Engineering+mechanics+solutions+manual

Solutions Manual For

Engineering Mechanics Manoj Kumar Harbola IIT Kanpur

1

Chapter 1 1.1

Rotational speed of the earth is very small (about 7  10 5 radians per second). Its effect on particle motion over small distances is therefore negligible. This will not be true for intercontinental missiles.

1.2

The net force on the (belt+person) system is zero. This can be seen as follows. To pull the rope up, the person also pushes the ground and therefore the belt on which he is standing. This gives zero net force on the belt. For the person, the ground pushes him up on the feet but the belt pulls him down when he pulls it, giving a zero net force on him.

1.3

A vector between coordinates (x1, y1, z1) and (x2,y2,z2) is given by ( x 2  x1 )iˆ  ( y 2  y1 ) ˆj  ( z 2  z1 ) kˆ . Thus (i), (ii) and (iv) are equal.

1.4

The vectors are (i)

2iˆ  3 ˆj  5kˆ

4iˆ  3 ˆj  kˆ

(ii)

(iii)

2iˆ  9 ˆj  5kˆ

(iv)

 3iˆ  3 ˆj  2kˆ

1.5

4iˆ  6 ˆj  10 kˆ

(i) The resultant vectors are 6iˆ  6kˆ , (ii) The resultant vectors are

1.6

2iˆ  6 ˆj  4kˆ

 A  B

,

2iˆ  6 ˆj  4kˆ

  A B

  B A

and  iˆ  3kˆ

 A

and 7iˆ  3kˆ

 B

1.7 On each reflection, the sign of the vector component perpendicular to the reflecting mirror changes. z

y 1.8

 v 2

O

x

The fly is flying along the vector from (2.5, 2, 0) to (5, 4, 4). This vector is 2.5iˆ  2 ˆj  4kˆ .

The unit vector in this direction is

2.5iˆ  2 ˆj  4kˆ 2.5iˆ  2 ˆj  4kˆ  . 6.25  4  16 26.25

The velocity of the fly is therefore 0.5 

2.5iˆ  2 ˆj  4kˆ 26.25

 0.25iˆ  .20 ˆj  0.39kˆ .





1.9 After time t, the position vectors rA and rB of particles A and B, respectively, are   rA  l sin t iˆ  l cos t ˆj rB  l sin t iˆ  l cos t ˆj   Their velocities v A and v B are given by differentiating these vectors with respect to

time to get   v A  l cos t iˆ  l sin t ˆj v B  l cos t iˆ  l sin t ˆj    Velocity v AB of A with respect to B is obtained by subtracting v B from v A    v AB  v A  v B  2l cos t iˆ

1.10

For rotation about the z-axis by an angle  v x '  v x cos   v y sin 

v y '   v x sin   v y cos 

v z'  v z

It is given that   30  . Therefore v x' 



1 vx 3  vy 2



v y' 

3



1  vx  v y 3 2



v z'  v z

1.11



Component of a vector A along an axis is given by its projection on that axis. This is obtained by taking the dot product of the vector with the unit vector along that axis. Thus   Ax  A  iˆ  A cos  1

  Az  A  kˆ  A cos  3

  Ay  A  ˆj  A cos  2

Also 2 A  Ax2  A y2  Az2

Substituting the expression for Ax, Ay and Az completes the proof. 1.12





(i) Dot product of two vectors A and B is   A  B  Ax B x  Ay B y  Az B z

This gives the dot product of the first vector of problem 1.4 each of the other vectors to be 4, 2 and 25. 



(ii) Cross product between two vectors A and B is r r A  B  (Ay Bz  Az B y )iˆ  (Az Bx  Ax Bz )ˆj  (Ax B y  Ay Bx )kˆ 



Taking A to be the fourth vector and B to be the first, second and the third vector gives the cross products to be  9iˆ  11 ˆj  3kˆ  9iˆ  5 ˆ j  21kˆ ˆ  33iˆ  11 ˆ j  24 k

1.13

If the angle between two vectors is , the cosine of this angle is given by

cos  

  A B   A B

. Thus

Between (i) and (ii) cos   Between (i) and (iii) cos   Between (i) and (iv) cos   Between (ii) and (iii) cos   Between (ii) and (iv) cos  

4

 0.127    82.7 

38 26 2

38 110  25 38 22

 0.037    88.2   0.865    149.8 

40  0.748    41.6  26 110 5 26 22

4

 0.209    102.1

11

Between (iii) and (iv) cos  

1.14

 0.380    67.6 

38 22

Vector A can be written as A  A cos  1iˆ  cos  2 ˆj  cos  3 kˆ  





Similarly B  B cos 1iˆ  cos  2 ˆj  cos  3 kˆ  



Taking the dot product between the two vectors and using the formula

cos  

  A B   A B

,

where  is the angle between the two vectors, we get the answer.

1.15

Magnitude

  A B 

  AB 

Similarly

2   2 A  B  2 A  B cos  2   2 A  B  2 A  B cos 

Equating the two gives

  A  B  0 which

implies that the two vectors are perpendicular

to each other.

1.16

  B  C   B y C z  B z C y iˆ   B z C x  B x C z  ˆj   B x C y  B y C x  kˆ





   A  B  C  Ax  B y C z  B z C y   A y  B z C x  B x C z   Az  B x C y  B y C x 



1.17

From the expression for







      C  A  B and B  C  A    A  B  C it is clear that it is equal

This comes out to be equal to





Ax

Ay

Az

Bx

By

Bz

Cx

Cy

Cz

to the determinant

Interchange of two rows in a determinant changes the sign of the determinant. This implies

Bx

By

Bz

Bx

By

Bz

Ax

Ay

Az

Cx

Cy

C z   Ax

Ay

Az  B x

By

Bz

Ax

Ay

Az

Cy

Cz

Cy

Cz

Cx

thereby proving the equalities in problem 1.16. 5

Cx

1.18

  B  C   B y C z  B z C y iˆ   B z C x  B x C z  ˆj   B x C y  B y C x  kˆ

Therefore





   A  B  C   Ay  B x C y  B y C x   Az  B z C x  B x C z  iˆ

  Az  B y C z  B z C y   Ax  B x C y  B y C x  ˆj

  Ax  B z C x  B x C z   Ay  B y C z  B z C y  kˆ

The x-component of

A B C y

x

y



   A B  C



 B y C x   Az  B z C x  B x C z   B x ( Ay C y  Az C z )  C x ( Ay B y  Az BZ )

On the right hand side above, add and subtract Ax B x C x to get

A B C y

x

y

 B y C x   Az  B z C x  B x C z    B x ( Ax C x Ay C y  Az C z )  C x ( Ax B x Ay B y  Az BZ )      B x ( A  C )  C x ( A  B)

Do the same manipulation for the other components to get



 







         A B C  AC B  A B C

1.19

( i)



 

     

AB  aiˆ  aˆj  aiˆ  akˆ  a ˆj  kˆ AC  aˆj  akˆ  aiˆ  akˆ  a ˆj  iˆ BC  aˆj  akˆ  aiˆ  aˆj  a kˆ  iˆ

 

(ii)

   











Body diagonal from the origin to the opposite corner is a iˆ  ˆj  kˆ . This vector is perpendicular to the vectors AB, AC and BC in the plane, because its dot product with each one of them vanishes. This shows that the diagonal is perpendicular to the plane.

6

Another way to see this is to find a vector perpendicular to the plane by taking cross product of any of the two vectors from AB, AC or BC, and show that it is parallel to the body diagonal. For example taking the cross product of AB and AC gives



AB  AC  a 2 kˆ  iˆ  ˆj



which is parallel to the body diagonal. (iii) If the angle between OA and AB is  then cos  



   

OA  BA aiˆ  akˆ  akˆ  aˆj 1   OA BA 2 a 2 a 2



This gives   60 . Note that we have taken dot product with the vector BA rather than AB because we wish to keep  less than 90 . In the same manner we get the angle between OA and AC also to be 60 . 1.20

The problem is to be solved exactly in tha same manner as done in example 1.3 by replacing the position vector by the velocity vector and the velocity vector by the acceleration.

1.21

Position vector R(t )  R  cos t iˆ  sin t ˆj  

  dR (t )  R  sin t iˆ  cos t ˆj Velocity vector v (t )  dt



  dv (t )   2 R cos t iˆ  sin t ˆj Acceleration a (t )  dt



1.22





(i) In reference frame 2, the components V x 2 , V y 2 , V z 2 of vector

 V at

time t are

given in terms of its components V x1 , V y1 , V z1 in frame 1 by formula (1.10) as V x 2 (t )  V x1 cos t  V y1 sin t V y 2 (t )  V x1 sin t  V y1 cos t V z 2  V z1

Here V x1 , V y1 , V z1 are time-independent because vector Differentiating V x 2 , V y 2 , V z 2 with respect to time, we get

7

 V is

constant in frame 1.

dV x 2 (t )  V x1 sin t  V y1 cos t  V y 2 dt dV y 2 (t )  V x1 cos t  V y1 sin t  V x 2 dt dV z 2 0 dt

(ii) From (i) it is clear that   dV  kˆ  V dt

1.23

Done in later chapters

1.24

Magnitude of a vector quantity

 A(t ) is fixed. This means   A(t )  A(t )  constant

Differentiating both sides with respect to time we get   A(t ) A(t )  0 dt

Since

 A(t )

is not zero, the equation above implies that

 A(t )

 dA(t ) and are dt

perpendicular to each other. An everyday example is a particle moving in a circle. The magnitude of its position vector is a constant and therefore its velocity, which is the time-derivative of its position vector, is perpendicular to the position vector.

1.25

 



OA  R1 sin t iˆ  cos t ˆj OB  R2  sin t iˆ  cos t ˆj



The area of triangle OAB is given as RR 1 1 OA  OB  R1 R2  2 sin t cos t   1 2 sin 2t 2 2 2

This is maximum at 2t  1.26

 2

or t 

 4

Let the angle between the z axis and the vector be . Then the component OB is OA sin  .

Thus the magnitude of OB is

kˆ  OA

.

However, its direction is

perpendicular to the plane containing the z axis and the vector OA. To get the proper

8

direction we again take cross product of kˆ  OA with kˆ . The component in the z direction is given as kˆ  OA . Thus the vector OA is general kˆ can be replaced by nˆ .

9

kˆ  OAkˆ  kˆ  OA  kˆ .

In

Chapter 2 2.1

(i)

y

L

T(y)

y

T(y+y)+

(ii) Since the element of length y is in equilibrium, we have T ( y )  T ( y  y ) 

M yg L

Using Taylor series expansion for T(y+y), which gives T ( y  y )  T ( y ) 

dT 1 d 2T y  ( y ) 2   dy 2 dy 2

And taking limit y  0 leads to the differential equation for T(y). The equation is dT M  g dy l

Solution of this equation is T ( y)  

M gy  C L

where C is the integration constant. This is determined by using the fact that at the lose end (y=L) of the rope, the tension is zero. This gives C  Mg

and T ( y ) 

10

M ( L  y) g L

2.2



Torque   r  F It is given that of vector

 r  2iˆ  ˆj

3iˆ  2 ˆj

and the force has magnitude 50N and acts in the direction

. Thus the force is 50 times the unit vector in the direction of the

given vector. This gives   3iˆ  2 ˆj   F  50 13   50 ˆ 3i  2 ˆj   50 kˆ With this the torque is  2iˆ  ˆj   13

2.3

13

(a)

100N

100N

(b) The centre of the rod is at (3, 2) The right end is at position   F  100 

 r  8iˆ  2 ˆj

and the force at this end is

3 ˆ 1 ˆ i  j  2 2 

The left end is at position

 r  2iˆ  2 ˆj

and the force at this end is

11

  F  100 

3 ˆ 1 ˆ i  j  2 2 

Torque with respect to the origin =

8iˆ  2 ˆj   100 

 3 ˆ 1 ˆ  i  j    2iˆ  2 ˆj  100 2 2  





 3 ˆ 1 ˆ  i  j   10iˆ  100 2 2    500kˆ

3 ˆ 1 ˆ  i  j 2 2 

Torque with respect to the centre of the rod =  8iˆ  100 

 3 ˆ 1 ˆ i  j    2iˆ  100 2 2  





 3 ˆ 1 ˆ i  j   10iˆ  100 2 2   ˆ  500k

3 ˆ 1 ˆ i  j  2 2 

Torque with respect to the left end of the rod =  10iˆ  100  ˆ  500k

3 ˆ 1 ˆ i  j 2 2 

Torque with respect to the right end of the rod =   10iˆ  100  ˆ  500 k

3 ˆ 1 ˆ i  j 2 2 

(c) Torques about all the points are equal because the net force on the rod is zero.

12

2.4 (a)

TA

TB

3m A

B 1m 70N

(b)

F

y

0.5m

30N

120N

 0 gives

T A  TB  220

(i)

Net torque about A is zero, which gives 3TB  1  70  1.5  30  2.5  120  415

(ii)

Equation (ii) gives TB  138 N This substituted in equation (i) gives T A  82 N 2.5 Component of force in the plane perpendicular to the axis is F cos  at a distance of R from the axis. Therefore the torque about the axis is RF cos  2.6 Free-body diagram of the block

13

N1 N2

W

N1, N2 and W are three forces in a plane. Thus they must pass through one common point for equilibrium. So the equilibrium conditions are only the force conditions.

F

horizontal

 0 gives

N 1 sin   N 2 cos 

F

vertical

 0 gives

N 1 cos   N 1 sin   W

Solution of these two equations is N 1  W cos 

and

14

N 2  W sin 

2.7 Free-body diagram of the plank

N Ry

2m

1m

100N 0.2m Rx

Free-body diagram of the block N

F

Nground

If the rod makes angle  with the horizontal then sin   0.2 and cos   0.98

(a) To get the horizontal force F, we first calculate the normal force N on the rod. To do so, we calculate the total torque about the hinged end of the plank   1  N  3  100 cos   and equate it to zero. This gives N  3  100 cos   3  98  294 N

Now we balance the horizontal forces

 F

horizontal

15

 0  on the block to get

F  294 sin   0.2  294  59 N

(b) Force balance on the plank

F

horizontal

F

vertical

 0  R x  N sin   59 N  0  R y  N cos   100

This gives R y  100  288  188 N

Minus sign in front implies that the direction is opposite to that shown in the free-body diagram above. 2.8

Free-body diagram of the rod

N2

N1 F

W

Balancing the vertical forces gives N1 = W = 50N Balancing the horizontal forces gives N2 = F Balancing the torque about the centre of gravity gives F

8  50  0.5

leading to F 

2.9

25  8.8 N 8

Free body diagram of the painting

16

T

N1

N2

N1+N2 Fx W

W

Force balance equations give Fx  T and N 1  N 2  W

N1 and N2 are equal because the component of torque perpendicular to the wall must vanish. This gives N1=N2=25N Balancing the component of torque parallel to the wall taken about the lower end of the painting gives 20 3  T  10  50

giving T 

25 3

17

 14.4 N

2.10

We first calculate the forces at the ends of the rod. These forces are applied by the supports. After finding the forces on the rod, we then calculate the forces and the torques applied by the wall on the supports. Free body diagram of the rod

N1

N2

140cm 60cm 35N

Free body diagrams of the left and the right supports

F1 1

F2 2

5cm

5cm

N1

The forces on the rod satisfy

F

N2

y

 0 which gives

N 1  N 2  35

Taking torque about the left end and using



 0 gives

140  N 2  60  35  N 2  15 N

This gives N 1  20 N

Now balancing vertical forces and the torque on the supports gives

18

For the left support F1=20N and  1  0.05  20  1Nm For the right support F2=15N and  2  0.05  15  0.75 Nm 2.11 Ry

60cm

Rx 40cm

N 40N To find the force applied by the plastic block, we balance torque about the upper left corner. This leads to 40  N  30  40



N  30 N

Balancing the vertical forces gives Ry = 40N Balancing the horizontal forces gives Rx = 30N Negative sign means that the direction of Rx is opposite to that assumed in the free-body diagram above. Free-body diagram of the pole

19

30N 40cm

40N

30N

N

 Balancing the vertical forces on the pole gives N = 40N There is no net horizontal force and the two horizontal forces give a couple = 300.4 = 12Nm Balancing the torques on the pole about the ground gives  = 300.4 = 12Nm 2.12

Free-body diagram of the table

90cm

ˆ j

Rx

iˆ

Ry

Nx

Ny 20N

To find Ny, we balance the torque on the table about its left hand edge to get 90  Ny  45  20

20



Ny  10 N

By balancing the vertical forces, we get Ry  10 N . The negative sign tells us that the force is direction opposite to that shown above. Free-body diagram of one of the rods

Ry 2 30

Rx 2

Sx

Sy Free body diagram of the entire system 90cm Nx

30 Ny

20N

2Sy

2Sx

To get Nx, we balance the component of the torque coming out of the paper on the entire system about the lower hinge. This gives 30 3  Nx  45  20



Nx  10 3 N

The negative sign again tells us that the direction of the force is opposite to that shown. Balancing the horizontal component of the force on the table then gives 21

Rx  10 3N

Note: The net force on each rod on its upper end is 

Rx ˆ Ry ˆ i j  5 3iˆ  5 ˆj which is along 2 2

the rod as it must be for the equilibrium of a rod held at its ends. Balancing the horizontal and vertical components of forces on each rod gives Sx 

Rx  5 3N 2

and

Sy  5 N

Thus the net force on each rod is 10N compressive. 2.13 Free body diagrams of the two side portions and the portion AC over the pulley: N

TA

TA

TC

RM g L

TC

L2 M g L

L1 M g L

F

Tension TA and TC at both ends of the portion over the pulley is the same because the torque about the centre must vanish. This gives TA  TC  Free body diagrams of the portion AB and BC

22

L1 Mg L

Neffy

Neffy

Neffx

TB

Neffx

TB

RM g 2L

RM g 2L

TA

TC

Notice that Torque of the normal reaction about the centre of the cylinder vanishes because for each small portion of the rope over the cylinder, the normal reaction is radial. Thus T A (or TC) and TB cannot be equal because they together provide a torque to balance the torque due to the weight of the rope. Balancing the torque about the centre on AB gives R  TA 

R 2



RM g  R  TB 2L

 R  M  TB   L1  g  2 2 L 

Thus if the net force by the cylinder on the rope is Neff at an angle  from the horizontal then by force balance R  M  N eff sin    L1  g  2  L 

R  M  N eff cos    L1  g  2 2 L 

Note that Neff acts at a point different from the centre of BC because on different infinitesimal portions it is different. 2.14 The support does not apply any torque about the x-axis. All other components and torques are balanced by the support. 2.15 When forces are applied at two points of the rod, force balance demands that the force be equal and opposite. However two such forces acting at two different points will give rise to a couple moment. The couple moment is zero only if the forces point along the rod (see figure below)

23

Couple moment non-zero

Couple moment zero

2.16 Let cables OA and OC make angle 1 and OB and OD angle 2 with the vertical. Then balancing the vertical forces gives 2T (sin 1  sin  2 )  45000

The sine of the angles is easily calculated to be sin 1 

1 2

sin  2 

1 11 4



2 5

This gives T=14050N 2.17 The torque direction is given by the direction of cross product

 ˆF n

, which is

perpendicular to nˆ . This implies there is no component of the torque in the direction of ˆ. n

2.18 The net force on the plate is

50iˆ  50 ˆ j  70iˆ  120iˆ  50 ˆ j

Therefore the force that must be applied to the plate to keep it in equilibrium is  F  120iˆ  50 ˆ j

. Since there are only three forces acting on the body, they must all pass

through the same point so that their net torque is zero. This is shown in figure below.

24

A

B

O

D



The force F

C

 120iˆ  50 ˆ j

 50    22.6 from the line DC. Thus  120 

1 is at an angle   tan 

it does not pass through O and intersects with side AD and diagonal BD of the square. Therefore: (i)

It is not possible to keep the square in equilibrium by applying the third force at O.

(ii)

It is possible to keep the square in equilibrium by applying the third force at a point on BD. Equation of BD with O as origin is y  x Equation of line along which the third force acts is y   Solving the two equations gives y  x 

a 5  3a    x  2 12  2 

3 a 34

This gives the distance of point O = 0.125a And from B = (iii)

 1 3    a  0.58a  2 34 

2 

It is clear that for equilibrium, the force can be applied only on AD and BC.

25

Chapter 3 3.1 For the three trusses shown, m = 21, j = 12. Thus they all satisfy 2j-3 = m. Thus they are all simple trusses. 3.2 Showing that pin E is in equilibrium There are five forces acting on E of which two (FE and ED) are horizontal, two (CE and the external load) are vertical and one (BE) is at an angle. We wish to check if the horizontal and vertical forces add up to zero. It is solved in 3.1 that 5000 5000 2 N  Tensile  , FBE  N  Tensile  3 3 10,000 10,000  N  Tensile  , FDE  N  Tensile  3 3

FFE  FCE

Since all the forces are tensile, they all pull the pin. In addition there is the external load of 5000N vertically down. The net horizontal force is

F

x

  FFE  FBE cos 45  FDE 

5000 1 5000 2 10000   3 3 3 2

0

Similarly the net vertical force is

F

y

 5000  FBE sin 45  FCE  5000 

1 5000 2 10000  3 3 2

0

3.3

(i) The truss has 4 members and 4 joints. Number of force balance equations therefore is 8 (2number of joints).

On the other hand, number of forces available is only 7

(3+number of members), which implies that the truss will not be stable and will collapse. In terms of stability condition 2 j  3  m which implies that the truss will collapse. (ii) If we add one more member to the truss, i.e make m = 5, then 2 j  3  m is satisfied and the truss becomes stable and a simple truss. Let us add a member across AC. To find forces in each member we start by first finding the forces applied by the external supports. The free-body diagram of the truss is as follows:

26

B

C

NAy

NAx

5000N A

D ND

The direction of the forces applied by the external supports has been anticipated as shown. To find ND, we balance torque about A to get

2  N D  (2  1.5 cos 60)  5000

  N D  6875N

 2.75  5000 

Now balancing the vertical and horizontal forces on the truss gives NAx = 0

and

NAy = 1875N

The negative sign again tells us that the direction of the force is opposite to that shown. We begin to apply the method of joints from point D since at this point there are two unknown forces FAD and FCD. Assuming these forces to be tensile gives the free body diagram of joint D as follows FCD

60 FAD 6875N

27

Balancing the vertical forces on point D gives FCD sin 60  6875  0 

FCD  7939 N

The negative sign shows that the force FCD is compressive and not tensile as assumed. Balancing the horizontal forces on point D gives FCD cos 60  FAD  0 

FAD  3969 N

The negative sign again shows that the force FAD is compressive and not tensile as assumed. Next we go to point A and balance the forces there. The free body diagram of point A is FAB FAC 60



3969N 1875N

In drawing the figure above, we have shown the direction of FAD according to it being a compressive force. The length of rod AC is =

 2  1.5 cos 60 2

 1.5 sin 60

2

 3.04 m

So sin  

1.5 sin 60  0.427 3.04

and

cos  

2  1.5 cos 60  0.904 3.04

Now balancing the horizontal forces at A gives FAB sin 60  FAC sin   1875

and balancing the vertical forces at A gives FAB cos 60  FAC cos   3969

Solving these two equations gives FAB = 0 and FAC = 4390N Since the sign of FAC is positive it is in the same direction as assumed and therefore tensile.

28

Now we can easily see that force FBC will be zero because point B is under equilibrium under only two forces FAB and FBC and FAB has already been determined to be zero. Thus FBC = 0. Thus all the forces are now determined. They are FCD  7939 N (compressive)

FAB = 0

FAD  3969 N (compressive)

FAC = 4390N (tensile) and FBC = 0.

Finally to check our answer we make the forces at point C and see if they all balance. The free body diagram of point C is

7939N 60  5000N 4390N

Balancing the horizontal forces at C gives 7939 cos 60  4390 cos   3969  3969  0

Balancing the vertical forces at C gives 7939 sin 60  5000  4390 sin   6875  5000  1875  0

This indicates that our answers are correct. Note: We see that FAB and FBC both vanish. This implies that members AB and BC may not bee needed for the truss. This is true because with just three members AD, AC and CD (m=3) there are only three joints (j=3) and the truss satisfies the condition 2 j  3  m for it to be a stable structure. 3.4 Free body diagram of the truss

29

Ry 30cm

B

Rx

C 

20cm

30N N A

Since point C is in equilibrium under one known and two unknown forces, both unknown forces can be determined easily. The forces on C look as follows FAC 

FBC

30N Balancing the vertical forces at C gives FAC sin   30

30

With sin  

2 13

this implies FAC = 54N (compressive)

Balancing the horizontal forces at C gives FBC  FAC cos   54 

3  45 N (tensile) 13

The only force left is at AB. We calculate this by balancing forces acting on pin A, which look as follows.

