened in the bottom of the vessel. Neglecting the viscosity of the water, determine how soon all the water will pour out of the vessel?
er si ta rio
m
2gz towards right, taking out mo-
.ja
velocity
ar a
na .
co
Let us choose an inertial frame in which the container is at rest at time t. In this frame the velocity of the container at time t + dt is dv to the left, while the liquid moves out with
200 2g
m
z(t)
t=
∴
m 1/2 m 1/2 0 − 0 …(iii) ρA 4 ρA From equation (iii) and (ii), we get the speed of container as m 1/2 m 1/2 v = 2 2 g 0 − 0 ρA 4 ρA
V(t)
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t=
A a 2g
0
∫
H
dh h
A 2H a g https://t.me/universitarios
…(iii)
4.30
s_ in f
o
ita rio
ita
ve rs
)
) Solution
e/ un i
∴ t=
2h g
Total time back to the original position, 2h = t1 (given in the problem) g
1 ∴ h = g t12 8
2t = 2
From equations (i) and (ii), we have T – T0 = Vda or T = T0 + Vda
1 2 gt 2
h
ht tp
s: //t .m
ht tp
rio s
ni ve rs e/ u t.m
s: //
)
(i) If d < dL, obtain an expression (in terms of d, t1 and dL) for the time t2, the ball takes to come back to the position from which it was released. (ii) Is the motion of the ball simple harmonic? (iii) If d = dL, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.
(i) Equation of motion, h =
When the block is at rest, from Newton’s law B – mg + T0 = 0 ...(i) where, B = Buoyancy force, and mg = Weight of block. Let the volume of the block is V. It is given that in the problem density of the liquid is greater than the solid block so, let the density of liquid is ρ and the density of solid is d. ∴ Vρg – Vdg + T0 = 0 ...(i) The block is in rest with respect to the vessel which is accelerated with respect to the ground. Hence, the acceleration of the block with respect to ground is ‘a’ upward. T + Vρg – Vdg = Vda ...(ii)
a = T0 1 + g
an a. m ar .ja w w w tp s: // ht ) Solution
T0 a g
16. A ball of density d is dropped on a horizontal solid surface. It bounces elastically from the surface and return to its original position in a time t1. Next the ball is released and its falls through the same height before striking the surface of a liquid of density dL.
co m
a given byT0 1 + . g
or
T = T0 +
or
15. The tension in a string holding a solid block below the surface of a liquid (of density greater than solid) is T0, when the containing vessel, as shown in the figure is at rest. Show that tension T, where the vessel has an upward acceleration a, is
Hydrostatics and Hydrodynamics
…(i)
In the second case, body released the some height when it is striking the surface it falls in liquid.
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Hydrostatics and Hydrodynamics
(ii) No
Hence, t2 = Time taken, Fall height h + time taken when the body goes in liquid in zero velocity + come back to the original position where the block is released.
(d L − d ) g =0 d Thus, the upward acceleration is zero. So, thus the ball continuous goes down in the liquid.
(iii) a =
where, t3 = Time taken when velocity of solid is zero.
t4 = Return back to the surface. t5 = From surface to original position. Upward acceleration of body in liquid,
rio s
ta si er iv un
m //t .
ht tp
(d L − d ) g d 2h × = 2 gh d (d L − d ) g
) Solution
At extreme position, Weight of the rod = S L d1 g Buoyancying force B = S L d2 g
=
s:
t1 d d 2h = (d L − d ) g 2(d L − d )
d2
e/
tp s
ht
2h3 2hd ⋅ d = a (d L − d ).g (d L − d )
v4 = at4
in
s_
rs ve ni
e/ u
t.m
://
So, t4 = =
d1
v2 2 gh d hd = = 2a 2.g (d L − d ) (d L − d )
ht
h3 =
ita
w .ja m
//w
2 gh d t1 d v = = a g (d L − d ) 2(d L − d )
tp s:
t3 =
w
∴
rio
Velocity of the block before strike the liquid v = 2 gh
fo
om a. c an
g (d L − d ) d
)
=
V d L g − Vdg Vd
ar
a=
17. A thin rod of length L and area of crosssection S is pivoted at its lowest point P inside a stationary, homogeneous and nonviscous liquid. The rod is free to rotates in a vertical plane about a horizontal axis passing through P. The density d1 of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by a small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic motion and determine its angular frequency in terms of the given parameters.
2h + t 3 + t 4 + t5 g
t2 =
2h t1 = g 2
t5 =
Hence, t2 =
4.31
t1 t1 d t1 d t + + + 1 2 2(d L − d ) 2(d L − d ) 2 o
dL d = t1 1 + = t1 (d L − d ) dL − d
θ
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4.32
Hydrostatics and Hydrodynamics
I = Moment of intertia at the rod about O. 1 = (S L d1 ) L2 3
Taking torque about a point O, we have L τ0 = (S L d2 g – S L d1 g) sin θ 2 L2 = Sg (d2 – d1) θ 2 (∵ θ is very small then sin θ ≈ θ) τ0 = Iα SL2 g (d 2 − d1 ) θ 2I Hence, angular acceleration is directly proportional to the angle so that the motion is simple harmonic. ∴
α=
3 d 2 − d1 g 2 d1 L
=
rio s_ in fo
na .c om
SL2 g (d 2 − d1 ) 1 2. (SL d1 ) L2 3
ω=
P
(c) the balance A will read less than 2 kg and B will read more than 5 kg (d) the balance A and B will read 2 kg and 5 kg respectively
=
Wa W Wa = a = =4 Wa − WW ∆W Wa / 4
er si ta rio s
un
m
[since weight of the body decreases by 25% of its original weight Wa (weight in air)] 2. The spring balance A read 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid as shown. In this situation (a) the balance A will read more than 2 kg (b) the balance B will read more than 5 kg
B
Ans. (b) and (c)
) Explanation When the mass is submerged )
A
t.m e/ s: //
Weight of the body in air Weight of the body in air − Weight of the body water
ht tp
=
ht tp s:
weight
iv
.m
//t
ht
)
) Explanation Relative density = Specific gravity
ita
e/
un iv
er s
w
s:
tp
//w
w
.ja
m
1. A body weights W in air and it loses its weight by 25% in water. The relative density of the body is 1 (a) 4 (b) 4 3 4 (c) (d) 4 3 Ans. (a)
ar a
SOLVED OBJECTIVE TY E QUESTIONS
in the liquid, a force of buoyancy act on it which reduces the reading of balance A. The reaction of this buoyancy is experienced by the liquid therefore the reading of balance B increases.
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3. A cube of mass m and density D is suspended form the point P by a spring of stiffness k. The system kept inside a beaker filled with a liquid of density d. The elongation in the spring, assuming D > d, is
⇒
A
a
ita rio s_
h
(b) lρa (d) hρg + lρa
ta
rio
s
er s
(a) hρg (c) hρg – lρa Ans. (d)
un iv
si
)
er
of width dx and area of cross-section ‘a’, at a distance x from the pont of the tank Mass of element = dm = (αdx) ρ Net force to the right on the element = (P + dP) α – Pα = αdP ∴ αdP = (Pαdx) α
iv
.m //t tp s:
.m e/
un
//t s:
tp
ht
or
C
C
A
A
∫ dP = ∫ Padx
ht
(i) spring force k x, where x = elongation of the spring, (ii) gravitational force w, weight of the cube = mg (iii) buoyant force Fb (or upward thrust) imparted by the liquid on the cube given as Fb = Vdg here V = volume of the immersed portion of the cube. For complete immersion, V = volume of the cube For equilibrium of the cube, kx + Fb = mg
dx
x
C
) Explanation Consider an element of the liquid
e/
s:
ht tp
mg
l
in fo
na .c
ar a
m
.ja
w
w
//w
)
α P
or PC – PA = Pal Also PB – PC = Pgh or PB – (PA + Pal) = Pgh or PB – PA = hPg + lPa
Fb
4. A sealed tank containing a liquid of density ρ moves with a horizontal acceleration a as shown in the figure. The difference in pressure between the points A and B is
B
) Explanation The cube is in equilibrium under
m
mg d 1 − k D
om
R
kx
x=
(d) none of these P
three forces
mg − Fb mg − Vdg = , k k
⇒
mg D 1 − k d
. . . . ... .. . . . ... . .. . . . . .. . . . . . . . . . . . . ... . .. . . . .. . ... . .. .. . . . ... . .. . . . . .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . .... . .. . . . . . . .. . . . . .. . .. . . . . . . . . . . . . . . ... . . .. . . . . .... ... . ..... .. . . . ... . .. . . . . .. . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . .. .. . . . . . . ... . .... ... . . ..... ... . . ..... .. . . . .... .. . . . . . . . . . . . . . .. .. . . . . . . .. . . . . .... .. . . . . . . . . . . .. . . . . .. .. . . . . .. .. . . . .. . . . . . . . . . .... ... . ..... .. . . m . . . . . . . . .. . . . . .. ... . . .. . . . . .... ... . . ..... .... . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .... . . .. . .. . . . . . . . . . . . .. . . .. . .. . . . . . .. .. . . . .. ... . . .. . . . . .... ... . . ..... .... . . . . . . .. .
4.33
m where V = D
mg d 1 + k D
(c)
(b)
mg d 1 − k D
(a)
x=
Hydrostatics and Hydrodynamics
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4.34
)
of mass, M = 1000 kg
∴
m
ta
h
si er iv
(a) l decreases and h increases (b) l increases and h decreases (c) both l and h increase (d) both l and h decrease Ans. (d)
decrease As density of coin is more than that of water, h will decrease
= (1.013 + 0.049) × 105 N = 1.062 × 105 N 6. A vessel contains oil (density = 0.8 g/cm2) over mercury (density = 13.6 g/cm3). A homogeneous sphere floats with half its volume immersed in mercury and the
8. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The column of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density ρ where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is
103 × 9.8 × 1 2 = 1.013 × 105 (1) N 2
) Explanation Due to decrease in weight, l will )
ht
tp
s: //
t.m
e/
un
//t
tp s:
ht
Pressure at C = PC =
ρh 2 F = Patm + ( ) 2
rio s
er e/
.m
tp
ht
PA + PB 2 = average pressure over the side wall ρg = Patm + 2 The force exerted on the side wall is
l
iv un
s: //
Given: Volume of the tank, V = 1m3 and Side of the tank = b = 1m Pressure at A = PA = Patm Pressure at B = PB = Patm + ρgh = Patm + ρgl ∴
Coin
si
.ja w w
w
B
o
ta rio
ar a
s_ i
nf
na .
A C
d1 + d 2 0.8 + 13.6 = = 7.2 g /cm 2 2 2
co m
1000 kg = = 1m3 1000 kg /m 2
h
d=
7. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there. After some time the coin falls into the water. Then
m ρw
Volume of water V =
)
4 3 2 2 πr dg = πr 3d1 g + πr 2 d 2 g 3 3 3
) Explanation The cubical tank contains water ⇒
) Explanation From the principle of floation
other half in oil. The density of the material of the sphere in g/cm3 is (a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 Ans. (c)
5. A cubical water tank is completely filled with water of mass m = 1000 kg. The force exerted by the water on the side wall of the tank, is (a) 1.062 × 105 N (b) 1.62 × 105 N 5 (c) 0.049 × 10 N (d) None of these Ans. (a)
Hydrostatics and Hydrodynamics
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Hydrostatics and Hydrodynamics
4.35
)
) Explanation Applying Bernoulli’s Principle
between the points 1 and 2, we have 1 1 ρgh1 + P1 + ρv12 = ρgh2 + P2 + ρv22 2 2 where, P1 = P0 + ρgh, P2 = P0, h1 = h2 = h1, P0 is the atmospheric pressure.
h
p 2R
h
m
in fo
a. co
(b) Mg – Vρg (d) ρg(V + πR2h)
(a) Mg (c) Mg + πR2 hρg Ans. (d)
h'
an
ta rio
ar
.ja
si
ta r
iv
si er
t.m
:// t.m
R = v2 × t t = time of fall can be given by
1 2 2h1 gt ⇒ t = 2 g
⇒ R = v2
ht tp s
h1 =
iv
Range where
e/ un
://
tp s
ht
9 AV 2 a <
1 ρgh = ρv22 , v2 = 2 gh (say ) 2
un
e/
tp s
ht
R
io s
er
w
w
:// w
Since,
m
and
9. A small hole is made at a height of 1 h′ = m from the bottom of a cylin 2 drical water tank and at a depth of h = 2 m from the upper level of water in the tank. The distance where the water emerging from the hole strikes the ground is
2h1 ;Putting v2 2 gh , we get g
R = 2 hh′
1 m and h = 2 m, 2 we get R = 2m
(c) 2 m Ans. (c)
(b) 1 m (d) none of these
(a) 2 2 m
R
10. In a cylindrical water tank there are two small holes Q and P on the wall at a depth of h1 from upper level of water and at a height of h2 from the lower end of the tank respectively as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of h1 is and h2 is
Putting h′ =
s_
cylinder – force at the top of cylinder = buoyant force on the cylinder i.e., F – πR2 ρgh = Vρg ⇒ F = ρg (V + πR2h)
)
) Explanation The Force on the bottom of the
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4.36
R1 = R 2 = R ⇒ v1t1 = v2t2 where, v2 = velocity of the efflux at
Q
Hydrostatics and Hydrodynamics
....(i)
2 = 2 g (H − h2 ) P
t1 = time of fall of the water stream through 2 (H − h1 ) g
(1) =
t2 = time of fall of the water stream through
2h2 g Putting these values in equation (i), we get (2) =
)
na .c om
) Explanation Since two streams strike at the
(H – h1) h1 = (H – h2) h2
or
ar a
v1
ita
h1
e/
un t.m e/ P
olved
ht tp s:
//t
iv
.m
tp ht un
OBJECTIVE TY E QUESTIONS of both the limbs are equal, what will be the difference in heights h of the liquid in the two limbs of U-tube, when the system is given an acceleration a in horizontal direction towards right as shown?
s: //
1. A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ0 (ρ0 rel="nofollow"> ρ0). The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle θ with the horizontal. Then
1 ρ0 2 ρ
(d) sin θ =
ρ0 ρ
ρ ρ0
(b) sin θ =
(c) sin θ =
1 ρ0 2 ρ
(a) sin θ =
ht tp
S
h1 =1 h2
a H
2. A liquid stands at the plane level in the U-tube when at rest. If area of cross-section
or
er si ta rio s
er s un iv
v2
s:
//w
w
P
w
.ja
H
[H – (h1 + h2)] [h1 – h2] = 0
Here, H – h1 + h2 is irrelevant because the holes are at two different heights ∴ h1 = h2
m
Q
h2
same point on the ground, therefore
rio s_ in fo
(b) 2 (d) < 1
(a) 1 (c) > 1 Ans. (a)
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L
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Hydrostatics and Hydrodynamics
La H
h
Hg (d) a
L2 a (c) H g
(b)
g L2 aH
(a)
4.37
si rio
s
er
iv
er
iv
un
8. An open tank containing non viscous liquid to a height of 5 m is placed over the ground. A heavy spherical ball falls from height 40 m over the ground in the tank. Ignoring air resistance, find the height to which ball will go back. Collision between ball and bottom of tank is elastic having e = 0.5. (a) 45 m (b) 35 m (c) 40 m (d) 20 m
9. A metallic sphere floats in immiscible mixture of water (ρw = 103 kg/m2) and a liquid (ρL = 13.5 × 103 kg/m3) such that
e/
.m
ht
tp
s:
//t
ht
(b) 20 cm (d) 100 cm
si
ta
un
e/
.m
tp
s:
//t
ht
s_ rio ta
w
//w w
s:
tp
(a) 10 cm (c) 50 cm
6. A conical block, floats on water with 90% height immersed in it. Height h of the block is equal to diameter, i.e., 20 cm. The mass to be kept on the block, so that the block just floats at the surface of water, is (ρwater = 1 g/cc)
fo
in
a.
an
ar
m
.ja
g −a (d) 2 g ρg g+a
5. The weight of a ballon is W1 when empty and W2 when filled with air. Both are weight in air by the some sensitive spring balance and under identical conditions. (a) W1 = W2 as the weight of air in the ballon is offset by the force of bouyancy on it (b) W2 < W1 due to the force of bouyancy acting on the filled balloon (c) W2 > W1 as the air inside is at a greater pressure and force has greater density than the air outside (d) W2 = W1 + weight of the air inside it
m
co
7. A block of mass 10 kg connected to another hollow block of same size and negligible mass, by a spring of spring constant 500 N/m, floats in water as shown in the figure. The compression in the spring is (ρwater = 1 × 103 kg/m3, g = 10 m/s2)
4. A beaker containing a liquid of density ρ moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is (a) hρg (b) hρ (g + a) (c) hρ (g – a)
(b) 980 g (d) 196 g
(a) 568 g (c) 112 g
3. There is a hole at the bottom of a large open vessel. If water is filled up to a height hm, its flow rate is ν. If water is filled to a height 4h, its flow rate will be (a) 4v (b) 3v (c) 4v (d) v
where, L = length of horizontal portion of the tube, H = initial height of liquid in both limbs
1 4 portion is in water and th 5 5 portion of the sphere is in the liquid. The density of the metal in kg/m3 is
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4.38
o
nf
s_ i
s
3g 8
si
ta
rio
rio
si ta
(d)
er iv un e/
.m //t
A
B
s:
16. A conical vessel is completely filled with a liquid of density s as shown in the figure. The force exerted by the liquid on the curved surface is
12. A wooden cube just floats inside water when a 200 g mass is placed on it. When the mass is removed, the upper surface of the cube is 2 cm above the water level. The size of the cube is
F
O
15. Air stream flows horizontally past an aeroplane wing of surface area 4m2. The speed of air over the top surface is 60 m/s and under the bottom surface is 40 m/s. The force of lift on the wing is (density of air = 1 kg/m3) (a) 800 N (b) 1000 N (c) 4000 N (d) 3200 N
ht
R
tp
P
m
er
un
e/
.m
//t
s:
tp
3 ρgR 3 2 (d) 0
(b)
ht
3g 4
(c)
iv
//w
s:
tp
ht
9 ρgR 3 2 (c) ρgR3
(a)
14. A unifrom wooden rod of length L and specific gravity S is free to rotate in vertical plane about the end a. The rod is released in the horizontal position inside liquid of specific gravity 1.5 S as shown in the figure. The instantaneous acceleration of the end B is g (a) g (b) 4
co a.