FAB N

 FAC This gives FAB  FAC sin   30 N

Additionally we can also solve for the normal reaction N and the forces Rx and Ry. These are N = 54N, Rx = 45N, and Ry = 30N 3.5 Rod AB provides a vertical force to hold pin A. However if it is removed and the vertical force is provided by a fixed pin joint, the structure will remain stable because we need 3j=6 forces for equilibrium of 3 joints; two of these are provided by the fixed supports and two by the two members. The forces in the members remain the same. So do the forces by the two support except that the fixed point at A also provides a vertical fore of 30N that was earlier provided by member AB. 3.6 Free-body diagram of the truss

31

Ry B

C

Rx

N A

D

100N Since pin D has only two unknown forces acting on it, we can start our calculations from this point onwards. The forces on D are

FCD

FAD 100N

It is immediately clear that FAD = 0 and FCD = 100 N (tensile) Next we go to pin C and balance the forces there. The forces acting on C are

32

FAC

FBC 100N

Balancing the vertical forces at C gives FAC  100 2 N (compressive)

Balancing the horizontal forces at C then gives FBC  100 N (tensile)

Next we go to pin A. The forces there are as follows

FAB N

100 2 N

Balancing the vertical forces at A gives FAB  100 N (tensile)

Balancing the horizontal forces at A gives N=100N Finally balancing forces at pin B will give the external forces Rx = 100N and Ry = 100N 3.7 Free body diagram of the truss is as follows

33

Ry Rx

C

D

B 500N

Ny Nx A (i)

E

There are 4 reaction forces at the supporting pins at A and B. In addition the forces generated by the members of the truss equal 6. This makes the total number of forces available = 10. The number of joints in the truss is 5 that require exactly 10 number of forces for equilibrium. Thus the truss is a stable one.

(ii)

It is also statically determinate since the number of forces available is equal to the number of equations to be satisfied for equilibrium.

(iii)

First we find Nx by balancing the torque about point B. This gives 0.75  Nx  1.5  500



Nx  1000 N

We now begin by balancing the forces at point D

34

FED

FCD 500N

Balancing the vertical forces at D gives FED  500 2 N (compressive)

Balancing the horizontal forces at D then gives FCD  500 N (tensile)

Nest we go to pin E because it has two unknown forces acting on it. The forces are as follows

FCE FAE

N

Balancing forces at E gives FAE  500 N (compressive) FCE  500 N (tensile)

35

Next we go to pin at A. The forces there are

Ny

1000N

500N FAC

Balancing the forces gives FAC  500 2 N (compressive)

and Ny  500 N

(tensile)

Finally we go to point C where only one force FBC is unknown. The forces on C are

FAC

FBC

500N

500N

Balancing the horizontal forces at C gives FBC=1000N As a final check, the value for FBC gives the horizontal force Rx by pin B on the truss to be 1000N and vertical force Ry to be zero. This is consistent with the overall equilibrium of the truss when it is treated as a system by itself.

36

3.8 Since each member of the truss weighs 50N, at each pin we take the load by each pin at that point to be 25N. The free body diagram of the truss is as follows; here each small arrow pointing down indicates the weight of the truss member, acting at its centre.

NBy

NBx

C

B

D

A

E

1000N

NE

We firs find NE. To do this we balance the torque about B. This gives l  NE 

l 3l  150  l  50   100  2l  1000  2 2

N E  2275N

This immediately gives, by balancing forces on the entire truss NBx = 0 and NBy = 925N The negative sign showing that the force I opposite to the direction assumed in the figure above. We begin at pin D as there are two unknown forces there. The force diagram on pin D is as follows (there are two members meeting at pin D that give a load of 225=50N there)

37

FCD

FDE 1050N

Balancing the forces gives FCD  1050 2 N  1485N (tensile)

and FDE  1050 N (compressive)

Next we go to pin E. The forces acting there are (including 325=75N from 3 members)

2275N

1050N

FAE FCE +75N

Balancing the forces gives FCE  2200 N (compressive)

and FAE  1050 N (compressive)

Next we go to pin A. The forces acting there are (including 325=75N from 3 members)

38

FAB

1050N

FAC

75N

Balancing the forces gives FAC  1050 2 N  1485 N (tensile)

and FAB   75  1050 N  975 N  FAB  975 N (compressive)

Net we move to point B where the forces are (including 225N=50N from two members) FAB

FBC

50N 925N

This immediately gives FAB = 975N (compressive) and FBC = 0 As a final check, one balances all the forces at C and sees that they all balance properly. This implies that the answers obtained by us are correct. 3.9 Free body diagram of the truss with the weight of each member included. The free body diagram is then as follows.

39

C

B

NA

A

F

Nx

ND

D

E 5000N

By balancing the torque about A we get

12  N D  2 1000  4  500  6 1500  8   500  5000  10 1000

  ND 

 67000 

16750 3

Balancing the vertical and horizontal forces on the truss, this gives NA 

11750 and Nx  0 3

For calculating force in members, we take the weight of each member shared equally at each joint. The forces on A are (including the weight of two members)

11750 N 3 FAF

500N

FAB Balancing the forces at point A

F

0

y

F

x

0

gives gives

FAB

11750  500 or FAB  4832 N (compressiv e) 3 2 F 10250 FAF  AB  N (tensile) 3 2 

Next we go to point F. The forces at point F are (including the weight of three members) 40

FBF

10250 N 3

FFE

750N Balancing the forces here gives FBF  750 N ( tensile) FFE 

10250 N ( tensile) 3

Next we go to point B since now there are only two unknown forces there. At point B the forces look as follows (including the weight of four members) 4832N

FBC 750N

FBE

1000N Balancing the forces

F

y

F

x

0

gives

FBE

0

gives

FBC 

2

 1000  750  4832 2



2357 2

4832 2



0 

FBE  2357 N (tensile) FBC  5083 N

Negative sign above means that the direction of the force is opposite to the one assumed. So FBC = 5083N (compressive)

41

We next consider point C and balance the forces there. The forces at point C are (including 750N form the weight of three members)

FCD 5083N 750N FCE Balancing the forces

F

x

F

y

 0 gives

FCD 2

 5083  FCD  7188 N (compressiv e)

 0 gives FCE  750 

FCD 2

 5083  FCE  4333 N (tensile)

Next we go to pin D where the normal reaction is

16750 N and balance the forces there. The 3

force diagram there is (including 500N form the weight of two members)

16750 N 3 FED 500N 7188N

It is easily seen that the vertical forces balance at this point. This points to the correctness of our calculations so far. Balancing horizontal forces then gives FED = 5083 N (tensile)

42

As a final check we should check whether all the calculated forces balance at pin E. The forces at pin E are (including the weight of four members)

2357N

4333N

N

5083N 1000N

5000N

All these forces balance as can be seen by calculating the net x and y components of the forces. Thus our calculations are correct. 3.10

Free body diagram of the truss 2000N

NBy C NBx

B

D

A

1000N

E NE

Taking torque about B we get 3  N E  4  2000 

NE 

8000 3

Balancing the horizontal and vertical forces now gives NBx = 1000 N

and

NBy = 

2000 N 3

Negative sign above means that the direction of the force is opposite to the one assumed.

43

We begin at pin D. The forces there are FED 

FCD

1000N

2000N

In the diagram above cos  

1  0.667 and sin   1.5

1.5 2  1  0.745 1.5

Balancing the vertical forces gives FED sin   2000 

FED  2683N (compressive)

Balancing the horizontal forces gives FCD  FED cos   1000  2789N

Next we go to pin E. The forces there are N

FAE 

 FCE

2683N Balancing the vertical forces gives

 2683  FCE  sin 



8000 3



FCE  895 N (compressive)

Balancing the horizontal forces gives FAE  2683 cos   FCE cos   1192 N (compressive)

44

Next we go to pin A. The forces there are FAC  1192N

 FAB

Balancing the vertical forces gives FAC = FAB. Balancing the horizontal forces gives

 FAB  FAC  cos  2 FAB cos  1192

 FAB  895N(compressive)

This also means that FAC = 895N (tensile) Now we go to pin C. The forces there are 895N 

2789N

FBC  895N

The vertical forces are already balanced here. Balancing the horizontal forces gives FBC  2  895  cos   2789 

FBC  1596 N(tensile)

To check our answers, we finally balance the forces at pin B and see if they all balance there. At pin B the forces are as follows

45

895N 

1596N

1000N N

It is easily seen that all the forces above balance. So our answers are all consistent. 3.11

The weight of the road = volume of the roaddensityg = 128.3200010 = 576000N This weight is divided equally between the two trusses on the sides. 288000N is supported by each truss. Weight of the members of the truss = 135000 = 65000N Total weight supported by each truss therefore is = 353000N Free body diagram of the truss is B

C

D

NAy

NAx

NE

A

H

G

F

353000N From the balance of forces, is clear that N Ax  0

N Ay  N E 

353000  176500 N 2

46

E

Thus only

Let us now consider forces at each pin one by one. Each pin has the following forces acting on it: The weight of the road divided over 5 pins, which is

288000  57600 N ; 5

the weight of the members at that pin; and the forces applied by the members. Let us now balance forces at point E. The forces on pin E are (including the weight of the members) 176500N

FEF  57600N

5000N

FDE

In the figure above sin  

3  0 .6 5

cos  

4  0 .8 5

Balancing the vertical forces in the figure above gives FDE cos   57600  5000  176500  FDE  142375 N (compressive)

Balancing horizontal forces then leads to FFE  FDE sin   85425N (tensile)

Next we consider pin D.

The forces on pin D are (including the weight of the

members)

47

142375N  FCD FDF 7500N

Balancing horizontal forces then leads to FCD  142375 sin   85425 N (compressive)

Balancing the vertical forces gives FDF  142375  0.8  7500  106400 N (tensile)

Next we look at pin F. The forces there are (including the weight of the members) 106400N

FGF

85425N  57600N

10000N

FCF

Balancing the vertical forces in the figure above gives FCF cos   57600  10000  106400 

FCF  48500 N (compressive)

Balancing horizontal forces then leads to FGF  FCF sin   85425 N  114525N (tensile)

By symmetry of the problem, forces on the members to the left of member CG will be exactly the same as on the corresponding members to its right. The only force that we now have to calculate is on member CG. For this we consider point G. Two horizontal

48

forces at G are by HG and GF and are equal to 114525N each. The forces at point G are then given as FCG

114525N

114525N  57600N

7500N

This gives FCG = 65100N Finally we check our answer at point C. The forces there (including the weight of 5 members meeting there) are 48500N

48500N

85425N

85425N

65100N

12500N

FCF

As is easily seen, the forces at C balance and therefore our calculations have been consistent throughout. 3.12

Free body diagram of the truss is as follows

49

2000N NA

1000N NE

C B

RA

D

 A

H

G

F

E

Here the distances and the angles are AH = HG = GF = FE = 2.5m tan  

BH = DF = 1.25m (similarity of triangles)

1.25  0.5    26.6  sin   0.447 and cos   0.894 2.5

Balancing torque about point A gives 10  N E  5  1000  2.5  2000 

N E  1000 N

Thus NA = 2000N and RA = 0 Now that the external reactions have been determined, we can go about calculating the forces in the members. We start with pin E because there are only two unknown forces there. The forces at E are 1000N

FEF

E

 FDE

Balancing the vertical forces in the figure above gives FDE sin   1000 

FDE  2236 N (compressive)

Balancing the horizontal forces gives FEF  FDE cos   2236 * 0.894  2000 N (tensile)

50

Nest we go to pin D. The forces there are FDF

2236N

D FCD Balancing the forces gives FCD  2236 N (compressive)

FDF = 0 Next we go to point F where the forces are as shown below. FCF 0N

FGF

F

2000N

Balancing the forces gives FGF = 2000N (tensile)

FCF = 0

Pin G is considered next. Since the forces there are only vertical and horizontal, even without making the forces there, we immediately can write FCG = 0 and FGH = 2000N (tensile) Next we consider point A where two members AB and AH meet and therefore there are two unknown forces. The forces there are

51

2000N



FAH

A

FAB Balancing the vertical forces gives FAB sin   2000 

FAB 

2000  4472 N (compressive) 0.4472

Balancing horizontal forces then gives FAH  4472 cos   4000 N (tensile)

Next we go to point B. Here there are four forces acting and each pair (FBH and 2000N; and FBC and 4472N) has two forces in opposite directions. Thus without solving the detailed force balance equations, we can directly write FBH = 2000N (compressive) and FBC = 4472N (compressive) Next we go to point H and balance the forces there. The forces there are as follows FCH

45 4000N

2000N

H

2000N By balancing the forces at H, it becomes clear that FCH  2000 2 N (tensile) Finally the answers are checked at C. The forces at C are as shown below

52

2236N

4472N

D

1000N

2000 2 N

As is easily seen, the horizontal a n vertical forces all balance at C. Thus our answers are all correct. To calculate the forces by method of sections, we make a cut through the truss so that it passes the concerned members. In the present case we take the following section of the truss and show various forces on that section. 2000N 2000N FCB

B  A

H

FCH

FGH

In the figure above, FCH is determined easily by taking torque about A since the torque due to FCB and FGH both vanish about A. This gives AH 

FCH 2

 AH  2000 

FCH  2000 2 N (tensile)

To find FCB, we balance the vertical component of the forces to get FCH 2

 FCB sin   0  FCB  

2000   4472 N sin 

Negative sign here means that the force is opposite to the direction assumed and therefore is compressive in nature. Finally, balancing the horizontal forces leads to

53

FGH 

3.13

FCH 2

 FCB cos   0 

FGH  2000  4472  0.894  2000 N (tensile)

The free-body diagram of the truss on one side is as follows (Notice that the weight of the truck is equally divided between the two trusses) C

B

D

NA RA

A

NE E

H

G

F

50kN We first calculate NE by balancing torque about A. This gives 16  N E  12  50 kN 

N E  37.5 kN

This gives NA = 12.5 kN and RA = 0 To find forces in members CD and DG, we make a cut through CD, DG and GF. This looks like the following D

FCD

37.5 kN

E FGF

F FGD 50kN

To find FGF, we balance torque about point D about which the torques due to FCD and FGD vanish. This gives

54

4  37.5  5  FGF



FGF  30kN (tensile)

To obtain FCD, we take torque about point where FDG and FGF intersect, which is point G. This leads to 5  FCD  8  37.5  4  50



FCD  20kN (compressive)

Now we balance the horizontal and vertical forces on the truss. Balancing horizontal forces gives FGD 

4 41

 20  30  0 

FGD  16kN

Negative sign here means that the force is opposite to the direction assumed and therefore is compressive in nature. To find the forces in the members BC and BG, we make a cut through the members BC, BG and HG as follows and then calculate the forces. B

FBC

12.5 kN A

FGH

H

FBG To obtain FBC, we take torque about point where FGH and FBG intersect, which is point G. This leads to 5  FBC  8  12.5 

FBC  20kN (compressive)

Next we find FGH by taking torque about B. We get 5  FGH  4  12.5 

FGH  10kN (tensile)

Finally we get FBG by balancing vertical and horizontal forces. Horizontal force balance gives FBG 

4 41

 10 

FBG  16kN (tensile)

Finally to find FCG, we make the following cut through the truss

55

FCG

B

D

12.5kN A

37.5kN

E H

G

F

50kN Since the vertical forces all balance, this implies FCG = 0. Further, the horizontal forces are also balanced.

56

Chapter 4 4.1

N s

Frictional force

F

4.2

Applied force max

Since the block is in equilibrium under three forces, the three forces must pass through the same point. Thus the normal reaction will be at the point where the arrow showing the weight meets the inclined plane. This is shown below. f

N

mg θ

4.3

The free body diagram of the block is as follows

57

N 

F 

f mg Balancing the horizontal forces gives f  F sin 

Balancing the vertical forces gives N  mg  F cos 

Since the maximum frictional force f max  N , for equilibrium we should have F sin     mg  F cos  

4.4



F

 mg sin    cos 

We consider two different situations when the weight on the table is about to move to the left or to the right. When it is about to move to the left, its free body diagram will look as follows N

10g

mg f

50g By equilibrium conditions, we have N = 50g f + mg = 10g 58

Since f  N

We have f  10 g  mg  0.1  50 g



5m

Thus the minimum value of m is 5kg when the frictional force is at its maximum pointing to the right. As m is increased above 5kg, frictional force becomes less and less, eventually changing direction and attaining its maximum value pointing left. In that situation, the free body diagram of the block on the table is as follows. N

10g

mg f

50g In this situation, the equation for horizontal force balance is f + 10g = mg This coupled with f  N leads to f  mg  10 g  0.1  50 g



m  15

Thus 5  m  15

4.5 Taking the x axis along the plane and the y axis perpendicular to the plane, the free-body diagram of the block looks as follows.

59

Y

N

F

X

30º

30º

F

100g The equations describing equilibrium in the X and the Y directions are

F

0

x

F

y



0



F cos 30  F   100 g sin 30  0 N  F sin 30  100 g cos 30  0

The first equation implies that F    F cos 30  100 g sin 30

Taking g = 9.8ms2, the value of F for different values of F is

4.6

F  600 N



F  500 N



F   29.6 N F   57.0 N

F  100 N



F   403.4 N

Free body diagram of the box when it is about to move (i.e. the frictional force is at its maximum) is shown below N

F h

b a

N mg

60

When the box is about to move, the friction is at its maximum and is equal to N. The force F also equals N at this point. This creates a couple that is counterbalanced by the couple formed by the weight of the box mg and N (=mg). This is the reason that N shifts towards the direction of the push. However, the maximum couple moment that can be created by mg and N is

a mg . 2

Thus for the box not to topple, the couple created by F and the friction should remain less than a mg . Thus implies 2 h  mg 

a mg 2



h

a 2

4.7 suppose each break show makes an angle  at the centre as shown below



The force F is assumed distributed uniformly over the shoe. Then the torque due to the frictional force will be b

   a

Fr

 r dr 

 2 b  a 2  2

2 F  b 3  a 3  2 F  a 2  ab  b 2   3 b 2  a 2  3  a  b

With two shoes therefore, the torque would be 4 F  a 2  ab  b 2   3  a  b

4.8 It is given that mass M is balanced by mass m. The contact angle is π. Since each time the string is wound once more around the rod, the mass M that can be balanced by m becomes twice as large, we have

61

M  m exp( )   2M  m exp(3 )  2  exp(2 ) 4M  m exp(5 ) This gives  = 0.11

4.9 Neglecting the length of the rope passing over the pulley, we have mass

side of the pulley that is balanced by mass

L2 M on the other side. Thus we have L1  L2

L1 L2 Mg  exp(  ) Mg L1  L2 L1  L2

4.10

L1 M on one L1  L2



L1  exp(  ) L2

As the weight is put, it has a tendency to move down. Hence the frictional force will be in the counterclockwise direction. Thus if the tension in the rope on the spring balance side is T1 and that on the weight side is T2 then   T2  T1 exp   2 

Now it is given that T1 = 5g and T2 = mg. Thus we get m  5 exp(0.2   / 2)  6.85kg

An interesting possibility exists if a person had pulled the weight down and then slowly brought it to equilibrium. In that case the tension will work in the other direction and m  5 exp( 0.2   / 2)  3.65kg

However we have not considered this possibility. 4.11 There is a range of M2 that exists because frictional force can act with its maximum value in one direction to the maximum in the other direction.

Largest value of M2 is when the

mass M1 is about to slide up the plane. The free body diagram of M1 in that case is as follows

62

N

T

f 

M1g

When the mass M1 is about to slide up, we have T  M 1 g sin   1 M 1 g cos   M 1 g (sin   1 cos  )      2  

The contact angle between the rope and the pulley is 

Since the rope has a tendency to move clockwise, the frictional force due to the pulley will be acting counterclockwise. Thus we have      M 2 g  T exp  2       2    Thus

      M 2 g  M 1 g  sin   1 cos   exp  2       2   

     M 2  M 1  sin   1 cos   exp  2       2   

The other extreme is when the mass M1 is about to slide down the plane. In that case the free body diagram of M1 is

N

T

f 

M1g

Thus we have T  M 1 g sin   1 M 1 g cos   M 1 g (sin   1 cos  )

63

Now the rope has a tendency to move counterclockwise, the frictional force due to the pulley will be acting clockwise. Thus we have      M 2 g exp   2       T  2    Thus

      M 2 g  M 1 g  sin   1 cos   exp    2       2   

     M 2  M 1  sin   1 cos   exp   2       2   

4.12 Free body diagram of the tire when it is loaded and is about to roll is as follows

F

f W

N

Balancing the vertical forces gives N  abP  W

 a

W bP

Balancing the horizontal forces gives F = f Balancing torque about the centre of the wheel gives FR 

a W 2

 F

64

W2 2bPR

Chapter 5 5.1

Consider a composite surface of total area A made up of N different surfaces. Then the coordinates  X C , YC  of the centroid satisfy

 xdA   ydA

AX C  AYC

Ai If the area of each surface is Ai (i  1 N ) then A   i

Now in the definition of the centroid, the integrals can be performed separately over each surface so that we can write AX C    xdA   Ai X Ci i

AYC 

i

i

  ydA   A Y i

i

Ci

i

i

This immediately gives

XC 

5.2

 Ai X C i A

  Ai X C i   Ai

 

 

and YC  

 Ai X C i A

  Ai YC i   Ai

 

  

By symmetry it is clear that XC = 2. We are nevertheless going to prove it below. We first calculate the area of the surface. It is 4

A

 ydx  0

 4   x  2 dx 4

2

0

4

 16    x  2  dx 2

0

Substituting z  x  2 we get 2

A  16   z 2 dz  16  2

Y

16 32  3 3

To calculate Xc, we take vertical strips of width dx on the surface at distance x from the origin and then calculate XC 

O x

X 65

 xdA A

Thus

 x 4   x  2  dx 4

XC 

2

0

32

2

3

Similarly to calculate YC, we take horizontal strips of width dy at height y and calculate YC 

Y

 ydA A

At height y, the strip extends from x1 to x2. These points are given by the equation y  4   x  2

y 1 O thusxhave We

X

x2

2



x1  2 

4 y

x2  2 

4 y

Therefore dA  ( x 2  x1 )dy  2 4  y dy

4

32 YC  2  y 4  y dy 3 0

Substituting y  4 sin 2  so that dy  8 sin  cos  d , we get  2

1

32 YC  2  4 sin 2   2 cos   8 sin cos  d  128 sin 2  cos 2  d  cos   3 0 0 1

 128  cos 2   cos 4   d  cos   0



256 15

This gives YC  

8 5

8 5

Thus the centroid is at  2,  

5.3

One curve (call it curve 1) y  4  ( x  2) 2 in this problem is the same as that in the problem above. The other curve (curve 2) is



y  16  4( x  2) 2  4 4   x  2 

66

2



The y-axis of curve 2 is thus 4 times curve 1. The area of curve 2 is therefore

128 . Similarly 3

the x coordinate the centroid of curve will remain at 2 but the y coordinate will be 4 

8 32  . 5 5

Thus we have A1 

32 3

 X C1 , YC1   

8 2,   5

;

A2 

128 3

 X C 2 , YC 2    

2,

32   5 

The area A for which we wish to obtain the centroid  X C , YC  is obtained by removing surface formed by curve 1 from the surface formed by curve 2. We thus have 128 32   32 3 3 AX C  A2 X C 2  A1 X C1 A

AYC  A2YC 2  A1YC1

5.4

 32 X C   A2  A1   2  32YC 



128 32 32 8    3 5 3 5



XC  2 YC  8

Trapezoidal loading is shown in the figure below

w2

f(x) w1

X1

X2

X

The total force on the beam will be equal to the area under the curve. Thus the total force is equal to

 w1  w2   X 2

2

 X1 

This load will be acting at the centroid of the area. Thus it acts at 67

X2

XC 



 x

X1

w1 

 w2  w1  x  X 1   dx X 2  X1 

 w1  w2  X 2  X 1 

X

2

  w





 w1  X 12  X 22  X 1 X 2  w2  w1  X 1  X 2  X 1 w1  3 2   w1  w2  X 2  X 1  2 2 2 2 2 3w1 X 2  3w1 X 1  2w2 X 1  2w2 X 2  2w2 X 1 X 2  2w1 X 12  2 w1 X 22  2 w1 X 1 X 2  3w2 X 12  3w2 X 1 X 2  3w1 X 12  3w1 X 1 X 2  3 w1  w2  X 2  X 1  2 2

X 2

2 1

2



w1 X 22  2 w1 X 12  w2 X 12  2 w2 X 22  w2 X 1 X 2  w1 X 1 X 2 3 w1  w2  X 2  X 1 

X 22  w1  2 w2   X 12  2 w1  w2   w2 X 1 X 2  w1 X 1 X 2  3 w1  w2  X 2  X 1 

Adding and subtracting w2 X 1 X 2 and w1 X 1 X 2 in the numerator we get

5.5



X 22  w1  2 w2   X 12  2 w1  w2   2 w2 X 1 X 2  w1 X 1 X 2  2 w1 X 1 X 2  w2 X 1 X 2 3 w1  w2  X 2  X 1 



 2w1  w2  X 1  X 2  X 1    w1  2w2  X 2  X 2  X 1  3 w1  w2  X 2  X 1 



1  2w1  w2  X 1   w1  2 w2  X 2  w1  w2  3

From figure 5.14, for a plate of width w w1  gh1 w

w2  gh2 w

Similarly, from figure 5.15 X 1  Y1 

h1 cos 

X 2  Y2 

68

h2 cos 

This gives from formula 5.9 X C  YC 



1 (2h1  h2 )h1  (h1  2h2 ) h2 3 (h1  h2 ) cos 



2 h12  h1 h2  h22 3  h1  h2  cos 



which is equivalent to the depth given by formula (5.17) 5.6

Loading on the tank door is triangular as shown below

N1 NA 0.25m 0.5m 153.125N NB

19.6N The average pressure is the pressure of water at the centroid of the submerged part. Thus the average pressure will be ghcentroid ( plate )  1000  9.8  0.125  1225 Nm 2

Thus the total force due to the water pressure is F  0.5  0.25  1225  153.125 N

This force acts at the centroid of the loading that is triangular in this case. Thus it is at a distance 1  0.25  2  0.50  0.417 m 3

below point A. We now apply equilibrium conditions to the door. This leads to 69

N 1  19.6 N 0.5  N B  0.417  153.125  N A  N B  153.125 

5.7

N B  127.7 N

25.4 N

The rectangular surface looks as follows Y

b X

a

To find Ixx, we take a strip (see figure above) of width dy parallel to the x-axis and calculate b 2

2  y  ady  

I xx 

b 2

ab 3 12

Similarly to find Iyy, we take a strip (see figure above) of width dx parallel to the x-axis and calculate a 2

I yy 

2  x  bdx  

a 2

a 3b 12

To find Ixy, we take a small square (see figure above) of size dxdy parallel and calculate a 2

I xy 

b 2

  xy dxdy   0

 a 2 b 2

by symmetry of the inegrand.