an ar m .ja
w w
11. The figure shows a semi-cylindrical massless gate (of width R) pivoted at the point O holding a stationary liquid of density ρ. A horizontal force F is applied at its lowest position to keep it stationary. The magnitude of the force is
P
D
P
13. A boat carrying iron balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water? (a) It will remain unchanged (b) It will rise (c) It will fall (d) First it will rise and then fall
10. The minimum horizontal acceleration of the container so that the pressure at the point A of the container becomes atmospheric is 3 4 (a) g (b) g 2 3 3 (c) g (d) g 4
(b) 10 cm (d) 20 cm
(a) 5 cm (c) 15 cm
(b) 4.0 × 103 (d) 3.0 × 103
(a) 4.5 × 103 (c) 3.5 × 103
Hydrostatics and Hydrodynamics
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4.39
Hydrostatics and Hydrodynamics
R
H2
H1
R
(d) 10 W
W + ρgHS 10
iv
.m
W −ρgHS 10
(b)
W
s: //
t.m e/
un
//t
ht tp s:
W 10
er si ta rio s
(c)
un iv
(a)
e/
s:
tp
ht
Liquid of density p
H
18. Water flows from a large reservoir through a pipe of cross sectional area S into a big stagnant lake. The level of water in the lake is at a height H1, above the end of the pipe as shown in the figure. The flow rate of water under steady state is
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(d) S 2 g (H1 − H 2 )
(c) S 2 g (H 2 − H1 )
20. A cylindrical drum open at the top contains 30 litre of water. It drains out through a small opening at the bottom 10 litres of water comes out in time t, the next 10 litres in further time t2 and the last 10 litres in further time t3. Then (a) t1 = t2 = t3 (b) t1 > t2 > t3 (c) t1 < t2 < t3 (d) t2 > t1 = t3
(b) S 2 gH 2
F
(a) S 2 gH1
ht tp
rio s_ in fo
ita
w w
//w
17. A piece of wood floats in water kept in a beaker. If the beaker moves with a vertical acceleration a, the wood will (a) sink deeper in the liquid if a is upward (b) sink deeper in the liquid if a is downward, with a < g (c) come out more from the liquid if a is downward with a > g (d) remain in the same position relative to the water
m .ja
πR 2 σg (↓) 3
(d) 2
19. A heavy load W is supported on a platform of area S by applying a force F on a small piston of area S/10. The value of F for equilibrium is
πR 2 σg (c) (↑) 3
er s
πR 2 σg (↓) 3
Lake
na .c om
(b)
ar a
πR 2 σg (↑) 3
(a)
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4.40
Hydrostatics and Hydrodynamics
ANSWERS 4. (b) 12. (d) 19. (b)
1. A cubical block of iron 5 cm on each side is floating on mercury in a vessel. (a) What is the height of the block above mercury level? (b) Water is poured into the vessel so that it just covers the iron block. What is the height of the water column? Density of mercury = 13.6 gm/cm3, density of iron = 7.2 gm/cm3. [IIT, 1973] [Ans. (a) 2.35 cm, (b) 2.54 cm]
8. (d)
4. An ornament, which weighs 36 g in air, weighs only 34 g when immersed in water. It is known that some copper is required to be mixed with the gold to prepare the ornament, find the amount of copper added. Take the density of gold to be 19.3 g/cm3 and that of copper to be 8.9 g/cm3. [Ans. m = 2.2 g]
er si ta rio s
ita
5. A tube bent at right angle is lowered into a water stream, as shown in the figure. The velocity of the stream relative to the tube is v = 3 m/s. The closed upper end of the tube situated at a height h0 = 15 cm from the water surface has a small orifice. Find the height h upto which the water jet will spurt. [Ans. 30.9 cm]
iv
un iv
e/
.m
V
h1
[Ans. 3.00 m/s]
7. The density of air in atmosphere decrease with height and can be expressed by the relation d = d0 e–αh
3. A large block of ice 5 m thick has a vertical hole drilled through it and is floating in the middle of the lake. What is the
6. A solid cylinder of radius R = 10 cm and of mass M = 2 kg floats in water with its axis vertical. Show that if it is slightly depressed and released, it will execute simple harmonic motion and find its period. [Ans. T = 0.5 s]
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minimum length of the rope required to scoop up a bucket full of water through the hole? [IIT, 1975] [Ans. h = 0.5 m.]
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2. A wide vessel with a small orifice in its bottom is filled with water and kerosene of densities ρ1 = 1 g/cc and ρ2 = 0.8 g/cc, respectively. Find the velocity of water flow if the height of the water layer is h1 = 30 cm and that of the kerosene layer is h2 = 20 cm. Neglect viscosity.
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8. A thin tube, sealed at both ends, is 1.0 m long. It lies horizontally, the middle 0.1 m containing mercury and the two equal ends containing air at standard atmospheric pressure. If the tube is now turned to vertical position, by what amount will mercury be displaced? Standard atmospheric pressure = 0.76 m of Hg.
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12. Two solid uniform spheres, each of radius 5 cm are connected by a light string and is totally immersed in a tank of water. If the specific gravities of the spheres are 0.5 and 2, find the tension in the string and the pressure between the bottom of the tank and the heavier sphere. [Ans. R = 2.566 N] 13. The side wall of a wide vertical cylindrical vessel of height h = 60 cm has a narrow vertical slit running all the way down to the bottom of the vessel. The length of the slit is l = 40 cm and the width b = 0.1 cm.
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10. Two identical cylindrical vessels with their bases at the same level, each containing a liquid of density ρ. The height of the liquid in one vessel is h1 and that in the other vessel is h2. The area of either base is A. What is the work done by gravity in equalizing the level when the two vessels are connected? [IIT, 1981] 2 h − h2 Ans. Aρg 1 2 11. A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water upto a height of 0.5 m. The specific gravity
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9. A vertical spring of force constant K = 500 N/m supports a cylinder of radius r = 0.05 m, height h = 0.2 m, and density ρ = 800 kg/m3. This cylinder is half dipped in water, as shown in the figure. (a) What is the elongation of the spring in the equilibrium position? (b) If the cylinder is slightly depressed and released, show that the resulting motion will be simple harmonic and find its period. [Ans. (a) x = 23.1 cm, (b) T = 0.93 s]
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[IIT, 1978]
of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. [Exclude the case θ = 0] [IIT, 1984]
the atmospheric pressure at sea level? Assume g to be constant. The numerical values of the constants are g = 9.8 m/s2, do = 1.3 kg m–3, α = 1.2 × 10–4 m–1 [Roorkee, 1977] [Ans. 1.06 × 105 N/m2]
4.41
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16. Calculate the rate of flow of glycerine of density 1.25 × 103 kg/m3 through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 newton/m2. [Roorkee, 1991] –4 3 [Ans. Q = 6.44 × 10 m /s]
With the slit closed, the vessel is filled with water. What is the resultant force of reaction of the water flowing out of the vessel immediately after the slit is opened?
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18. A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross section of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. [IIT, 1997]
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14. Water is flowing out of a tank through a tube bent at right angle. In the figure radius of the tube is r = 0.5 cm and the length of its horizontal section is l = 20 cm. The rate of water is Q = 0.5 litre/second. What is the moment of reaction forces of flowing water, acting on the tube’s wall, relative to the point O?
[Ans. F = 3.14 N]
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17. A cylindrical tank 1 m in radius rests on a platform 5 m high. Initially, the tank is filled with water to a height of 5 m. A plug, whose area is 10–4 m2, is removed from an orifice on the side of the tank at the bottom. Calculate: (a) initial speed with which the water flows from the orifice, (b) initial speed with which water strikes the ground, (c) the time taken to empty the tank to half its original value. Given, g = 10 m/s2 [Roorkee, 1995]
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15. A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev. per sec, find the difference in height of the liquid at the centre of the vessel and at its sides. [Roorkee, 1987] [Ans. 0.01m]
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[Ans. τ = 0.7 N-m]
[Ans. = 29025 J]
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Hydrostatics and Hydrodynamics
4.43
QUESTION BANK
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2. A large tank is filled with two immiscible liquids of densities ρ1 and ρ2 as shown in figure. A small hole is made at the bottom of tank from which liquid of density ρ2 starts flowing out. Bernoulli’s equation can be applied between points
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8. Explain why detergents should have small angle of contact?
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7. Explain why the surface tension of a liquid is independent of the area of the surface.
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1. A cylindrical vessel filled with water is released on an inclined surface of angle θ as shown in figure. The friction coefficient of surface with vessel is µ (< tan θ). Then the constant angle made by the surface of water with the incline will be
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4. Atmospheric pressure at a height of about 6 km decreases to nearly half the value at the sea level, through the height of the atmosphere is more than 100 km. Explain.
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3. The blood pressure in humans is greater at the feet than at the brain. Explain.
12. The size of the needle of a syringe controls flow better than the thumb pressure exerted by a doctor while administering an injection. Why?
15. Does it matter if one used gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
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13. A fluid flowing out of a small hole in a vessel, results in a backward thrust on the vessel. Explain.
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14. Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
2. The force required by a man to move his limbs immersed in water is smaller than the force for the same movement in air. Explain.
1. A balloon filled with helium does not rise in air indefinitely but halts after a certain height. Explain.
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3. Shape of the meniscus formed by two liquids when capillaries are dipped in them are shown. In I, it is hemispherical whereas in II, it is flat. Pick correct statement regarding contact angle formed by the liquids in both situations.
6. A square box of water has a small hole located in one of the bottom corner. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed v0, as shown in figure. When the box is half empty, it is tilted by 45º so that the hole is at the lowest point. Now the water will flow out with a speed of
(b) A and B (d) A, B and C
(a) A and C (c) B and C
Hydrostatics and Hydrodynamics
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8. A cylindrical vessel filled with water upto a height h is connected to a light inextensible string. The string is wound on a solid uniform disc of radius R. With what velocity will the first water particle be ejected from the vessel through a small hole made at a depth y = (3/10) R, just after the disc is spined with an angular speed ω = g /R and released? Assume that mass of the disc is equal to the sum of mass of vessel plus water. Ignore friction anywhere and avoid slipping of the string with the disc.
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7. A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative density of silver is 10, then tension in the string will be [take g = 10 m/s2] (a) 37.12 N (b) 42 N (c) 73 N (d) 21 N
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9. A block of iron is kept at the bottom of a bucket full of water at 2ºC. The water exerts buoyant force on the block. If the temperature of water is increased by 1ºC the temperature of iron block also increases by 1ºC. The buoyant force on the block by water (a) will increase (b) will decrease (c) will not change (d) many decrease or increase depending on the values of their coefficient of expansion
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14. The area of cross-section of the two vertical arms of a hydraulic press are 1 cm2 and 10 cm2 respectively. A force of 10 N applied, as shown in the figure, to a tight fitting light piston in the thinner arm balances a force F applied to the corresponding piston in the thicker arm. Assuming that the levels of water in both the arms are the same, we can conclude
12. The figure shows a soap film in which a closed elastic thread is lying. The film inside the thread is pricked. Now the sliding wire is moved out so that the surface area increases.
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13. Consider the following statements: (i) Young’s modulus is numerically equal to the stress which will double the length of a wire. (ii) Viscosity of gases is greater than that of liquids. (iii) The surface tension of a liquid decreases due to the presence of insoluble contamination. The number of above statements that are true is (a) one (b) two (c) three (d) zero
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The radius of the circle formed by elastic thread will (a) increase (b) decrease (c) remains same (d) data insufficient
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Hydrostatics and Hydrodynamics
F = 100 N F = 50 N F = 25 N F, as applied, cannot balance the effect of the force on the first piston.
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15. A space 2.5 cm wide between two large plane surfaces is filled with oil. Force required to drag a very thin plate of area 0.5 m2 just midway the surfaces at a speed of 0.5 m/sec. is 1 N. The coefficient of viscosity in kg-sec/m2 is
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(a) the pressure in the water is higher at lower points in the stream (b) the pressure in the water is lower at lower points in the stream (c) the pressure in the water is the same at all points in the stream (d) pressure variation will depend upon density and exit speed of the water.
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20. A rain drop of radius r has a terminal velocity v m/s in air. The viscosity of air is η poise. The viscous force on it is F. If the radius of drop be 2r and the drop falls with terminal velocity in same air, the viscous force on it will be (a) F (b) F/2 (c) 4 F (d) 8 F
23. A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by
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4.48
Hydrostatics and Hydrodynamics
to pull a thin wire ring up (refer figure) if it is initially resting on a horizontal water surface. The circumference of the ring is 20 cm and its weight is 0.1 N
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28. A capillary tube of radius 0.20 mm is dipped vertically in water. The height of the water column raised in the tube, will be (surface tension of water = 0.075 N/m and density of water = 1000 kg/m3. Taking g = 10 m/s2 and contact angle 0º). (a) 7.5 cm (b) 6 cm (c) 5 cm (d) 3 cm
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7. A beaker is filled in with water is accelerated a m/s2 in + x direction. The surface of water shall make an angle (a) tan–1 (a/g) backwards (b) tan–1 (a/g) forwards (c) cot–1 (g/a) backwards (d) cot–1 (g/a) forwards
3. Water is flowing through a long horizontal tube. Let PA and PB be the pressures at two points A and B of the tube (a) PA may be equal to PB (b) PA may be greater than PB (c) PA may be smaller than PB (d) PA = PB only if the cross-sectional area at A and B are equal
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5. Statement 1: The velocity of flow of a liquid is smaller when pressure is larger and vice-versa. Statement 2: According to Bernoulli’s theorem, for the stream line flow of an ideal liquid, the total energy per unit mass remains constant.
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6. Statement 1: As wind flows left to right and a ball is spined as shown, there will be a lift of the ball. Statement 2: Decrease in velocity of air below the ball, increases the pressure more than that above the ball.
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10. In a peculiarly shaped tube as shown in the diagram the lower part is a quarter circle having radius of curvature R. The vertical limbs of the tube are separated by a distance R. A liquid of density ρ is poured into the tube of two height R and 2R in left and right limbs respectively. A valve fitted in the circular part maintains this liquid height difference. The valve is suddenly opened. If the gauge pressure at the lowest point is P and acceleration of the liquid is ‘a’ just after opening the valve, choose the correct choice (a) P = 2ρgR 4 + 2π (b) P = ρgR 4+π (c) P = πρgR 2g (d) a = 4+π
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8. Statement 1: In the three cases shown in the figure, force exerted by liquid on three vessels is same. Statement 2: Pressure at the bottom in each case is same.
1. An arrangement of the pipes is shown in the figure. The flow of water (incompressible and nonviscous) through the pipes is steady in nature. Three sections of the pipe are marked in which section 1 and section 2 are at same horizontal level, while being at a greater height than section 3. Correctly match order of the different physical parameter with the options given. 5
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(r) Total energy per unit volume
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Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
2. Bernoulli’s equation can be written in the following different forms (column I). Column II lists certain units each of which pertains to one of the possible forms of the equation. Match the unit associated with each of the equations.
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to a height h in each. Neglect atmospheric pressure.
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Column I (a) If the radius of a soap bubble A is four times that of another soap bubble B, then the ratio of excess pressure (PB/PA) will be (b) If two small drops of mercury, each of radius R coalesce to form a single large drop, the ratio of the total surface energy before and after change will be (c) The energy required to below a bubble of radius 4 cm and 3 cm in the same liquid is in the ratio of (d) Two soap bubbles are blown. In the first bubble excess pressure in 4 times that of the second soap bubble. The ratio of radii of first to second soap bubble
4. Match the column I with column II.
2 (b) Time to acquire (q) V = 2 r g (ρ − σ) 9 η terminal speed (c) Radius of drops (r) Less viscous force of water < 0.01 mm (d) Greater veloc- (s) Independent of ity of flow of a the density of liquid liquid.
Column I Column II (a) The net force exert- (p) 2hρgA2 ed by liquid on side walls is (b) The force exerted by (q) zero the liquid on the base of the vessel is (c) The weight of the (r) hρg (A2 – liquid is A1) on the liquid (d) The walls of the (s) Less then vessel at the level X 2hρgA2 exert a downward force is
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Hydrostatics and Hydrodynamics
ASSAGE BASED QUESTIONS
PASSAGE-1 A liquid of density ρ is filled in a beaker of cross section A to a height H and a cylinder of ice of mass m base are ‘a’ is floating in it as shown in the figure.
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PASSAGE-2 If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquids gets changed. In case of horizontally accelerated liqa uid (ax), the free surface has the slope x . In case g of vertically accelerated liquid (ay) for calculation of pressure, effective g is used. A closed box with horizontal base 6 m by 6 m and a height 2 m is half filled with liquid. It is given a constant horizontal acceleration g/2 and vertical downward acceleration g/2.
7. Maximum value of water pressure in the box is equal to (a) 1.4 MPa (b) 0.14 MPa (c) 0.104 MPa (d) 0.014 MPa
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1. Pressure due to the weight of ice will be maximum at (a) 1 (b) 2 (c) both 1 and 2 (d) neither 1 or 2
Take atmospheric pressure as P0.
6. Water pressure at the bottom of centre of box is equal to (atmospheric pressure = 105 N/m2, density of water = 1000 kg/m3, g = 10 m/sec2) (a) 1.1 MPa (b) 0.11 MPa (c) 0.101 MPa (d) 0.011 MPa
5. Length of exposed portion of top of box is equal to (a) 2 m (b) 3 m (c) 4 m (d) 2.5 m
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14. A penguin floats first in a liquid of density ρ0 then in a fluid of density 0.95ρ0 and then in a fluid of density 1.1ρ0. Rank the densities according to the magnitude of buoyant force on the penguin greatest first. (a) 1.1 ρ0, ρ0, 0.95 ρ0 (b) 0.95 ρ0, ρ0, 1.1 ρ0 (c) ρ0, 0.95 ρ0, 1.1 ρ0 (d) All tie
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Hydrostatics and Hydrodynamics
1. A cube with a mass m = 20 g floats on the surface of water. Each face of the cube a is 3 cm long. What is the distance between the lower face of the cube and the surface of the water? (S.T. of water = 70 dyn/cm]
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2. The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
4. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The
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SUBJECTIVE QUESTIONS
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Hydrostatics and Hydrodynamics
9. A manometer reads the pressure of a gas in an enclosure as shown in figure (a) When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers in mercury and the atmospheric pressure is 76 cm of mercury. (a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury. (b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
5. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 Nm–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kgm–3 (g = 9.8 ms–2).
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8. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 ºC) is 2.50 × 10–2 Nm–1 If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
11. A long, square wooden block is pivoted along one edge as shown. The block is in equilibrium when immersed in water to the depth shown. Assuming friction in the pivot is negligible, evaluate the density of the wood (in SI unit, up to two significant digit). /
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length of the slider is 30 cm. What is the surface tension of the film?
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2. A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in figure. The lower density liquid is open to the atmosphere having pressure P0. [1995] (a) A homogeneous solid cylinder of length L (L < H/2), cross-sectional area A/5 is immersed such that if floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid. Determine
(b) 4d/5 (d) d/5
(a) 5d/4 (c) 4d
12. In the incompressible flow through the device shown in figure, velocities may be considered uniform over the inlet and outlet sections.
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(i) the density D of the solid. (ii) the total pressure at the bottom of the container. (b) the cylinder is removed and the original arrangements is restored. A tiny hole of area s(s << A) is punched on the vertical side of the container at a height h (h < H/2). Determine (i) the initial speed of efflux of the liquid at the hole. (ii) the horizontal distance x travelled by the liquid initially and (iii) the height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm. (Neglect the air resistance in these calculations)
1. A homogeneous solid cylinder of length L (L < H/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure P0. Then, density D of solid is given by [1995]
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tube is kept inclined in the vertical plane as shown in the figure. The area of crosssection of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. [1997]
string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is [2001]
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5. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to [2000]
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4. Water from a tap emerges vertically downwards with an initial speed of 1 ms–1. The cross-sectional area of tap is 10–4 m2. Assume that the pressure is constant throughout the steam of water and that the flow is steady, the cross-sectional area of stream 0.15 m below the tap is [1998]
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8. A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm3 and 1.2 g/cm3, respectively. The height of liquid
(b) 2πL (d) L/2π
6. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and mass M. It is suspended by a
(a) L / 2π (c) L
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A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm.
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11. A container of width 2a is filled with a liquid. A thin wire of weight per unit length λ is gently placed over the liquid surface in the middle of the surface as shown in the figure. As a result, the liquid surface is depressed by a distance y (y << a). Determine the surface tension of the liquid. [2004]
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10. A liquid of density 900 kg/m is filled in a cylindrical tank of upper radius 0.9 m and lower radius 0.3 m. A capillary tube of length ℓ is attached at the bottom of the tank as shown in the figure. 3
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9. A soap bubble is being blown at the end of very narrow tube of radius b. Air (density ρ) moves with a velocity v inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is T. After sometime the bubble, having grown to radius r separates from the tube. Find the value of r. Assume the r >> b so, that you can consider the air to be falling normally on the bubble’s surface. [2003]
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[2002] (a) Find the total force exerted by liquid A on the cylinder. (b) Find h, the length of the part of the cylinder in air (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.
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Hydrostatics and Hydrodynamics
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PASSAGE-1 The density of block is ρ/3 and is used to block a hole H.
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18. If water level is lowered below h2, then (a) the block will not rise (b) the block will start rising again at h2 = h/3 (c) the block will start rising again at h2 = h/2 (d) the block will start rising again at h2 = h/5 [2006]
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14. A U-shaped tube contains a liquid of density ρ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column. [2005]
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(c) |P0πR2 + Rρgh2 – 2RT| (d) |P0πR2 + Rρgh2 – 2RT| 20. Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. [2008] and Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. (a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1 (b) Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True
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Hydrostatics and Hydrodynamics
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6. (b) 14. (d)
7. (b) 15. (b)
8. (a)
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7. Surface tension of the liquid is the force per unit length of an imaginary line drawn in the free surface of liquid. So, it does not depend on the area of the surface.
4. The density of air is maximum near the surface of earth and decreases rapidly with height up to 6 km. and becomes nearly half of its value at the sea level at a height of 6 km. So, atmospheric pressure at a height of 6 km. decrease to nearly half of its value at sea level. Beyond 6 km height density of air changes very slowly.
8. Detergents having small angle of contact dissolve easily in water. 9. In the absence of external forces the liquid drop experiences only surface tension force and tries to minimise its surface area. For a given volume the minimum surface area is of a sphere.
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3. The pressure exerted by a liquid column is directly proportional to its height. The height of blood column in the human body is more at feet than at the brain. Therefore, the blood pressure in humans in greater at the feet than at the brain.
10. If we blow under the paper, the pressure on the lower side will decreases. The pressure above the paper is atmospheric pressure which will bend the paper downward. 11. When we try to close a water tap with our fingers, it reduces the area of outlet of water
6. For water-glass surface the angle of contact is acute. To achieve this acute valve of angle of contact the water spreads out an a clean glass surface. On the other hand for mercury-glass, angle of contact is obtuse. In order to achieve this obtuse valve of angle of contact, the mercury form drops.
5. The pressure is transmitted equally in all directions. In other words, there is no fixed direction for the pressure due to liquid. Hence, hydrostatic pressure is a scalar quantity.
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2. The density of water is much greater than air, so the upthrust on the limbs of a man is more. As a result of which the effective weight of limbs of a man decreases in water. So, it is easier for him to move his limbs in water than in air.
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1. Initially, the weight of the displaced air (upthrust) is greater than the weight of helium filled balloon. We know that the density of air and acceleration due to gravity decreases with height. Due to which upthrust also decreases. Therefore, the balloon halts at a height where weight of balloon just equals to the weight of air displaced.
P
CONCE TUAL QUESTIONS
ar a
HINTS AND SOLUTIONS
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Hydrostatics and Hydrodynamics
SUBJECTIVE QUESTIONS 1. Downward force = Buoyant force
1 12. Bernoulli’s equation P + ρv 2 + ρgh = con2 stant, shows that P occurs with power one and v occurs with power two. Therefore, influence of velocity is more than pressure. Moreover needle of syringe has very small cross sectional area compared to the thumb. That is why a needle controls the flow rate better.
Mg + 4 Ta = a2 x pg
13. The fluid flowing out of a small hole acquires a large velocity and hence, large momentum. As no external force is acting on the vessel, according to principle of conservation of momentum, the vessel gets a backward momentum.
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a.
an
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2. 0.637 ms–1
3. 6.2 × 106 Nm–2
4. 2.5 × 10–2 Nm–1
5. For the first bore, pressure difference (between the concave and convex side) = 2 × 7.3 × 10–2/3 × 10–3 = 48.7 Pa. Similarly for the second bore, pressure difference = 97.3 Pa. Consequently, the level difference in the two bores is [48.7/(103 × 9.8)] m = 5 mm. The level in the narrower bore is higher. 6. v = 0.33 m3
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P
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0.2 + 0.0084 0.2084 = = 0.02315 m 9 9 = 2.315 cm.
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1 2 mv ρv 2 2 is is energy per unit weight = 2 (m /ρ) 1 2 mv v2 2 = is energy per unit volume and m 2 energy per unit mass
20 × 10−3 × 10 + 4 × 70 × 10−3 × 3 × 10−2 9 × 10−4 × 103 × 10
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=
=
1. Use equation of continuity and concept “pressure is greater at lower and broader section.” 1 2 mv 1 2 v 2 2 = 2. Kinetic energy is mv ∴ 2 2g (mg )
Mg + 4Ta a 2ρg
x=
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15. No, one can use gauge pressure instead of absolute pressure in applying Bernoulis’ theorem, if the pressure at the two points where Bernouli’s equation is applied are significantly different.
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14. No, Bernouli’s theorem is used for streamline flow only.
MATCH THE COLUMN TY E QUESTIONS
[
7
jet and hence, velocity of water increase. (av = constant)
4.62
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7. 48.2º or tan–1(1.118)
R
PR
1. (a)
2. (i)
h(3H − 3h)
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8. (a) zero (b) 0.25 cm. (c) g/6 9. 4T/ρv2
13. Horizontal 81 × 104, Vertical 33.9 × 104
s
dQ ∝ r5 dt
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12.
13. 2 m
14. H =
ω2 L2 2g
15. (a)
16. (b)
17. (a)
18. (a)
19. (b) 20. (a) As the stream falls down, its speed will increase and cross-section area will decrease. Thus it will become narrow. Similarly as
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11.
12. 2500 + 999 cos(4πt)kg/s
360 B + 4320 B – 8.64 B ρ = 0 4680 ρ= = 540 kg/m3 8.64
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10. (1/720) Ns/m2
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Torque due to upthrust force = (0.6 × 1.2 × B) × 103 × B × 10 × 0.6 = 4320 B Torque due to weight = ρ × (1.2)2 × B × q × 0.6 = 8.64 Bρ
3H 3 , H 8 4
4. (c)
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× 10 × (x) (0.6 – x)B.dx = 360B
0
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3
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∫ 10
(iii) At h =
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(ii)
g 2
3. 29025 J/m3, 29400 J/m3
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2
H H H ρgb × × 2 2 3 = 4 H = 32 cm. = 2 H 9 ρgbH 2 + ρgb 2 11. Torque due to liquid pressure 0.6
(3H − 4h)
(b) (i)
10. Force due to atmospheric pressure = ρ0A = ρgH × bH F1 = ρgbH2 This force acts at a height H1 = H/2 from bottom Force due to mercury on side wall, ρgbH 2 F2 = 2 This force acts at a height H2 = H/3 from bottom. ∴ distance of net force from bottom F H + F2 H 2 = 1 2 F1 + F2 ρgbH 2 ×
5d dg (6 H + L) (ii) P = P0 + 4 4
9. (a) absolute pressure = 96 cm of Hg, gauge pressure = 20 cm of Hg for (a) absolute pressure = 58 cm of Hg, gauge pressure = 18 cm of Hg for (b) mercury would rise in the left limb such that the difference in its levels in the two limbs becomes 19 cm.
8. 20.0 Nm–2, 1.06 × 105 Nm–2.
4.63
EVIOUS YEA S’ IIT-JEE QUESTIONS
Hydrostatics and Hydrodynamics
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4.64
Hydrostatics and Hydrodynamics
P2 = pressure just inside the bubble at the end 1 = P0 +
21. (b) P1 = pressure just inside the bubble at
⇒ Air will flow from end 1 to end 2
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the end 2 = P0 +
4T R
R > r ⇒ P2 < P1
the stream will go up, speed will decrease and cross-section area will increase. Thus it will become broader. Hence statement 1 is correct and statement 2 is correct explanation also.
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c h a p t e r
5
m
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5.1 SIMPLE HARMONIC MOTION
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Simple Harmonic Motion
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Simple harmonic motion is defined as a type of motion in which a particle moves to and fro about fixed point in such a way that its acceleration is directly proportional to displacement and it is always directed toward mean position.
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A and B is extremes position O is the mean position.
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Example Motion of simple pendulum is simple harmonic motion. Motion of large mass droped in the tunnel passing through the centre of the earth is simple harmonic motion.
A
B
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C
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5.2
Simple Harmonic Motion
5.2 ANALYSIS OF SIMPLE HARMONIC MOTION Consider a particle of mass m executing simple harmonic motion along x-axis with mean position at origin. From above definition, if displacement of particle is positive x-direction then the acceleration is negative x-direction.
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O
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ma = – k x
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m
So a∝–x or F ∝ –x Thus, a is directly proportional to x then force is also directly proportional to x. F=–kx ...(i) where, k is constant From equation (i), if mass of the particle is m, then
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d 2 x -k = x ...(ii) dt 2 m This equation is called as the differential equation of simple harmonic motion The general expression for x (t) satisfying the equation is x (t) = A sin (ωt + φ) ...(iii) k [from equations (ii) and (iii)] m
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where, ω =
Amplitude The amplitude of a particle executing simple harmonic motion is its maximum displacement. From equation (iii), the amplitude of particle is A. x (t) = A sin (ωt + φ) Since maximum value of sin θ = 1, therefore – A ≤ x ≤ A
Time Period Time period of the particle in simple harmonic motion is the time taken by the particle to complete one cycle of motion and is denoted by T. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.3
Consider the equation of simple harmonic motion, i.e., x (t) = A sin (ωt + φ) 2π , then ω
If t is increased by
2π x (t ) = A sin ω t + +φ ω = A sin (ωt + 2π + φ) = A sin (2π + ωt + φ) = A sin (ωt + φ)
Thus, the function x (t) repeats itself after on interval of
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Frequency
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k 2π m ..... since ω = = 2π m ω k
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T=
⇒
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Time period, T=
2π . Therefore, ω
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ω 1 k = 2π 2π m For solution of equation of simple harmonic motion a = – ω2 x dv ∴ = -ω2 x dt dv dx or × = -ω2 x dx dt vdv or = -ω2 x dx v(nu) =
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The frequency of a particle executing simple harmonic motion is equal to the number of oscillations completed in one second. It is measured in cycle per second (Hertz) and is denoted by
∫ vdv = -ω ∫ x dx 2
or
v2 = – ω2x2 + C …(v) 2 2 2 v =C–ω x It is clear that v increase as x decrease and at a particle, when v = 0 and x = A (amplitude) or
v2 = ω2A2 – ω2x2
v = ω A2 - x2
or
Again
…(vi)
C = ω 2 A2
then,
dx = ω A2 - x2 dt
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Simple Harmonic Motion
Integrating,
or or
sin -1
dx A2 - x2
= ∫ ω dt
x = ωt A
x = sin (ωt + φ) A x = A sin (ωt + φ)
…(vii)
∴
dx v= = ωA cos (ωt + φ) dt
and
a=
na .c om
dv = -ω2 A sin (ωt + φ) = -ω2 x dt
5.3 CONCEPT OF PHASE
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Phase means state of motion which described by velocity displacement and acceleration.
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m
x = A sin (ωt + φ) dx v= = Aω cos (ωt + φ) dt a = – ω2 A sin (ωt + φ)
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Let or or
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Initial Condition
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v, x and a depends on (ωt + φ), i.e., (ωt + φ) is controller of phase. Hence, (ωt + φ) is phase angle on simple pendulum. θ = ωt + φ At t = 0, θ = φ known as initial phase or epoch. φ represent initial condition and actual equation of simple harmonic motion depends emit a condition.
s: //
5.4
x = 0 at t = 0 x = A sin (ωt + φ) 0 = A sin φ sin φ = 0 φ = 0 and π φ=0 x = 0+ t = 0 φ=π x = 0– t = 0 A
B
Actual equation of simple harmonic motion: For x = 0+ at t = 0 https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
Let or or
x = A sin ωt for φ = 0 x = A and at t = 0 x = A sin (ωt + φ) A = A sin (ωt + φ) sin φ = 1
π 2 Actual equation ∴
φ=
π x = A sin ωt + = A cos ωt 2
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Concept of Reference Circle
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Pt=0 x Reference line
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tp ht x = A cos (ωtt + φ) y = A sin (ωt + φ)
P t = t
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Projection of a particle in uniform circular motion perform simple harmonic motion.
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dx = A ω sin (ωt + φ) dt aθ = – ω2 A cos (ωt + φ) = – ω2x
vx =
Physical Signification of Phase Physically phase represents the mood. Mood of the particle: (i) Equation of projective wave x = A sin (ωt – kx) indicate that mood (phase) of particle is function of time. (ii) Equation of stationary wave x = 2a cos kx × sin ωt indicate that mood (phase) depends on time only. https://t.me/universitarios_infohttps://www.jamarana.com
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5.5
5.6
Simple Harmonic Motion
5.4 ENERGY CONSIDERATION IN S. H. M. Simple harmonic motion’s defined by the equation F = – kx The work done by the force F during a displacement from x to x + dx is dW = Fdx = – kx dx The work done in a displacement from x = 0 to x is x
W = ∫ (-kx) dx
0
in fo
an a. c
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1 = - kx 2 . 2 Let U(x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force.
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ω=
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As
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1 and U( x) = kx 2 . 2
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1 U( x) - U(0) = - W = kx 2 2 Let us choose the potential energy to be zero when the particle is at the centre of oscillation x = 0 Then, U(0) = 0
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tp
t.m e/
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1 k = mω2 - U( x) = m ω2 x 2 2 The displacement and the velocity of a particle exceeding a simple harmonic motion are given below. x = A sin (ωt + φ)
tp
v = Aω cos (ωt + φ)
ht
and
Thus potential energy at time t is 1 U = mω2 x 2 2 1 = mω2 A 2 sin 2 (ωt + φ) 2
And the kinetic energy at time t is
1 K = mv 2 2
1 = mA 2 ω2 cos 2 (ωt + φ) 2
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Simple Harmonic Motion
5.7
The total mechanical energy at time t is E=U+K 1 = m ω2 A 2 2 [sin2 (ωt + φ) + cos2 (ωt + φ)] 1 = m ω2 A 2 2 We see that, the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected. Hence, force is conserved as mechanical energy is conserved. Graph of potential energy
na .c om
1 U = mω2 A 2 2
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v2 x2 + =1 ω2 A 2 A 2
w
⇒
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m
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Graph between velocity vs displacement v2 = ω2 (A2 – x2) ⇒ v2 = ω2A2 – ω2x2 ⇒ v2 + ω2x2 = ω2A2
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x=A
u=0 x=0
x=+A
This is the equation of ellipse.
5.5 CHARACTERISTICS OF SIMPLE HARMONIC MOTION (i) It is a periodic motion, i.e., motion repeats itself after second interval of time. All periodic motions are not simple harmonic motion but all simple harmonic motions are periodic. e.g., motion of the earth, round is periodic but not simple harmonic motion (ii) It is an oscillatory motion, i.e., to and fro motion (i.e., vibrating motion). All oscillatory motions are not simple harmonic motions but all simple harmonic motions are oscillatory motions.
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Simple Harmonic Motion
Example A ball of density (d) is dropped in the liquid of density ρ (ρ > d) from height h. What type of motion the ball will perform, is it simple harmonic motion? Find time period of motion (viscous force on resistance has to be negligible).
) Solution
Weight of liquid, W = Vρg Displaced liquid = B Weight of the body = Vdg = Mg
Hence, acceleration inside liquid is upward
ita
v = 2gh
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h
C
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B
V ρg
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m
A
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g (ρ - d ) d
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=
B - Mg Vρg - Vdg = M Vd
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a=
ht
The motion of ball is to and fro, i.e., oscillatory but not simple harmonic motion because acceleration is not directly proportional to displacement. Time period of motion:
∴ t1 =
1 2 g t1 2
s: //
h=
ht tp
Time from A to B
t.m e/
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5.8
2h g
Time from B to C vc = vB – at2 = 0 ∴ t2 =
2 gh VB = a a
Total time = 2 (t2 + t1) 2 gh 2h = 2 + a g https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.9
Example Find time period of motion of ball on the given inclined plane.
) Solution
Motion of ball on the inclined plane is oscillatory but not simple harmonic motion because a ∝ x. C
m
ar a
α
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1 g sin α t12 2
iv
.m
2h cosec α 2h = cosec α g sin α g
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s: //
t.m e/
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un
//t
⇒ t1 =
B
er s
w
w
//w ht
tp
s:
h cosec α =
β
ita
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Time from A to B 1 s = ut + at 2 2
h
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h
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na .c om
A
elongation = d
Block in equilibrium
Similarly, t2 =
2h cosec β g
Hence, time period of oscillation = 2 (t1 + t2) (iii) Acceleration is directly proportional to displacement. (iv) Acceleration is always opposite to displacement from mean position which is always considered as initial position for the sake of measuring displacement. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
Note: Direction of displacement is not distance by direction of velocity of simple harmonic motion Here, a∝–x a = – ω2 x 2π ω= = 2πn T
⇒ ⇒
∆K + ∆U = 0
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Example
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na .c om
m
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∆K = – ∆U F · ∆x = ∆U -∆U ⇒ F= ∆x Change in kinetic energy, ∆K = work-done = F ∆x ∂u f = ∂x
(v) The force acting on the particle is conservative i.e., (K + U) = constant
)
//t
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) Solution
iv
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The equation of a particle exceeding simple harmonic motion is x = (5 m) sin π −1 ( π s )t + 3 . Write down the amplitude time period and maximum speed. Also find the velocity at t = 1s.
t.m e/
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Comparing with equation, x = A sin (ω t + φ) We see that amplitude = 5 m 2π 2π = = 2sec. ω πs -1 Maximum speed = Aω = 5 m × πs–1 = 5π m/sec. dx Velocity at time t = = Aω cos (ωt + φ) dt
ht tp
At t = 1s
s: //
And, time period =
π −5π m/sec v = (5 m) (πs -1 ) cos π + = m / sec. 3 2
Example A particle of mass 0.2 kg undergoes simple harmonic motion according to the equation
(i) What is the time period of oscillation? (ii) What are initial condition?