70

5.8

Y’ X’



b O a

From the figure sin  

b

cos  

a2  b2

a a2  b2

2ab a  b2 a2  b2 cos 2  1  sin 2 2  2 2 a b

sin 2  2 sin  cos  

2

From the formula for transformation of area moments (taking X and Y axis as in the problem above) we get I x'x' 

I xx  I yy





2



ab a 2  b 2 24



a 3b 3 6 a2  b2





I xx  I yy

cos 2  I xy sin 2 2 ab b 2  a 2 a 2  b 2  24 a2  b2





Similary

71

 

 

I y' y ' 

I xx  I yy



2





ab a 2  b 2 24



ab a 4  b 4 12 a 2  b 2

 

  

I xx  I yy

cos 2  I xy sin 2 2 ab b 2  a 2 a 2  b 2  24 a2  b2



 

 

and I x'y'  

I xx  I yy 2



sin 2  I xy cos 2



ab b 2  a 2 2ab 2 24 a  b2







a b a b 12 a 2  b 2 2

2



2

2







5.9

Y

b

X

a

To calculate IXX we take a horizontal strip of width dy at y (see figure) for dA and calculate

72

b

I XX   y 2 dA 2  y 2 b

a 2 b  y 2 dy b

Taking y  b sin  , we get  2

I XX  2



b 2 sin 2 

 2

 2

a 2 sin 2 2 ab 3 b cos 2 d  2ab 3  d  b 4 4  2

Similarly for IYY, we take a vertical strip at x for dA and calculate a

I YY   x 2 dA  2  x 2 a

b a 2  x 2 dx a

Taking x  a sin  , we get I YY 

a 3 b 4

And by symmetry I XY  0

We now calculate the moments and product of inertia about a set of axes rotated by an angle with respect to the original one and with the same origin. Thus I x'x' 

I xx  I yy



I xx  I yy

cos120   I xy sin 120 

2 2 2 2 ab(b  a ) 1 ab(b 2  a 2 )   8 2 8 ab  3a 2  b 2  16

Similarly I y'y ' 

I xx  I yy



I xx  I yy

cos 120   I xy sin 120 

2 2 2 2 ab b  a  1 ab b 2  a 2    8 2 8 2 2 ab a  3b   16

Product of inertia is calculated using the formula

73

 3

I x' y ' 

I xx  I yy

sin 120   I xy cos120 

2 3 ab b 2  a 2   2 8 ab  3a 2  b 2  16

5.10 Y

y O

x2

x1

X

To calculate IXX we take a horizontal of width dy strip at y (see figure) for dA and calculate I XX 

y

R

2

dA  2  y 2 R 2  y 2 dy 0

To evaluate the integral, we substitute y  R sin  so the integral is transformed to  2

I XX  2 R 4  sin 2  cos 2  d 0



4  2

R 2

 sin

2

2 d

0

Now substituting z  2 , we get I XX  



R4 sin 2 z dz 4 0

 R4 8

74

Y

x X O

Similarly for IYY, we take a vertical strip of width dx at x for dA (see figure) and calculate I YY   x 2 dA 2R



x

2

R 2   x  R  dx 2

0

Taking  x  R   R cos  we get 0

I YY  R 4  1  cos   sin    sin  d 2





 R 4  1  2 cos   cos 2   sin 2  d 0



2 Now  sin  d  0

 , 2





0

0

2  cos  sin  d  0 and

I YY 

 sin

2

 cos 2  d 

 . This gives 8

5 4 R 8

5.11 Product of area about the origin O is given as I xy 

 x y A i

i

i

i



 xy dA

If the centroid is at O’ which has the coordinates  x0 , y 0  and the coordinates of a point with respect to O’ are  x , y  then x  x0  x

y  y0  y

75

Y

Y’

X’

(x0,y0)

X

O Therefore I xy 

 xy dA    x

0

 x  y 0  y  dA  x 0 y 0 A  x 0  y  dA  y 0  x dA   x y  dA

However, by definition of the centroid

 xdA  0

 y dA  0

Thus I xy  x 0 y 0 A 

 x y  dA

5.12 Consider the moments and product of inertia of a square about a set of axes parallel to its sides and passing through its centre.

Y

X

a For this set of axes

76

 a4   I xx  I yy    12 

I xy  0

Now by the formula I x'y' 

I xx  I yy 2

sin 2  I xy cos 2

Ix’y’ will always remain zero because of the equality of Ixx and Iyy irrespective of the angle of rotation of the new set of axes x’y’. Thus any set of axes passing through the centre is the principal set of axes. 5.13 The formulae for the moments of inertia in rotated frames are

I x'x' 

I y'y' 

I xx  I yy 2

I xx  I yy 2





I xx  I yy 2

I xx  I yy 2

cos 2  I xy sin 2

cos 2  I xy sin 2

Taking the second derivative of these expressions with respect to  , we get

I xx  I yy d 2 I x 'x '  4 cos 2  4 I xy sin 2 2 2 d d 2 I y'y' d 2

4

I xx  I yy 2

cos 2  4 I xy sin 2

Thus the two derivatives have opposite signs. This implies if one of them is a maximum, the other one will be a minimum.

77

Chapter 6 6.1 (i) (a) and (d) are the virtual displacements because these are the only ones consistent with the constraint that the block can move only in the vertical direction. (ii) Suppose the strech is y0. In that situation, the forces on the block are as shown

ky0

mg

Now a virtual displacement gives a displacement of y in the vertical direction. Taking it in the vertically up direction gives the virtual work to be

W   ky 0  mg y Equating this to zero gives y0 

mg k

6.2 Figure below shows the students and the plank on a wedge and a possible virtual displacement of the system.

x

(3-x)

 40g

30g 50g

It is clear that as long as the point on the wedge does not move, the only possible displacement is the rotation of the plank about this point and the system has only one degree of freedom. For the

78

virtual displacement shown, the displacement and the virtual work done by various forces is as follows: 40kg :

virtual displacement  x ;

Center of gravity of plank : 50kg :

virtual work  40 xg

virtual displacement   x  1.5 ;

virtual displacement   3  x  ;

virtual work  30 x  1.5 g

virtual work  50 3  x  g

Since the net virtual work must vanish for equilibrium, we have

 50 3  x   40 x  30 x  1.5 g

 0   120 x  195  0 

x

195  1.625 120

6.3 (i) Since the number of parameters required to describe the system is 1, the number of degrees of freedom is 1. We choose it to be the angle , the rod makes from the vertical. A virtual displacement will be to change  by .

1.5 m



2kg

T 20kg

(ii) To apply the principle of virtual work, we need to calculate the virtual work done by various force when  is changed to  +.

For this we first write the coordinates (x, y) of the tip of the

rod and yCG of the centre of gravity of the rod with respect to the pivot point. These are

79

x  1.5 sin 

y  1.5 cos 

y CG  0.75 cos 

As  is changed to  +, these coordinates change and the changes are given by x  1.5 cos  

y  1.5 sin  

y CG  0.75 sin  

Thus the total virtual work done by the external forces – 2g at the CG, 20g at the tip, both in the positive y direction, and T at the tip in the positive x direction – is W  1.5 cos   T  1.5 sin   20 g  0.75 sin   2 g 

Equating this to zero gives T  21g tan 

6.4 To find the forces applied by the bricks, we treat these forces as external. For only vertical motion, there are two degrees of freedom. One is the vertical displacement of the centre of gravity and the other the rotation of the plank about the CG. Equivalently, we can take the vertical displacements of the ends A and B as the virtual displacements. We choose the second option because this is related directly to the forces applied by the bricks. The plank in equilibrium and virtually displaced is shown below

1.5m

y2

1m

y1

NA

100N 2m

NB 500N

Now the vertical virtual displacement of different points is Point A :

y1

Point B :

y 2

Centre of gravity of the plank :

y1  y 2 2

80

  y 2   y1  y1  1.5   0.25 y1  0.75 y 2 2  

Athlete : Thus the total virtual work done is

 y  y 2  W  N A y1  N B y 2  100 1   500 0.25y1  0.75y 2 2     N A  50  125y1   N B  50  375y 2

Equating the virtual work to zero and therefore the coefficients of each independent displacement (y1 and y2) to zero gives NA=175N and NB = 425N 6.5 (i) Constraints on the system: length of the strings holding the masses is fixed (ii)

There is only one degree of freedom. This can be understood as follows. There are three variables needed to describe the system: The distance of two masses and one pulley from the ground. However there are two constraints: To of the strings have fixed lengths. Thus only one variable is left to change freely.

(iii)

In terms of the lengths shown in the figure

h1

M1 M2

y1

h2

y2

The constraint that the longer string has fixed length is expressed as

 h1  y1   2 h1  h2   3h1  2h2  y1  const.

81

The constraint that the shorter string has fixed length is expressed as h2  y 2  const.

(iv)

The constraint forces are the tension in the two strings. It is by these tensions that the constraints are maintained.

(v)

To apply the method of virtual work, we make a virtual displacement y1 of mass 1. The corresponding displacement of mass 2 is then y 2 . However, since there is only 1 degree of freedom, we will finally express y 2 in terms of y1 to apply the method of virtual work. The virtual work done in these processes is W   M 1 gy1  M 2 gy 2

Notice that the two virtual displacements are not independent. Therefore we should not conclude that M1=M2=0 for equilibrium. To apply the method, we first express the virtual work in terms of only y1 . From the first constraint equation, since h1 if constant,  2h2  y1  0  h2  

y1 2

From the second constraint we have h2  y 2  0  y 2  h2  

y1 2

Substituting these in the expression for the virtual work gives W   M 1 gy1  M 2 g

y1  M     M 1  2  gy1 2 2  

Equating this to zero then gives, for equilibrium M 2  2M 1

6.6 (i) Since the two blocks are free to move in one dimension without any constraint, the number of degrees of freedom for the system is 2. (ii)

The system in equilibrium is shown below. At equilibrium spring on the left is stretched by x1 and the total stretch of the two springs together is by x2. Thus spring on the right is stretched by (x2x1).

Thus force on mass m1 is k1 x1 to the left and

k 2 ( x 2  x1 ) to the right. These two forces must be equal for equilibrium but we wish

82

to obtain this by applying the principle of virtual work. Similarly the two force acting on m2 are F and k 2 ( x 2  x1 ) to the left. Again these two forces must be equal for equilibrium and we will obtain this by applying the principle of virtual work. k1

m1

k2

m2 F

x1

x2

As the mass m2 is moved by a virtual displacement  x 2 , let us assume that mass m1 moves by  x1 . Then the virtual work done will be W   k1 x1x1  k 2 ( x 2  x1 )x1  Fx 2  k 2 ( x 2  x1 )x 2

Equating this to zero and therefore the coefficients of  x1 and  x 2 to zero gives k1 x1  k 2 ( x 2  x1 ) and F  k 2 ( x 2  x1 )

This gives x1 

F  1 1    . and x 2  F  k1  k1 k 2 

6.7 (i) There is only one degree of freedom. Although the piston moves in the vertical direction and the wedge in the horizontal direction, their movements are connected because the piston moves on the surface of the wedge. (ii) Normal reactions on all the surfaces are the constraint forces. In the present context, the constraint force specific to the piston’s movement on the surface of the wedge is the normal reaction of the wedge surface on the piston. (iii)

if the distance of the middle line of the piston is at a distance a from the origin, its height is y and the distance of the left edge of the wedge is x (see figure) then the constraint is expressed as

83

m

x



F

y

a

y   a  x  tan 

  y   x tan 

Now the method of virtual work is applied by considering x as the free variable, varying it by x, and equating the total virtual work to zero. The virtual work is W   F x  mg y    F  mg tan   x

Equating the coefficient of x to zero gives F  mg tan  6.8 When the equilibrium angle is , the distances of various points (see figure) , taking A as the origin are as follows x B  l sin

 2

x D  l sin

 2

y C  2l cos

The force on the two points B and D due to the spring is   k  l 2  2l sin  2 

to the left on B and to the right on D

84

 2

A

x B

D

y



C W It is clear that we need only one parameter  to specify the system. Thus there is only one degree of freedom. Now let us make a virtual displacement by changing  by  . In that situation l  x B    cos 2 2

l  x D   cos 2 2

y C  l sin

 2

Thus the total virtual work done as  is changed by  is    l   l    W   k  l 2  2l sin  cos  k  l 2  2l sin  cos  mgl sin   2 2 2 2 2 2 2   

Vanishing of the virtual work then implies    l   l     k  l 2  2l sin 2  2 cos 2  k  l 2  2l sin 2  2 cos 2  mgl sin 2   0      

85

Or equivalently 2  2 sin

If

  mg     tan 2  kl  2

mg  mg   0 then the solution is   . For very small value of therefore we have    x kl 2 kl 2

, where x is very small.

Substituting in the equation above, we get x   mg  x    2  2 sin  45      tan  45   2   kl  2  

Since

x is small, we get by Taylor series expansion 2 x x 1  x  sin  45    sin 45   cos 45    1  2 2 2 2  x x  tan  45    tan 45   sec 2 45   1  x  2 2 

Thus the equation for equilibrium is 2 2

1  x   mg   1     (1  x) 2   kl  2



 1

x 

2



mg  mg   kl  kl

Or to a good approximation x

mg kl

2

Thus 

 mg  2 2 kl

6.9 (i) Even if we consider only one dimensional motion, we would require two variables, one the angle that the bar makes with the vertical and the other describing the position of the mass. However, the two are connected by a rope. Hence the two variables cannot vary independently since the length of the rope is foxed (constraint). Therefore there is only one degree of freedom in the system. (ii)

The constraint is the length of the rope remaining constant. It is enforced by the tension that develops in the rope. This tension makes the movement of the bar and the mass restricted. Thus it is the tension in the rope that is the force of constraint.

86

(iii)

A virtual displacement would be to displace the bar by an angle  from its equilibrium position. Since the length of the rope is fixed, the midpoint of the bar moves by the same distance as the mass connected to the rope. Therefore the tension does positive work at one end of the rope and exactly equal but negative work at the other end. The sum of the work done by the tension then vanishes.

(iv)

Let us say we make a virtual displacement of the bar by turning it counterclockwise by an angle from the vertical. Then the end of the rod moves by l in the direction of the force. At the same time, the virtual displacement of the mass is equal to the movement of the midpoint of the bar. Thus the mass moves by

l  2

opposite to the gravitational force (see figure) F 

m

The net virtual wok therefore is l W  Fl  mg  2

Equating this to zero then gives F

mg 2

6.10 Initially the bars are at 90. The weight has a tendency to fall down so the torque applied is such that it tends to pull the weight up. Thus is a virtual displacement of angle  is made in the direction of the torque, the weight will be lifted up by the corresponding distance y.

87

The corresponding virtual work done by the torque  is  and the corresponding work by the gravity is Wy. By the method of virtual work then we have   Wy  0

This gives  W

y 

Next we calculate the relationship betweeny and . As the rod is rotated by angle , the length of the horizontal diagonal decreases by

p  . If the corresponding angle at the 2

vertical corner changes from 90 to 90+α, then we have (see figure)

(90+α)



W

Change in the length of the horizontal diagonal 

= 2 l sin 

This should equal

90    l    l  l   2 l sin 45   l  cos 45      2 2 2 2 2 

p  so we have 2 l p    2  2

As the angle changes the length of the vertical diagonal changes by  90    l    l  l   2 l cos 45   l 2 l cos  sin 45      2 2 2 2 2  

Thus we have

88

y 

l 2

 

p  2

y p   2



This gives  

6.11 (a)

pW 2

For the motion in a plane, the orientation of the rod can be described by the

displacement of its centre of mass and the angle it has rotated by about its CM. Or equivalently the displacement of its two ends is sufficient to describe its orientation. Thus the degrees of freedom is 2. (b) We take the vertically down displacement of the two ends as the virtual displacement as shown in the figure.

yCM y1 y2

Let the centre of mass be at a distance

2L from the left hand end of the rod. In that case, 3

the centre of mass moves down by y CM  y1 

2  y 2  y1   1 y1  2 y 2 . The virtual work 3 3 3

done by the springs in such a virtual displacement is  ky1y1 and  ky 2y 2 respectively while

89

that by the weight of the rod is Wy CM 

W 2W y1  y 2 . Thus equating the net virtual work to 3 3

zero gives W 2W y1  y 2  ky1y1  ky 2y 2  0 3 3

Now equating the coefficient of each independent displacement to zero gives y1 

W 3k

y2 

90

2W 3k

Chapter 7 7.1

For Cartesian coordinates (x,y), the planar polar coordinates are

r

x2  y2

and

 y  . If y<0 and x<0, 270>>180 and if y<0 and x>0, >270.  x

  tan 1 

Unit vectors are given by rˆ 

7.2

xiˆ  yˆj  yiˆ  xˆj and ˆ  r r   At each point the velocity v is given as  v .rˆ  rˆ   v .ˆ  ˆ . Thus for (iv)



 v  2iˆ  3 ˆj

for a particle at (-2,-3), the velocity is

  2iˆ  3 ˆj     2i  3 j  rˆ    2iˆ  3 ˆj   3i  2 j  ˆ   5rˆ  12 ˆ

ˆ

ˆ

13

7.3

ˆ

ˆ

13

The two points with polar coordinates

13

 r1 , 1  and  r2 ,  2 

and the corresponding

vectors are shown in the figure below

 r2

ˆ2

  r1  r2

ˆ1 2

 r1

(21)

1

(i) As is clear from the figure, the angle between the vectors r1 and r2 is   2  1  . This also 



the angle between unit vectors rˆ1 and rˆ2 and unit vectors ˆ1 and ˆ2 . Therefore rˆ1  rˆ2  ˆ1  ˆ2  cos  2  1 

91



Similarly, angle between r1 and ˆ2 is

     2  1  and that between r2 and ˆ1 is 2

    2  1  . Therefore 2  rˆ1  ˆ2  cos   rˆ2  ˆ1  cos 

(ii) (iii)

7.4

    2  1    sin   2  1  2     ( 2  1 )   sin   2  1  2 

  Similarly, from the figure r1  r2  sin   2  1  zˆ   r1  r2 

r12  r22  2r1 r2 cos  2  1 

The trajectory of the projectile is shown schematically in the figure below. The horizontal distance from the origin to the point of highest elevation (height 1.25m) is 2.5 3m .

The vertical component of the velocity at this point vanishes while the

horizontal component is 5 3ms -1 .

Similarly, the vertical component of the

acceleration is 10ms-2 vertically down. Also shown in the figure are unit vectors in the radial and the tangential directions at the highest point and on the ground.

ˆ

rˆ ˆ

1.25m



rˆ

2.5 3m From the figure it is clear that for the point of highest elevation tan  

1.25 2.5 3



1 2 3

;

sin  

1 13

;

cos   2

3 13

Therefore the radial and tangential components of the velocity at the highest point are

92

v r  5 3 cos   5 3  2

3 30  ms 1 13 13

v  5 3 sin   5

3 ms 1 13

Similarly the radial and tangential components of the acceleration are a r  10 sin   

10 13

a  10 cos   20

ms  2

3 ms 2 13

On the ground, the radial unit vector points towards positive x direction and the tangential unit vector is in the negative y direction.

Thus the radial component of the

velocity is its x component and the tangential component is its y component. Therefore v r  5 3ms 1

v  5ms 1

Similarly, since the gravitational acceleration on the ground is in the negative y direction, its radial and tangential components are a  10ms 2

ar  0

7.5

The position of the particle at time t is shown in the figure below.

2ms-1

1

 2t The polar coordinates of the particle at time t are given as  1   2t 

  tan 1 

r  4t 2  1

Therefore r 

4t

  

4t  1 2

This gives the velocity in polar coordinates as

93

2 4t 2  1

 v  rrˆ  rˆ 

4t 4t  1 2

rˆ 

2

ˆ

4t 2  1

Differentiating the velocity vector gives  dv  dt

4 4t  1 2

Now substituting rˆ  ˆ  

rˆ 

 4t

16t 2 2



1

32

4t

rˆ 

4t  1 2

rˆ 

 4t

8t 2



1

32

ˆ 

2 4t  1 2

 ˆ

2 2  ˆ and ˆ   rˆ  2 rˆ , we get the acceleration above to 4t  1 4t  1 2

be zero. 7.6

(i) Kepler’s second law states that the rate of the area swept (

) A

by the radius vector is

constant. This can be expressed as (see figure for the symbols used)   1 r 2  constant A 2

 r 

Sun

Now differentiating the equation above with respect to time gives 1 2  r   rr  0 2



r 2  2rr  0

The expression on the right is the tangential acceleration. Thus Kepler’s second law gives tangential acceleration to be zero. (ii)

Since the tangential acceleration is zero, the force is only in the radial direction. Acceleration in the radial direction is

94



a r  r  r 2 We differentiate the orbit equation r  r 



r0 with respect to time to get 1  e cos 

r0 e sin 

1  e cos   2



r 2e sin  2 A e sin   r0 r0

 is constant, differentiating the equation above once more with respect to time, we Since A

get

 r  2 A  0  1  r 2 A e cos  4 A 2  r0    r     1  2 r0 r0 r0 r  r  This gives 4 A 2  r0   1  r 2  2 r  r  2 4 A  r0 4 A 2    1  r   r0 r 2  r r4   4 A 2  1  2   r  0  r

ar 

This shows that the force is proportional to r2. 7.7

It is given that mx   F sin  t my  F cos  t

which gives F  my sec  t

Now write y  x tan t and differentiate it twice with respect to t to get y  x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t

This gives





F  my  m x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t sec t

Substitute this in mx   F sin  t to get





x   x tan t  2 x  sec 2 t  2 2 x sec 2 t tan t sin t sec t

Multiplying and rearranging terms gives

95

x  2 x tan t  2 2 x tan 2 t

The equation does not appear to be easily ingrable. 7.8 It is given that r  2t and   1rad s 1 Since r (0)  1 , integrating the equation for r gives

(i) r (t )

t

1

0

2  dr   2t ' dt '  r (t )  1  t or equivalently r (t )  t 2  1



(ii) v  rrˆ  rˆ  2trˆ  (t 2  1)ˆ If the velocity vector is at 45 to the radius vector, we have

2trˆ  (t

  v .rˆ  v cos 45 

2



1  1)ˆ .rˆ  2



4t 2  (t 2  1) 2



This gives 2t 

4t 2  t 4  1  2t 2 t 4  6t 2  1  2 2

Squaring both sides gives 8t 2  t 4  6t 2  1 OR t 4  2t 2  1  0

This is equivalent to

t (iii)

2



2

 1  0  t  1s

Radial acceleration zero implies





a r  r  r 2  0 r  2t  r  2 which leads to





2  t 2  1  0 giving t  1s

This also gives the distance from the origin to be r (t  1)  1  1  2m 7.9 Free body diagram of the bead when its radius vector is making an angle  (increasing as the particle slides down) from the vertical is shown below

96

N R



 mg

Taking the components of the forces in the radial and tangential directions and equating these to mass times the radial and tangential components of the acceleration, respectively, gives



In the radial direction

N  mg cos   m r  r 2

In the tangential direction

mg sin   m r  2r







Since the particle moves on a path of constant radius r  R , we have r  r  0 When substituted in the equations above, this gives N  mg cos    mR 2 g sin   R

1 d 2 Now using   , we get from the second equation above 2 d

R 

R d 2  g sin  2 d

which gives upon integration 

R 2   g  sin  d  g 1  cos    R 2  2 g 1  cos   2 0

When substitutes in the equation N  mg cos    mR 2 , this gives N  m 3 g cos   2 g 

7.10

The free body diagram of the bead at equilibrium is shown in the figure below.