π x (t) = 3 sin π t + 4
5.10
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Simple Harmonic Motion
5.11
(iii) What is the total energy of the particle if potential energy is zero at mean position? (iv) What are the kinetic and potential energies of particle at time at t = 1s? (v) At what time instants is the particle’s energy purely kinetic? )
) Solution
Comparing the given equation with x (t) = A sin (ωt + φ)
m
fo io iv
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si
ta
iv
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From the position x = 1.5 2 m
tp
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s:
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//t
1 With the velocity = (3π ) m/sec. 2
.m
1 (iii) Kinetic energy = mω2 A 2 2 1 = (0.2) (π) 2 (3) 2 2 = 0.9 π2
ht
π 3 m (iv) At t = 1s, x(t ) = 3sin π + = 4 2 π v(t ) = 3π cos π + 4 3π =m/sec. 2
tp
s:
//t
ht
s
w
Hence, the particle starts moving at t = 0
s:
tp ht
si
3π m/sec. 2
⇒ v(0) =
//w
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w
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m
dx π = v(t ) = 3π cos πt + dt 4
ta r
ar
s_
= 1.5 2 m
rio
an
in
π 4
(ii) x (0) = 3sin
a. co
we get A = 3 m ω = π rad/sec. π φ= 4 2π = 2 sec. (i) T = ω
1 Kinetic energy = mv 2 2
9π 2 1 = (0.2) = 4.44 J 2 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
1 U = kx 2 2 1 = m w2 x 2 2 1 9 = (0.2)(π) 2 2 2 = 4.44 J (v) Energy is purely kinetic at mean position i.e., when x = 0 π x (t ) = 3sin πt + 4
π = 0, π, 2π, 3π ...... 4
ar a
πt +
ita
m
1 3 7 11 t=- , , , 4 4 4 4
⇒
.ja
w
rio s_ in fo
π 0 = 3sin πt + 4
we have
na .c om
Using
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w
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At these time instants, particle crossing origin and hence its energy is purely kinetic.
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//t
iv
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tp
s:
(i) If particle starts at t = 0 from mean position, its equation is x(t) = A sin ωt, if initial velocity is positive. x(t) = – A sin ωt, if initial velocity is negative. (ii) If the particle starts at t = 0 from right extremes, its equation is
ht
t.m e/
x(t) = A cos ωt.
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(iii) If the particle starts at t = 0 from left extremes, its equation is x(t) = – A cos ωt.
ht tp
5.12
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Simple Harmonic Motion
5.13
Example The friction coefficient between the two blocks shown in the figure is µ and the horizontal plane is smooth. (i) If the system is slightly displaced and released, find the time period. (ii) Find the magnitude of frictional force between the blocks when the displacement from the mean position is x. (iii) What can be the maximum amplitude if the upper block does slip relative to the lower block?
ar a
) Solution
m M
rio s_ in fo
na .c om
k
er si ta rio s
.m
M+m k
iv
(ii) Acceleration of the blocks at displacement x from the mean position is -kx a = -ω 2 x = M+m
t.m e/
ht tp s:
un
//t
ht
er s
un iv
e/
And, time period, T = 2π
tp
s:
//w
w
w
k M+m
ω=
ita
.ja
m
(i) For small amplitude the two blocks oscillate together. The angular frequency is
Thus resultant force on the upper block is
s: //
-m kx M+m
ht tp
ma =
This force is provided by the friction in the lower block. Hence, magnitude of the frictional force =
mk | x | . M+m
(iii) Maximum force of friction required for simple harmonic motion of the upper mk A block is at the extreme positions. But the maximum frictional force can M+m mk A only be µ mg. Hence, = µ mg M+m or A =
µ ( M + m)g k
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Simple Harmonic Motion
Example A particle is moving in a straight line with simple harmonic motion. Its velocity has the values 3 m/sec and 2 m/sec when its distance from mean positions are 1 m and 2 m respectively. Find the length of its path and period of its motion. )
) Solution
At x = 2 m,
v1 = 3 m/sec. v2 = 2 m/sec.
At x = 1 m,
From the formula
v2 = ω2 (A2 – x2) ...(i)
9 = ω2 (A2 – 1)
w .ja
s
iv
un
si ta
5 = 1.29 rad/s 3
ω=
t.m
tp
And
e/
s: //
w
w
er si
A = 6.4 = 2.53m
∴
un
2π = 4.875 ω
ht
t.m
tp
Time period, T =
e/
s: //
iv
Hence, length of the path = 2 A = 5.06 m
er
5 A2 = 32
or
ht
ta rio
9 A2 – 36 = 4 A2 – 4
m
or
rio
ar
an
9 A2 -1 = 4 A2 - 4
s_ in fo
co m
...(ii)
a.
and 4 = ω2 (A2 – 4) From equations (i) and (ii), we have
s: //
Example
ht
tp
A particle of mass m is located in a undimensional potential field where the potential energy of the particle depends on the co-ordinate x as U(x) = U0 (1 – cos ax); U0 and a is constants. Find the period of small oscillations that the particle performs about the equilibrium position.
) Solution )
∴
As U(x) = U0 (1 – cos ax)
Fx = -
dU dx ...(i)
= – U0 a sin ax
or
Fx = – U0 a · ax
(because for small angle of oscillations sin ax ≅ ax) or
5.14
Fx = – U0 a2x
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Simple Harmonic Motion
5.15
But for small oscillation Fx = – mω2x U0a2 m
ω=a
or
U0 m
Hence, time period, T =
2π m a U0
na .c om
=
2π ω
= 2π
m a U0 2
rio s_ in fo
∴ ω2 =
er si ta rio s
er s
w
w
) Solution
ita
.ja
m
ar a
Example Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.
e/
d 2x = mg - 2 F θ dt 2
un t.m e/
ht tp s:
m
iv
.m
By Newton’s Second law
//t
ht
tp
s:
un iv
//w
At that position, the force on the ball = mg – 2 F sin θ
s: //
(because θ is very small then sin θ ≅ θ) θ
ht tp
θ
F
x
F
mg
or
m
d 2x x = mg - 2F dt 2 l/2 4 Fx = mg l
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5.16
Simple Harmonic Motion
4 Fx l
a=g-
Thus,
4F mgl x ml 4F
=-
x′ = x -
Putting,
mgl , we get 4F
-4 F x′ ml 4F ω2 = ml
a=
Thus,
2
m
ml F = 0.2 sec.
T=π
ta si rio
s
er
//w w
w
.ja m
or
in fo
ar
rio s_
an
a.
A π2 4 F = T2 ml
or
co
or
4F 2π = T ml
ita
un iv
tp s:
5.6 GENERAL APPROACH FOR ANALYSING SIMPLE HARMONIC MOTION
ve
ht
.m
rs
e/
The following steps must be carefully followed while analysing simple harmonic motion for a particle:
e/ un i
//t
.m
tp
ht
tp s:
//t
ht
s:
(i) Displace the particle mean position (equilibrium position) by a distance x along the line of simple harmonic motion. (ii) Analyse the forces on the particle and try to show that (a) The magnitude of net force is proportional to x. (b) The force is directed towards the mean position, i.e., it has a restoring nature. (iii) Find the value of proportionality constant from the expression of the magnitude of force. m (iv) Find the time period using, T = 2π k
5.7 SIMPLE PENDULUM A simple pendulum is an idealized body consisting a point mass, suspended by a light inextensible cord. When pulled to one side of its equilibrium position and then released, the pendulum oscillates. In figure, the string of a simple pendulum of length l attached to a bob of mass m is shown making an angle θ with the vertical.
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Simple Harmonic Motion
θ
5.17
l
T
na .c om
mg sinθ
θ mg sinθ mg
rio s_ in fo
ar a
The forces acting on m are mg or W (gravitational pull of the earth) and T (Tension in the cord).
er si ta rio s
ht
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
⇒ Choose the axis along tangent to the arc of motion and along the radius. Resolve the forces along the two axes. ⇒ The radial component of magnitude mg cos θ balances the force and tension in the string. ⇒ The tangential component of magnitude mg sin θ is the restoring force acting on m to return it to the equilibrium state, so the magnitude of restoring force is F = mg sin θ
arc ..... θ = radius mg x So F = mg θ = …(i) l ⇒ F∝x So the simple pendulum executes simple harmonic motion. The relation between magnitudes of force and displacement is F = kx ...(ii) From equations (i) and (ii), we get mg ⇒ k= l ⇒ θ=
x l
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
This force is acting towards mean position. If the angle θ is very small, then sin θ is very close to θ (sin θ → θ) in radians. So sin θ ≅ θ or F ≅ mg θ (here θ is angular displacement) Now, the magnitude of displacement along the arc is x (but for small θ, are can be taken as a straight line segment).
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5.18
Simple Harmonic Motion
So, the time period of a simple pendulum, when its amplitude is small, is given as m k m mg / l
= 2π
l g
Also, Frequency, v = =
1 k 2π m 1 2π
g l
ar a
5.8 LIQUID OSCILLATING IN A U-TUBE
rio s_ in fo
= 2π
na .c om
T = 2π
//t
er si ta rio s iv
.m
tp ht
D
2x
A
B
Displaced liquid
t.m e/
ht tp s:
un
In equilibrium
C
x
x
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
Consider a liquid column of total length 2l inside a U-tube as shown:
ht tp
s: //
In the U-tube, liquid is displaced from state of equilibrium by distance x. It is clear from diagram that the total difference in liquid levels between limbs is 2x. Consider equal cross-section A on both sides (tube is uniform throughout). The restoring force acting on the liquid is the thrust of excess pressure due to the column ABCD. The magnitude of restoring force is F = pressure × area = (2x) ρgA F = (2 A ρg) x or
...(i)
F∝x
Hence the liquid executes simple harmonic motion in the tube Comparing equation (i) with F = kx, we get k = 2Aρg Hence, the time period is https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
T = 2π = 2π
5.19
m k m 2A ρg
T = 2π
2A lρ 2A ρg
or
T = 2π
l g
na .c om
∴
rio s_ in fo
But mass of oscillating liquid = m = A (2l) ρ
ar a
5.9 OSCILLATION OF A CYLINDER FLOATING IN LIQUID
iv
er si ta rio s
e/
L
.m ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Consider a cylinder of mass m and density d floating on the surface of a liquid of density ρ. The total length of a cylinder is L and the length immersed in the liquid is l. Taking A as the area of cross-section, we apply the law of floatation on the cylinder.
Cylinder in equilibrium
Cylinder displaced downward by a distance x https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
B
mg
Weight of cylinder = Buoyant force mg = Vρg
or
Ld = lρ
rio s_ in fo
(AL) dg = (Al) ρg
ar a
∴
na .c om
where, V is the volume inside the liquid. ...(i)
un iv
e/
.m
tp
s:
//w
F = B – mg = (l + x) A ρg – LA dg = Aρgx + l A ρg – LA dg
er si ta rio s
w
er s
w
ita
.ja
m
If the cylinder is displaced from this position, it oscillates up and down. Let, it displace through a distance x and we analyse the net force acting on it. As the volume occupied inside the liquid increases, the buoyant force B becomes greater than weight and hence, net force acts upwards. Magnitude of net force is
ht
un
//t
iv
From equation (i), F = (Aρg) x Hence, the magnitude of the net force is proportional to x and the direction of force is towards the equilibrium or mean position. Thus, the cylinder will executes simple harmonic motion Computing with F = kx, we get k = A ρg
ht tp
s: //
t.m e/
ht tp s:
5.20
Hence, time period, T = 2π = 2π
m k m Aρg
Using m = (AL) d and Ld = lρ, we can further simplify for T as T = 2π
ALd Alρ = 2π Aρg Aρg
T = 2π
l g
Hence, the time period depends only on the length of the immersed portion. https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
5.21
5.10 OSCILLATIONS OF A BLOCK CONNECTING TO A SPRING (i) Horizontal spring
na .c om
Elongation = d +x
rio s_ in fo
Block displaced downwards by x
er si ta rio s
.m
e/
F = ksx
iv
i.e., F is proportional to x. Block executes simple harmonic motion with k = ks m ks
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
Let a block of mass m be placed on a smooth horizontal surface and rigidly connected to a spring of force constant ks whose other end is permanently fixed. When the spring is in its underformed condition, block is in equilibrium. Let it displace the block by a distance x as shown. The net force acting on the block has a magnitude ksx and is directed towards its mean position. Hence, magnitude of the force is
s: //
Time period, T = 2π
ht tp
(ii) Vertical spring If the spring is suspended vertically from a fixed point and carries the block at its other end as shown, the block will oscillate along the vertical line.
ksd elongation = d
Undeformed spring
Block in equilibrium
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mg d = mg ksd=
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Simple Harmonic Motion
In its equilibrium position, the block elongates the springs by a distance d. In this position the weight of block and elastic force of spring balance each other. If the block displaced downwards by a distance x, the elongation in springs becomes (d + x) and the elastic force becomes greater than the weight of the block. The net force now acts upwards and has a magnitude.
er si ta rio s
er s
un iv
e/
ks((d+x)
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
s:
From equation (i), F = ksx
ita
.ja
//w
w
w
F = ks (d + x) – mg F = ks d + ks x – mg
rio s_ in fo
Block displaced downwards by x
m
ar a
na .c om
Elongation =d+x
ht
5.22
net force upwards
mg
Hence, the magnitude of net force is proportional to x and its direction is towards the mean position (equilibrium position). The block executes simple harmonic motion and k = ks. Time period = 2π
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m ks
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Simple Harmonic Motion
5.23
5.11 COMBINATION OF SPRINGS (i) Springs in series: When two springs of force constants k1 and k2 are connected in series as shown, they are equivalent to a single spring of force constant k which is given by 1 1 1 = + k k1 k2 k1k2 k1 + k2
na .c om ar a
≡
er s
un iv
k1k2 k 1 + k2
un
t.m e/
ht tp s:
//t
Block displaced downwards by x
k=
iv
.m
e/
s: tp ht
ita
m .ja //w
w
w
k2
k
rio s_ in fo
k1
er si ta rio s
k=
ht tp
s: //
(ii) Springs in parallel: For a parallel combination as shown, the effective spring constant is
k1
k = k1 + k2
k2
≡
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k = k1 + k2
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5.24
Simple Harmonic Motion
k1
k2
k1+k2
) Solution
rio s_ in fo
ar a
na .c om
Example A block of mass M, at rest on a horizontal, frictionless table, is attached to a rigid support by a spring of force constant k. A bullet of mass m and speed v strikes the block as shown in the figure. The bullet remains embedded in the block. Determine the amplitude of the resulting simple harmonic motion in ter ms of m, M, v and k.
0
N
ht tp
s: //
t.m e/
Y
un
iv
er si ta rio s
e/
ht tp s:
//t
ht
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
Velocity of block and bullet after collision. mv = (m + M) v′ mv ∴ v′ = (m + M)
For the maximum amplitude 1 1 1 (m + M)v′2 = k x 2 = k A 2 2 2 2
…(i)
From equation(i), m+M A= v′ k = =
(m + M) mv 1 . = mv k (m + M) k(m + M )
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Simple Harmonic Motion
5.25
Example A massless spring of force constant 3.60 N/cm is cut into halves.
ar a
ita
er s
w
.ja
m
(i) Force constant = k1 + k2 = 3.60 N/cm + 3.60 N/cm = 7.20 N/cm
un iv
e/
.m
M k
iv
1 1 k = f 2π M
or
M= =
k f2 4π 2
t.m e/
1 1 k = ⋅ f 2 4π 2 M
ht tp
or
k
s: //
where, f = frequency
un
//t
or
T = 2π
ht tp s:
ht
tp
s:
//w
w
(ii) Time period of single spring,
er si ta rio s
) Solution
M
rio s_ in fo
na .c om
(i) What is the force constant of each half? (ii) The two halves, suspended separately support a block of mass M shown in the figure. The system vibrates at a frequency of 2.87 Hz. Find the value of the mass M.
7.20 × (2.87) 2 = 1.48 kg. 4 × (3.14) 2
Mg
Example A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is a0 and the length of the pendulum is l, find the time period of small oscillations about the mean position.
) Solution
For mean position, the acceleration of the bob with respect to the car should be zero. If θ be the angle made by the string with the vertical, the tension, weight and the pseudo force will add to zero in this position.
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Simple Harmonic Motion
α
T
θ
ma0 (θ + α) mg
na .c om
Suppose, at some instant during oscillation, the string is further deflected by an angle α so that the displacement of the bob is x. Taking the components perpendicular to the string,
m
ar a
rio s_ in fo
Component of T = 0 Component of mg = mg sin (α + θ) Component of ma0 = – ma0 cos (α + θ). Thus, the resultant component
ita
.ja
F = m [g sin (α + θ) – a0 cos (α + θ)]
er si ta rio s
un iv
//w
w
er s
w
Expanding the sine and cosine and putting cos α ≈ 1, sin α ≈ α ≈
x , we get l
tp
e/
s:
x F = m ρg sin θ - a0 cos θ + g cos θ + a0 sin θ l
…(i)
iv
.m
At x = 0, the force F on the bob should be zero, as this is the mean position. Thus, by equation (i) 0 = m [g sin θ – a0 cos θ] ...(ii) a Solving, we get tan θ = 0 g
cos θ =
s: //
a0
…(iii)
ht tp
Thus sin θ =
t.m e/
ht tp s:
un
//t
ht
5.26
(a0 ) 2 + g 2 a0
…(iv)
(a0 ) 2 + g 2
Putting equations (ii), (iii) and (iv) in equation (i), we get F = m g 2 + a02 where,
ω=
x l
or
F = mω2x
g 2 + a02 l
This is an equation of simple harmonic motion with the time period t=
2π l = 2 ω ( g + a02 )1/4
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Simple Harmonic Motion
5.27
mao
mg
An easy working rule may be found out as follows. In the mean position, the tension, the weight and the pseudo force balance. From figure, the tension is
rio s_ in fo
T = a02 + g 2 m
ar a
or
na .c om
T = (m a0 ) 2 + (mg ) 2
l l = 2π 2 T/m [ g + a02 ]1/4
er si ta rio s
ita
er s
w
w
.ja
t = 2π
m
This play the role of effective g. Thus, the time period is
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
Example Determine the period of oscillations of mercury of mass m = 200 g poured into a bent tube seen the figure whose right arm forms an angle θ = 30º with the vertical. The cross-sectional area of the tube is A = 0.50 cm2. The viscosity of mercury is to be neglected.
θ Equilibrium position
θ x
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5.28
Simple Harmonic Motion
) Solution
Total displaced liquid = x (1 + cos θ) Weight of liquid = ρAx (1 + cos θ) g ...(i) If the mercury rises in the left arm by x, it must fall by a slanting length equal to x in the other arm. Total pressure difference in the two arms will then be given by equation (i) This will give rise to a restoring force – ρg Ax (1 + cos θ)
...(ii)
or ma + ρg A x (1 + cos θ) = 0 ρg A (1 + cos θ) x m This equation in simple harmonic motion
m ρgA (1 + cos θ)
er si ta rio s
er s
un iv
//w
w
w
Putting the value, we get T = 2.94 sec.
ita
.ja
m
∴ T = 2π
2
2π ρ g A or = (1 + cos θ) m T
rio s_ in fo
ρg A (1 + cos θ) m
ar a
ω2 =
na .c om
or a = -
e/
tp
s:
5.12 ANGULAR ACCELERATION OF A RIGID BODY
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
Linear simple harmonic motion is the right-left (or up-down) oscillations of the particles. A rigid body can also execute clockwise and anti-clockwise oscillations. These oscillations are called as angular oscillations. A very simple example of angular oscillations is a thin disc suspended by a wire attached to the centre of the disc. If the disc is rotated in the horizontal plane, the wire will be twisted. The twisted wire will exerts a torque on the disc tending to return it to its original position. This is a restoring torque (compare it with a restoring force in linear oscillations). For small twist angles θ, the restoring torque is proportional to the angular displacement θ (Hook’s law).
Wire
Disc
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Simple Harmonic Motion
5.29
Hence τ = – C θ ...(i) Here negative sign indicates that if θ is anti-clockwise, torque is clockwise and vice versa i.e., torque is opposite to θ. The equation (i) is the condition for angular simple harmonic motion. It can easily be compared to F = – kx Torque = – Cθ Iα = – Cθ or
I
d 2θ = - Cθ dt 2 d 2θ C =- θ dt 2 I
⇒
ar a
d 2θ k =- x dt 2 m
rio s_ in fo
na .c om
This differential equation for angular simple harmonic motion can also be compared with the same equation for linear simple harmonic motion
m
Following the same comparison we can see that
I and the variation of θ with time C θ (t) = θmax sin (ωt + φ)
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
Time period, T = 2π
e/
.m
tp
s:
where, θmax is the angular amplitude or the maximum angular displacement from the mean position.
iv un
//t
ht
5.13 PHYSICAL PENDULUM
ht tp
s: //
t.m e/
ht tp s:
A rigid body is suspended from a fixed point O. It can perform angular oscillations in a vertical plane about the horizontal axis through O perpendicular to the plane of oscillations. This rigid body can be called as a physical pendulum.
O θ
d C
mg
Let C be the centre of gravity of such a rigid body suspended from O. The distance of C from point of suspension O is d. If we displaced the body anti-clockwise through a small angle θ, the weight of the body creates a clockwise restoring torque about axis through O. Magnitude of restoring torque is https://t.me/universitarios_infohttps://www.jamarana.com
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Simple Harmonic Motion
τ = mg (d sin θ) = mg dθ (if θ is small) Comparing with τ = C θ, we have C = mgd Let I be the moment of inertia of the body about the axis through O. Therefore
Time period = 2π
I mgd
co m
Example A body makes angular simple harmonic motion of amplitude π/10 rad. and time period 0.05 sec. If the body is at a displacement θ = π/10 rad. at t = 0, write the equation giving the angular displacement as a function of time. θ = θm sin (ωt + φ)
ar an a.
m
s
ni v
//t s: ht tp
sin φ = 1
π 2
e/ u
φ=
.m
or
or
ht tp s: //t
.m
e/
s:
At t = 0, θ =
…(i)
er
//w
w
π θ = sin[(40πs -1 )t + φ] 10
π rad. 10 Putting in equation (i), we get π π = sin φ 10 10
ht tp
ita
un iv er s
w
∴
2π 2π = = 40πs -1 T .05
.ja
ω=
π rad 10
si ta rio
Here θm = amplitude =
rio s_ in fo
)
) Solution Let the required equation be
π π π By equation (i), θ = sin (40π s -1 )t + = cos [(40πs −1 ) t ] 2 10 10 Example
A solid cylinder is attached to a mass less spring with constant k = 3 Nm–1 and it rolls without slipping. If the system is stretched by 0.25 m and released. (i) Show that under these condition the centre of mass of cylinder executes 3m simple harmonic motion with period T = 2 π 2k (ii) Find the translational kinetic energy and rotational kinetic energy as it passes through equilibrium position.
5.30
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na .c om
Simple Harmonic Motion
) Solution
5.31
er s
e/
…(i)
un
//t
iv
.m
er si ta rio s
un iv
//w
w
1 where, v = ωR and I = mR 2 for solid cylinder 2 1 mω2 R 2 + mω2 R 2 + kx 2 = constant or 2 Differentiating equation (i) with respect to t, we get
s: tp ht
rio s_ in fo
ita
.ja
1 2 1 2 1 2 mv + Iω + kx = constant 2 2 2
w
i.e.,
m
ar a
(i) First method: From the conservation of energy Total energy of cylinder and spring system is constant
ht tp
s: //
t.m e/
ht tp s:
dω dx dω mR 2 2ω +ω + 2kx = 0 dt dt dt dx Here, = v = ωR dt or x = Rθ 3α m + 2 kθ = 0 2k α=θ or 3m This equation is the simple harmonic equation because angular acceleration is directly proportional to the angular displacement. T = 2π
3m 2k
Second Method: Let the cylinder rotates in antic-lockwise direction. In this rotation if angle rotates is θ, then the spring shifts atleast a distance Rθ from mean position. The friction force opposed this force.
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Simple Harmonic Motion
Rθ θ Mean position
na .c om
Thus, net force = (k Rθ – f) towards mean position. Applying τ = Icm α, and F = ma
rio s_ in fo
f R = I α, and k Rθ – f = ma
∴ k Rθ – f = m α R f = (k Rθ – m α R)
w
er s
w
or
un iv
kR 2 θ Icm + mR 2
kRθ
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
e/
⇒ α=
tp
s:
//w
or (k Rθ – m α R) R = Icm α
er si ta rio s
ita
.ja
m
a = αR
ar a
where, a is the linear acceleration and α is the angular acceleration. For rolling without slipping.
ht
5.32
f
The magnitude of the angular acceleration is proportional to the angle θ. Hence, the cylinder performs simple harmonic motion ω2 =
kR 2 Icm + mR 2
2
k R2 2π or = 1 T m R 2 + mR 2 2 or
T = 2π
3m 2k
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5.33
1 1 (ii) Translational kinetic energy = mv 2 = m (Aω) 2 2 2 1 2π A = (0.25) = m A 2 ⋅ 2 T
Here,
Simple Harmonic Motion
2
1 (2π) 2 2k = m A2 ⋅ ⋅ 2 (2π) 2 3m 1 3N × 2 = × 0.25 × 0.25 × = 2 3 = 6.25 × 10–2 J
na .c om
ar a
m
11 = mv 2 2 2
er si ta rio s
ita
er s
un iv
e/
s:
The rotational energy is half of the translational kinetic energy.
iv
ht tp s:
un
//t
ht
.m
tp
//w
w
w
.ja
1 = × 6.25 × 10-2 J 2 = 3.1 × 10−2 J
rio s_ in fo
1 1 1 = Iω2 = ⋅ mR 2 ω2 2 2 2
Rotational kinetic energy
t.m e/
)
s: //
ht tp
) Solution
Obviously for small β, the ball excutes parts of simple harmonic motion. Due to the perfectly elastic collision, the velocity of ball simply reversed. As the ball is in simple harmonic motion (| θ | < α on the left). Its motion law is ma + mg sin θ = 0
1. A ball is suspended by a thread of length l at the point O on the wall, forming an angle α with the vertical. Then the thread with the ball was deviated through a small angle β (β > α) and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum.
WORKED OUT EXAMPLES
O αβ
θ l
mg sinθ
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mg cosθ mg
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5.34
Simple Harmonic Motion
) Solution
θ θ = – g sin θ = - g l = – ω2 θ ……. (i) If we assume that the ball is released from the extreme position θ = β at t = 0, the solution of differential equation would be taken in the form θ = β cos ωt
)
or
When the hydrometer is displaced, then the net upward force
F = ρπ r2 (h + x) g – mg
...(ii)
From equations (i) and (ii) F = ρπ r2 xg ρπ r 2 xg ∴ a= x = ω2 x m
……. (ii)
g t l
= β cos
...(i)
mg = ρ π r2hg
Let at equilibrium, h is length in the liquid. Then at equilibrium
o
nf s_ i io ta r si rio
s
er
er si ta
iv
m ρπ r 2 g
T = 2π
e/ un
tp
t.m
ht
4πm ρg r 2
3. A cylinder of mass m and radius r rolls without slipping on a circular surface of radius R as shown in the figure. Find the frequency of oscillation when the cylinder is displaced slightly from its equilibrium position.
tp
2. Calculate the period of small oscillations of a hydrometer shown in the figure, which was slightly pushed down in the vertical direction. Mass of the hydrometer is m = 50 g, the radius of its tube is r = 3.2 mm, the density of the liquid is ρ = 1.00 g/cm3. The resistance of the liquid is assumed to be negligible.
=
s: //
ht
m
a. iv e/
s:
//t
l π -1 α + sin g2 β
π -1 -1 becausesin x + cos x = 2
(h + x)
Hence acceleration is directly proportional to the displacement
.m
tp
ht
=2
un
-1 α π - cos β
s:
l g
//w
time period, T = 2t′ =2
a∝x
an
-1 α π - cos β
∴
m .ja
l g
=
or
ar
-α l cos -1 g β
t′ =
w w
or
co
g -α = β cos t′ l
If t′ be the time taken by the ball to go from the extreme position θ = β to the wall, i.e., θ = – α, then from equation (ii)
R m
r
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Potential energy = mgh
)
) Solution
v = velocity of centre of mass of the cylinder, and ω = angular velocity of rotation of the cylinder For pure rolling,
= mg (R – r) (1 – cos θ) E = Total energy of the system at B
1 1 = mv 2 + Iω2 + mgh 2 2 2
co m
fo
3 d θ m(R - r ) 2 + mg (R - r ) (sin θ) = 0 2 dt
in
s_ io
ta r
io
s
er iv
ta r
si
un
iv
d 2θ 2 g =θ dt 2 3 R -r
er
e/
un
//t .
e/
Hence the motion is simple harmonic motion. Hence angular frequency of oscillation,
1 I ωc 2 2 2
1 R - r 2 = I θ 2 r
ω=
2 g . 3 R -r
ht tp
s:
//t .
m
s:
ht tp
….. [For small θ, sin θ. θ] ⇒
m
Reference position
=
d 2θ 2 g + θ=0 dt 2 3 R - r
si
.ja w w ht tp
o
A
The total energy of the system consists of kinetic energy (rotational and translational) and potential energy should remain the same all the time. 1 Translational kinetic energy = mv 2 2 1 = m(R - r ) 2 θ 2 2 1 Rotation kinetic energy = I ωc 2 2
⇒ ∴
R
dE =0 dt
an m
(R – r) cos θ
ar
θ
s: //w
B
Now,
a.
C R –r
3 = m(R - r ) 2 θ 2 + mg (R – r) (1 – cos θ) 4
…(ii)
V R -r ωc = = θ r r
⇒
mg (R – r) (1 – cos θ)
…(i)
v = (R - r ) θ
∴
1 11 R - r 2 = m(R - r ) 2 θ 2 + mr 2 θ + 2 2 2 r
Let θ be the angular displacement of the line OC as the cylinder passes through the position B.
4. A certain mass of a perfect gas is enclosed in a cylinder of volume V0 fitted with a smooth heavy piston of mass m and area A. The piston is displaced through a small distance downwards so as to compress the gas isothermally, and then it is let go. Show that the resultant motion is simple harmonic motion and find its period. Take the atmospheric pressure as Patm.
v = r ωc
Let
5.35
Simple Harmonic Motion
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5.36
weight and hence magnitude of net upward force is
)
) Solution
F = PA – mg – Patm A
PatmA
xA F = P0 1 + A - mg - Patm A V0 From equation (i),
⇒ PO A
mg
Po
F=
Hence, magnitude of the force is proportional to x and its direction is towards equilibrium position. Comparing with F = kx, we get
co m
Piston in equilibrium
mg
in
s
iv
si
ta
un
er
e/
//t
tp s:
ht
⇒
T = 2π
= 2π
mV0 mg 2 Patm + A A
mV0 Patm A 2 + mg A
5. A uniform rod is placed on two spinning wheels as shown in the figure. The axis of the wheels are separated by a distance l = 20 cm the coefficient of friction between the rod and the wheels is µ = 0.18. Demonstrate that in this case the rod perform harmonic oscillations. Find the period of these oscillations.
.m
e/ u
//t
s:
tp
ht
ni v
.m
ht
-1
m V0 P0 A 2
xA P = P0 1 V0
m k
rio
w //w s: tp
Let A be the area of croass-section of the cylinder. When, the piston is in equilibrium, forces on it must balance. ∴ P0 A = Patm A + mg ...(i) Let us now displace the piston through a small distance x in downward direction. As the compression of the gas is isothermal, the final pressure P is given by P0 V0 = P (V0 – xA) PV P= 0 0 V0 - xA
= 2π
er s
w
P
ita rio s_
.ja m
PA
P0 A 2 V0
Time period, T = 2π
ar
x
k=
an a.
PatmA
P0 A 2 x V0
fo
Simple Harmonic Motion
xA = P0 1 + V0 [∵ (1 + t)n ≈ 1 + nt for small t.]
Considering the forces on the displaced piston, thrust of the gas pressure now exceeds its https://t.me/universitarios_infohttps://www.jamarana.com
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From equations (i), (ii) and (iii), we get
)
) Solution
In the equilibrium position, the centre of mass of the rod lies midway between the two rotating wheels. Let us displaced the rod horizontally by some small distance and then released it. Let us depict the forces acting on the rod when its centre of mass is at distance x from its equilibrium position. Since, there is no net vertical force acting on the rod, Newton’s second law gives
d 2x 2g =-µ x 2 dt l Hence, time period, T = 2π
fo
co m
in
a.
s_
an
rio
ar m
si
s rio ta si iv er
mg x
e/
...(i)
)
The physical system consists with a pulley and the block. Earth and spring are considered as external object. Here the physical phenomenon consists in the oscillatory motion of system. Choosing on inertial frame, let us direct the x-axis as shown in the figure. Initially the system is in equilibrium position. Now from the condition the translation equilibrium for the block, T0 = mg ...(i) Similarly, for the rotational equilibrium of the pulley k ∆l R = T0 R or
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T0 = k ∆l
…(iii)
l + 2x l - 2x N1 = N2 2 2
As the rod experience no net torque about an axis perpendicular to the plane of the figure through the centre of mass of the rod
…(ii)
ht
tp
For the translational motion of the rod from the equation
v
un
x
s: //t .m
ht N1 + N2 = mg
Fx = max md 2x µN1 - µN 2 = dt 2
x
) Solution First Method: Shown in figure (b).
er iv un
e/
µN2
mg
O
:// t.m
x
T
T
ta
.ja w c
tp s
ht tp
s:
µN1
//w
o
k∆ l
T 0`
N2
w
N1
2l µg
6. Find the time period of the system shown in the figure. The radius of the pulley is R, its moment of inertia relative to the rotation axis is I, the mass of the body is m and the spring constant is k.
l 2 c
l 2µg
T=π
l 2
5.37
Simple Harmonic Motion
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...(ii)
5.38
econd method: The mechanical energy of the system (pulley + block) is the joint field of spring and gravity is conserved.
s
From equations (i) and (ii) mg ∆l = ...(iii) k Now let us disturb the equilibrium of the system, no matter in which way to analysis its motion at an arbitrary shown in the figure. Now from Newton’s second law of motion for the block Fx = max d 2x mg - T = m 2 ...(iv) dt
Therefore, ∆T + ∆U = 0
2 2 1 x 1 2 1 v0 1 2 m x + I mv + ( ) I 0 2 R 2 2 R 2
1 1 + k (∆l + x) 2 - k (∆l ) 2 - mgx = 0 2 2
Similarly, for the pulley for its rotational motion from the equation. τz = I αz (where z-axis is the axis of rotation) d 2θ TR - k (∆l + x) R = I 2 ...(v) dt
m
Here, v0 = 0 at x = 0 which is constant.
in
rio
ta
s
rio
ta
iv
er si
un e/
Hence,
.m s:
I m+ 2 R 2π ω
= 2π
or
tp
k
Hence, the time period T =
ht
ω2 =
k x = - m + I2 k
//t
ht
d x k x = - 2 dt m+ I R 2 2
I x + kx = 0 m + 2 R
[because at equilibrium position kAl = mg]
//t
s:
tp
I d 2x -kRx = mR + 2 R dt
or
x x (Al + x) - mg = 0 R R
er si or
Solving equations (iv) and (vii) and using the initial condition of the problem, we get
or
1 I x 1 x+ k 2 m + 2 2 2 R R 2
iv un
...(vii)
e/
I d 2x R dt 2
.m
ht
tp
TR - k (∆l + x) =
s:
From equations (v) and (vi)
s_
ar an
w
...(vi)
//w w
.ja
2
or
Differentiating with respect to time, we get
m
a = αR d x d 2θ = R dt 2 dt 2
fo
co
a.
But
2
1 1 x 1 m( x ) 2 + I + k (∆l + x) 2 - mgx = 0 2 2 R 2
m+ k
I R2
T = 2π
x
m+ k
I R2
7. Two springs, each with unstretched length 0.20 m but having different force constants k1 and k2 are attached to opposite ends of a block with mass m on a level, frictionless surface. The other ends of the springs are now attached to two points P1 and P2, 0.10 m from the original position of the ends of the spring. Let k1 = 1.0 N/m, k2 = 3.0 N/m and m = 0.20 kg.
Simple Harmonic Motion
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Simple Harmonic Motion
5.39
Determine the length of each spring when the block is in its new equilibrium position after the springs have been attached to the pins.
R M
C θ α
P2
P1
A
)
) Solution
fo
s
rio
si ta
er
iv
un
mA gR – R mg sin α = 0 mA = m sin α
e/
s: //t .m
ht tp
∴ or
//t .
s:
ht tp
| F1′ | = | F2′ | k1 (0.1 + x) = k2 (0.1 – x) 0.1 + x = 0.3 – x 4.0 x = 0.2 m x = 0.05 m length of spring = 0.35 m and 0.25 m
8. A uniform cylindrical pulley of mass M and radius R can freely rotate about the horizontal axis C. The free end of thread tightly wound on the pulley carries a dead weight A. At a certain angle α if counter balances a point mass m fixed at the rim of the pulley, find the angular frequency of small oscillation of the arrangement.
si ta
e/
m
ht tp
Since, | F1 | ≠ | F2 |, the mass will move towards the right so as to attain equilibrium position at which these forces become identical. If x is the distance moved by the mass towards right side, we have
i.e., or or or
er
un
| F2 | = k2 (0.1 m) = (3.0 N/m) (0.1) = 0.3 N
iv
| F1 | = k1 (0.1 m) = (1.0 N/m) (0.1 m) = 0.1 N
s: //w w
w
.ja
m
rio
ar a
When the springs are attached to the pins, each spring will under go elongation by 0.1 m. The elongation will exerts forces on the mass which are given by
s_
in
na
)
) Solution
If we get the system in motion, no matter in which way, from the conservation of mechanical energy of oscillation at an arbitrary position (as shown in the figure). 1 1 1 MR 2 2 mA (θ ) 2 R 2 + m(θ ) R 2 + (θ) + mg 2 2 2 2 [R cos (α + θ) – R cos α] – mA g Rθ = constant
R2 M 2 m sin α + m + (θ) + mg R[cos 2 2 (α + θ) – m cos α] - m sin α Rθ = constant or
0.2 m
In this problem, the weight of A is restoring force for half oscillation and for the other half oscillation the weight of point mass m is restoring. Hence, the solution of this problem from energy conservation is more meaningful. It is also because the restoring and conservative forces are responsible for the conservation of mechanical energy of oscillation. Before writting the equation of energy, let us find the mass of the dead weight A. From the condition of the problem, at the equilibrium position: τC2 = 0
om
0.1 m
.c
0.1 m
0.2m
Differentiating with respect to time, we get R2 M m sin α + m + 2 ⋅ θθ + mg R (α + θ) 2 2 θ – mg R sin α = 0
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5.40
or
M R m sin α + m + θ + mg 2 [sin α cos θ + cos α sin θ] – mg sin α = 0
or
m in rio
iv
)
un
) Solution
e/
(i)
un e/
s:
X1
ht
tp
//t
ht
.m
tp
s:
//t
iv
.m
tp
ht
s
er si
ta
.ja
w
s: //w w
9. A rod AB of mass M is attached as shown below to a spring of constant k. A small block of mass m is placed on the rod at its free end A. (i) If end A is moved down through a small distance d and released, determine the period of oscillation. (ii) Determine the largest allowable value of d if the block m is to remain at all times in contact with the rod.
si
2mg cos α [M + 2m(1 + sin α)]R
er
ω=
rio
2mg cos α θ [M + 2m(1 + sin α)]R
s_
an
ar
θ=-
m
∴
fo
M R m sin α + m + θ + mg cos α θ = 0 2
ta
a. co
[for small θ, cos θ ≃ 1 and sin θ ≃ 0]
or
10. In the given figure, the spring has a force constant 5000 N/m. The pulley is light and smooth. The spring and the string are light, the suspended block is of mass 1 kg. If the block is slightly displaced vertically down from its equilibrium position and released, find the period of its vertical oscillations.