97

N

 mg



The horizontal components of the normal reaction N provides the centripetal force while the vertical component balances the weight of the bead. Thus N sin   mx 2 N cos   mg

Dividing the first equation by the second one gives tan  

x 2 g

The slope of the curve is also equal to tan  . Thus x 2 dy   4cx 3 g dx

This gives 2   4cg 2 cg

x

7.11

Since  

4 and y  cx 

4 16cg 2

1 2 t , we have 2

r  2  t 2 ,

r  2t





and

r  2 ;   t

and

  1

2 2 2 Thus a r  r  r 2  2  t 4 and a   r  2r   t  4t  5t

7.12

We know that rˆ  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ

ˆ  cos  cos  iˆ  cos  sin  ˆj  sin  kˆ

ˆ   sin  iˆ  cos  ˆj

Therefore 98

 ˆ   sin  cos  iˆ   cos  sin  iˆ   sin  sin  ˆj   cos  cos  ˆj   cos  kˆ







  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ   cos   sin  iˆ  cos  ˆj   rˆ   cos  ˆ

Similarly  ˆ    cos  iˆ  sin  ˆj 



  sin  rˆ  cos  ˆ

7.13



Since m3 y 3  T2  m3 g

We get from the solution y3 

4m1 m2   m1  m2  m3 4m1 m2   m1  m2  m3

T2  m3



g

4m1 m2   m1  m2  m3 g  m3 g 4m1 m2   m1  m2  m3

8m1 m2 m3 g 4m1 m2   m1  m2  m3

From T2  2T1  0 , we get

T1 

4m1 m2 m3 g 4m1 m2   m1  m2  m3

Now from m1 y1  T1  m1 g we get

y1 

3m2 m3  4m1 m2  m1 m3 g 4m1 m2   m1  m2  m3

And from m2 y 2  T1  m2 g we get

y 2 

3m1 m3  4m1 m 2  m2 m3 g 4m1 m 2   m1  m2  m3

7.14

99



m

l

 T2

T1

(i) The tension in the outermost string provides the centripetal force for the outermost bead. Let the tension in this string be T1. Then T1  Nml 2

Similarly, centripetal force for the second bead from outer end is provided by the difference in the tension T2 in the second string and tension T1 in the first string. Thus T2  T1  ( N  1) ml 2

 T2   2 N  1 ml 2

T3  T2  ( N  2)ml 2

 T3   3N  3 ml 2

T4  T3  ( N  3) ml 2

 T4   4 N  6 ml 2

Extending further

and so on. In general we can write for the ith string from the outside, Ti  Ti 1   N  (i  1) ml 2

 Ti  2   N  (i  2) ml 2   N  (i  1) ml 2  

This is then easily seen to be Ti   iN   (i  1)  (i  2)  (i  3)    2  1  0  ml 2 i (i  1)     iN  ml 2  2   2 iml   2 N  i  1 2

(ii)

Now we generalize the result of part (i) to a rope of length L and mass per unit length λ. To make the transformation from the problem in part (i) to this problem, we consider the mass of each bead to be distributed over the connecting string whose length we take to be vanishingly small i.e. l  0 . Thus we have L  Nl , x  ( N  i )l ,  L  x   il and m  l

Substituting this in the expression for Ti above, we get

100

( L  x) 2 (2 Nl  il  l ) 2  2  ( L  x )(2 L  L  x  l ) 2  2  ( L  x )( L  x  l ) 2

T ( x)  Ti 

On taking limit l  0 , we then get T ( x) 

 2 2 (L  x 2 ) 2

To get this answer by considering the rope directly, we take a small portion of the rope of length x at a distance x from the centre O. The centripetal force to it is provided by the difference in the tension at its two ends, as shown in the diagram below T(x+x)

T(x) Then T ( x)  T ( x  x)  x 2 x 

T ( x)  T ( x  x)   2 x x

This gives the differential equation 

dT   2 x dx

The solution of this equation is T ( x)  C 

 2 x 2 2

where C is the integration constant. With the condition that the tension vanishes at x=L, we then get C

 2 L2 2

and

T ( x) 

 2 2 L  x2  2

Consider a small section of the rope making a small angle  at the centre of the

7.15

loop. The force at its two ends due to the tension in the rope is shown in the diagram below

101

/2  T

T Tsin(/2)

For small angle the forces at the ends give a net force towards the centre which is equal to      2T  T 2  2

2T sin 

This provides the required centripetal force for the segment. Therefore T  R  R 2

7.16

 T  R 2  2

The relevant coordinates and the free body diagrams of the two masses are shown below

r T m

 y

T

mg

M Mg

102

We treat the mass m using the polar coordinates as shown in the figure. The total number of unknowns in the problem are : y coordinate of mass M, polar coordinates of mass m and tension T in the string. The corresponding equations are M y  Mg  T m r  r 2  mg cos   T m r  2r  mg sin 

 





And the constraint equation r  y  constant

 r  y  0 and r  y  0

7.17 (i) After time t, the end of the rod that was at the origin has moves by a distance . Thus the relationship between the x and y coordinates will be y  tan  1 2 x  At 2



y cot   x 

1 2 At 2

(ii) Free body diagram of the bead is shown below

θ θ

N

Thus the equations of motion are m x  N sin  m y   N cos 

From the constraint equation A  y cot   x

Thus N sin   m A  y cot  



N  mA cos ec  my cot  cos ec

which gives N cos   mA cot   my cot 2 

103

1 At 2 2

Thus my   mA cot   my cot 2 

 my cos ec 2   mA cot 

Or y   A cos  sin   

A sin 2 2

Integrating the equation above with the condition y (0)  0 and y (0)  d gives

(iii)

y (t )  d 

Thus the bead will take time t 

A sin 2 t 2 4

4d to reach the lower end. A sin 2

Suppose the lower corner with angle  is at the origin at t=0. Then if the position of

7.18

the mass is (x, y) at time t, the relationship between these coordinates is y  tan  1 2 x  At 2



y cot   x 

1 2 At 2

if the acceleration of the wedge is A. This implies A  y cot   x

(i)

Now if the mass falls vertically down, y   g which gives A  g cot  i.e. the wedge accelerates to the right with acceleration g cot  .

The interpretation is

quite simple: when the wedge moves by horizontal distance should move vertically down by

minimum acceleration is (ii)

1 At 2 , the mass 2

1 2 gt and the relationship between the two at 2

1 gt 2 2  tan   A  g cot  . 1 At 2 2

For the particle not to move with respect to the wedge, we have y  0

which implies x  A .

104

The free body diagram of the mass is as follows Y N

θ

X

Thus we have by the conditions above N cos   mg  N sin   mA

which gives A   g tan 

7.19

From example 7.9 mx2  N 1 sin  M x1   N 1 sin  y 2  ( x1  x2 ) tan 

N1 

mg cos  m   2  1  sin   M  

This gives (assuming x 1 (0)  x1 (0)  0 ) m g cos  sin  x1   M m   sin 2    1 M  

Similarly x2 

g cos  sin  m   sin 2    1 M  

This gives

105



m 2  g sin  M m  1 sin 2   M 

 1

y 2   

  

Thus if mass m starts from height h, it will take time 2h (m  M cos ec 2 ) (m  M ) g

2h  y 2

t

And at this time the speed of the wedge will be x1t  m cot 

2 gh (m  M )( m  M cos ec 2 )

7.20 The coordinates used in solving the problem are shown in the figure below P1 P2 T h1

T m

T T

P3

y1

h2

m y2

y3

Let the tension in the string be T. The equations of motion for the two masses are my1  T  mg my 2  T  mg

And since the pulley P3 is massless, we have 2T  k  y 3  h   0

Thus we have four unknowns y1, y2, y3 and T but only three equations. One more equation is provided by the constraint equation. The constraint is that the length of the string is a constant. If the heights of the two fixed pulley are h1 (for P1) and h2 (for P2), the constraint is expressed as

106

 h1  y1    h1  y3    h2  y3    h2  y 2   const. or equivalently as y1  2 y 3  y 2  const. 

y1  y 2  2 y 3

Now adding the equations of motion for the two masses gives

m y1  y 2   2T  2mg On substituting y1  y 2  2 y3 and 2T  k  y 3  h  from the equations above, we get  2my 3  ky 3  kh  2mg

This gives y 3 

k kh   y3   g   2m 2m  

The general solution of the equation above is the sum of its homogeneous solution yh(t) and the particular solution yp(t). We have y h (t )  A cos t  B sin t y p (t )  h 

2mg k

Here A and B are two constants to be determined by the initial conditions and   the general solution is 2mg   y 3 (t )  A cos t  B sin t   h   k  

Now the initial condition is y 3 (t  0)  h

y 3 (t  0)  0

These give A

2mg k

B0

Thus y 3 (t )  h 

2mg  1  cos k 

k 2m

This immediately gives through the equation 2T  k  y 3  h   0  T  mg  1  cos 

107

k 2m



t  



t  

k . Thus 2m

This gives k t 2m

y1   g cos

This is easily integrated. Since it is gives that y 1 (t  0)  0 and y1 (t  0)  h , we get y1 (t )  h 

2mg   1  cos k 



k 2m

t 

k 2m

t 



In exactly tha same manner we get y 2 (t )  h 

2mg k



 1  cos  

 

7.21 (i) Wedge m3 is free to move only in the horizontal direction; m1 and m2 move both horizontally as well as vertically. Thus we would have had 5 degree of freedom. However, there are three constraints: The length of the string is fixed, mass m 1 moves only in the vertical shaft and mass m 2 moves on the plane of the wedge.

These

constraints reduce the degrees of freedom to 2. Thus there are only two degrees of freedom. (ii) The origin and the coordinate axes chosen to describe the motion are given in the figure below. Also shown are the free-body diagrams of the three masses

m1 m2

y1

m3

x1 x2

x3

108

y2



T

T

N2

T N1

N1

T

N2

m3

m1g m2g

m2g N3

We see that there are in total 8 unknowns: x1, y1, x2, y2, x3, N1, N2, N3 and T. The equations of motion are: mx1  N 1 (i) my1  T  m1 g (ii) m3 x3   N 1  N 2 sin   T cos  m2 x2  N 2 sin   T sin  (iv) m 2 y 2  N 2 cos   T sin   m2 g

N 3  T 1  sin    N 2 cos   m3 g

(iii) ( v) ( vi)

These are the equations of motion. In addition there are three constraint equations.

y 2   x3  x 2  tan 

y 2   x3  x2  tan  ( vii)  h  y1    x2  x1  sec  const.   y1   x2  x1  sec  0  x3  x1   const.  x1  x3 (ix ) 

( viii)

In the above, h is the height of the wedge. There are a total of nine variables and nine equations. Of these equation (vi) is not relevant for the motion since the wedge moves only in the horizontal direction. Equations (i), (iii), (iv) and (ix) give m3 x1   m1 x1  N 2 sin   T cos   m1 x1  m2 x2

which gives

 m1  m3  x1  m2 x2  0 This is the equation expressing momentum conservation in the horizontal direction. Similarly equation (iv) and (v) along with (ii) give

109

m2 x2 cos   m2 y 2 sin   T  m2 g sin   m1 y1  m1 g  m2 g sin 

Now substituting for y1 from (viii) and y 2 from (vii) and using (ix), we get

m2 x2 cos   m2  x1  x2  tan  sin   m1  x2  x1  sec   m1 g  m2 g sin  Which is equivalent to m2 x2 cos 2   m2  x1  x2  sin 2   m1  x2  x1    m1  m2 sin   g cos  

 m1  m2  x2   m1  m2 sin 2   x1   m1  m2 sin   g cos 

Now using  m1  m3  x1  m2 x2  0 , we eliminate x2 from the equation above to get



 m1  m2  m1  m3  





x1  m1  m2 sin 2  x1   m1  m2 sin   g cos 

m2

This gives

x1 

m2  m1  m2 sin   g cos   x3 m12  2m1m2  m1m3  m2 m 3  m22 sin 2 

And using  m1  m3  x1  m2 x2  0 , we then get

x2  

 m1  m3  m1  m2 sin   g cos  m  2m1 m2  m1m3  m2 m 3  m22 sin 2  2 1

Now using equation (viii) we get

y1  

 m1  m2  m3  m1  m2 sin   g m  2m1 m2  m1 m3  m 2 m 3  m22 sin 2  2 1

Using equation (vii), we get

y 2 

 m1  m2  m3  m1  m2 sin   g sin  m12  2m1 m2  m1 m3  m2 m 3  m22 sin 2 

As m3   , we get

x1  x3  0 x2  

 m1  m2 sin   g cos  m1  m2 

y1  

 m1  m2 sin   g  m1  m2 

y 2 

 m1  m2 sin   g sin  m1  m2 

7.22 Consider two parts of the rope x and (L-x) in length. The free body diagrams (showing only the horizontal forces) of the rope and these two parts are shown below

110

x

F

Friction

T(x)

F Friction

T(x) Friction

Since the rope is moving with constant speed, there is no net force on the rope or any part of it. Thus from the free body diagram of the rope F  Mg

From the free body diagram of the left portion of the rope F  T ( x)  

 L  x  Mg M xg  0  T ( x)   L L

Or from the free body diagram of the right portion of the rope T ( x)  

 L  x  Mg L

It is also instructive to solve the problem by considering a small portion of length dx at distance x from the left and balancing the forces there to get a differential equation for T(x). The free body diagram of such a portion is as given below dx T(x)+dT(x)

T(x) Friction T ( x )   T ( x )  dT ( x )  

dx Mg L



dT ( x ) M  g dx L

With the condition that T(L) = 0, the equation above can be integrated to get

111

T ( x)

x

M 0 dT   L g L dx



T ( x) 

 L  x  Mg L

7.23 The force applied should be such that the frictional force on mass m is sufficient to balance its weight. Free body diagrams of the two blocks are shown below

friction F

N

M

mg

N friction Mg

If the entire system is moving with acceleration a then N  Ma F  N  ma

This gives a

F M m

N

and

M F M m

If mass m is not falling then friction  N  

 M  m  mg M F  mg  F  M m M

Thus Fmin 

 M  m  mg M

For m=16kg, M=88kg, =0.38 and g=9.8ms-2, we get Fmin 

104  16  9.8  487.7 N 0.38  88

112

Normal reaction

7.24 The forces on the bead are: Its weight, normal reaction NV in the vertical direction and normal reaction NH in the horizontal direction. Thus the free body diagram of the bead is as follows (assuming the rotating arm is going into the page)

NV friction

rˆ

NH mg

These force provide the radial and the tangential acceleration given by









 a  r  r 2 rˆ  r  2r ˆ

Since the rod is rotating with a constant angular speed , we have





 a  r  r 2 rˆ  2rˆ

(i) If the bead is stationary at r=R, we have  a   R 2 rˆ

This gives N V  mg

NH  0

friction  mR 2

Since friction    N , we get for stationary bead mR 2  mg   02 

(ii)

g R

If    0 and negligible weight of the bead, we have N V  mg

N H  2mr

N  m g 2  4r 2 2  2r

friction   N  2 mr

The minus sign for the friction shows that since the bead slides outwards, the frictional force in inwards. The equation describing the motion of the bead is then r  r 2  2 r 

OR

r  2 r   2 r  0

Assuming a solution of the form r (t )  e t and substituting it in the equation above we get 2  2    2  0

Solution of this equation gives

113



1   1   2  



and





 2   1   2  

Thus the general solution is

 

r (t )  A exp t



 

1   2    B exp  t

1  2  



Here A and B are to be determined by the initial conditions that r (t  0)  R and r (t  0)  0 . Thus A B  R



 

1  2   A 



1  2   B  0

The solution is A

R  2 

1   2    1  2

B

and  

R  2

1   2    1  2



 

Thus the distance of the bead from the center is given as

r (t ) 

R 2 1 

2



  

1   2   exp t

 

1  2   

  

1   2   exp  t

1  2  

 

7.25 The solution for the distance travelled by the particle is x (t ) 

F k





m  k m  t   t  k 1  e  

For k=0, we can’t substitute this directly in the formula since we are dividing by k in the expression above. Thus we take the limit k  0 . In this limit, the first nonzero term we get is  m k 1 k2 2  1  1  t  t  higher order terms  2 k  m 2m  1 k 2  t 2m

t

This gives x (t ) 

7.26 The equation of motion for the particle is 114

1F 2 t 2m

y   g 

k y OR m

y 

k y   g m

The solution of the homogeneous equation y 

k y  0 m

is k   y  A exp  t   m 

Here A is a constant to be determined by the initial conditions. The particular solution is y  

mg k

Therefore the full solution is k  mg  y  A exp  t   k  m 

Now the initial condition is y (t  0)  v0 . This gives A  v0 

mg k

Thus mg  k  mg   y (t )   v 0   exp  t   k  k   m 

This is easily integrated to get y(t) also. Thus y (t )  

mg m mg  t   v0   k k k 



 k    1  exp  m t      

When the ball reaches the highest point, its speed is zero. If this time is tup, then mg    k  mg 0   v0   exp  t up   k  k   m 

mg k 1  k   exp  t up    mg    kv   m   v0    1  0  k  mg   

OR t up 

kv  m  ln 1  0  k  mg 

This gives the height h to be kv0  mg m  m mg   h  ln 1   v0  k k mg k k   

115

 1    1 kv0   1  mg  









 

mv0 kv 0  m2 g    ln 1  k mg  k2 

If the total time of flight is T then T=tup+tdn, where tdn is the time taken to come down from the highest point. At time T, y(T)=0. Therefore 0

mg  t up  t dn   m  v0  mg  k k k 



 k   k    1  exp  m t up  exp  m t dn        

Thus tdn will be given by solving 

mg mg t up   t dn k k

mg k m mg   k     v0  exp  t dn    1 mg  k  k   m     v0    k   

     

OR t up  t dn 

1 mg  k      1  exp  t dn     v0  g k  m      t dn 

v0 m  k      1  exp  t dn   g k  m   

To understand whether tup is larger or tdn is larger, let us see the time change in the limit of very small k. In that case (up to order k) t up 

 kv  m  kv 1 k 2 v02 v 0 1 kv02 m  ln 1  0    0   ......    k  mg  k  mg 2 m 2 g 2 g 2 mg 2 

And tdn is given by solving t up  t dn 

v0 m   k     1  exp  t dn   g k   m  

v0 m  k 1 k2 2    1  1  t dn  t dn  g k  m 2 m2  v 1 k 2  0  t dn g 2m  t dn 

This gives 2mv0 2m 2  kv  t   2 ln 1  0  kg mg  k   2mv0 2m 2  kv0 1 k 2 v02 1 k 3 v03   2     ....  2 2 3 3 kg 3m g k  mg 2 m g  2 dn



v02 2 kv03  g 2 3 mg 3

Therefore

116

t dn

v  0 g



2 kv 0   1   3 mg  

12



v 0 1 kv02  g 3 mg 2

A comparison shows that tup is smaller than tdn. This makes sense because the while coming down, the average speed is smaller since the particle has lost energy due to viscosity and continues to do so. This also gives the total time of flight to be (up to order k) 2v0 5 kv02   g 6 mg 2

T  t up  t dn

However, this approximation will be valid only if

kv02  1 mg 2

1 5.27 It is given that v 0  100ms and   45 . Therefore v 0 sin   v0 cos   50 2ms 1 . It is

also given that g  10ms 2 . v02 sin 2   250m (i) Height = 2g

Range =

2v 0 sin  v cos   1000m g

(ii) When k  0 , from the expression derived in the problem above, we get with the initial vertical speed v 0 sin  mv0 sin  m 2 g  kv sin   2 ln 1  0 Height = k mg k    

Substituting the values , we get For k=0.1, Height = 202m For k=0.2, Height = 172.3m To find the range, we first fine the total time of flight and then use formula derived in example 7.13 to find the horizontal distance travelled. From the solution of the previous problem, we know that the time of flight T is given by solving mg m mg  T   v0 sin    k k k 



 k    1  exp  m T      

For k=0.1, this gives 200T  54141  exp(0.05T )

OR

117

T  27.071  exp(0.05T )

Since the time of flight without drag is

2v 0 sin   14.14 s , and as the result of the problem g

shows the time of flight becomes smaller for k  0 , we tabulate T and the right hand side of the equation above to find T for T  14

It is clear from the Table above that the expression on the right was smaller than T till T=13s and becomes larger at 12.5s. Thus the time of flight will be between 13 and 12.5s. A little more tabulation gives T=12.8s. Substituting this in x(T )  Since t up 





mv o cos   k / m  T 1 e , we get Range = 669m k

kv sin  m  ln 1  0 k  mg 

  6.05s , we get tdn= 6.75s. This confirms numerically that time 

taken to come down is greater. For k=0.2, the equation to determine the time of flight is 100T  1071  exp(0.1T )

Making a table like above

x (T ) 



T 14 13.5 13 12.5

OR

T  17.071  exp(0.1T )

27.071  exp(0.05T )

13.62 13.29 12.93 12.58



mv o cos   k / m  T 1 e  490m k

118

gives T=11.8s and Range

(iii)

k=0

k=0.1

k=0.2

(iv) When drag is introduced, it is the range that is affected much more than the height. (7.28) In the problem above, we have already found the range for   45 . Let us now take the case of k=0.1 and find the range for   40 and   50 . For   40 , the equation

119

mg m mg    k   T   v0 sin     1  exp  T   k k k    m  

beomes 200T  52861  exp  0.05T   or T  26.431  exp  0.05T  

and gives T=11.7s. This gives a range of x (T ) 





2  100 cos 40 1  e 0.0511.7 = 678m. 0 .1

Thus for   40 , the range increases. For   50 , the equation mg m mg  T   v0 sin    k k k 



 k    1  exp  m T      

beomes 200T  55321  exp  0.05T   or T  27.661  exp  0.05T  

and gives T=13.8s. This gives a range of x (T ) 





2  100 cos 50 1  e 0.0513.8 = 641m. 0.1

Thus for   50 the range decreases. The two calculations above show that for maximum range, the projectile should be launched at an angle less than 45 . The reason for this is as follows. Since the horizontal speed reduces as the projectile moves, it should cover a larger distance in the initial part of the flight. For this it is better to have a relatively larger horizontal component of the velocity compared to the case when there is no drag. Thus the angle should be smaller than 45 . (7.29) In this case the drag force is proportional to the square of the speed. So the equation of motion will be given as follows for the motion up and motion down (taking vertically up direction to be the positive y direction)

120

Motion up

my   mg  ky 2

Motion down

my   mg  ky 2

2 Since we are only interested in height, we change y to 2 dy  y  to get the speed as a

1 d

function of the vertical distance of the ball from the ground. The first equation in that case is

 

1 d k y 2  y 2   g 2 dy m

The solution of this equation is the sum of the solution

 

1 d k y 2  y 2  0 and a particular solution y 2 2 dy m

p

( y)

2 y

h

( y ) of

the homogeneous equation

. These solutions are

mg  2k  y 2 ( y )  A exp  y  and y 2 ( y )   h p m  k 

Here A is a constant to be determined from the initial conditions. The full solution therefore is mg  2k  y 2 ( y )  A exp  y  m  k  2 2 The initial condition is that y ( y  0)  vi . This gives

A  vi2 

mg k

This the speed of the ball as it moves up is  2k y 2 ( y )  vi2 exp  m 

 mg y  k 



 2k  1  exp  m  

  y   

At the maximum height h, the speed becomes zero. Therefore  2k  mg   2k   0  vi2 exp  h  1  exp  h  m  k  m    

This gives

h

kvi2  m  mg k  m     ln  ln 1  2k  mg k  vi2  2k  mg 

Now we consider the motion for downward motion. This can be rewritten as

 

1 d k 2 y 2  y   g 2 dy m

121

2 y

Again the solution of this equation is the sum of the solution 1 d k 2 2 2 equation 2 dy  y   m y  0 and a particular solution y

p

( y)

h

( y ) of

the homogeneous

. These solutions are

mg  2k  y 2 ( y )  A exp y  and y 2 ( y )  . Thus the complete solution is h p k  m  mg  2k  y 2 ( y )  A exp y  k  m 

Now the initial conditions are y ( y  h)  0 . This gives, with h 

kvi2  m    ln 1  2k  mg 

 kvi2  mg mg k    0  A 1   A mg  k kv 2  1 i mg This gives y 2 ( y )  

mg k  2k  mg exp y  2 k kvi  m  1 mg

If the final speed is vf, then v 2f  

vi2 mg k mg   k kv 2 kv 2 1 i 1 i mg mg

2 2 Note that of k=0 then v f  vi The answer can also be written as

1 1 k  2  2 mg v f vi

122

Chapter 8 8.1 Since there is no external force on the system in the horizontal direction, the total momentum in the horizontal direction is conserved. Initial momentum in the horizontal direction = momentum of the carriage + momentum of rain = Mv  0 = Mv Final momentum of the system after time t

=  M  mt  v f

Here vf is the final velocity. Equating the two moment gives vf 

Mv  M  mt 

8.2 Since the water leaking out of the carriage still has a horizontal velocity equal to the velocity of the carriage, total momentum of water after it came out for time t is = mtv If the initial amount of water in the carriage was m0, then the initial momentum of the system (carriage + water in it) =  M  m0  v If the aped of carriage (with left over water in it) after time t is vf, then by momentum conservation

 M  m0  mt  v f

 mtv   M  m0  v  v f  v

8.3 Exactly like in problem 8.2, there will be no change in the speeds of the two bicyclists. This is easily done by considering the momentum of the two friends before and after the books are given by one of them to the other person. Consider the person giving the books. Her momentum before transferring the books is Mv . After she gives the books, let her speed by vf. Then by momentum conservation Mv  mv   M  m  v f

 vf  v

Similarly, for the person receiving the books Mv  mv   M  m  v f

 vf  v

8.4 Conserve momentum after the first bullet has been fired. Initial momentum is 0. Let the velocity (since the motion is one dimensional, we write only the symbol for it, the direction is taken care of by the sign) of the gun after the bullet is fired be v1. Since the relative velocity of the bullet when it leaves the gun is u, and the bullet leaves the gun when the gun

123

is already moving with v, bullet’s speed ug with respect to the ground is calculated as follows: u  u g  v1

 u g  u  v1

Therefore momentum conservation gives

 M 0   N  1 mv1  m u  v1   0

 v1  

mu M 0  Nm

Now the momentum of the gun and  N  1 bullets in it is   M 0   N  1 m 

mu  M 0  Nm 

Now let the speed of the gun after the second bullet is fired be v2. Then momentum conservation gives

 M 0   N  2 mv2  m u  v2    M 0   N  1 m  v2  

mu  M 0  Nm 

mu mu  M 0  Nm M 0   N  1 m

Similarly one can now show that if the speed after the third bullet is fired is v3 then v3  

mu mu mu   M 0  Nm M 0   N  1 m M 0   N  2  m

Generalizing this we get after N bullets have been fired

v final 

k  N 1

 M k 0

0

mu  ( N  k )m 

8.5 (i) Momentum of the system = sum of the momentum of each particle = 0.2iˆ  0.4 ˆj kg ms1 (ii)

velocity of the centre of mass = total momentum/total mass



1 0.2iˆ  0.4 ˆj 0.3 2 4 ˆ  iˆ  j ms 1 3 3 

8.6 (i) acceleration of the CM

124



 Fnet  total mass iˆ  ˆj  0.3 10 ˆ ˆ  i j 3



(ii)



No, the acceleration is not in the same direction as the momentum of the CM.