M R m sin α + m + θ + mg 2 [sin α ⋅ cos θ + cos α ⋅ sin θ] – mg sin α = 0
or
M R 2 m sin α + m + θ + mg R 2 sin (α + θ) – mg R sin α = 0
Simple Harmonic Motion
At equilibrium X2 T
T B
or
)
) Solution Try yourself.
l
or
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b
m
A
Initially, there is no elongation of spring. When it goes in equilibrium position 2 T = Kx1 T = mg 2 mg x1 = k
k
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...(i)
5.41
and x2 = 2x1 4 mg = k mg x2 = k/4
Simple Harmonic Motion
O
…(i)
(ii) Similarly, 2T = mg T = k x1
l
m k/4
Hence, time period, T = 2π
)
co m
) Solution
in
si
s
rio
ta
si
er iv
e/
1 ml 2 2 l 1 (θ) + mg (1 - cos θ) + k (l θ) 2 = 2 3 2 2 constant
un
//t
e/
s:
m
tp
//t .
Hence, time period, T = 2π
1 2 l 1 I(θ) + mg (1 - cos θ) + k (l θ) 2 = 2 2 2 constant
Differentiating with respect to time, we get 1 ml 3 l 1 g θθ + mg sin θθ + kl 2 2θθ = 0 2 3 2 2
m 4k
θ θ
11. Find the frequency of small oscillation of a thin uniform vertical rod of mass m and length l hinged at the point O shown in the figure. The combined stiffness of the springs is equal to k. The mass of the springs in negligible.
s:
mg 4k
tp
x2 =
ht
T 2k
ht
=
or
.m
tp
x x2 = 1 2
or
ht
mg
or
un
iv
//w
s:
2T
w
er
w
At equilibrium
X2
1 2 1 Iω + mgh + kx 2 = constant 2 2
ta
.ja
m
T
rio
ar
s_
an
X1
fo
Let us locate the rod at the position when it makes angle θ from the vertical and in this problem both, gravity and spring forces are restoring conservative force. From the energy conservation
a.
T
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Simple Harmonic Motion
Thus, for very small θ
)
) Solution The electro-magnetic field induced in the loop
3g kl θ = - 1 + θ 2l mg
ε = l (v × B) = v . Bl
...(i)
5.42
L
12. A loop is formed by two parallel conductors connected by a solenoid with inductance L and a conducting rod of mass m which can freely (without friction) slide over a conductor. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction B. The distance between the conductors is equal to l. At the moment t = 0 the rod is imparted an initial velocity v0 direction to the right. Find the law of motion x(t) if the electric resistance of the loop is negligible.
di - iR = ε - vBl dt
Applying Kirchhoff’s law
3g kl 1 + 2l mg
ω=
…(ii)
L
…(iv)
rio s_ in
fo
ar an
ta
-
…(v)
rio
si ta
d 2 v B2 l 2 + v=0 dt 2 mL
iv er
v = A sin ω t
un
∴
v (t) = v0 sin ω t
e/ where,
ω=
Hence,
x(t ) =
Bl ml v0 cos ωt ω
13. A body of mass m fell from a height h onto the pan of a spring balance. The masses of the pan and the spring are negligible, the stiffness of the later is k. Having struck to the pan, the body starts performing harmonic oscillations in the vertical direction. Find the amplitude and the energy of these oscillations.
ht
.m //t s: tp
Lm d 2 v = vBl Bl dt 2
s
er si iv
e/
un
or
//t l
…(iii)
From equations (iii) and (v), we get
s: tp B
dv dt
co m
a.
i Bl = - m
di m d 2v =dt Bl dt 2
or
ht L
di = vBl dt
Ampere’s force oppose the velocity
+z
.m
l
B
R=0
But
∴
m
.ja
w w //w
Vo
s: tp
L
ht
Hence,
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Simple Harmonic Motion
5.43
because at the extreme position kinetic energy is zero. Although the weight of body m is conservative force, it is not restoring in this problem, hence Uextreme is only concerned with the spring force. Therefore
m
h
)
m
) Solution
14. A body A of mass m1 = 1.00 kg and a body B of mass m2 = 4.10 kg are interconnected by a spring as shown in the figure. The body A perform free vertical hurmonic motion with amplitude a = 1.6 cm and frequency ω = 25 s–1. Neglecting the mass of springs, find the maximum and minimum values of force that this system exerts on the bearing surface.
1 m2 g 2 E = U extreme = k a 2 = mgh + 2 2k
ta si er iv
tp
mg k
ht
∆x =
e/ un
.m
s:
While the body A is at its upper extreme position, the spring is obviously elongated by the mg amount a - 1 k If we indicate y-axis in vertically downward direction, then by Newton’s second law of motion mg m1 g + k a - 1 = m1 ω 2 a k
The mechanical energy of oscillation which is conserved equals, E = Uextreme’
or
m g k a - 1 = m1 (ω 2 a - g ) k
mg 2 hk 1+ k mg
a=
2
mg 2 mgh a = x - ∆x = + k k
Hence, amplitude,
) Solution
//t
ht
∴
rio
iv un
//t
s:
At the equilibrium position, there is no force. Thus mg = k ∆x
B
)
2
mg mg 2 mgh + + k k k
tp
x=
.m e/
ht
Neglecting negative sign, we get
s
er si
w
//w
s:
2mg ± (2mg ) 2 + 8 mg hk 2k
tp
x=
o
ta rio
ar m
kx – 2 mg x – 2mgh = 0
w
or
A
.ja
1 -mg (h + x) + kx 2 = 0 2 2
s_
an
in f
a.
co
From energy conservation, for the maximum contraction of the spring, the total energy is conserved. Loss in potential energy is gain in the spring energy. Thus
…(i)
(Because at any extreme position the magnitude of acceleration of an oscillatory body equals ω2 a and is restoring in nature.)
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5.44
Simple Harmonic Motion
passing through the middle of bar. Find the period of small oscillations of the sleeve. At what value of ω will there be no oscillations of the sleeve?
If N be the normal force exterted by the floor on the body B, while the body A is at its upper extreme position, then from Newton’s second law for the body B A
B
M
mg N + k a - 1 = m2 g k mg N = m2 g - k a - 1 k
or
From equation (i),
)
om
) Solution
ita rio s x
.m From Fx = maximum for the sleeve in the frame of rod – kx + mω2x = m x k x = - - ω2 x or …(i) m 2π = 0.7sec ∴ time period, T = k 2 -ω m It is obvious from equation (i) that the sleeve will not perform small oscillations if
ω≥
k = 10rad/sec. m
15. In the arrangement shown in the figure, the sleeve M of mass m = 0.20 kg is fixed between two identical springs whose combined stiffness is equal to k = 20 N/m. The sleeve can slide without friction over a horizontal bar AB. The arrangement rotates with a constant angular velocity ω = 4.4 rad/s about a vertical axis
B
//t ht tp s:
N′ = m1 (g + ω2a) + m2g N′ = (m1 + m2) g + m1 ω2a
From Newton’s third law, the magnitude of the forces are N′ and N respectively.
M
Hence,
mω x 2
kx
rs
t.m
://
tp s
ht
From equation (i),
A
ve
…(iii)
In this, case if N′ be the normal force exerted by the floor on the body B, then from Newton’s second law for body B, we get
mg N′ = k a + 1 + m2 g k
fo
ita
e/ un iv
mg k a + 1 = m1 ( g + ω 2 a ) k
ht tp s:
or
//w
w
er s
w
mg m1 g - k a + 1 = m1 (-ω 2 a ) k
e/ un i
.ja
Fy = may for the body A at this state
Obviously, the sleeve performs small oscillations in the frame of rotating rod. In the rod’s frame, let us depict the force acting on the sleeve along the length of the rod while the sleeve is at a small distance x towards right from its equilibrium position. The free body diagram of block does not contain Coriolis force because it is perpendicular to the rod.
rio s_ in
m
ar an a. c
When the body A is at its lower extreme position, the spring is compressed by the distance m1 g . a + From Newton’s second law in k projection from, i.e.,
N = m2g – m1 (ω2a – g) = (m1 + m2)g – m1 ω2a
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Simple Harmonic Motion
5.45
)
iv
) Explanation The motion will SHM only if
un
t.m e/
k ( x - a)2 , 2
dU = k ( x - a) dx
F = –k (x – 1)
This is the condition for SHM about point x=+a Other function do not satisfy this condition.
5. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downwards push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is
2. Which of the following functions represent S.H.M.? (a) t + sin ωt (b) sin2 ωt (c) sin ωt + 2cos ωt (d) sin ωt + cos 2 ωt d2y acceleration or 2 ∝ - y . dt This is true for (c)
⇒
s: //
F1 = mat = mg sin θ
ht tp
mv l
2
dU dx
er si ta rio s
)
un iv e/ .m
//t
ht tp s:
T – mg cos θ =
ita er s
w //w s:
)
For U( x) =
mg cosθ
) Explanation Considering the F.B.D. of the
rio s_ in fo
na .c om
ar a m .ja
(d) U(x) = constant
) Explanation F = -
tp
ht
mg sin θ
k (x - a2 ) 2 (b) U(x) = k1x + k2x2 + k3x3 (a) U( x) =
(c) U(x) = Ae–bx
w
T
and
4. The potential energy U(x) of a particle executing SHM is given by
θ
bob
3. The time-period of a simple pendulum of small amplitude (a) is directly proportional to the square root of its length (b) is inversely proportional to the square root of acceleration due to gravity (c) is independent of the amplitude (d) all of these Ans. (c)
1. A simple pendulum of length l and mass (bob) m is oscillating in a plane about a vertical line between angular limits –φ and + φ. For an angular displacement θ (where | θ | < φ), the tension in the string and the velocity of the bob are T and v respectively. Then (a) T cos θ – mg mv 2 (b) T – mg cos θ = l (c) magnitude of the tangential acceleration of the bob is |aT| = g sin θ (d) T = mg cos θ
t
SOLVED OBJECTIVE YPE QUESTIONS
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5.46
a height 3 Re from the earth’s surface, where Re is earth’s radius. The period of the pendulum will be
1/2
1 k - Aρg 2π M
1/2
1 k + Aρg (b) 2π M
8. A simple harmonic motion has an amplitude A and time period T. The time required by it to travel from x = A to x = A/2 is T T (a) (b) 4 6 T T (d) (c) 4 3
1/2
in
ta
s
rio
ni
π 2π sin = sin t 6 T T t= ⇒ 12 Hence, time required to travel from x = A to A x= will be 2 T T T - = 4 12 6
⇒
9. A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period
Phase difference
7. A second’s pendulum is placed in a space laboratory orbiting around the earth at
A 2π = A sin t 2 T
tp ht
s:
//t
ht
.m
e/ u
t.m e/
tp
s: //
ht
)
π 2π y1 = a sin + Q and y2 = b sin + Q 2 3 2π π π = + Q - + Q = 3 2 6
fo
un iv er
tp s:
π 2π 2 t + Q and y2 = b sin 3 t + Q . The phase difference between these after 1 sec is π (a) π (b) 2 π π (c) (d) 4 6
∴
T and time required to travel A to x = 0 is 4 A from x = to x = 0 is equal to the time 2 A required to travel from x = 0 to x = which 2 is equal to 2π y = A sin t T
si
w .ja
w
//w
6. Two SHM are given by y1 = a sin
) Explanation t = 1 sec
rio s_
) Explanation Time required to travel from x =
m
L Fr = Mg - k ( x + y ) + A + y ρg 2 = -(k + Aρg ) y
)
ar an a
released, then
om
.c
L Mg = kx + A ρg 2 Let it be given a small downward push y and
)
) Explanation In equilibrium,
k + Aρg M
)
and its time period will be infinite.
1 k + Aρg (d) 2π Aρg
ω2 =
) Explanation The pendulum will not oscillate
1/2
1 k + AρgL2 (c) 2π M
∴
(b) 2 3s (d) infinite
(a) zero (c) 4 s
(a)
ve rs ita
Simple Harmonic Motion
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(c) 2π
L g
L g sin α
(d) 2π
L g tan α
)
) Explanation mg = kx
(b) 2π
L g cos α
(a) 2π
1 (c) T = π mg / x(M + m) 2 (M + m) (d) T = 2π mgx
x(M + m) mg
10. The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined pane of inclination α, is given by
(b) T = 2π
5.47
mg x(M + m)
(a) T = 2π
Simple Harmonic Motion
mg M +m and T = 2π x k
) Explanation Effective gravity inside the vehicle )
k=
is g cos α.
rio s_ in fo
na .c om
⇒
er si ta rio s
ita
iv
(d)
1 s 4
(b)
1 s 3 1 s 6
6. A block of mass m is tied with a spring of length l0 fixed at A and placed on a smooth horizontal surface as shown. The block is compressed and a smooth vertical wall is placed at l1. Now if the block is further compressed by length a, the period of oscillation of block will be (assume the collision between the block and the wall to be elastic)
1 s 2
(c)
ht tp
s: //
(a)
un
t.m e/
ht tp s:
3. A small bob is tied to a light string and is suspended in a lift. The bob is set into oscillations. The time period of oscillations is maximum, when the lift is (a) at rest (b) moving upward with acceleration (c) moving downward with acceleration (d) time period will be same for all conditions
5. A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?
un iv
e/
.m
//t
(d) none of these
4. For a particle undergoing SHM, the velocity is plotted against displacement. The curve will be (a) a straight line (b) a parabola (c) a circle (d) an ellipse
er s
w
w
//w
s:
tp
ht
2m B
2. As the acceleration of a vibrating particle executing simple harmonic motion increases, its time period (a) increases (b) decreases (c) remains unchanged (d) first increases, then decreases
(c) π
.ja
1. If potential energy of body of mass m is given by u = A + Bx2, then find the time period of each cycle of motion will be B 2B (a) π (b) π m m
m
ar a
UNSOLVED OBJECTIVE TYPE QUESTIONS
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5.48
fo
s
rio
ta
un iv er
//t
come to rest (c) undergo vertical SHM with a time
e/ .m //t tp s:
ht
m2 g and k
m1 k (d) undergo vertical SHM with a time period of 2π
period of 2π
(m1 + m2 ) k
A 3 k 2 M
(d) A
k M
er s
iv
un .m e/
(a) remain undisturbed
M k
8. The displacement y of a particle executing certain periodic motion is given by y = 4 1 cos2 t sin (1000t). This expression 2 may be considered to be the superposition
(c) A 3
(b)
k M
(a) A
10. Two blocks of masses m1 and m2 are attached to the lower end of a light vertical spring of force constant k. The upper end of the spring is fixed. When the system is in equilibrium, the lower block (m2) drops off. The other block (m1) will (b) move up through a distance
s: tp ht 0–60°
(d) all of these
ita rio s_
ar
m
.ja
w
w
//w
tp s: ht M
an a. co m
7. An insect of negligible mass is siting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude A in front of a plane mirror placed as shown. The maximum speed of insect relative to its image will be
in
l -l k sin -1 0 1 m l0 - l1 + a
si
(d) 2π
(a) speed of the particle is maximum at x = 0 π (b) amplitude of oscillations is 2a (c) time period of small oscillation is, m T = 2π 2 a U0
l -l m k sin -1 o 1 k m l0 - l1 + a
(c) 2π
Then (for the small oscillations)
m k l -l sin -1 o 1 k m a
(b) 2π
9. A particle of mass m is moving in a potential well, for which the potential energy is given by U(x) = U0(1 – cos ax), where U0 and a are constants.
(lo - l1 ) m k sin -1 k m (lo - l1 + a )
(a) 2π
A
of n independent harmonic motions. Then n is equal to (a) 2 (b) 3 (c) 4 (d) 5
l1
Simple Harmonic Motion
M
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11. A coin is placed on a horizontal platform, which undergoes horizontal simple hormonic motion about a mean position O. The coin does not slip on the platform. If the force of friction acting on the coin is F, then (a) F is always directed towards O (b) F is directed towards O when the coin is moving away from O, and away from O when the coin moves towards O (c) F = 0 when the coin and platform come to rest momentarily at the extreme position of the harmonic motion (d) F is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion
m
k2 k1
o
nf
s_ i
io
er si
17. A simple pendulum with a solid metal bob has period T. The metal bob is now immersed in a liquid of density one-tenth that of the bob. The liquid is non-viscous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be 9 (a) T (b) T 10
(d)
10 T 9
9 T 10
ht tp
s: //t .m
e/ un
iv
ta
rio
s
ta r
er si
iv
e/ un
t.m
s: //
ht tp
(d)
16. A simple pendulum with angular frequency ω oscillates simple harmonically. The tension in the string at lowest point is T. The total acceleration of the bob at its lowest position is T (a) -g (b) zero m g T (c) g (d) T + m m
(c)
18. A block of mass m compresses a spring of stiffness k through a distance as shown 2 in the figure. If the block is not connected with the spring and the impact of the block with the vertical wall is elastic, the period of motion of the block is
(b) 2 (d) 2
14. Two S.H.M.’s are represented by the π equations, y1 = 10 sin 3 πt + and 4 y2 = 5 (sin 3πt + 3 cos 3πt). Their amplitudes are in the ratio of
k2 k1
(c)
a. co
an
ar
.ja m
w
w
w
s: //
ht tp
13. A simple pendulum has a time period T1 when on the earth’s surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is (a) 1 (c) 4
15. Two bodies M and N of equal masses are suspended from two separate massless springs of spring-constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, then ratio of the amplitude of vibration of M to that of N is k k1 (b) (a) 1 k2 k2
12. In the previous question, the angular frequency of the simple harmonic motion is W. The coefficient of friction between the coin and the platform is µ. The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time (a) at the extreme positions of oscillations µg (b) for an amplitude of W2 (c) at the mean position g (d) for an amplitude of µ W2
5.49
(b) 1 : 2 (d) 1 : 4
(a) 2 : 1 (c) 1 : 1
Simple Harmonic Motion
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5.50
(b) the time period of oscillation will decrease (c) the time period of oscillation will remain unaffected (d) the mean position of simple hormonic motion will shift to the right
Simple Harmonic Motion
k
(d) none of these
7Q
m k
19. A spring-block system undergoes simple harmonic motion on a smooth horizontal surface. The block is now given some positive charge and uniform horizontal electric field to the right is switched on. As a result (a) the time period of oscillation will increase
E
s ita rio
7. (c) 15. (d)
8. (b) 16. (a)
ve
rs
6. (c) 14. (c)
un i e/
s:
.m
tp
//t
tp s:
p
ht
RACTICE EXERCISE 1
1. A particle executes harmonic oscillations along the x axis about the equilibrium position x = 0 with frequency ω = 4/s. At a certain instant, the particle has a position coordinate x0 = 25 cm and velocity v0 = 100 cm/s. Find the position x and the velocity v of the particle after a time t = 2.4 s. [IIT, 1978] π Ans. x = 35.36sin ωt + 4 ; v = 81.9cm/s
of elasticity for the wire, obtain the frequency of oscillation for simple harmonic motion along the vertical line. [MNR, 1990] 1 YA Ans. 2 π mL
2. A point mass m is suspended at the end of a massless wire of length L and crosssection A. If Y be the Young’s modulus
3. A certain spring, when stretched by 0.25 m, stores 5 joule of energy. It is kept vertical by fastening its lower end to a rigid support. Find the mass of a block which, if attached to the free end of this spring, will execute 5 oscillations per second. [Ans. 0.162 kg]
ht
e/ un i
5. (d) 13. (d)
.m
4. (d) 12. (a, c) 20. (b)
//t
3. (c) 11. (a, d) 19. (c, d)
ita rio
ve rs
w //w
s:
2. (c) 10. (c) 18. (c)
ht tp
1. (c) 9. (d) 17. (c)
a.
an
ar
w .ja
m
ANSWERS
s_ in fo
m
20. The time period of a spring-mass system is T in air. When the mass is partially immersed in water, the time period of oscillation is (a) T (b) < T (c) > T (d) ≤ T
co
(c) (1 + π)
m k
(b) (π + 4)
m k
l/2
l/2
l
(a) 2π
m
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o
nf
s_ i
rio
s
si ta
rio
α l cos.-1 - Ans. 2 g β
12. Consider the situation shown in the figure. The block on the left moves with a velocity v towards block on the right which is placed in equilibrium. Assume all the collisions to be elastic and all the surfaces to be smooth. Neglecting the dimensions of the blocks, prove that the motion of the blocks are periodic and find their periods.
ht tp
s: //
t.m
e/ un
iv
er
11. Two linear simple harmonic motions of equal amplitudes and frequencies ω and 2ω are impressed on a particle along the axes of X and Y respectively. If the initial π phase difference between them is , 2 find the resultant path followed by the particle. [Roorkee, 1985] [Ans. Parabola]
s: //t .m
ht tp
Mg 2 γP + A 1 A Ans. n = 2π MV0
si ta
er
iv
un
e/
s:
ht tp
10. A ball is hung by a thread of length l from a point P of the wall that makes an angle α with the vertical. The thread with the ball is then deviated through a small angle β (β > α) and set free. Assuming the collision of the ball with the wall to be perfectly elastic, find the period of such a pendulum.
a. co m
ar
m
.ja
//w
w w
an
5. A mass M attached to a spring oscillates with a period of 2 sec. If M is increased by 2 kg, the period increases by 1 sec. Find the initial mass M assuming that Hooke’s law is obeyed. [IIT, 1979] [Ans. 1.6 kg]
7. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of the mass M. The piston and cylinder have equal cross sectional area A. Atmospheric pressure is P and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from the equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT, 1981]
Determine the maximum velocity of the point. [Ans. 2.73 aω]
6. The potential energy of a particle of mass M situated in a unidimensional potential field varies as U(x) = U0 (1–cos ax) where U0 and a are constants. Find the time period of small oscillations of the particle about the mean position. m Ans. T = 2π 2 a U0
5.51
8. Find the time period of small oscillations of a ball suspended by a thread of length l = 0.20 m if it is placed in a liquid whose density σ is n = 3.0 times less than the density ρ of the ball. The resistance of the liquid is to be neglected. [Ans. T = 1.1 sec.] 9. A block is resting on a piston which is moving vertically with simple harmonic motion of period 1.0 sec. At what amplitude of motion will the block and piston separate? What is the maximum velocity of piston at this amplitude? [Roorkee, 1985] [Ans. 0.25 m; 1.57 m/s]
4. A point executes two harmonic oscillations simultaneously along the same direction: x1 = a cos ω t and x2 = a cos2 ω t.
Simple Harmonic Motion
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5.52
Simple Harmonic Motion
er si
ht
tp
s:
s
rio
iv
M A
B
15. Two particles execute simple harmonic motion of same amplitude and frequency along the same straight-line. They pass one another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them? [MNR, 1986] [Ans. 120º]
//t
.m
e/
un
//t
s:
ht tp
er s
e/
.m
ht
18. A ball of mass m and radius r rolls on a rough concave surface of large radius R. It performs small oscillations about the mean position. Assuming the slipping to be absent, find its time period. [MNR, 1990] 7(R - r ) Ans. T = 2π 5g 19. In the given figure, a sleeve of mass m is fixed between two identical springs, each having a force constant equal to k. The sleeve is free to slide without friction over a horizontal bar AB, rotating with a constant angular velocity ω, about a vertical axis passing through the middle of the bar. Find the time period of small oscillations of the sleeve. For what value(s) of ω, the sleeve will not oscillate at all?
ita
un
iv
w
//w
14. A uniform plate of mass M is placed horizontally and symmetrically on two wheels, separated by a distance L and rotating in opposite directions as shown in the figure. The coefficient of friction between each wheel and the plate is µ. If the plate is slightly displaced along its length and released, show that it executes simple harmonic motion. Find its time period.
tp s:
w
(kl = 1.8 N/m, k2 = 3.2 N/m, m = 200 gm) [IIT, 1985] [Ans. 2.83 sec]
L Ans. T = 2π 2µg
ta rio
.ja
m
s_
ar an
in
a.
fo
co
m
16. A horizontal spring-block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass k m is put on it and the two move together. ν m m Find the new amplitude and frequency of L vibration. [Roorkee, 1988] Ans. Both the amplitude and frequency change by a factor 2L m +π Ans. v k M Ans. Both the amplitude and frequency change by a factor 13. Two light springs of force constants k1 M+m and k2 and a block of mass m are in one 17. A simple pendulum is made by attaching line AB on a smooth horizontal table a 1 kg bob to a 5 m copper wire of diamsuch that one end of each spring is fixed eter 0.08 cm and it has certain period of on rigid supports and the other end is oscillation. Next, a 10 kg bob is substifree as shown in the figure. The distance tuted for 1 kg bob. Calculate the change CD between the free ends of the springs in period if any. is 60 cm. If the block moves along AB (Young’s modulus of copper = 12.4 × 1010 with a velocity 120 cm/sec in between the N/m2) [Roorkee, 1988] springs, calculate the period of oscillation [Ans. 0.004 s] of the block.
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rio s_ in fo
iv
er si ta rio s
un iv
e/
.m
ht tp
s: //
22. One end of each of the two identical springs, each of force constant 0.5 N/m, are attached to the opposite sides of a wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched and are collinear in a horizontal plane. To the wooden block is fixed a pointer which touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s, find the equation of path traced by the pointer on the paper
t.m e/
un
//t
ht tp s:
24. A uniform cylinder of length L and mass M having cross-sectional area A is suspended with its length vertical from a fixed point by a massless spring such that it is half submerged in a liquid of density σ at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with small amplitudes. If the force constant of the spring is k, calculate the frequency of oscillation of the cylinder. [IIT, 1990] 1 k + Aσg Ans. n = 2 π M
er s
w
//w
s:
tp
ht
M Mm . x0 ,(b)T = 2π Ans. (a) m k M + (M + m)
ita
.ja
w
m
ar a
21. A circular ring hanging from a nail in a wall undergoes oscillations of amplitude θ = 2º and period T = 2 s. Find (a) radius of the ring, (b) speed of the particle opposite to the point of suspension as it passes through the position of rest, (c) acceleration of this particle when it crosses the mean position, and (d) acceleration of this particle when it is at one of the extreme positions. Given, g = π2m/s2. [Ans. (a) 50 cm, (b) 10.996 cm/s, (c) 1.19 cm/s2, (d) 34.4 cm/s2]
na .c om
2 2 Ans. T1 = 3 s; T2 = 3 s
23. In the given figure, the mass of the trolley is M, that of the block is m, and the force constant of the spring is k. Initially, the entire system is at rest. The spring is now stretched through x and the system is released. Find: (a) amplitude of the resulting simple harmonic motion of the block and the trolley, as seen by an observer from the road, and (b) periods of the two simple harmonic motions.
25. Find the frequency of small oscillations of a thin vertical rod of mass m and length l hinged at the point O, as shown in the figure. The equivalent force constant of the springs is equal to k. Assume the springs to be massless and the rod to be uniform.
Find out the ratio of their amplitudes. What are the time periods of the two motions? [IIT, 1986]
20. Two simple harmonic motions are represented by the following equations: π y1 = 10 sin (12t + 1) 4 y2 = 5 (sin 3πt + 3 cos 3πt)
5.53
and distance between two consecutive maxima of this path. [Roorkee, 1990] [Ans. x = 0.02 cos 10 (t – 10 z), ∆z = 6.28 cm]
1 k ;ω ≥ Ans. T = 2π k m - ω2 m
Simple Harmonic Motion
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5.54
Simple Harmonic Motion
in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 metre. Each spring has a natural length 0.06π metre and spring constant 0.1 N/m. 26. A point particle of mass 0.1 kg is executInitially, both the balls are displaced by ing S.H.M. of amplitude 0.1 m. When the an angle θ = π/6 radian with respect to particle passes through the mean posithe diameter PQ of the circle (as shown tion, its kinetic energy is 8 × 10–3 joule. in figure) and released from rest. Obtain the equation of motion of this (a) Calculate the frequency of oscillation particle if the initial phase of oscillation of ball B. is 45º. [Roorkee, 1999] (b) Find the speed of ball A when A and B are at the two ends of the diameter PQ. A B (c) What is the total energy of the system? π/6 [IIT, 1993] P π/6 Q 1 Ans. (a) n = π cycle/sec,(b) v = 0.0628m/s, (c), (c)3.947 × 1 1 π Ans. (a) n = cycle/sec,(b) v = 0.0628m/s, (c), (c)3.947 × 10-4 J Ans. 0.1sin 4 t + π 4 29. Find the time period of the system shown in the figure. The radius of the pulley is 27. A uniform rod AB of mass m = 1.5 kg is R, its moment of inertia about the axis of suspended by two identical threads of rotation is I, mass of the body is m and equal length l = 0.9 m, as shown in the figthe spring constant is k. ure. When turned through a small angle about a vertical axis passing through its middle point C, the threads are deviated by an angle θ = 5º The rod is then released to allow it to perform small oscillak∆l TO tions. Find the time period of oscillation of the rod. m O [Ans. 1.1 s]
rio
ita
.ja
s
rio
ita
rs
ve
ht
tp
s:
28. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed
//t
ht
.m
tp
e/ u
s:
ni
//t
ht
.m
tp
e/
s:
//w w
un iv er s
w
m
ar
s_
an
in
a.
fo
co
m
1 3k 3 g + Ans. n = 2 π m l
x
mg
1 m+ 2 R Ans. T = 2π k
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Simple Harmonic Motion
5.55
QUESTION BANK
1. A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod’s other end is supported by a spring of force constant k. The rod is rotated (in vertical plane) by a small angle θ from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is
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2. Suppose that a grandfather clock (a simple pendulum) is running slowly. That is, the time it takes to complete each cycle is longer than it should be. Should one shorten or lengthen the pendulum to make the clock keep the correct time? Why?
3. The amplitude of a simple pendulum decreases with time. But there is no decrease in the amplitude of a pendulum clock. Why?
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2. A particle of mass m is attached to three identical springs A, B and C each of force constant k as shown in figure. If the particle of mass m is pushed slightly against the spring A and released, then the time period of oscillation is:
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7. A particle is oscillating in simple harmonic motion. The time required for the particle to travel through one complete cycle is equal to the period of the motion, no matter what the amplitude is. But how can this be, since larger amplitude mean that the particles travel farther? Explain.
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8. A pendulum clock is taken to moon. Will it gain or lose time?
9. Time period of a particle in SHM depends on the force constant k and mass m of the
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m 2k
(d) 2π
m 3k
m k
(b) 2π
(c) 2π
2m k
(a) 2π
m . A simple pendulum 2k executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? particle: T = 2π
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6. Distant explosion can break glass windows. How?
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5. What is the period of oscillation of a simple pendulum if its bob is made of ice?
(a)
4. A car travels over a road that contains a series of equally spaced bumps. Explain why a particularly jarring ride can result if the horizontal velocity of the car, the bump spacing, and the oscillation frequency of the car’s suspension system are properly “matched”.
1. Can pendulum clocks be used in artificial satellite? Explain.
ONLY ONE OPTION IS CORRECT
CONCEPTUAL QUESTIONS
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5.56
3. A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude? (a) s/2 (b) s/6 (c) s/4 (d) s/3
2πt π + (c) x(t ) = a sin 4 4
πt π (d) x(t ) = a cos + 3 2
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8. The uniform stick of mass m length L is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle θ, it executes SHM. The frequency of the motion is
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6. The following figure depict a circular motion. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P can be shown as
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1 3K 2π m
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(b)
1 3K 2π 2m
(d) None of these
(c)
πt π (b) x(t ) = a cos + 4 4
1 6K 2π m
(a)
.
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5. A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is T T 4 cos -1 (a) (b) 2π 20 5
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7. A block of mass ‘m’ is suspended from a spring and executes vertical SHM of time period T as shown in figure. The amplitude of the SHM is A spring is never in compressed state during the oscillation. The minimum force exerted by spring is never in compressed state during the oscillation. The minimum force exerted by spring on the block is
4. A system is shown in the figure. The time period for small oscillations of the two blocks will be
Simple Harmonic Motion
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9. On a smooth inclined surface a body of mass M is attached between two spring. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is
1/2
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(c)
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14. The friction coefficient between the two blocks of masses 1 kg and 4 kg shown in figure is µ and the horizontal plane surface is smooth. If the system is slightly displaced from the mean position and released, it will execute SHM. The maximum amplitude for which the upper block does not slip relative to the lower will be – (K is spring constant)
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11. A uniform cylinder of length L and mass M having cross-sectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is halfsubmerged in a liquid of density d at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the
(c)
(a)
13. Three simple harmonic motions in the same direction having the same amplitude A and same period are superposed. If each differs in phase from the next by 45º then (a) The resultant amplitude is (1 + 3) A (b) The phase of the resultant motion relative to the first is 90º (c) The energy associated with the resulting motion is (3 + 2 2) times the energy associated with any single motion. (d) The resulting motion is not simple harmonic.
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10. A simple pendulum suspended in a vehicle has a time period T when the vehicle travels along a frictionless road AB. What will be its time period when the vehicle travels due to its weight only along the frictionless path BC?
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12. A particle is moving on x-axis has potential energy U = 2 – 20x + 5x2 Joules along x-axis. The particle is released at x = –3. The maximum value of x will be [x is in meters and U is in joules] (a) 5 m (b) 3 m (c) 7 m (d) 8 m
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5.58
(b) 2π
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15. The spring block system as shown in figure is in equilibrium. The spring connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A is
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18. If the displacement (x) and velocity v of a particle executing simple harmonic motion are related through the expression 4v2 = 25 – x2 then its time period is (a) π (b) 2π (c) 4π (d) 6π
19. A large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizontal circular path. If ℓ is the length of the string from m to the top end of the tube and θ is angle between
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16. A long uniform rod of length L and mass m is free to rotate in a horizontal plane about a vertical axis through a centre O of the rod. A spring of length l0 with force constant K is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. If the rod is rotated by a small angle θ and, then released, the time period of the oscillation is
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17. A loop consists of two cords of length ℓ and 2ℓ, and their masses per unit length are their masses per unit length are µ and 2µ. It is placed in stable equilibrium over a smooth peg as shown in the figure. When slightly displaced, it executes SHM. The period of oscillation is
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Simple Harmonic Motion
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4 2h sin α g
(d) None of these
23. A condenser of capacity 20µF is first charged and then discharged through a 10 mH inductance. Neglecting the resistance of the coil, the frequency of the resulting vibrations will be (a) 356 cycles/sec (b) 133 cycles/sec (c) 233 cycles/sec (d) 1333 cycles/sec
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this part and vertical part of the string as shown in the figure, then time taken by m to complete one circle is equal to
Simple Harmonic Motion
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24. A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is
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25. A solid sphere of mass 1 kg and diameter 0.3 m is suspended from a wire. If the twisting couple per unit twist for the wire is 6 × 10–3 N-m/radian, then the time period of small oscillations will be (a) 0.7 sec (b) 7.7 sec (c) 77 sec (d) 777 sec
22. A particle of mass m slides back and forth between two inclined frictionless planes as shown. If h is initial height, then time period of motion is
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21. A block of 4 kg produces an extension of 0.16 metre in a spring. The block is replaced by a body of mass 0.50 kg. If the spring is stretched and then released the time period of motion will be (a) 0.283 sec (b) 0.0283 sec (c) 2.83 sec (d) 28.3 sec
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20. A loaded vertical spring executes simple harmonic oscillations with period of 4 s. The difference between the kinetic energy and potential energy of this system oscillates with a period of (a) 8 s (b) 1 s (c) 2 s (d) 4 s
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5.60
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frequency given by
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(b) System performs SHM with angular
frequency given by
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32. A massless rod is pivoted at point O. A string carrying a mass m at one end is attached to the point A on the rod. The period of small vertical oscillation of mass m around its equilibrium position is
(c) Tension in string will be 0 when the system is released. (d) Maximum displacement of m1 will be m2 g k
29. A 1 kg body when suspended from the lower end of a light spring produces a vertical extension of 9.8 cm in it. The time period of the oscillations of the spring will be
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28. Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then (a) Time period of the simple pendulum will be more than that of the springmass system. (b) Time period of the simple pendulum will be equal is that is of the springmass system. (c) Time period of the simple pendulum will be less than of the spring-mass system. (d) Nothing can be said definitely without observation.
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31. m1 and m2 are connected with a light inextensible string with m1 lying on smooth table and m2 hanging as shown in figure. m1 is also connected to a light spring which is initially unstretched and the system is released from rest.
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2π cycles/sec 100
30. Two SHMs are respectively represented by y = a sin(ωt – kx) and y = b cos(ωt – kx). The phase difference between the two is (a) π/2 (b) π/4 (c) π/6 (d) 3π/4
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27. Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of the system.