8.7 If the base of the cylinder is in the xy plane, as shown in the figure, the x and y coordinates of the CM are (0, L/2). We thus have to calculate the z coordinate of the CM.

z y L x R

To obtain the z coordinate of the CM, consider a rectangular sheet of thickness dz at height z, as shown in the figure below.

z

R

If the density of the material that the cylinder is made of is , the z coordinate of the CM, by definition, is

125

R

z CM 

  2 L R 2  z 2 zdz 0

L R 2 2



4 R 2

R



R 2  z 2 zdz

0

To evaluate the integral, we substitute z  R sin  and dz  R cos d . This gives z CM 

4 R 2

 2

 R cos   R sin   R cosd 0

1



4R cos 2 d  cos    0



4R 3

8.8 (i) CM of a cone shown in the figure below

h

r z

R 2

The CM is on the axis of the cone by symmetry. To calculate its height, we take a thin disc of thickness dz at height z. By similarity of triangles, it radius r is given by r R  hz h

 r   h  z

R h

If the density of the material that the cone is made of is , then the position of the CM is given by h

z CM 

  z   r 2 dz 0

1 2 R h 3

h





3 R2 2 h  2  2 h  z 2  2hz zdz  4 R h0h

It is reasonable that the location of the CM is more towards the base since larger mass of the cone is concentrated there. 126

(ii)

Hemispherical bowl of radius R is shown in the figure below.

r R

z



The CM will be on the line passing through the centre of the base. To calculate its height zCM, we take a ring of height dz at height z. According to the figure z  R sin  and dz  R cos  d ,

z , R

sin  

cos  

r  R

R2  z2 R

If the mass per unit area for the shell is , then the mass dm of the ring is dm  2 r  R d  2 R 2  z 2 

dz  2 R  dz cos 

Thus zR

z CM 

 zdm

z 0 zR

2R 

zR

 zdz

z 0 2

2R 

 dm



R 2

z 0

N



8.9 Given N particles of masses mi (i=1-N) with total mass M   mi at positions ri (i=1-N), i 1

position of their CM

 RCM

is given as N

 RCM 

  mi ri i 1 N

m i 1

i

127

N





m r i 1

M

i i

Now let us make m subsystems of these masses with number of particles N1, N2, N3……Nm in Ni

them. Then we have the mass of each subsystem as M i   mi . The position of the CM can i 1

then be written as N1

N

 RCM 

  mi ri i 1 N

m i 1



 N2  m r  i i   mi ri  ....... i 1

i 1

M

i

By definition of the CM we have for the position

 RCMi of

Ni   each subsystem M i RCMi   mi ri . i 1

Thus the expression above can be written as N

 RCM 

  mi ri i 1 N

 mi i 1



m



M 1 RCM 1  M 2 RCM 2  ..... m

Mi



M R i

i 1

i 1

CM i

m

M i 1

i

This shows that the CM of the system can be calculated by treating each susbsystem as a point particle of mass Mi located at the CM of each subsystem. 8.10

To find the CM, we will treat the cone and the hemisphere as two subsystems. It is also clear by symmetry that the CM will be on the extended axis of the cone. Taking the axis as the z direction with z = 0 at the base of the cone, we have z CM 

mass of the sphere  z CM ( sphere)  mass of the cone  z CM (cone) total mass

Assuming the entire system is made of a material of uniform density, we get

z CM 



3R1 2R13 H R22 H    4 2 2 8 3 4 3   3R1  R2 H 2R13 R22 H 8R13  4 R22 H  3 3

8.11Since there is no external force on the system in the horizontal direction, the position of the CM will remain unchanged as the small block moves from one side to the other. Taking

128

horizontal direction to be the x-direction, let the position of the CM when the block in on the left be X1 and let it be X2 when the block is on the right. Then the poison of the CM of the block is (X1R) and (X2+R), respectively. Since the CM does not move, we have X CM  m X 1  R   MX 1  m X 2  R   MX 2

This immediately gives

 X 2  X1   

2mR m  M 

8.12 When the ball is compressed, it looks like shown in the picture below

R

R x

The radius r of the circular area of contact for x<
R 2   R  x

2



2 Rx

Thus, if the pressure in the ball remains unchanged, the force that the ball applies on the wall is F   r 2 p  2 pRx

8.12 As a photon hits the surface, it gives it an impulse proportional to its momentum. If it gets absorbed, the impulse J 

 2 and if it is reflected then J  . And the force by the c c

stream of photons hitting the surface will be nJ where n is the number of particles hitting the surface per second. If the cross-sectional area of the parallel beam of light is a, then n  acN

129

Thus the pressure P (i)

when the light is completely absorbed P 

(ii)

when light is perfectly reflected P 

acN    N  a c

acN 2   2 N a c

8.14 By equation (8.42b), the force on the planar surface will be equal to momentum transfer per unit time. On hitting the surface, the component of momentum perpendicular to the plane becomes zero while that parallel to the plane remains unchanged. Thus all the momentum that water stream carries perpendicular to the surface is transferred to it. The momentum carried by the stream of water per second is Its component perpendicular to the surface is

d 2 d 2 v  v  v 2 4 4

d 2 v 2 4 2

When the water stream hits the surface, it makes an elliptical cross sectional area on the surface because the surface is slanted. The major axis of the ellipse is = And the minor axis remains the same as

d 2 d  2 2

d 2

Thus the cross-sectional are of the ellipse is =

d 2 2 v 4 2 Thus the pressure on the surface =

d 2 2 2

d 2 2 2



v 2 2

8.15 At steady state flow let the mass flow rate from the upper portion of the hour-glass be  . If the height through which it falls before hitting the lower surface is h then the m

amount of mass in the air is the rate at which the mass is falling and the time it takes  for it to reach the bottom. Thus it is m

2h g

and its weight is

 m

2hg

. Thus the

hour glass should have weighed less by this amount. However as the sand hits the

130

bottom, it transfers momentum to the hour glass that exactly compensates for the weight in the air. This is shown as follows. As the mass hits the bottom, its speed is 2hg

. Thus the momentum it transfers to the bottom per second is

 m

2hg

.

8.16 Consider equation (8.43) for the rocket.     ( M  m) v  m  u  v   Fext t



Now since the mass coming out leaves the rocket with u rel , we have     u rel  u  (v  v )

This is because when the mass leaves the rocket, it has already achieved velocity   (v  v ) . This gives in the equation above    Mv  m u rel  Fext t

8.17 Example (8.9) using equation derived above. The example is solved exactly as done 







in the text except that in applying the equation derived above, u rel  u  (v  v ) each time the bullet is fired. This immediately leads to equation (8.50) and the rest is the same as done in the example. 8.18 (i) Force needed to hold the chain is equal to the force required to hold the part of the chain hanging vertically. This force is = gh (ii)If the chain is to be pulled at a constant speed v, its mass increases at the rate of v . This gives momentum change per unit time = v 2 . This is the additional force required to provide momentum. Thus the total force is gh  v 2 . This is also seen easily by the rocket equation M

  dv dM   u rel  Fext dt dt

Now it is given that

 dv  0, dt

dM  v, u rel  v . dt

equation immediately gives the result derived above. (iii)

The rocket equation, with

dM  v, u rel  v is dt M (t )

dv  v 2  F dt

131

This substituted in the rocket

If at a time t, the length of the chain on the table is x then M (t )   ( h  x) and dv dv dx 1 dv 2   . Thus the rocket equation can be rewritten as dt dx dt 2 dx

 dv 2 (h  x)  v 2  F 2 dx

This is integrated as V2

x

dv 2 2 dx ' 0 F  v 2   0  h  x'

Upon integration this gives 1  F 2  h x  ln  ln  2    F  V    h 



F  h x   2 F  V  h 

2

Upon solving this gives V 

1  h  x



F 2 x  2hx 



8.19 Since the peg is frictionless and the length of the portion of chain passing over the peg is negligible, the other portion has length (Lx) and the tension in the chain is the same throughout. Taking the tension to be T, the equation of motion for the two portions is (see figure)

132

T

T

( L  x) x(t)

M ( L  x) g L

M xg L  M   M  x  x   x g T  L    L  M   x)  M ( L  x) g  T ( L  x )( L L L



M M ( L  x) x  T  ( L  x) g L L



x 

Adding the two equations to eliminate T gives M x 

2Mx g  Mg L

2g x  g L

This equation is also obtained by direct application. Since the total mass being moved is M, the net force is

M 2 Mx g  x  ( L  x)  g  Mg and the acceleration is x . L L

The solution for the equation above is a sum of the solution of the homogeneous equation x 

2g L x  0 and the particular solution x p  . The solution of the homogeneous part is L 2 

x h (t )  A exp 

 2 g  t   B exp  l  

2 g  t L 

Where A and B are two constants to be determined by the initial conditions. The full solution is

133

 x(t )  A exp 

The initial conditions are x(t  0) 

 2g  t   B exp  l  

2g  L t  L  2

3L L and x (t  0)  0 . This gives A  B  . This 4 8

give the solution for x(t) to be x(t )  

 L exp 8 

 2g  L t   exp  l  8 

 L 1  1  cosh  2  2 

2g  t l 

2g  L t   L  2

  

8.20 (i) Since the pressure inside the box is p, and the force is unbalanced over an area S, the force on the box will be pS. pS M

(ii)

The acceleration of the box will be

(iii)

In the simplest calculation, the rate at which the molecules are coming out in one second will be those contained in a cylinder of height v x , where v x is the average speed in the x direction in the rms sense. Thus the rate at which the gas will be leaking out is Snmv x where n is the number density of molecules and m the mass of each molecule.

(iv)

Equation (8.45)   dv dM  M  u rel  Fext dt dt

In our case

M

 dM   Snmv x ; u rel  v x ; dt

 Fext  0 so

dv 1  Snmv x2  Snmv 2  pS by equation (8.40) dt 3  Mi   .  M f  

8.21 By the rocket equation v f  u ln

If the mass of the fuel is M, we have

M i  1000  M , Mf = M. Thus we have

 1000  M   1000  

6  5 ln

134

This gives 1 

M  6  exp  1000  5

 M  2320kg

8.22 In this case the rocket equation becomes (assuming vertically up direction to be positive) m

dv  mu  mg  bmv  dt

dv  bv   u  g  dt

Solution to the equation above is given by the sum of the solution for the homogeneous part and the particular solution. This gives v (t )  A exp( bt ) 

u  g b

A is determined by the initial condition v(t  0)  0 . This gives A   v (t ) 

u  g 1  exp(bt ) b

135

u  g . Thus b

Chapter 9 9.1

Consider a frame the origin O and another frame with origin O’. Frame O’ is moving with velocity V in the x-direction with respect to frame O. Let there be a force F act on a particle; the force is the same in the two frames.

Then by the work energy

theorem in frame O, we have xf

1 2 1 2 mv f  mv i   F ( x) dx 2 2 xi Now in the frame O’, the corresponding velocities are related to the velocities in frame O as follows v f  V  v 'f

vi  V  vi'

and x  x 'Vt



dx  dx'Vdt

and t  t '  dt  dt '

Substituting this in the work energy theorem gives



1 m v 'f 2



2

 

1  m v i' 2

x' f



2

 mV v

' f

v

' i

tf

   F ( x' ) dx'  V  Fdt  ti

x' i

tf

Now

 Fdt  is noting but the momentum change m v

f

 vi   m v 'f  vi'  , which is the

ti

tf

same in both the frames. Thus

 Fdt   m v

' f



 vi' . This, when substituted in the

ti

equation above leads to



1 m v 'f 2



2

 

1  m v 'i 2

2

x' f



Which is the work energy theorem in frame O’. 9.2

Kinetic energy of each particle is 1  0.1  12  0.05 J 2

and

1  0.2  2 2  0.4 J 2

136

 F ( x ' )dx'

x' i

Thus the total kinetic energy of the system is = 0.45J Momentum of the system is = 0.1  0.4  0.5kgms 1 Velocity of the CM therefore is =

0.5  1.667 ms 1 0.3

2 PCM 0.25   0.417 J Thus the kinetic energy of the CM is = 2 m1  m2  2  0.3

Thus the kinetic energy of then particles in the CM frame is = 0.450.417 = 0.033J This can also be checked by calculating the kinetic energy of each particle in the CM frame, which is 1 2 2  0.1  1.0  VCM   0.05  1.0  1.667   0.022 J 2 1 2 2  0.2   2.  1.667   0.1   2.0  1.667   0.011J 2

9.3

Since the momentum of the system is zero, the kinetic energy of the CM = 0

9.4

Masses, velocities, moment and kinetic energies o feach particle are shown in the table below mi(kg) 1 2 3 4 5

vi(ms1) 1 2 3 4 5

pi(kgms1) 1 4 9 16 25

Thus the total mass is = 15kg Total momentum is = 15kgms1 Velocity of the CM =

 15  1ms 1 15

Total kinetic energy = 112.5J 2 PCM 15  15   7.5 J Kinetic energy of CM = 2  total mass 2  15

Kinetic energy about the CM = 105J

137

KEi(J) 0.5 4.0 13.5 32 62.5

This is easily checked by calculating the kinetic energy of each particle about the CM and adding them all up. Thus KE about the CM =



1 2 2 2 2 2 1    1  1  2   2  1  3    3  1  4   4  1  5    5  1 2



= 105J 9.5

From equation (9.37), we have v1  VCM  v 2 V CM 

m2 (v1  v 2 ) m1  m 2

m1 (v1  v 2 ) m1  m 2

Thus the kinetic energy of the system

1 1 1 1 m1 m22 2 2 2 m1v1  m2 v 2  m1VCM   v1  v2  2  m1m2 VCM  v1  v2  2  m1  m2  2 2 2 2  m1  m 2  1 1 m 2 m12 2  v1  v2  2  m1m2 VCM  v1  v2   m2VCM  2 2 2  m1  m 2   m1  m2  Since v rel  v1  v 2 , we get 1 1 1 1 m1 m2 2  v1  v2  2 m1v12  m2 v 22   m1  m2 VCM  2 2 2 2  m1  m2  

9.6

(a)

We

will

be

F ( x)  

using

U ( x )    F ( x ) dx  C

1 1 2 2 MVCM  v rel 2 2 dU ( x ) dx

to

calculate

the

force.

Thus

, where C is a constant chosen according to the reference

point x0, where U  x0   0 . U ( x)    kxdx  C  

(i) F ( x )  kx (ii) F ( x) 

1 x  a2 2

1 2 kx  C 2

To calculate the potential energy, substitute x  a sinh z so that

dx  a cosh zdz to get U ( x )   sinh 1

x C a

138

(iii) F ( x )  5 sin 2 x

U ( x )    5 sin 2 x dx 

5 cos 2 x dx  C 2

(iv) F ( x)  5 sin 2 2 x U ( x)    5 sin 2 2 x dx  

(b)

5 1  cos 4 x dx   5  x  sin 4 x   C  2 2 4 

The plots of force and potential in each of the cases above are as follows. The force is shown on the left and the corresponding potential on the right in each case. (i)

We have taken k = 1.5. We have chosen the constant such that U(0) = 0.

(ii)

The plots are drawn for a = 5

(iii) We have chosen the constant such that U(0) = max.

139

(iv)We have chosen the constant such that U(0) = 0. Notice that the average force is always positive, i.e. pointing in the positive x-direction. Therefore the potential energy curve keeps going down as x increase.

9.7 We use U ( x)    F ( x) dx  C and keep the potential continuous everywhere. In the plots the force is shown on the left and the corresponding potential on the right in each case. (i)

  k if x  0 F ( x)     k if x  0

140

x  0 U ( x)   (k )dx  C  kx  C 

  U ( x)  k x  C

x  0 U ( x)   kdx  C  kx  C   If we choose

U (0)  0,

U ( x)  k x

For the plots drawn below we have chosen k = 2.

(ii)

F ( x)  k x

1 2 kx  C 2 1 x  0 U ( x)    ( kx) dx  C  kx 2  C 2 x  0 U ( x)    kx dx  C  

If we choose U(x) = 0, C=0. For the plots below, we have chosen k=2.

141

(iii)

 0 if x  a

F ( x)  

  kx if x  a

The nature of force implies that the potential is a constant for  a  x  a x  a U ( x)   kx dx  C  x  a U ( x) 

1 2 kx  C 2

1 2 ka  C 2

1 2

2 Choosing U(0) = 0 gives C   ka and

x a x a



1 k x2  a2 2 U ( x)  0 U ( x) 



For the plots below, we have chosen k=2 and a =15.

(iv)

 0 if x  a

 F ( x)   C  2 if x  a x The nature of force implies that the potential is a constant for  a  x  a x  a U ( x)    x  a U ( x) 

C C dx  const   const 2 x x

C  const a

142

Choosing U(0) = 0 gives const 

C C for x  a and const   for x  a a a

Thus



C C  x  a U ( x)  x  a  C C   x  a U ( x)   x a   x  a U ( x)  0   For the plots below, we have chosen C=250 and a =20.

9.8

We use F ( x )  

dU ( x ) . This gives dx

(i) U ( x)  kx

(ii)

U ( x)  k x

(iii) U ( x) 

1 x  a2 2

(iv) U ( x)  

F ( x)  

dkx  k dx

 d kx   dx  k x  0 F ( x)     d   kx  k x  0  dx F ( x)  

d  1  2x   2 2  2 dx  x  a  x  a2





2

d  1   2x 1 F ( x)     2   2 dx  x  a   x 2  a 2  2 x2  a2

143

(b) Plot for part (i) is straightforward; it is a constant force and linearly varying potential Plots for part (ii) are similar to that of part (i) in problem 9.7. This has been given above. (iii)

We have chosen a = 5. The force is shown on the left and the corresponding potential on the right.

Notice that the force is negative for negative x and positive for positive x. Thus it is a force that pushes a particle away from x = 0 and the position at x = 0 is an unstable equilibrium point. Thus when a particle is displaced slightly from this point, it will runaway to infinity. (iv)

We have chosen a = 5. The force is shown on the left and the corresponding potential on the right.

Notice that the force is positive for negative x and negative for positive x. Thus it is a restoring force and the position at x = 0 is a stable equilibrium point. Further the force is varying almost linearly with x near x = 0. Thus when a particle is displaced slightly from this point, it will perform simple harmonic oscillation.

144

9.9 9.10

Since all the potentials except (i) are time-dependent, only (i) is conservative. For the block not to fall off the track, its speed v at the top of the loop should be such that v2  g  v 2  Rg R

Now by the conservation of energy we have mgh  mg (2 R ) 

1 mgR mv 2  2 2

 h

5 R 2

9.11 For the potential energy U ( x)  C  x  a  3 we have the potential energy curve (with parameters C = 0.5 and a =2)

It is clear from the curve that a particle will always experience a force in the negative direction except at a = 2. From this it is clear that the point x = a is a point of unstable equilibrium. For the potential energy U ( x)  C  x  a  3 we have the potential energy curve (with parameters C = 0.5 and a =2)

145

This potential energy curve gives force opposite to the displacement, and the force is zero at x = a. Thus this point is a point of stable equilibrium.

Notice that unlike the potential energy for a

 1 2 kx  , this potential energy is quite flat near the equilibrium point (because  2 

spring 

 x  a  4   x  a  2 for

x  a  1.

For larger displacements from x = a, the curve rises much faster

than the spring potential energy. 9.12

Let the velocities of the balls be v1 and v2 after the collision (since this is a one dimensional motion, we are not putting vector signs on top of the velocities). Then by momentum conservation mV  m v1  v 2   V  v1  v 2 1 1 mV 2  m v12  v 22   V 2  v12  v 22 2 2

The first equation gives V 2  v12  v 22  2v1v 2 This and the second equation implies that 2v1v 2  0 Thus either v1 is zero, i.e. the first ball is not moving. This is the situation after collision. Or v2 is zero, i.e. the second ball is not moving. This is the situation before collision. 9.13

According to the situation give, the two balls are undergoing an elastic collision (no loss in energy) when they are moving in the opposite directions with equal speed V. Let

146

the mass of the lighter ball be m and that of the heavier one be M. If the velocity of smaller ball is v1 after the collision, then by equation 9.36 (taking vertically up direction to be positive) v1 

 M  m V M m



M 3MV  mV ( 2V )  M m M m

In the extreme limit, when M>>m, we have v1  3V If the balls are dropped from a height h then V



2 gh

If after the collision the smaller ball goes to height H then

3V 

2 gH

These two equations give H  9h 9.14 By momentum conservation we have

 m1  m2  v  m1v0 Totla initial energy = Final energy =

 v

m1v o m1  m2

1 m1v 02 2

m1 1 m12 v 02    initial energy  2 m1  m2 m1  m2

As is clear, the final energy is less than the initial energy. Therefore some energy is lost in the process. Now consider a pile of beads, each of mass m, connected with strings of length l between them. As the first bead falls over the edge, its initial speed is zero bu by the time the string connecting it to the second bead is taut, it gains a velocity of

v

2 gl

. However, as soon

as it pulls the second bead, by momentum conservation at that instant (neglecting the impulse due to gravity) the speed of two beads is v

m 2 gl  2m

gl 2

Now the two beads fall together and as the string connecting the second bead to the third bead becomes taut, the energy of hse two beads is 1 gl 5  2m   2mgl  mgl 2 2 2

This gives the speed of the two beads to be

147

5gl 2

Thus when the third bead is pulled over, momentum conservation at that instant gives the speed of the three beads to be =

2m 5 gl 2 3m



2 3

5 gl 2

One can go on like this and build up the solution up to p th bead falling by recognizing a pattern. We can also do the problem in the following way. Suppose when the (p-1)th bead has just fallen of the edge, the speed of the beads becomes v p 1 .

Then when these beads fall and the string between the (p-1) th and pth string becomes

taut, the energy of the system will be 1  ( p  1) m  v 2p 1  ( p  1) mgl 2

Thus the speed of these beads just before the pth bead falls off the edge is given by 1 1 ( p  1) mv 2   ( p  1) m  v 2p 1  ( p  1) mgl 2 2



v 2  v 2p 1  2 gl

By momentum conservation, therefore, we get the speed when the pth bead falls off as follows pmv p  ( p  1) mv



pv p  ( p  1) v 2p 1  2 gl



p 2 v 2p  ( p  1) 2 v 2p 1  ( p  1) 2 2 gl

We know that v1  0 i.e. when the first bead just falls off the edge of the table, its speed is zero. Now we can write ( p  1) 2 v 2p  ( p  2) 2 v 2p 1  ( p  2) 2 2 gl

And so on so that continuing in this manner we get ( v1  0 )





p 1

p 2 v 2p  v12  ( p  1) 2  ( p  2) 2  ( p  3) 2  ..........  1 2 gl  2 gl  i 2  i 1

( p  1) p( 2 p  1)  2 gl 6

This gives vp 

9.15

( p  1)(2 p  1) gl 3p

From problem 8.18, the total force required to move the chain is gh  v 2 . Thus the power delivered by the person pulling the chain is = ghv  v 3 On the other hand, the energy of the chain is = kinetic energy + potential energy If the length of the chain on the table is x then

148

dx  v and dt

Kinetic energy =

1 ( x  h)v 2 2

Thus Total energy E = This gives

potential energy =

1 gh 2  xgh 2

1 1 ( x  h)v 2  gh 2  xgh 2 2

dE 1 3  v  vgh dt 2

Thus the rate of change of the energy of the chain is not equal to the power delivered. As shown in the previous problem, the energy is lost in the inelastic collision. One may ask a question: what if instead of the chain, it s a rope or a strin that is being pulled? Where does the energy go in that case? The answer is that the energy difference is then used in stretching the rope to generate the required force. 9.16 Let the length of the chain be l. If its length y is hanging from the table, the force pulling it down is =

Mg y L

Taking the vertically down direction to be positive, the coordinate of the end of the hanging portion of the chain, when its length is y , is also y. Thus the equation of motion of the chain is M y 

Mg y l

or

y 

g y0 L

The equation can also be derived by considering the two protions, the one hanging from the table and the other on the table, separately. 1 dy 2 Writing y  we get from the equation above 2 dy

1 dy 2 Mg M  y0 2 dy l

v



1 Mg M  d y 2  2 0 l

y

 y' dy'  0

y1

This gives 1 1 Mg 2 Mv 2   y  y12   0 or 2 2 l

1 1 Mg 2 1 Mg 2 Mv 2  y  y1 2 2 l 2 l

The second term in the equation above is the change in the potential energy of the chain as the length of its hanging portion changes from y1 to y. Thus the total energy of the chain

149

remains unchanged as it slips off or the energy is conserved. This can also be seen as follows When length y1 is hanging, kinetic energy = 0

potential energy = 

1 Mg 2 y1 2 l

Total energy = 

1 Mg 2 y1 2 l

When length y is hanging, kinetic energy =

1 Mv 2 2

potential energy = 

Total energy =

1 Mg 2 y 2 l

1 1 Mg 2 Mv 2  y 2 2 l

It is clear from the above that the total energy remains unchanged. 9.17 As the chain unfolds, the mass of the hanging (and moving) portion keeps on changing. Thus the problem is like the variable mass problem. We take the mass per unit length of the chain to be λ. Taking the vertically down direction to be positive, the coordinate of the end of the hanging portion of the chain, when its length is y , is also y. The external force on the chain is yg ,

dm  y , and the relative velocity of the dt

mass that is added to it is  y . Thus the equation of motion my 

dm u rel  Fext of dt

the chain is yy  y 2  yg

OR

yy   y 2  yg

1 dy 2 Using y  we write this equation as 2 dy

1 dy 2 y 2  g 2 dy y

A

2 We now solve this equation for y 2 as the sum of the solution y  y 2 of the homogeneous

part and the particular solution

2 yg to get 3 y 2 

A 2  yg y2 3

150

Here A is a constant. Using the condition that y 2 ( y  y1 )  0 , we get A  

2 3 y1 g and 3

therefore

2 y 3  y13  y  g 3y 2 2

The same result can be obtained by letting p   and l  0 such that pl  y in the solution for a coiled set of beads in problem 9.14. Now let us compare the energy of the chain at the beginning o fthe motion and agter it has slipped so that y of its length is hanging. At the beginning, when length y1 is hanging, kinetic energy = 0

potential energy =  

1 gy12 2

Total energy =

1 gy12 2

When length y is hanging, Kinetic energy = Potential energy = 

 y 3  y13  g  1 y 2 g  1 y13 g 1 yy 2   2 3y 3 3 y

1 gy 2 2

Total energy = 

 2 y1  1 y13 1 1   Initial total energy  gy 2  Initial total energy g  gy 2   3 y 6 6  3y 

Thus we see that the energy is lost during the motion. This happens due to the inelastic collision between the chain links as each new link is pulled by the moving chain (see problem 9.14).

151

Chapter 10 10.1 (a) The fields are shown on the axses. They are the same all over the place. y 

(i) F ( x, y )  y iˆ x

y 

(ii) F ( x, y )  x ˆj x

(b) If we think of the force arrows as velocity of a fluid, we see that a matchstick thrown in that fluid will tend to rotate clockwise in (i) and counterclockwise in (ii). Thus the curl of both the fields is not zero; it is negatve for (i) and positive for (ii). (c)

curlf of force field (i) iˆ     F  x y

ˆj  y 0

kˆ    kˆ z 0

Curl of force field (ii) iˆ     F  x 0

152

ˆj  y x

kˆ   kˆ z 0

10.2

Y (2, 1)

O

X



(i) Force field F ( x, y )  y iˆ . Path 1: Work done is zero as the particle moves along the x-axis. Similarly as it moves from (2,0) to (2,1), work done is again zero since the particle is moving perpendicular to the force field. Thus alongh path 1, the net work is zero. Path 2: Work done is zero as the particle moves along the y axis. However, as it moves from (0,1) to (2,1), it is under a constant force of 1 unit in the x direction. So the work done by the field is 2 units. Path 3: Along path 3 y 

 x x . Thus F ( x, y )  iˆ . The displacement along path 3 is given as 2 2

 1 dl  dxiˆ  dy ˆj  dxiˆ  dx ˆj . Thus 2

  12 dW  F .dl   xdx  1 unit 20

(ii)



Force field F ( x, y )  x ˆj

Path 1: Work done is zero as the particle moves along the x-axis since the particle is moving perpendicular to the force field. Similarly as it moves from (2,0) to (2,1), Field is constant and

153

equal to 2 units. Thus the work done by the field will be 2 units. Thus alongh path 2, the net work is 2 units. Path 2: Work done is zero as the particle moves along the y axis because the field is zero. As it moves from (0,1) to (2,1), it is is moving perpendicular to the force field so the work done is zero again. So the net work done by the field is zero. Path 3: Along path 3 y 

 x . Thus F ( x, y )  x ˆj . The displacement along path 3 is given as 2

 1 dl  dxiˆ  dyˆj  dxiˆ  dx ˆj . Thus 2

W 

  12 F  .dl  2 0 xdx  1 unit

In both the cases the work done depends on the path. It is therefore consistent with the curl of both the fields not being zero. r 10.3 Force field is F  2 y iˆ  3 x ˆj . The paths are shown below

Y

(1, 1) Path2 path1

O

X

We will calculate the work done along path 1 in two parts: first when the particle mones along the x-axis and secondly when it is moving from (1,0) to (1,1). Along the x-axis, the force is

 F  3 x ˆj

and the displacement is dxiˆ . Thus the work done is zero when the particle

moves along the x-axis.  F  y iˆ  3 ˆj

When the particle moves from (1,0) to (1,1), the force is

and displacement is

dy ˆj .

Thus the total work done is 1

W   3dy  3 units 0

154

This then is the total work done along path 1. Along path 2 y = x so that dy = dx. The work done is given by the integral 1 1 1   1 5 W   F .dl   2 ydx   3 xdy   2 xdx   3 ydy  units 2 0 0 0 0

Since the work done along two paths is different, the force field is NOT conservative. 10.4 The figure for the path si shown below. The way we have set up the transformations Y C

(2,1)

A



(0,1)

B X

x  (1  cos  ) and y  (1  sin  ) , the corresponding angle is as shown in the figure and

varies from π to 2π. Thus the integral W ACB 

F

x

dx 

ACB )

F

y

dy

ACB

from A to B over the

semicircular path is W ACB  A

2  x ydx  A

ACB ) 2



ACB

x3 dy 3

  A  1  cos   1  sin   sin  d  2



A 3

2

 1  cos  

2

cos  d

The integral can be carried out easily and gives W ACB 

8A 3



10.5 F   k1 y iˆ  k 2 x ˆj . The force field will be conservative if its curl vanishes. The curls of it is    F 

iˆ  x  k1 y

ˆj  y  k2 x

If the curl has to vanish, we should have k1  k 2 . 155

kˆ    k 2  k1 z 0

For the force field

 F   y iˆ  x ˆj

it is clear by inspection that the potential should be

U ( x, y )  xy . However, to derive it formally we put 

U ˆ U ˆ i j   y iˆ  x ˆj x y

This gives U  y x

Which is integrated to U ( x, y )  xy  f ( y ) U

Differentiating this with respect to y and writing y  x gives df  0 OR dy

f  const.

By the condition U (0,0)  0 , the const. is zero. Thus U ( x, y )  xy 10.6 (i) Curl of the field iˆ     F  x 3A

ˆj  y Az

kˆ    Aiˆ z 0

is not zero. Therefore the field is not conservative. (ii) The curl of the field    F 

iˆ  x Axyz

ˆj  y Axyz

kˆ    A x y  z  iˆ  y  z  x  ˆj  z  x  y  kˆ z Axyz





is not zero. Therefore the field is not conservative.

10.7 (i)



The function f ( x, y )  2 exp   

( x  2) 2  ( y  1) 2   will be a constant where its 4 

argument ( x  2) 2  ( y  1) 2 is a constant. Thus the contours are given by ( x  2) 2  ( y  1) 2  const.

These are concentric circles with centre at  2,  1 . (ii)

The gradient of the function is given by 156

  x  2  2   y  1 2    x  2  2   y  1 2   ˆ     i  2 exp    j   exp     y  4 4  x    

 x  2 iˆ   y  1 ˆj 

10.8 Kinetic energy of the system of N particles with masses mi  i  1 N  and positions  ri  i  1 N  is

 1 N mi ri 2  2 i 1 

Let the position of the CM be R so that

   V  R.

If the positions and velocities of these

  particles with respect to the CM are given respectively by ric  i  1 N  and ric  i  1 N 

then    r  R  ri

and

      r  R  ric  V  ric

Substituting this in the equation for kinetic energy above, we get  2 1 N  2 N   1 N 1 N 2 m r  m V  m r  m r  i i 2  i ic  i i ic  V 2 i 1 2 i 1 i 1 i 1 M

Denoting the total mass

 mi as M and using the property of the CM that i 1

N



m r i 1

i ic

 0,

we get  2 1  1 N 1 N 2 mi ri  MV   mi ric2  2 i 1 2 2 i 1 This is the desired result. 10.9 When some energy is released during collision, the equations during the collision are     p1  p 2  p1  p 2 (momentum conservation)     p12 p 22 p1'2 p 2'2   E   (energy conservation) 2m1 2m 2 2m1 2m 2 Here the unprimed quatities are before collision and the primed quantities are after collision. Further E  0 . The easiest way to prove that in the bove situation, the paricles cannot move stuck together is to write all quatities in the with respect to the CM. In that case (referring to quantities in the CM frame with subscript c)       p1c  p 2c  p1c  p 2 c  0  p1c   p 2 c and

157

  p1c   p 2 c

And using the splitting of the kinetic energy as in the problem above and the fact that     p1c   p 2c and p1c   p 2c     p12c p 22c p1' 2c p 2' 2c   E   2m1 2m2 2m1 2m2



 p12c 2

  1 p1' 2c  1 1  1     E      2  m1 m2   m1 m2 

Now if the particles are moving together, their momenta after the collision with respect to the CM will be zero. This is also seen mathematically as follows. By momentum conservation   p1c   p 2 c And if they are moving together, then   p1c  p 2 c

  The above two equation give p1c  p 2c  0

Putting this in the energy conservation equation gives   p12c  1 p12c  1 1  1     E  0  E       2  m1 m2  2  m1 m2  However that is not possible because two positive quantities cannot add up to zero. Thus the particles cannot move stuck together. 10.10

The picture of the carom coin and the striker is shown below Y

Direction of impulse on coin

2.05cm

1.55cm

2ms-1

2cm X

The coin will move in the direction of the impulse that it receives from the striker. This direction is perpendicular to their common surfaces and theefore along the line joining their centres. Thus the coin moves at an angle  such that

158

sin  

2 5  (2.05  1.55) 9



cos  

56 9

and   33.75

We will work in cgs units. Initial momentum of the system is that equal to the initial momentum of the striker which is

 p  3000iˆ

. 

If after the coliision, the velocity of the striker is V  V x iˆ  V y ˆj and the speed of the coin is  v 56 ˆ 5v ˆ v then its velocity would be v  v cos  iˆ  v sin  ˆj  i j and the momentum 9

9

conservation will lead to v 56  15V x  3000 9 5v 5  15V y  0 9 5

Since the collision is elastic, the total energy is conserved and we get 1 1 1  5  v 2   15  V x2  V y2    15  200 2 2 2 2

These equations simplify to v 56  3V x  600 9 5v  27V y  0 v 2  3V x2  3V y2  120000

These are 3 equations for 3 unknowns so all the answers can be obtained from these equations. Substituting V x  200 

56 v 27

and V y  

5v 27

In the energy conservation equation gives  4v 2993    0 9   3

v

This gives v  249.4cms 1  2.5ms 1 after collision. The trivial answer v  0 of course refers to the situation before collision. This then leads to V x  1.30ms 1 and Vy  0.46ms 1 . This gives the angle of deflection  0.46    19.5 below the x-axis.  1.30 

1 of the striker to be tan 

159

(ii) KE o fthe CM remains unchanged during the collision. It is therefore 3000 2  10  7  0.0225 J . 2 5  15

(iii)

Velocity of the CM =

15  2  1.5 ms 1 15  5

Therefore the striker is moving with velocity 0.5 ms1 and the coin with -1.5 ms1 along the xaxis in the CM frame.

The magnitude of momentum of each body is 0.0075 kgms1 in

directions opposite to each other (see figure). In the CM frame the magnitude of velicities does not change during an elastic collision. Thus the momentum and the velocity vectors of the striker and the coin just rotate. Further, the impulse J is in the direction (see figure) such that it makes and angle 

2    33.75  3.6 

  sin 1 

from the x-axis. As is clear from the figure, the magnitude of J must be such that the final momentum has the same magnitude as the initial momentum. This is seaily done by drawing a circle of radius equal to the magnitude of the momentum with its centre at the tail of the initial momentum vector. We then draw the impulse vector from the head of the initial momentum vector at the angle  and take its length such that its head touches the circle. The vector from the centre to this point gives the final momentum. immediately shows that    p f  pi  J

It is also clear from the figure that  J  2  0.0075  cos 33.75  0.0125kgms 1

It is also clear from the figure that the angular momenta vector srotate by  CM  180  2  33.75  112 .5

160

The construction

J 0.0075 kgms1

33.75 0.0075 kgms1

 CM

As a check we now calculate the velocities of the striker and the coin in the lab frame from the information above and compare our answer. As is clear from the figure, the x component of the striker’s velocity in the CM frame is V xCM  0.5 cos 112 .5  0.187ms 1

This then gives Vxlab as follows (keep in mind that the CM is moving in x direction only) V xCM  V xlab  VCM

 V xlab  VCM  V xCM  1.5  0.187  1.313ms 1

And we get from the figure (the CM is moving in x direction only) V ylab  V yCM  0.5 sin 112 .5  0.462ms 1

Thus V striker ,lab 

2 2 V xlab  V ylab  1.39ms 1

Similarly for the coin (see figure) V xCM  1.5 cos180  112 .5  0.574ms 1

This gives V xlab  VCM  V xCM  1.5  0.574  2.074ms 1

and we get from the figure V ylab  V yCM  1.5 sin 180  112 .5  1.386ms 1

This gives Vcoin ,lab 

2 2 V xlab  V ylab  2.49ms 1

These match with the previously obtained answers. 161

10.11 As in the problem above, the stationary balls will move in the direction of impulse after the collision.

This is going to be along the line joining the centre of the striking ball

and the centres of the stationary balls. At the time of colliding, the three balls will look as shown below. Direction of impulse on ball 30

Direction of impulse on ball Since the net impulse (due to the two stationary ball) will be along the line of its initial motion by symmetry (see figure), it will continue to move along its original direction. This is also seen by noticing that the statinary balls have a net momentum in the direction of v0. Thus there will be no component of the momentum to be balanced in the direction perpendicular to that.

Thus the striking ball will continue to move along its original

direction. Let the speed of the striking call be v1 after the collision and the speeds (equal by symmetry) of the other balls be v. Then by momentum conservation mv0  mv1  2  m  v

3 2

 v 0  v1  v 3

Since the collision is elastic, we have by energy conservation 1 2 1 2 1 mv0  mv1  2  m  v 2 2 2 2

 v 02  v12  2v 2

There are two unknowns v1 and v and two equations so we can get their values. Squaring the momentum conservation equation and subtracting it from the energy conservation equation gives





v v  2v1 3  0

One of its solutions is the trivial solution v  0 referring to the situation before collision. Theother solution is

162

v  2v1 3

Substituting this in the the momentum conservation equation gives v1  

v0 5

and v 

163

2v 0 3 5

Chapter 11 11.1 Disc rotating about a point on its periphery. Consider first a rotation about a point on the periphery by an angle. This is shown on the left in the figure below. The dashed circle shows the initial position of the disc while the solid circle shows the position after rotation.

 

On the right we show the same rotation carried out by translating the CM of the disc first to its new position, as shown by the straight arrow and then carrying out the rotation about the CM. We have made the disc in this position by dotted circle. As is evident from the figure, the magnitude and the sense of rotation is the same as in the figure on the left. 11.2 Now we generalize the result of the problem above. On the right in the figure below, we show the same rotation as in the figure on the left carried out by translating the opposite end of the diameter of the disc first to its new position, as shown by the straight arrow and then carrying out the rotation about the this point. We have made the disc in this intermediate position by dotted circle. As is evident from the figure, the magnitude and the sense of rotation is the same as in the figure on the left. Similarly, it can be shown about any other point.

164

 



11.3 The angular momentum L about the origin is given by

   L  mr  v .

Thus the answers

in different cases are 

(i) L  2iˆ  2iˆ  0

(ii)

(iii) L  2 ˆj   2iˆ  ˆj   4kˆ 

 L  2 ˆj  2iˆ  4kˆ

(iv) L   iˆ  2 ˆj    2iˆ  ˆj   kˆ  4kˆ  3kˆ 

11.4 The position of the wheel and the stone stuck to its periphery is shown in the figure below.

Y

Vt x(t) y(t)

t X

165







Thus we have for the position r (t ) , velocity v (t ) (obtained by differentiating r (t ) with 



respect to time once) and acceleration a (t ) (obtained by differentiating r (t ) with respect to time twice) of the stone  r (t )  Vt  R sin t  iˆ  R cos t ˆj

 v (t )  V  R cos t  iˆ  R sin t ˆj

 a (t )   2 R sin t iˆ   2 R cos t ˆj

As is evident, the acceleration is towards the centre of the wheel. It is the centripetal acceleration.

The force for this is provided by the force that keeps the stone stuck to the

wheel. This force is   F  ma (t )  m 2 R sin t iˆ  m 2 R cos t ˆj

The angular momentum

 L (t ) of

the stone with respect to the origin is



 

   L (t )  mr (t )  v (t )  m  Vt  R sin t  iˆ  R cos t ˆj   V  R cos t  iˆ  R sin t ˆj  mVRt sin t  R 2  VR cos t  kˆ



This gives  dL  mV 2 Rt cos tkˆ dt

The angular momentum changes due to the torque provided by the force that keeps the stone stuck to the wheel and is equal to the centripetal force. Thus the torque 

 m

 Vt  R sin t  iˆ  R cos t ˆj   

2

R sin t iˆ   2 R cos t ˆj



 mV Rt cos t kˆ 2

Thus it is equal to the rate of change of the angular momentum. 11.5 We assume that at time t = 0, the paricle is on the x axis. Thus after time t, its position vector is give as (see figure)  r (t )   R  R cos t  iˆ  R sin t ˆj

166

Y

R

t

O

X



Thus its velocity and acceleration are (obtained by differentiating r (t ) with respect to time once for the velocity and twice for the acceleration)  a (t )   2 R cos t iˆ   2 R sin t ˆj

 v (t )  R sin t iˆ  R cos t ˆj

As expected, the acceleration is the centriprtal acceleration. It is provided by the external force towards the centre. The angular momentum of particle with respect to the origin is



 

   L  mr (t )  v (t )  m  R  R cos t  iˆ  R sin t ˆj   R sin t iˆ  R cos t ˆj  mR 2 1  cos t  kˆ



Its time derivative is  dL   m 2 R 2 sin tkˆ dt

The torque due to the external force is



 

     mr (t )  a (t )  m  R  R cos t  iˆ  R sin t ˆj    2 R cos t iˆ   2 R sin t ˆj



  m 2 R 2 sin t kˆ

This is the same as the rate of change of angular momentum. 11.6 The angular momentum for a collection of particles about an origin O is    LO   mi ri  vi i



Now let is choose a different origin O’ such that the position vector of O’ from O is R (see figure) so that    ri  R  ri '





and vi  vi'

167

Z’ Z

 ri '

 ri

 R

O

O’

Y

Y’

X’

X Substituting the expressions above in the formula for



 LO



, we get

          LO   mi ri  v i  mi R  ri '  v i  R   mi vi   mi ri '  vi' i

i

i

i

 ' where we have used the fact that vi  vi .

Now the angular momentum about O’ is  LO ' 

'

m r

i i

  vi'

Thus if the total momentum of the system is zero, then 

'

m v  m v i

i

i

i

i

i

 0 and L  L O O'

11.7 At the maximum distance Rmax and the minimum distance Rmin from the sun, the velocity vector is perpendicular to the radius vector. If the speed of the earth at these distances is Vmax and Vmin, respectively, then the angular momentum of the earth at these two points is MearthRmaxVmax and MearthRminVmin , respectively. By the conservation of angular momrntum we have MearthRmaxVmax = MearthRminVmin This gives Vmax R 1.47  min   0.967 Vmin Rmax 1.52

168





11.8 Initial momentum p i , the final mpmentum p f and the change in the momentum    p  p f  pi are shown in the figure below.

 p

 pf

   

1 

2

Y

    X

 pi 

As is evident from the figure, the magnitude of p is 2 p sin

 where 2

  p  pi  p f

and

it is in the direction bisecting the angle      between the incident and the scattered direction.

Mathematically it can be seen as follows. Choose the direction of incoming

particle to be the x direction. Then  pi  piˆ and

 p f  p cos iˆ  p sin ˆj

 p  p cos   1 iˆ  p sin ˆj     2 p sin 2 iˆ  2 p sin cos ˆj 2 2 2      2 p sin   sin iˆ  cos ˆj  2 2 2  

Thus the magnitude of p is 2 p sin

    and it is in the direction   sin iˆ  cos 2 2 2 

ˆj  . 

(ii) Since the paticle is moving parallel to the x axis at a distance d, its angular momentum is mvd going into the plane of the paper. This can be seen as follows.

The position and

velocity vectors of the particle are  r  xiˆ  dˆj



and v  viˆ

This gives for the angular momentum    L  mr  v   mvd kˆ

Further, since the force is central, the angular momentum about the origin is a constant.

169

(iv)

  dp   F , we have p  Since dt





 Fdt . Therefore  Fdt is in the same direction as

 p .

To calculate



 Fdt , we change the integration over time to integration over the angle by

using the angular momentum conservation. Using polar coordinates, we write the angular 

2 momentum of a particle moving in the xy plane as L  mr

problem, it is a constant equal to

 mvd kˆ

mr 2

d ˆ k . Since in the present dt

, we have

d   mvd dt

 dt  

r2 d vd



The force F between the two charges is in the radial direction and is equal to  Qq F  k 2 cos  iˆ  sin  ˆj r





Thus  2  Qq ˆ  sin  ˆj   r d F dt  k cos  i   r2 vd 



k









Qq cos  iˆ  sin  ˆj d vd 





Qq  sin iˆ  1  cos   ˆj vd Qq    k  2 cos   sin iˆ  cos vd 2 2 2 k

Thus the magnitude of



 Fdt is 2k

ˆj  

Qq  cos . Comparison of this expression with that for vd 2

the momentum change shows that the two are in the same direction and 2 p sin

 Qq   2k cos 2 vd 2

Or d k

Qq  Qq  cot  k cot 2 pv 2 2 mv

This is the Rutherford formula.

170

11.9 (i) At angle  1 the potential energy of the girl is MgL1  cos 1  

1 MgL12 with 2

respect to the equilibrium point, i.e. when the swing is in the vertical position. The kinetic energy of the system at the lowest point , if the angular speed is 1 , is 1 ML212 . By energy conservation 2 1 1 ML212  MgL12 2 2

(ii)

 1  1

g L

Whe the child stands up at the lowest point, all the external forces on the her are passing through the pivot so her angular momentum remains unchanged. However 2 as she stands up her moment of inertia about the pivot is M  L  d   I CM where ICM

is her moment ofinertia about her CM. Neglecting it gives the moment of inertia to be M  L  d  2 . Then by conservation of angular momentum M  L  d   2  ML21 2

keeping only linear terms in (iii)

 2 

L2 2d   1   1   1 2 L  L  d 

d since d<
The final kinetic energy of the girl is 2

1 1 2d  2 2 I 2  M  L  d   1   1 2 2 L   1 4d  2   M  L2  2dL  1   1 2 L   1 2d   4d   ML212  1    1  2 L   L   1 2d    ML212  1   2 L   

keeping only linear terms in

d since d<
system has increased in comparison with the original energy. Threfore the swing would go higher on the other side.

171

(iv)

Suppose the swing goes up to angle  2 on the othwr side, we have by energy conservation and the fact that

1 1 ML212  MgL12 2 2

1 2d  1  2 ML212  1    Mg  L  d  2 2 L  2 



L    Ld 

  22  



 1 

2d  2  1 L 

Now we have the approximation d<
L  2d   d  2d  3d   1    1   1   1 L   L  L  L  Ld   

keeping only linear terms in

d . Thus L 

3d  2   1  L  

1

2

1

3d    2   1  1 2L  

This gives the increase in the angle of swing in half a period. While going back, the new amplitude 1 by the time the swing reached its starting point will be. 3d  3d    1   2  1    1  1   2L  2L   

2

3d    1  1   L  

In the full period, the net increase in the amplitude will therefore be  1  1 

172

3d 1 L

Chapter 12 12.1 Moment of inertia of a ring. Let the mass of the ring be m and its radius R. Since all the points on the radius are at the same distance from the axis, each small mass m on the periphery gives the same contribution mR 2 to the moment of inertia. Thus the total moment of inertia is I   mR 2  mR 2

A disc can be thought of made of many rings of width dr at different radii (see figure). M

2 The moment of inertia of each ring is r   R 2 2 rdr . Thus

the total moment of inertia is I 

M

R

2

2r 3 dr 

1 MR 2 2

For the moment of inertia about the diameter of a ring, consider a small portion of it of extent d at and angle  It moment of inertia about the diameter is

M  R cos   2 Rd . 2R



MR 2 Thus the total moment of inertia is 2

2

 cos 0

2

 d 

1 MR 2 2

12.2 Moment of inertia of a rectangular sheet. The sheet is shown in the figure below

173

Z Y y b

X

a

We first calculate its moment of inertia about the x axis. For this, let us take a thin strip of width dy parallel to the x axis and at a distance y from it. Its mass is

M M a dy  dy . ab b

Since the perpendicular distance of all its points is y from the x-axis, its moment of inertia about the x axis is y 2

M dy . Thus the total moment of inertia about the x-axis is b

M Ix  b

b

Mb 2 b y dy  12  2

2

2

Similarly by taking a thin strip parallel to the y axis, we will get the moment of inertia Iy about the y axis to be

M Iy  a

a

2

2  y dy 

a

2

Ma 2 12

Let us now calculate the moment of inertia about the z axis. Consider again the strip parallel to the x axis.