(b)
(b) 102 - 52 ω (d) 15ω
(c)
(a) 102 + 52 ω (c) 5ω
(a) 200π
26. A particle is subjected to two simple harmonic motions given by: y1 = 10 sin ωt, and y2 = 5 sin(ωt + π) The maximum speed of the particle is
Simple Harmonic Motion
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(c) Mg 1 +
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41. A body of mass 0.1 kg is attached to two springs of force constants 6 N/m and 4 N/m and supported by two rigid supports. If the body is displaced along the length of the springs, the frequency of vibrations will be (a) 5 vibrations/sec (b) 10 vibrations/sec (c) 5/π vibrations/sec (d) π/5 vibrations/sec
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40. The potential energy of a particle executing SHM changes from maximum to minimum in 5 s. Then the time period of SHM is (a) 5 s (b) 10 s (c) 15 s (d) 20 s
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37. A particle is executing simple harmonic motion along a straight line 8 cm long. While passing through mean position its velocity is 16 cm/s. Its time period will be
36. A 5 kg. weight is suspended from a spring. The spring stretches by 2 cm/kg. If the spring is stretched and released, its time period will be (g = 10 m/s2) (a) 0.628 sec (b) 6.28 sec (c) 62.8 sec (d) 0.0628 sec
2
39. A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period (a) 42 minutes (b) 1 day (c) 1 hour (d) 84.6 minutes
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35. The height of liquid column in a U tube is 0.3 metre. If the liquid in one of the limbs is depressed and then released, then the time period of liquid column will be (a) 1.1 sec (b) 19 sec (c) 0.11 sec (d) 2 sec
2 (d) Mg 1 + 2 L
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34. What will be the percentage change in the time period of a simple pendulum if its length is increased by 6%? (a) 3% (b) 9% (c) 6% (d) 1/9%
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33. The angular velocity and the amplitude of a simple pendulum are ω and A respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is U, then the ratio of T to U is (a)
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38. A simple pendulum with length L and mass M of the bob is vibrating with amplitude a. Then the maximum tension in the string is (a) Mg
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(a) 0.157 sec (c) 15.7 sec
5.61
42. The time period of oscillation of a block of mass m attached to a light spring of spring
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Simple Harmonic Motion
m in the absence of all k other forces. Suppose that a block of mass m is attached to a fixed spring on one side and rests (i.e., is not attached to) against an identical spring on the other side. It is now allowed to oscillate. The frequency of small oscillations is
constant k is 2π
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5.62
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48. A uniform rod of mass m and length ℓ is hinged at its mid point in such a way that it can rotate in the vertical plane about a horizontal axis passing through the hinge. One of its ends is attached to a spring of spring constant k which is unstretched when the rod is horizontal. If this end is now given a small displacement and released angular frequency of the resulting motion is
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47. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest, its time period is T. With what acceleration should lift be accelerated upwards in order to reduce its time period to T/2? (a) g (b) 2g (c) 3g (d) –3g
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44. A bar magnet is oscillating in the earth’s magnetic field with time period T. If its mass is increased four times then its time period will be (a) 4T (b) 2T (c) T (d) T/2
.
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(d)
(b)
2 K /m g /
49. The displacement of a particle is given at time t, by: x = A sin(–2ωt) + B sin2 ωt Then,
K /m
(c)
(a)
45. A man of mass 60 kg is standing on a platform which is oscillating up and down with frequency 2 oscillations/sec and amplitude 50 cm. A machine on the platform indicates weight of the man with respect to time, then
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43. A block of mass ‘m’ is connected to two springs of spring constants ‘2k’ and ‘3k’ respectively, as shown in the diagram, with both the springs being vertical and fixed at their other ends. Both springs are initially unstretched and the mass m is given an initial velocity ‘u’ in the downward direction. The block oscillates. The amplitude of oscillation of the block is given by
(2 + 2) k π m
(d)
46. The bob in a simple pendulum of length ℓ is released at t = 0 from the position of small angular displacement θ. Linear displacement of the bob at any time t from the mean position is given by
1 k (2 - 2) π m
(4 - 2 2) k π m
(b)
(c)
1 k π m
(a)
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the maximum reading of the machine will be (g = 10 m/s2) (a) 10 kg (b) 532.8 kg (c) 103 kg (d) 104 kg
Simple Harmonic Motion
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52. Two particles P and Q describe SHM with the same amplitude A and the same frequency f. The maximum distance separating the particles is observed to be A. The phase difference between the particles is
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51. The displacement x of a particle in motion is given as a function of time by x(x – 4) = 1 – 5 cos ωt then (a) particle execute SHM (b) particle execute oscillatory motion which is not SHM. (c) the motion of particle is not oscillatory (d) the particle is not acted upon by force when x = 4
3 mω2 A sin ωt (d) µ ( M + m) g + 2
55. A particle of mass m and charge –q moves along a diameter of uniformly charged sphere of radius R and carrying a total charge +Q. The frequency of the SHM of the particle, if the amplitude does not exceed, is
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54. A particle executes S.H.M. starting from its mean position at t = 0. If its velocity is 3bω, when it is at a distance b from the mean position, when ω = 2π/T, the time taken by the particle to move from b to the extreme position on the same side is (a) 5π/6ω (b) π/3ω (c) π/2ω (d) π/4ω
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50. A block of mass m, attached to a fixed position O on a smooth inclined wedge of mass M, oscillates with amplitude A and linear frequency f. The wedge is located on a rough horizontal surface. If the angle of the wedge is 60º, then the force of friction acting on the wedge is given by (coefficient of static friction = µ).
53. The displacements y of a particle executing a certain periodic motion is given by
B 4 (b) the motion of the particle is not SHM, but oscillatory with a time period of T = π/ω (c) the motion of the particle is oscillatory with a time period of T = π/2ω (d) the motion of the particle is aperiodic. A2 +
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5.63
(b) π/2 (d) 2π/3
2
(a) zero (c) π/3
(a) the motion of the particle is SHM with
Simple Harmonic Motion
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(c) the energy associated with the resulting motion is (3 + 2 2) times the energy associated with any single motion. (d) the resulting motion is not simple harmonic
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5. A block is placed on a horizontal plank. The plank is performing SHM along a vertical line with amplitude of 40 cm. The block just loses contact with the plank when the plank is momentarily at rest. Then (a) the period of its oscillation is 2π/5 sec. (b) the weight of the block on the plank is double its actual weight, when the plank is at one of the positions of momentary rest.
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(a) At time t2, its position lies in between – Xm and O. (b) The phase difference ∆φ between points A and B must be expressed as 90º <∆φ < 180º. (c) At time t1, its speed is decreasing (d) At time t1, it is going towards Xm.
(a) The displacement of the particle at t = 0
3. Three simple harmonic motions in the same direction having the same amplitude ‘a’ and same period are superposed. If each differs in phase from the next by 45º, then (a) the resultant amplitude is (1 + 2)a (b) the phase of the resultant motion relative to the first is 90º
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2. A 20 gm particle is subjected to two simple harmonic motions x1 = 2 sin 10 t, x2 = 4 sin π 10t + 3
of the particle will be 200 7 m/s 2 (d) Energy of the resultant motion will be 28 J
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4. A particle is executing SHM between points – Xm and Xm, as shown in figure-I. The velocity of V(t) of the particle is partially graphed and shown in figure-II. Two points A and B corresponding to time t1 and time t2 respectively are marked on the V(t) curve.
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1. A conservative force has the potential energy function U(x) as shown by the graph. A particle moving in one dimension under the influence of this force has kinetic energy 1.0 J when it is at position x1. Which of the following is/are correct statement(s) about the motion of the particle?
ption
ONE OR MORE THAN ONE O MAY BE CORRECT
Simple Harmonic Motion
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(c) the block weights 1.5 times its weight on the plank halfway down from the mean position. (d) the block weights its true weight on the plank, when velocity of the plank is maximum.
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9. Two blocks of masses 3 kg and 6 kg rest on a horizontal smooth surface. The 3 kg blocks is attached to a spring with a force constant k = 900 N/m which is compressed 2 m from beyond the equilibrium position. The 6 kg mass is at rest at 1 m from mean position. 3 kg mass strikes the 6 kg mass and the two stick together.
11. The amplitude of a particle executing SHM about O is 10 cm. Then (a) When the K.E. is 0.64 of its max. K.E. its displacement is 6 cm from O. (b) When the displacement is 5 cm from O its K.E. is 0.75 of its max. P.E. (c) Its total energy at any point is equal to its maximum K.E. (d) Its velocity is half the maximum velocity when its displacement is half the maximum displacement. 12. A horizontal plank has a rectangular block placed on it. The plank starts oscillating
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(a) We can find time period of oscillation. (b) We cannot find time period of oscillation. (c) We cannot find initial phase of oscillation. (d) We cannot find frequency of oscillation.
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Simple Harmonic Motion
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.
(b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1.
(c) Statement 1 is True, Statement 2 is False.
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ASSERTION AND REASON QUESTIONS Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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2. Statement 1: In simple harmonic motion the total energy is proportional to the square of the amplitude. Statement 2: Total energy is given by 1 2 kA 2
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1. Statement 1: The time period of a simple pendulum is independent of its amplitude of vibration. Statement 2: If the amplitude of a simple pendulum is doubled; its vmax is also doubled.
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(d) Statement 1 is False, Statement 2 is True.
vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then (a) the period of oscillation is (2π/5) (b) the block weighs double its weight, when the plank is at one of the positions of momentary rest. (c) the block weighs 0.5 times its weight on the plank halfway up (d) the block weighs 1.5 times its weight on the plank halfway down
6. Statement 1: The percentage change in time period is 2%, if the length of simple pendulum increases by 4%. Statement 2: Time period is directly proportional to length of pendulum.
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Column I The block’s mass in kg The block’s displacement at t = 0 in metres The block’s acceleration at t = 0.10 s in m/s2 The block’s maximum kinetic energy in joule
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8. Statement 1: Two simple harmonic π motions are given by y1 = 10sin 3πt + 4 y = 5(sin 3 π t + 3 cos3 π t ). and 2 These have amplitudes in the ratio 1 : 1 Statement 2: y2 represents two waves each of amplitude 5 and so total amplitude is 10, same as that of y1. 9. Statement 1: The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. Statement 2: The potential energy and kinetic energy of a particle in SHM, don’t vary linearly with position.
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7. Statement 1: When a body executing SHM passes through equilibrium it has zero acceleration. Statement 2: When a body executing SHM passes through equilibrium it has maximum energy.
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3. For a particle under going linear SHM about x = 0, choose the correct possible combination. Symbols have their usual meaning.
PASSAGE BASED QUESTIONS
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PASSAGE–2 You know very well that a block attached to an elastic spring performs SHM. Consider such a spring block system lying on a smooth horizontal table with a block of mass m = 2 kg attached at one end of a spring (k = 200 N/m) whose other end is fixed. The block is pulled so that the spring is extended by 0.05 m. If at this moment (t = 0), the block is projected with a speed of 1 m/s in the direction of increasing extension of the spring.
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2. In this projection φ is (a) π/2 (b) π (c) 3π/2 (d) 0
1. In this projection ω is (a) v/R (b) m2/R (c) R/v (d) None of these
4. A particle of mass 2 kg is moving on a straight line under the action of force F = (8 – 2x)N. The particle is released at rest from x = 6 m. For the subsequent motion match the following (All the values in the column II are in their S.I. units)
(b) velocity is (q) mean position negative (c) acceleration is (r) 0 < x < A negative (s) –A < x < 0 (d) v × a = 0
Column II (p) extreme position
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PASSAGE–1 An object of mass m is moving in uniform circular motion in xy-plane. The circle has radius Rand object is moving clockwise around the circle with speed v. The motion is projected onto the x-axis where it appears as simple harmonic motion according to x (t) = Rcos (cot ωt + φ). The motion starts from point of coordinates (0, R).
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(d) None of these
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10. The speed of block P just before P and Q are about to collide for the first time
PASSAGE–3 Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequency of 2 cm and 1 Hz respectively. Equilibrium positions of the particles A and B are at the co-ordinates (3 cm, 0) and (0, 4 cm) respectively. At t = 0, B is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin.
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(a) 0.268 s (c) 0.47 s
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4. A heavy ball of mass m is tied to a weightless thread of length ℓ. The friction of the ball against air is proportional to friction of the ball against air is proportional to its velocity relative air : Ffr = µv. A strong horizontal wind is blowing at a constant velocity v. Determine the period T of small oscillations, assuming that the oscillations of the ball attenuate in a time much longer than the period of oscillations.
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6. An A-shaped wire is made of bent wire ABC and a straight wire DE. Rings are provided which can slide over the bent wire as well as on straight wire. The arrangement is dipped into a soap solution and is then taken out and is placed in a vertical plane. The wire DE hangs in equilibrium at a distance L = 4.9 cm from the bent B as shown. When DE is displaced vertically parallel to itself it starts oscillating. Find its period of oscillation in milli-seconds. Surface tension of soap solution is 0.07 N/m. (g = 10 m/s2)
3. Determine the period of oscillations of mercury of mass m = 200 g poured into a bent tube whose right arm forms an angle θ = 30º with the vertical. The cross-sectional area of the tube S = 0.50 cm2. The viscosity of mercury is to be neglected. [
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1. A thin rod of length L and uniform crosssection is pivoted at its lowest point P inside a stationary homogeneous and nonviscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density d1 of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.
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PREVIOUS YEARS’ IIT-JEE QUESTIONS
12. A smooth horizontal disc rotates about the vertical axis O with a constant angular velocity ω0. A thin uniform rod AB of length l performs small oscillations about the vertical axis A fixed to the disc at a distance a from the axis of the disc. Find the period of these oscillations.
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Find the period of its oscillations if it rolls without slipping.
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11. The pulley of radius r shown in figure has a moment of inertia I about its axis and mass m. Find the time period of vertical oscillations of its centre of mass. The spring has spring constant ‘k and the string does not slip over pulley.
14. A uniform semicircular cylinder of radius r and weight W is displaced through a small angle θ from its equilibrium position as shown in the figure.
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The free end of a thread tightly wound on the pulley carries a deadweight A. At a certain angle α it counterbalances a point mass m fixed at the rim of the pulley. Find the frequency of small oscillations of the arrangement.
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same period are superposed. If each differ in phase from the next by 45º, then [1999] G
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2. A particle of mass m is executing oscillation about the origin on the x-axis. Its potential energy is U(x) = k |x| 3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is [1998]
6. The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination α, is given by [2000]
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9. A solid sphere of radius R is floating in a liquid of density ρ with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing simple harmonic motion. Find the frequency of these oscillations. [2004] 10. A mass m is undergoing SHM in the vertical direction about the mean position y0 with amplitude A and angular frequency ω. At a distance y from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m. Find the distance y, (measured from the mean position) such that the height h attained by the block is maximum (Aω2 > g). [2005]
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PASSAGE A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity V0 = V0iˆ. The coefficient of friction is µ. [2008]
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If EI, EII and EIII are the percentage errors in
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(a) (a) (c) (b) (d) (c) (a)
w
1. 9. 17. 25. 33. 41. 49.
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ONLY ONE OPTION IS CORRECT
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A
nswers
(d)
2k M
2k 3M
(c)
(b)
k M
(a)
14. The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is (a) –kx (b) –2kx (c) –2kx/3 (d) –4kx/3
Simple Harmonic Motion
MATCH THE COLUMN TYPE QUESTIONS 1. (a) → r (b) → p (c) → q (d) → s
2. (a) → r, s (b) → r, s (c) → s (d) → q
3. (a) → r, s (b) → r, q, s (c) → p, r (d) → p, q, r, s 4. (a) → r (b) → t (c) → q (d) → r (e) → q 5. (a) → r (b) → s (c) → p (d) → q
PASSAGE BASED QUESTIONS 1. (a) 9. (b)
2. (c) 10. (a)
3. (b) 11. (b)
4. (c) 12. (a)
5. (a)
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6. (a)
7. (b)
8. (d)
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Simple Harmonic Motion
5.75
HINTS AND SOLUTIONS
1 Maximum energy = mVm2 = 4 J 5 1 2 or kA = Emax 2 1 1 = × 200 × 0.04 = 4 J = Vm2 = 4 J 2 2
. Inside the satellite, the g effect of g is zero. So pendulum will not oscillate.
3. In the case of simple pendulum, amplitude decreases because energy is lost in overcoming air resistance, friction etc. In the case of pendulum clock, this energy lost is balanced by the energy supplied by the wound spring.
m
ita
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iv
s: //
ht tp
At t = 0.1 sec, acc. is maximum = –ω2A = –200 m/s2
2. (a) T = 2π /3 g = 1.1 sec; (b) E =
3. T = 0.8 sec 4. It can easily be seen that the ball attains the equilibrium position at an angle α of deflection of the thread from the vertical, which is determined from the condition µv tan α = . mg
m T 2k ⇒ m = 2 = 0.2 kg k 4π
1. 3/2 sec (mgℓα2)/2 = 0.05 J
T = 2π
un
t.m e/
ht tp s:
//t
SUBJECTIVE QUESTIONS
MATCH THE COLUMN TYPE QUESTIONS V 2π 1. Vm = Aω ⇒ A = m = × (0.2) = 0.20 m; ω 2π
er si ta rio s
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e/
.m
1 1 = kA2 = × 2 × 4 N-m = 4 j 2 2 As particle has started its motion from positive extreme ∴ Phase constant = π/2
8. The period of oscillation of a simple pendu-
. The value of g is less in g moon than that on the surface of earth. So T increases. The pendulum takes more time to complete one oscillation. Hence it will lose time.
Time to go from x = 2 m to x = 4 m (i.e., from extreme position to mean position) T π = = 4 2 Energy of S.H.M.
er s
w
w
//w
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tp
s:
6. Explosion produces sound waves of large amplitude in all directions. Sound waves are pressure waves and carry mechanical energy. These waves make glass windows to execute forced vibrations. Glass it brittle so it breaks.
lum is T = 2π
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5. In the case of an ice pendulum, the period depends on the radius of ice bob. As the ice melts radius decreases. So period decreased. But if the centre of mass of ice remains constant then there will be no change in the period.
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4. F = 8 – 2x = –2(x – 4) At equilibrium position, F = 0 ⇒ x = 4 m As particle is released at rest from x = 6 m, i.e., it is one of the extreme position. Hence, amplitude A = 2 m. Here, force constant k = 2 N/m ⇒ mω2 = 2 or ω = 1 2π ∴ Time period, T = = 2π ω
1. No. T = 2π
CONCEPTUAL QUESTIONS
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5.76
2
13. ω0 =
2mg cos α MR + 2mR (1 + sin α)
14. T = π
(9π = 16)r 2g
om
µv = g 1+ mg
(mg ) 2 + (µv) 2 g g′ = =g cos α mg
2 3ω02 α
12. T = 2π
force F = (mg ) 2 + (µv) 2 and a small drag force (Fig. 148). Consequently, the motion of the ball will be equivalent to a weakly attenuating motion of a simple pendulum with a free-fall acceleration g' given by
1 2 + m r 11. 2π 4k
During the oscillatory motion of the ball, it will experience the action of a constant large
Simple Harmonic Motion
e/
11. (a)
12. (a, b, c)
∆T ∆g ∆ = +2 13. (b) g = 4π2 2 T T g ∆ ∆T ⇒E= +2 , greater the value of t, T lesser the error. Hence, fractional error in the Ist observation is minimum.
14. (d) 2kx – f = ma ⇒ f.R = Iα a = Rα ⇒ ma = 4kx/3 (directed opposite to displacement) 15. (d) –(2kx) R = IPα
10. g/ω2
9. 150
(b) T = 2π ( R /g ) ≈ 84.6 min
s
1 3g 2π 2 R
un
9.
g + v 4 /R 2
10. (a) T = 2π ( R /2 g ) = 1 hour,
iv e
8. (a)
1 2
3g 3k + rad/sec 2 M
rs ita
7. (a)
.m
//t
tp ht
8. ω =
7. T = 2π
6. 440 ms.
m(k1 + 4k2 ) k1k2
5. T = 2π
s:
( g /) + 1 + (µv /mg ) - µ (4m )
2
rio
6. (a)
//t
s:
tp
ht
2
fo
5. (a, c)
e/
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tp
ht
2π
2
rio
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4. (d)
un iv
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s:
The period of small (but still damped) oscillations of the ball can be determined from the relation 2π T= g ′ / l - µ 2 (4m 2 ) =
3 g (d 2 - d1 ) 2d1 L
3. (b)
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2. (a)
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1. ω =
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a. c
PREVIOUS YEARS’ IIT-JEE QUESTIONS
α=-
4 KR 4k ( Rθ) = θ 2 3mR 3m
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