By parallel axis theorm the moment of inertia of this strip about the

z axis is = (moment of inertia of its CM with respect to the z-axis + its moment of inrtia about the axis passing through its CM and parallel to the z-axis) dI z  y 2

M 1 M dy  a 2 dy b 12 b

which upon integration leads to Iz 

M 2 a  b2  12

174

12.3 The disc and a strip on it perpendicular to the axis of rotation (y-axis) is shown in the figure below. The strip is of width dy and at a distance y fron the centre. Y

R

y X

Length of this strip is

2M

and its mass is  R 2

2 R2  y2

R 2  y 2 dy . Thus its moment of

inertia about the y-axis is

dI y 







1 2M 2M  4 R2  y2  R 2  y 2 dy  R2  y2 2 2 12 R 3 R



3

2

dy

The integration is easily carried out by substituting y  R cos  and gives the resuly Iy 

1 MR 2 4

12.4 This problem can be done exactly in the same manner as above. We now divide the shell inti thinn rings at an angle  (see figure).



175

M

M

The radius of the ring is r cos  and therefore its mass is 4 r 2 2 r cos   rd  2 cos  d and its moment of inertia about the rotation axis is dI  r 2 cos 2  

M cos  d 2

Thus the total moment of inertia is 

Mr 2 2 3 Mr 2 4 2 2 I cos  d    Mr 2  2 3 3 2

12.5 In this problem, we divide the sphere into thin discs and add the moments of inertia of all discs. Thus for a disc of thickness dy at height y (see figure), its radius is

R2  y2

mass is









M 3M  R 2  y 2 dy  R 2  y 2 dy . 4R 3 3 4R 3

Y

R

y X

Therefore, the moment of inertia of the disc is dI 

2 1 3M 2 3M  3  R  y 2    R 2  y 2  dy  3  R 2  y 2  dy . 2 4R 8R

Integrating it from R to R gives the total moment of inertia I

3M 8R 3

R

 R

R

2



2

 y 2 dy 

3M 8R 3



 2 R 5  

176

2R 5 4R 5  3M 16 R 5 2      MR 2 5 3  15 5 8R 3

, its

12.6 For the moment of inertia about the y-axis, the problem is done exactly in the same way as in 12.5 by taking a disc at height y. The only difference is that the mass M is now distributed over half the volume and the y integration runs from 0 to R. Thus the mass of the disc is









M 3M  R 2  y 2 dy  R 2  y 2 dy 3 3 2R 3 2R

And its moment of inertia is dI 

2 1 3M 2 3M 2  3  R  y 2    R 2  y 2  dy   R  y 2  dy 3 2 2R 4R

Thus the total moment of inertia is given as I

3M 4R 3

R

 R

2



2

 y 2 dy 

0

3M 4R 3

 

 R 5 

R 5 2R 5  3M 8 R 5 2      MR 2 5 3  5 4 R 3 15

Similarly, to calculate the moment of inertia about the x axis, we make a cylindrical shell as we did in example 12.3 (see figure) Y

R

y X



Thus the moment of inertia will be calculated exactly like in example 12.3. The mass of this shell is given by 1 M   2 y dy  2  2 2 R 3 3 3M  3 R 2  y 2 y dy R

dm 

177

Th factor of

1 comes because there is only half the shell in a hemisphere. 2

Therefore the

moment of inertia of the hemisphere is I   y 2 dm 

3M R3

R

y

R 2  y 2 dy 

3

0

2 MR 2 5

12.7 A conical shell is shown below. To calculate the moment of inertia, we take a ring at a distance x from the base extending from x to x+dx. Then by similarity fo triangles, its

radius is y  r

 h  x h

dx

h

. Similarly the length of its side is cos   2 2 dx h r Y

x

r



X

h

Thus the mass of the ring is dm 

m

 r h2  r 2

 2 y 

h2  r 2 2m dx  2  h  x  dx . Its moment of h h

inertia therefore is 2m 2mr 2  h  x  3 dx dI  y  2  h  x  dx  4 h h 2

The moment of inertia of the shell is then given by integrating the expression above over x from 0 to h and gives 2mr 2 I h4

h

2mr 2   h  x dx  0 h4 3

h

 h

3

0



 3h 2 x  3hx 2  x 3 dx 

mr 2 2

Now we calculate the moment of inertia about the y-axis. To do this we use the parallel axis theorem and write the moment of inertia of the ring considered above as that of its CM plus its moment of inertia about the axis parallel to the y axis and passing through the CM. Thus

178





2m 1 2 2m 2m mr 2 3 2 3 dI  x  2  h  x  dx  y  2  h  x  dx  2 hx  x dx  2  h  x  dx 2 h h h h 2

Integrating this we get I

mr 2 mh 2  4 6

We now do the calculations for a solid cone. The calculations proceed in exactly the same manner as for the shell bu now instead of a ring, we take a disc at distance x. The mass of the m

3m

2 2 disc is dm  r 2 h 3 y dx  h 3  h  x  dx . Thus its moment of inertia is

dI 

1 2 3m 3mr 2 2  h  x  4 dx y  3  h  x  dx  5 2 h 2h

Thus the total moment of inertia is I 

3mr 2 2h 5

h

  h  x

4

dx 

0

3 mr 2 10

For the moment of inertia about the y axis, we again consider the same disc and use the parallel axis theorem to write the moment of inertia of the ring considered above as that of its CM plus its moment of inertia about theaxis parallel to the y axis and passing through the CM. So dI  x 2 

3m 1 2 3m 3m 2 3mr 2 2 2 2        h  x  4 dx h  x dx  y  h  x dx  x h  x dx  3 3 3 5 4 h h h 4h

Now the integration over x from 0 to h gives 1 3 mh 2  mr 2 10 20

12.8 (i) The total angular momentum = angular momrntum of large disc + angular momentum of small disc L  I 1 (large disc)  I 2 (small disc) I1 

1 MR 2 2 1 2

I2 is calculated using the parallel axis theorem and its value is md 2  mr 2 Thus L 

1  MR 2  mr 2    md 2  2

179

(ii)

Let the large disc start rotating with angular speed  in the opoosite direction. Then by conservation of angular momentum 

(iii)

1 1 mr 2  MR 2  mr 2    md 2   mr 2  0    2 2 MR 2  m  r 2  2d 2 

For M  0 and d  0 , we have  

This keeps the total angular momentum zero since the large disc rotates in the opposite direction with exactly the same angular speed as given to the small disc. (iv)

In a helicopter also, as the large rotor starts, the body of the helicopter will tend to rotate in the opposite direction. To prevent this, a tail rotor is fitted that provides the counterbalancing torque.

12.9 The position of the person (represented by the filled circle) after time t is shown on the platform in the figure below.

u O

B

ut A

As the person moves on the platform, the platform starts rotating clockwise with angular speed  so that the total angula momentum remains zero. As the platform rotates, the  person also moves with it. The net velocity v of the person with respect to the ground is   equal to (velocity u of the person with respect to the platform + rotational velocity v rot of the person due to the rotation of the platform). This is expressed as    v  u  v rot

Thus the angular momentum of the person is 180

    l p  M r   u  v rot 

 Here r is the position vector of the person from the origin O on the axis. The position, various distances and the velocities, as seen from the top, are shown in the figure below. B

 u a 2

O

 r  v rot

ut

A

It is clear from the figure that   a r     2

2

 a   ut   2 

2

  a   r  v rot   r 2        2 

  a r u  u 2



2

 a   ut   2 





2

 

Here the sign takes care of the ddirection of angular momentum along the aixs of rotation; it is positive for counterclockwise rotation and negative for clockwise rotation. Thus the angular momentum of the perso is lp 

  a Mau  M    2   2 

2

 a   ut   2 





2

 

The angular momentum of the platform is l plat  

ma 2  6

Now equation the total angular momentum to zero (by conservation of angular momentum) gives

181

  a Mau  M    2   2 

2

 a   ut  2  



2



  

ma 2 0 6

This gives  (t ) 

Ma u 2 2  ma  a  a   M       ut  6  2    2 

Total time that the person takes to move from A to B is



2

2

 

a u

Thus the angle  through which the platform rotates by the time the person reaches B is 

au

au

0

0

  (t )dt 



Ma u 2   a ma 2 M   6   2 

2

 a     ut   2 

dt 

2

 

This integration can be done by substituting z

a  ut 2

and

dz  udt

So we get a 2

a 2

Ma 2 a 1   dz  dz  m M 2 1 2 2 a 2  m 2 2 a 2     a z   a  Mz  4   6  6M 4  1 1  m  m   tan  which gives dz  a    sec 2  d to get Now substitute z  a   6M

4



where

1 m 1 tan  0     2  6M 4



1 2

 6M

0

a 2  0



m 1   sec 2   6M 4  d 1 2  m 2   a sec    6M 4 

a 

. This gives 

0 1  m     6M 4 

182

4

If m = 0, we have  0 

  and   . Thus although on the platform, the person has 4 2

moved from A to B but overall he remains at his original position since the platform has rotated back by  

 . 2

For m  M , we have 

1 m 1 tan  0     2  6M 4

Thus  0 

1 2

2 4m    1  2 6M 



1 2

 1

m 3M

   , where   1 . We can therefore write 4          tan  sec 2    1  2 4 4  4 

tan  0  tan

Thus  

m 6M

This gives m   1 m           4 6M   4 6M 

12.10



1 2

m   m   m         1   1    3M  2 3M  2  2 3M   



This is similar to the variable mass problem except that in this problem we will get the answer by applying the principle of conservation of angular momentum because this is an extended body and if we try to apply Newton’s second law to each point particle, it is impossible to solve the problem because of a large number of internal forces involved. Let a small mass m come out at a given time t during the time interval from t to t+t from one of the containers. Let the angular speed of the display be  at t. Then the speed of gases coming out with respect to the ground is  u  l  , as shown in the figure

183

u-l u-l 

If in time t the angular speed of the display becomes    , then by conservation of angular momentum 4  m l 2  4 m  m  l 2       4ml  u  l 

Here factor of 4 comes because of 4 containers. Here m is the mass of each container at time t. Obviously m  M  m   t . Neglecting the second order term m , we get m l 2   mlu



m l  mu

Thus d dm u  dt dt m l

This equation can not yet be integrated because that. Obviously

dm should be written in terms of m for dt

dm dm  . This gives dt dt d dm u  dt dt m l

Integrating this gives, starting from  = 0 at t = 0  (t )  

u l

M  m t



M m

dm u  M m    ln m l  M  m  t 

Thus after the entire gubpowder is exhausted, we have 

u  M m ln  l  M 

184

12.11 This problem is exactly the same as the problem above except that the moment of inertia of the display is different because of different positioning of the containers of gunpowder. Conservation of angular momentum now gives 2

 l  l 2    4 m  m  l       4 m  m     2  2 l l  4ml  u  l   4m  u    2 2

2

4 m l 2   4m 

    

Neglecting second-order terms and simplifying leads to 

 m l 2  m 

l2 4



lu      m lu   2  



5 3 m l 2   mlu 4 2

Following the steps in the solution of problem 12.10, this leads to  (t ) 

12.12

6u  M m   ln 5l  M  m  t 

and  

6u  M  m  ln  5l  M 

As the water comes out of the sprinkler pipes, it carries angular momentum with it. If the opposing torque were not there, the system will rotate in the opposite direction with increasing angular speed to conserve the angular momentum. However it does not happen because of the opposing torque which is the external torque on the system. Taking the direction of rotation of the rods to be positive, we have L t  t   L(t )  external torque  t

Here external torque is   with the minus sign indicating that it is in the direction opposite to the rotation. Since the mass of the rod remains unchanged, if the mass of water coming out from each end in time interval t is m , we have L t  t   2  

ml 2       4  m l  u  l  12 2 2 

ml 2       2ml  u  l  6 2  

and L(t )  2 

ml 2 ml 2   12 6

Thus we have ml 2 l     2ml  u     t 6 2  

185

This gives ml 2 d dm  l  2 l u     6 dt dt  2 

dm  Su , where  is the density of water. In dt

Since the area of each hole is S, we have

steady state, the sprinker system rotates with constant angular speed so that Substituting

d  0. dt

dm d  Su and  0 in the equation above gives dt dt



2u   l Sul 2

Assuming that at the beginning of turning the spinker on, all pipes are filled, we also solve how the angular speed changes with time until it approaches the above vaue.

The

differential equation for  can be written as d  6 S u 12 Su 2 6     2 dt m ml ml

This equation has the solution   6 Su   2u  (t )  A exp  t     m   l Sul 2    

Since  (t  0)  0 , we get  2u  A    Sul 2  l   

This gives  2u   Sul 2  l

 (t )  

 6Su     1  exp  t m       

 Let the angle between the unit vector nˆ and vector r be 

12.13 (see figure).

186

ˆ n 



 r

 r

O

 It is clear from the figure that the magnitude of r is r sin   and it is in the direction     of nˆ  r . Since the magnitude of nˆ  r is also r sin   , we get r  nˆ  r .

12.14

As the disc hits the rough surface, let the surface apply an impulse J on it (see figure).

 V

V1

J As a consequence the disc moves with a smaller velocity and also starts rotating because the impule applies a torque on it about its CM. By rolling condition we have V1  R . By the equation for the CM, we get J  mV  mV1

And by the torque equation JR 

This gives V1 

1 mR 2 2



2 V 3

187

J 

1 1 mR  mV1 2 2

12.15

Free body diagram of the disc is given below.

F f Here f is the frictional force on the disc. By the equation for the motion of its CM we get F  f  ma

Taking torque about the CM of the disc we get fR 

1 mR 2 2



f 

1 mR 2

For pure rolling we have a  R . Substituting this in the above two equations we get a

2F 3m

12.16 (i) Free body diagram of the cylinder is shown below. J

d Jf

Here Jf is the frictional impuse on the disc. By the equation for the motion of its CM we get J  J f  MV

Similarly the angular momentum equation about the CM gives MR 2  2 MRV  2

Jd  fR 

Where the second equality follows from the rolling condition V  R . Solving the above two equations leads to

188

 3MV   1 R 2 J  

d 

(ii)

For a sphere the equations are the same as for a cylinder except that its I about the CM is different. Thus the equations become J  J f  MV

2MR 2  5 2MRV  5

Jd  fR 

Their solution gives  7 MV   1 R .  5J 

d 

The rod in example 12.11 bounces back with angular speed 

12.17

and speed V. Before it impacts the ground, its angular momentum about the point of impact is equal to the angular momentum of its CM because it is not rotating, and it is pointing into the page (see figure). l/2  V



h

Thus Linitial 

ml 2gh cos  2

After the impact, the rotation about CM gives angular momentum coming out of the page and the translational motion gives angular momentum going into the page. Thus L final 

ml 2 mlV  cos  12 2

189

By conservation of angular momentum, we have ml 2 gh ml 2 mlV  cos   cos  12 2 2



l  6V cos   6 2 gh cos 

By energy conservation we have 1 ml 2 2 1   mV 2  mgh 2 12 2



l 2 2  12V 2  24 gh

These are two equations for two unknowns. Substituing l



 6 cos  V 

2 gh



from the

first equation into the second equation, we get the equation

 









V 2 1  3 cos 2   6 2 gh cos  V  2 gh 1  3 cos 2   0

This is a quadratic equation in V, and its solution is   1  3 cos 2    V   2gh  2  1  3 cos  

The plus sighn in front of 1 above gives the trivial solution that the velocity of the rod before impact is

2 gh

downward. And after the impact its velocity is gives by the second

root which is V 

From l



 6 cos  V 

2 gh

 1  3 cos 2    2  1  3 cos  

2gh 

 , this gives 

12 2 gh cos  l 1  3 cos 2  

It is evident from the equations above, for  

12.18

 , the rod rebounds without any rotation. 2

This problem is similar to the problem 12.17 . However, the CM and the moment of inertia about the CM are different in this problem. In this case the CM is at distance

3ml  l from mass m. Thus the moment of inertia about 3m

the CM I CM  2ml 2  m  4l 2  6ml 2

190

Like the previous problem, before the system impacts the ground, its angular momentum about the point of impact is equal to the angular momentum of its CM because it is not rotating, and it is pointing into the page. Its value is Linitial  3ml 2 gh cos 

Let the system bounce back with angular speed  and speed V after the impact. Then L final  6ml 2  3mlV cos 

Thus angular momentum conservation about the point of impact gives 6ml 2  3mlV cos   3ml 2 gh cos 



2l  V cos  

2 gh cos 

Similarly energy conservation gives 1 1 6ml 2 2   3mV 2  3mgh 2 2

Again we substitute l 





2l 2 2  V 2  2 gh



1 V  2 gh cos  in the second equation to get 2



1   V 2  1  cos 2    2  

Its solution gives one trivial solution



1   2 gh cos  V  2 gh 1  cos 2    0 2  

and for speed after impact

2 gh



1  2  1  cos   2  V  2gh  1  1  cos 2     2   and 

12.19

2 2 gh cos  1   l  1  cos 2   2  

The figure below shows the rod and the stone just before the impact.

191

3l 4

(i)

l

Right after the impact the angular momentum about the pivot is conserved. Thus, if the rod and stone stuch with it move with angular speed  after the impct, we have (moment of inertia of the rod and the stone together is m0 v 0

With m0 

m and v0  3

3l   3l    m0   4  4 



 3l    4

2



 m0 



ml 2   3 

)

ml 2    3 

gl , we get 

(ii)

2



12 25

g l

The kinetic energy of the rod and the stone system after the impact is KE 

1  3l    m0  2   4

2



ml 2  2 1 25 2  12  ml        3  2 48  25 

2

g 3  mgl l 50

If the rod rises by an angle , the potential energy of the system at this point is PE  mg

l 1  cos    m0 g 3l 1  cos    3mgl 1  cos   2 4 4

Equating the KE and PE, we get cos   0.92

(iii)

   23

The impulse by the pivot. As soon as the stone hits the rod and gets stuck with it, the horizontal momentum of the system changes. This change is brought about by

192

the impulse that the pivot applies on the rod. Thus by calculating the change in the momentum of the system, we obtain the impulse that the pivot applies on the rod. Initial momentum of the system = m0 gl 

m gl 3

Final momentum of the system = momentum of the CM of the rod  momentum of the stone l  m   m0 2 3ml 12   4 25 9  m gl 25

3l  4 g l

Thus after the stone gets stuck, the momentum of the system changes by 2  9 1   m gl  m gl  75  25 3 

in the same direction that the stone was moving initially. Thus the impulse given by the pivot is 2 m gl 75

or

2 m0 gl 25

in the same direction that the stone was moving initially. It is also instructive to solve this problem by angular momentum consideration by calculating the change in the angular momentum of the system and attributing it to the torque provided by the impulse. For this we apply the equation dLCM   applied ,CM dt

Here CM is the centre of mass of the entire system (rod+sone). Note that this equation can be applied only about the CM of the system and NOT about the CM of the rod alone.

Distance of the CM from the pivot point =

1  l m 3l  9l  m     4m 3  2 3 4 16

l  9l l     above the system CM. Similarly the stone is 16  16 2 

Thus the CM of the rod is 

m0 v 0 v 3l  3l 9l   0 in the     below the system CM. The velocity of the CM is 16 m  m0 4  4 16 

193

same directon as the stone. Thus befor the stone strikes the rod, it is moving with respect to the system CM with speed

with speeed 

3v 0 in the same direction and the CM of the rod is moving 4

v0 (in the opposite direction). This is shown in the figure below in the 4

system CM frame.

9l 16

v0 4 3v 0 4

Thus the angular momentum of the system in its CM before the impact is Linitial  Lrod  Lstone  ( L of CM of rod )  Lrod (about CM of rod )  Lstone  m 

l v0 3l 3v   0  m0   0 16 4 16 4

mlv 0 16

where we have used the fact that m0 

m . The sense of rotation is counter clockwise. 3

Immediately after the stone hits the rod, rod starts moving with speed 9l 9l 12v 0 27     v0 16 16 25l 100

With respect to the CM of the system, the CM of the rod and the stone are moving with speed (keep in mind that  is the same about any point and we are thinking of the motion of the system as the translation of the system CM plus rotation about the system CM. Thus with respect to the system CM, the motion is roational)

194

rod vCM 

l l 12v 0 3v 0     16 16 25l 100

and

v stone 

3l 12v 0 9v 0   16 25l 100

Thus the angulat momentum of the system about the system CM after the impact L final  Lrod  Lstone  ( L of CM of rod )  Lrod (about CM of rod )  Lstone l 3v 0 ml 2 12v 0 3l 9v     m0   0 16 100 12 25l 16 100 19mlv 0  400  m

where we have used the fact that m0 

m . The sense of rotation is counter clockwise. 3

Thus the impuse gives a torque equal to 3mlv 0 1  19    200  400 16 

L final  Linitial  mlv 0 

The negative sign here shows that the change is in clockwise direction. Its magnitude is the torque impulse which is equal to the impulse J provided by the pivot times the distance fo the pivot from the system CM. For clockwise sense, J is in the same direction as the firction of stone’s velocity. Thus J

12.20

9l 3mlv 0  16 200

 J

2mlv0 75

or

2m0 lv 0 25

Let the distance of the sweet spot be ls. The ball hitting the bat and its swing is shown in the figure below.

195

F

ls

0 The point where the bat is held is the pivot point of the bat. Further, let the ball come horizontally with speed vi and retun alonh the same line with speed vf. Considering clockwise rotation to be represented by the positive direction, we have: Initial angular momentum Li  I 0  mvi l s Final angular momentum L f  mv f l s Since there is no external torque about the pivot, we have Li  L f







I 0  mvi l s  mv f l s or m v f  vi 

I 0 ls

Finally the momentum change of the system is caused by the force applied on the bat. Initial momentum is the sum of the momentum of the bat and that of the ball. The final momentum is that of the ball only. This gives  F  Pf  Pi  mv f  ( mvi  MLc  0 )

Since for the sweet spot force is zero, we have m v f  vi   MLc  0

This combined with the equation m v f  vi   ls 

(i)

I 0 gives ls I MLc

For M=2 kg, LC=40 cm and I=0.3kgm2, we get 196

ls 

(ii)

0.3 m  37.5cm 2  .4

From the momentum equation above, we get

0  (iii)

m v f  vi  MLc



0.154  60  11.55 rad s 1 2  0 .4

With the wrapping of the tape, the mass the moment of inertia of the bat and its CM all change. We have I new  0.3  0.050  0.65 2  0.321 kg m 2

and M new Lc new  MLc  0.05  0.65  2  0.4  0.05  0.65  0.8325 kg m

This gives l s new 

0.321 m  38.6cm 0.8325

Thus the sweet spot is lowered by 1.1 cm. 12.21

The problem is solved as example 15.2 in chapter 15.

12.22

For generality we first take the masses on the dumbel to be of mass M each and the mass striking to have mass m.

Since there are no external

forces or torques acting on the system, its momentum and angular momentum remains the same before and after the collision. For convenience we take angular momentum about the origin. We show various distances in the figure below. Y M l/2 v

O

m

X

a l/2 M

The momentum components of the mass m are as follows

197

a

p x  mv 

a l 2

2

4

l 2

p y  mv 

a l 2

2

4

2mva



4a 2  l 2 mvl



4a 2  l 2

Initial angular momentum about the origin arises only from the motion of mass m. When particle is on the x axis, only the y component of its momentum gives the angular momentum about the origin and it is Linitial  mv 

l 2 a l 2

2

4

a 

mvla

clockwise

4a 2  l 2

After the mass m gets stuck with M after hitting it, the CM of the system is at xCM  0

y CM 

 M  m l 2  M l  2M  m 

2



ml 2 2 M  m 

immediately after the impact. By momentum conservation the momentum of the system after the imact is the same as before it. Thus the angular momentum of the CM of the system after the impact is (see figure below) Y (M+m)

py px

yCM M

X

O

M

LCM  y CM  p x 

m 2 vla

 2M  m 

4a 2  l 2

Since angular momentum is conserved, the rest of the angular momentum comes from the angular momentum about the CM of the system. Thus the angular momentum about the CM is LaboutCM  Linitial  LCM 

2 Mmvla

 2M

 m  4a 2  l 2

This then gives the rate of rotation  through the equation I aboutCM   LaboutCM

198

For this we first calculate I aboutCM . It is  l  ml   2 2  2 M  m   

I aboutCM   M  m  

2

 l  ml   2 2  2 M  m   

 M 

2



M  M  m l 2  2M  m 

Thus M  M  ml 2 2 Mmvla   2M  m   2 M  m  4a 2  l 2



m  2va   M  m  l 4a 2  l 2

  

With these we easily calculate the velocity of mass (M+m) and mass M right after the impact. These will be given by adding the velocity of the CM to their velocity due to rotation about the CM. Thus px  l  ml Ml   v x ( M  m )  v xCM       2M  m  2M  m   2 2 2 M  m  

v y ( M  m )  v yCM 

px  l  ml  M  m  l   v xM  v xCM       2M  m   2M  m  2 2 2 M  m  

v yM  v yCM 

py

 2M  m py

 2M  m

For m  M these answers become Angular momentum of the system CM about the origin after the impact =  Angular momentum of the system about its CM 

Mvla 3 4a 2  l 2

2 Mvla 3 4a 2  l 2

and v x ( M m) 

va 4a  l 2

v xM  0

v y ( M  m )  v yCM 

2

vl 3 4a 2  l 2

vl

v yM  v yCM 

3 4a 2  l 2

Chapter 13

13.1

2 2 2 (i) I xx   mi  y i  z i   2mw i







I zz   mi xi2  y i2  2m l 2  w 2 i

I xz  I zx   mi xi z i  0 i





I yy   mi xi2  z i2  2ml 2 i



I xy  I yx   mi xi y i   mlw i

I yz  I zy   mi y i z i  0 i

199

Thus  2mw 2  mlw  0   2 I    mlw 2ml 0  2 2   0 0 2m l  w   

(ii)

To find the principal axes, one rotates the frame with respect to the z-axis by an angle  so that the coordinates for each particle transform by the formulae x'  x cos   y sin  y '   x sin   y cos 

and the resulting coordinates make I x ' y '  0 . Numbering the masses as shown in the figure, we get Y

Y’

4

3 X’

w 1

2



X

l x1'  0

x2'  l cos 

x3'  l cos   w sin 

x 4'  w sin 

y1'  0

y 2'  l sin 

y3'  l sin   w cos 

y 4'  w cos 

Substituting these in the formula for I x ' y ' and equating the resulting expression to zero gives tan 2 

lw l  w2 2

Substitution of new (x’, y’) also gives 





l

2

 w2



2



l 2 w2

I x 'x '  m  l  w    l 4  w4  l 2 w2 l 4  w 4  l 2 w 2   2

2

and





2  l 2  w2 l 2 w2 I y ' y '  m l 2  w 2   l 4  w4  l 2 w2 l 4  w4  l 2 w2 





200







13.2 The principal axes (1,2) of the rectangle passing through its CM are shown in the figure when it is in the plane of the paper. Axis (3) is coming out of the paper. 2



1

 a a

(i)

Components of the angular velocity are 1   cos   

(ii)

b

b a b 2

2

 2   sin   

a a  b2 2

Components of angular momentum along the principal axes are L1  I 11 

ma 2 b

L2  I 2  2 

12 a 2  b 2

mab 2

 L

ma 2 b 12 a  b 2

2

1ˆ 

mab 2 12 a  b 2

2

2ˆ 

L3  0

12 a 2  b 2

Thus if the unit vectors along the principal axes are denoted as

(iii)

3  0

ˆ, 2 ˆ 1

mab 12 a 2  b 2

ˆ then and 3

 a1ˆ  b2ˆ

Kinetic energy of the rectangle K .E . 

1   1 1 ma 2 b 2 2   L  I 112  I 2 22  2 2 2 12 a 2  b 2 

13.3 The system and its principal axes (1,2) are shown in the figure below. Principal axis (3) is coming out of the paper.

201

2





a

1 L

b

We have I 1  2  2m 

a2  ma 2 4

1   cos    L1  I 11 

b a2  b2 ma 2 b a2  b2

I 2  2  2m 



b2  mb 2 4

 2   sin    L2  I 2  2 

I3  m a2  b2 a



3  0

a2  b2

mab 2

L3  0

a2  b2

Rate of change of angular momrntum can be found easily by either the Euler’s equations or by taking components of angular momentum in directions parallel to and perpendicular to  the direction of  .  By Euler’s equations: Since the components of  in the direction of the principal axes remain unchanged, we have for the change in angular momentum    dL L dt ˆ  ( L   L )3 ˆ  ( 2 L3   3 L2 ) 1ˆ  ( 3 L1  1 L3 )2 1 2 2 1

With the components calculated above, we get   a2  b2 dL  (1 L2   2 L1 )3ˆ  mab 2  2 2 dt  a b

 ˆ  3 

 dL ˆ . Thus at Thus the vector is always in the direction opposite to that of principal axis 3 dt

the instant shown, it is pointing into the paper.

202

By taking components of angular momentum in directions parallel to and  perpendicular to the direction of  : If we decompose the angular momentum   components L|| in the direction parallel to  and L perpendicular to  , then as the body rotates, L|| remains unchanged and L rotates with angulae speed  and changes at the rate L and that gives the rate of change of angular momentum. From the figure above, it is

clea that  a2  b2   2 2   a b 

L  L1 sin   L2 cos   mab 

This is also shown in the figure above. It is then clear that at the instant shown this vector is going to rotate into the plane of the paper. Thus the direction of change of angular momentum is into the paper and its magnitude is  a2  b2   2 2   a b 

L  mab 2 

13.4 The picture of the system of eight particles is shown below. We also show the diagonal about which we consider it rotating. Y 

X

Z (i)

To show that the coordinate system chosen is is also the principal axes system, it is sufficient to show that all the off diagonal elements of its moment of inertia tensor vanish. This is easily shown. We do it for Ixy here.

203

 l  h h h h l  h h h h              0  2 2 2 2 2 2 2 2 2 2 

I xy  I yx   mi xi y i   m  i

Thus the set of axes chosen is the principal set of axes.

(ii)

For the diagonal shown, the unit vector is =

liˆ  hˆj  wkˆ l 2  h2  w2

liˆ  hˆj  wkˆ



Thus the angular velocity is given as   

l 2  h 2  w2

To calculate the angular momentum, we also need the moment of inertia about the principal axes. These are  h 2 w2    2m h 2  w 2  4 4  



I 1   mi ( y i2  z i2 )  8  m   i

 w2 l 2     2m w 2  l 2 4 4 i   l 2 h2    2m l 2  h 2 I 3   mi ( xi2  y i2 )  8  m    4 4 i  



I 2   mi ( z i2  xi2 )  8  m  









Thus we have L1 





2 m h 2  w 2 l l 2  h2  w2

L2 





2 m w 2  l 2 h l 2  h 2  w2

L3  





2 m l 2  h 2 w l 2  h 2  w2

The kinetic energy

K .E . 

1 1 1 2m 2  h 2 l 2  w 2 l 2  w 2 h 2  I 112  I 2 22  I 3 32  2 2 2 h 2  l 2  w2

13.5 As the rod rotates, its angular momentum also changes. The required torque is provided by the gravitational force pulling it down.

Thus the rod can be in

equilibrium at two positions. Absolutely vertical or at an angle  from the vertical. The first solution is trivial. For the second solution, we find  by equating the angular momentum change about the pivot (a staionary point) to the torque. Later we will check the answer by applying the torque equation about the centre of mass also. The rotating rod in the plane of the paper is shown in the figure along with its principal axes at the pivot; axis (3) is coming out of the plane fop the paper. Also shown is the free 204

body diagram of the rod. The pivot applies a vertical force N to balance the weight mg of the rod and a horizontal force F to provide the required centripetal accelerartion to the CM of the rod. 2

N

1



F 



mg

The moment of inertia of the rod about the principal axes at the pivot are I1 

ml 2 3

I2  0

I3 

ml 2 3

The angular velocity components are 1   sin 

 2   cos 

3  0

Thus the angular momentum components are L1 

ml 2 sin  3

L2  0

L3  0

Since the angular velocity and its components along the principal axes are constant, we have for the torque    dL ˆ  ( L   L )3 ˆ      L  ( 2 L3   3 L2 ) 1ˆ  ( 3 L1  1 L3 )2 1 2 2 1 dt 

Thus 1  0

2  0

3  

ml 2 2 sin  cos  3

Here the negative sign shows that the torque is pointing into the paper. This is proided by the weight of the rod. The torque of the weight about the pivot is pointing into the paper and its magnitude is 205

l sin  2

mg

Thus we should have l ml 2 2 sin  cos  mg sin   2 3



 3g  2   2l 

  cos 1 

 The same answer can also be obtained by taling horizontal (perpendicular to  ) LH and  vertical (parallel to  ) LV components of the angular momentum and then calculating the

rate of change of angular momentum as LH. Now we calculate F. This provideds the centripetal force so we have  l  F  m sin    2  2 

Check: To check our answers, we apply the torque equation about the CM and see if the forces calculated by us satisfy this. About the CM we have  dLCM    applied ,CM dt

Further, for the principal axes at the CM (parallel to those at the pivot) I1 

ml 2 12

I2  0

I3 

ml 2 12

So that (angular velocity components are the same) L1 

ml 2 sin  12

L2  0

L3  0



3g  2   2l 

Thus we have, with cos    1  0

2  0

3  

ml 2 2 sin  cos  mgl sin   12 8

Let us see if , N and F calculated by us give the same torque. This is in the direction perpendicular to the paper and its magnitude is

F

l l l2 l cos   N sin   m  2 sin  cos   mg sin  2 2 4 2



3g  2  we get the torque as  2l 

Substituting cos   

206

l2 2 3g l mgl sin  m  sin    mg sin    2 4 2 8 2l

Thus our answers also satisfy the torque equation about the CM and are therefore correct. 13.6 (i) Since the wheel is rolling without slipping, it rotates about the horizontal axis with an angular speed  H 

V R

Further, it rotates about the vertical axis with an angular speed V 

V . This can also be L

seen easily as follows. Since the wheel rotates a full 2 radians in time

2L , its angular V

V  2L    in counterclockwise direction looking at it L  V 

speed about the vertical is V  2  from the top.

(ii) Angular momentum about O = angular momentum of the CM + angular momentum about the CM Angular momentum of the CM about O = MLV in the vertically up direction Angular momentum about the CM has both the horizontal and the vertical components. Angular momentum about the CM =

1 1  V MR 2   in the vertically up direction + MRV 4 2  L

in the horizontal direction pointing towards O. Thus total angular momentum about O is Horizontal component LH =

1 MRV in the horizontal direction pointing towards O. 2

Vertical components LV= MLV  (iii)

MR 2V in the vertically up direction. L

As the wheel rotates, the vertical component of the angular momentum remains unchanged while the horizontal component rotates. therefoe equal to V L H 

1 MRV 2 2 L

If seen from the top, its direction is as shown in the figure below

207

Its rate of change is

O LH

LH

(iv) The free body diagram of the wheel, when it is going into the page is shown below. N1 O

F

L

mg

N

As the wheel rotates, the change in the angular momentum is such that at the position shown above, it will point out of the paper. On the other hand, torque due to the weight about O is pointing into the paper. Thus to generate torque coming out of the paper, the normal reaction of the ground has to be more than the weight of the wheel so that Torque  change in angular momentum L N  mg  

1 MRV 2 2 L



N  mg 

1 MRV 2 2 L2

1 MRV 2 By balancing the vertical forces, we get N 1  2 L2

Thus when the wheel is moving, it is pulled down by the vertical shaft at the centre. In turn, the shaft is pulled up. Thus with time it will tend to come out of the ground. The horizontal force applied by the shaft provided the centripetal acceleration. Thus 208

F

mV 2 L

The explanation for why the wheel presses theground harder has been given above. Another way to think about it is as follows. If there were no gravity, the horizontal shaft of the wheel will tend to turn clockwise about a horizontal axis as the wheel moves so that the change in the torque is zero. In turn the wheel presses the ground and generated enough torque so that the rate of change of its angular momentum is equal to the torque generated. 13.7 This problem is like the previous problem except for position of the centre of mass and the moment of inertia of a cone.

We again obtain the horizontal component of the

angular momentum of the cone and multiply it with the vertical component of the angular velocity of the cone to get the rate of change of the angular momentum. The horizontal and the vertical components of the angular velocity and the free body diagram of the cone are shown below.

V

N1

H

F mg

Since the speed of the centre of cone’s base is V and its radius is R,  H  V 

N

V . Similarly R

V . h

The principal axes of the cone at its vertex are its axis and two axes perpendicular to it. The moment of inertia about the axis of the cone is

3 mR 2 . Thus the horizontal moment 10

of inertia is LH 

3 V 3 mR 2   mVR 10 R 10

And its rate of change is

209

3 3 mV 2 R mVR  V  10 10 h

Coming out of the plane of the paper at the position of the cone shown. This shoud be equal to the torque about the pivot. Thus 3 3 mV 2 R hN  hmg  4 10 h

3 3 mV 2 R N  mg  4 10 h 2



And N1  mg  N

1 3 mV 2 R N 1   mg  4 10 h 2



Similarly, the force F provides the centripetal force. Thus F  m

13.8

3 3 mV 2 h  V2  4 4 h

This is a problem where we take the components of the angular velocity according

to the convenience of applying the rolling condition. So we split the angular velocity so that it has component  about the vertical and  about its axis.

We then find the

relationship between them by demanding that the speed of all the points on the ground vanishes so that the cone rolls without slipping. (i) Now a point at distance x from the vertex (pivot point) on the line touching the ground moves with speed x due to the rotation about the vertical axis through the vertex and speed x sin  (see figure below) in the opposite direction due to rotation about the axis of the cone. 



 x

 Thus for rolling we get    sin 

Now if the centre of the base moves with speed V, we have

210

V  h cos 

 

V h cos 

This also gives 

V h cos  sin 

As is clear from the components, the net angular velocity



of the cone is  cos  parallel

to the line touching the ground an in direction shown in the figure. Thus    cos  

(ii)

V V h2  R2  h sin  hR

To calculate the angular momentum of the cone, we calculate the angular momenta along the principal axes shown in the figure below. 2

1





From the figure it is clear that L1   I 1 cos   

3 V h2  R2 mR 2   10 hR

h h R 2

2



3 mRV 10

2 2 3 R 3  3  V h R  3  V mR 2  mh 2     mR 2  mh 2  2 2 20 80 hR 20 80     h h R

L2  I 2 sin   

In this case, the angular momentum changes because its vertical component rotates with angular speed . Thus the rate of change of angular momentum is given by  dL   LV    L2 cos   L1 sin   dt   3  3 h 3 R  V mR 2  mh 2    mRV     2 2 2 2 80 10  h h R h R    20 3 mRV 2 3 mhV 2   20 h 80 R 

V h2  R2 hR

and it is in the direction of axis 3. Its sign depends on the relation between h and R. Check: The answer can be checked easily by applying the Euler equations that give

211

L 1   2 3 ( I 3  I 2 )  0 L 2   31 ( I 1  I 2 )  0 3 3   L 3  1 2 ( I 2  I 1 )   2 sin  cos    mR 2  mh 2  80  20  2 V  3 3   mR 2  mh 2    hR  20 80 

which is the same answer as obtained above. 13.9 In the figure below, we show the horizontal angular momentum LH of the rotating disc, its change LH as the platform rotates and its free body diagram. N1 N2

 LH

LH mg

We have LH 

 dL 1  LH  mR 2 0  dt 2

1 mR 2 0 2

If the torque applied on the disc is zero, i.e., N1 and N2 are equal, there cannot be any change in the angular momentum. Hence the disc will tend to rotate clockwise in the position shown so that N1 will become larger than N2. Finally, N1 and N2 will be such that the torque is equal to change in the angular momentum. This gives N 1  N 2  mg

l  N1  N 2   1 mR 20  2 2

Solving these two equations gives N1 

mg mR 2 0   2 2l

N2 

mg mR 2 0   2 2l

13.10: It has been worked out in section 13.5.3 (see figure 13.18). Carry it out further for each axis step by step. 13.11 We show in the figure below the principal axes for the system at the pivot point and the forces that act on the vertical rod at the bearings. Principal axis 1 is along the rod

212

and 2 and 3 are perpendicular to the rod with axis 3 coming out of the plane of the paper at the instant the rod is shown. In the free body diagram of the rod, we have not shown the vertical forces. The cetre of mass of the system is also indicated in the figure. N1

2







1

LH

d

d

N2 The moment of inertia about the principal axes are as follows I1  0

I2  I3  m

l2 9l 2 5 2 m  ml 4 4 2

The angular velocity components are 1   cos 

 2   sin 

3  0

Thus the angular momentum components are L1  0

5 L2  ml 2 sin  2

L3  0

The torque required to keep the rod rotating can be calculated either by Euler’s equations or by the horizontal component LH of the angular momentum and multiplying it by . Euler equations give  1  ( 2 L3   3 L2 )  0  3  (1 L2   2 L1 ) 

 2  ( 3 L1  1 L3 )  0

5 2 2 ml  sin  cos  2

This is the same answer as obtained through LH as is easily checkd.

213

The torque is provided by the reactions of the bearings. The difference in these reactions also provides the centripetal force as the CM of then system is moving in a circle of radius l sin  . Thus 2

 N1  N 2  d  5 ml 2 2 sin  cos 



2

N1  N 2 

5 ml 2 2 sin  cos  2d

l N 1  N 2  2m  sin    2  ml 2 sin  2

Solving these two equations gives N1 

ml 2 sin  2

13.12

 

 1

5l cos    2d 

N2 

ml 2 sin  2 

 1 

5l cos    2d 

In the figure below we show the principal axes for all the three rigid bodies. Axes (1, 2) are in the plane of the paper and axis 3 is coming out of the the paper in all three cases. 

1



2





(a)

 1

(b)

(c)

In all the three cases we have 1   cos 

 2   sin 

3  0

and L1  I1 cos



2

2

L2  I 2 sin 

Thus we have

214

L3  0

1

 dL     dt   dL     dt 

 ( 2 L3   3 L2 )  0 1

 dL     dt 

 ( 3 L1  1 L3 )  0 2

 (1 L2   2 L1 )   I 2  I 1   2 sin  cos  3

Now in case (a) I1 

mb 2 12

I2 

ma 2 12

sin  

b a b 2

cos  

2

a a  b2 2

This gives  dL     dt 

 3

m 2 ab  a 2  b 2  12  a 2  b 2

  

In case (b) mb 2 I1  12

ma 2 mb 2 I2   . 12 12

This gives

 dL     dt 

 3

ma 2 2 sin  cos  12

In case (c) I1 

13.13

mR 2 4

I2 

mR 2 . 2

This gives

 dL     dt 

 3

mR 2 2 sin  cos  4

As the ring rotates, the tension in the string does two three things: it balances the weight of the ring, it provides the centripetal acceleration and it provides the torque for the angular momentum change of the ring. The figure below shows the principal axes of the ring at its CM and the free body diagram of the ring. Axes (1, 2) are in the plane of the paper while axis 3 is coming out of the paper.



O

2



A 1 B

A

B

mg 215

T

The components of the angular velocity are 1   cos

 2   sin 

3  0

Therefore components of the angular momentum along the principal axes at the CM are L1 

1 mR 2 cos  2

L2  mR 2 sin 

L3  0

Thus the change in the angular momentum is  times the horizontal component of the 



1 2

2 2 angular momentum =   mR  sin  cos   mR  cos  sin   





1 mR 2 2 sin  cos  2

And it points in the direction of axis (3) i.e. coming out of the page at the instant shown. Euler equations also give the same answer through  dL     dt   dL     dt 

 ( 2 L3   3 L2 )  0 1

 dL     dt 

 ( 3 L1  1 L3 )  0 2

 (1 L2   2 L1 )   I 2  I 1   2 sin  cos  3

However, at the position shown, the tension and the weight of the ring give no torque about the CM of the ring. Thus the internal forces will move the ring so that there is an opposing change in the angular momentum. This ring moves so that its B end moves up. This implies that the ring moves towards position 1. However, as it moves up, the tension gets misaligned with the CM and starts giving a torque about it and finally the ring stops at an angle such that the torque equals the angular momentum change. Now balancing the forces gives T cos   mg

T sin   m R  l  2 sin 

These equations give

216



T  m R  l   2

cos 



g  1  2  R  l  

  sin  cos 

g2 g2  sin   1  4 2  1  4 2   R  l  2  R  l  

g  2  R  l

If the ring moves up by an angle , The perpendicular distance of the tension and the CM is about  R . This gives a torque  3   RT   mR R  l  2

Equating this to the rate of change of angular momentum gives  mR R  l  2 

1 1 g mR 2 2 sin  cos   mR 2 2 2 2 2   R  l

This gives Rg



13.14

2  R  l  2

2

Following the notation of example 13.7, we have (using parallel axis theorem for calculating

I )

mR 2 50  10 3   2  10 2  I   10 5 kgm 2 2 2 mR 2 I   ml 2  5  10 6  50  10 3  25  10  4  1.3  10  4 kgm 2 4 2

Given  S  100 rads 1 (i) The presseion angular frequency will then be p 

mgl 50  10 3  9.8  5  10 2   7.8 rads 1  450s 1 5 I 10  100

Seeing the change in the angular momentum for the given sense of rotation, the top will be coming out of the paper. (ii) We see that  p   S . Therefore for the nutation motion we can write (   0 ) 

 p sin  0



1  cos  t  with 

217



I S I

The maximum difference between  and 

0

is therefore

2 p sin  0



.

Substituting all the

numbers, we get  max   0 

(iii)

2 p sin  0



 30  18.6  48.6

The frictional force is needed to provide the centripetal force. Thus F friction  ml sin 30 2p  50  10 3  5  10 2  0.5  7.8 2  0.076 N

13.15

The soulution of the Euler equations governing the motion is the same as in

example 13.7 but the initial conditions are different. Thus we have  2  A sin  0 t  B cos  0 t 

 sin  cos  S t  I S

 3   A cos  0 t  B sin  0 t 

Here  0 

 I  I   I

S

 sin  sin   S t  I S

. Let the initial angular speed given about the vertical be  0 . Then

the initial conditions for the subsequent motion are  2 (t  0)   0 sin  0

 3 (t  0)  0

These give A0

  B    0  I S 



 sin  0 

Thus we get    2    0  I  S     3    0  I S 



 sin  0 cos  0 t  

 sin  cos  S t  I S



 sin  sin   S t  I S



 sin  0 sin   0 t   

If the precession velocity of the top is  p then

218

 p sin    2 cos  s t   3 sin  s t

  sin  cos  S t  sin  0 cos  0 t  cos  s t I S   sin  sin  S t  sin  0 sin  0 t  sin  s t  I S

  0 sin  0 cos  0 t   S t  

 I     0 sin  0 cos  S t   I I  S   



    sin   sin  0 cos I  S t        I   

This gives  p  0

 I  sin  0  cos  S t   sin  I S  I 

 

 1 

 I  sin  0 cos  S t  sin   I 

   

And    2 sin  s t   3 cos  s t   sin  cos  S t  sin  0 cos  0 t  sin  s t I S   sin  sin  S t  sin  0 sin  0 t  cos  s t  I S

  0 sin  0 sin   0 t   S t  

      0  I S 



 I   S t   I 

 sin  0 sin  

Upon integration, the equation for  gives

 (t )    0 

I I S



   0  

 I S

  I    sin  0  1  cos  S t    I     

The variation of  (t ) with time is similar to that shown in figure 13.23. To get the angle through which the top has precessed, we substitute the value of  (t ) in the  (t ) and integrate it numerically. Finally to see the motion of the tip of the equation for  p

axis of the top,  is plotted against

 

p

(t ) dt

219

.

Chapter 14 14.1

For this spring  

(i)

(ii)

k  m

Asin   0.1      , A  0.1 Acos  0  2

25  3.54 rad s1 2



giving x(t )  0.1sin  3.54t  

  2

A sin   0.1    A  0.11, tan   2.38 or   1.17rad A cos  0.15  A cos  0.042

x (t )  0.11 sin  3.54t  1.17 

(iii)

Asin   0.1    A  0.104, tan   3.57 A cos  0.1  A cos  0.028 sin   0    cos  0      which gives   1.84rad 2 tan   0  x(t )  0.104 sin  3.54t  1.84 

(iv)

Asin   0.1    A  0.115, tan   1.76 A cos  0.2  A cos  0.0565 sin   0    cos   0      0 which gives   1.06rad 2 tan   0  x (t )  0.115 sin  3.54t  1.06 

220

leading

to

(v)

A sin   0.2

   A  0.202, tan   7.14 A cos  0.1  A cos  0.028 sin   0    cos   0      0 which gives   1.43rad 2 tan   0  x (t )  0.202 sin  3.54t  1.43

(vi)

A sin   0.2    A  0.208, tan   3.57 A cos  0.2  A cos  0.056 sin   0    cos   0       which gives   1.84rad 2 tan   0  x (t )  0.208 sin  3.54t  1.84 

14.2

It is given that x (t )  5 sin(2t   ) . Thus the velocity is v (t )  10 cos(2t   ) and the acceleration is x (t )  20 sin( 2t   ) . This is shown below for different values of . Displacement curve is the one with smallest amplitude of 5, the velocity curve has intermediate amplitude of 10 and the acceleration curve has the largest amplitude of 20. Notice that phase difference of π and π give identical curves.  0

221





 4

 3



 2

222

 

 4

 

 3

 

 2





2 3

3 4



 

2 3



3 4



223

14.3

Equilibrium points of the potential are given by

dV ( x )  0. dx

For V ( x )  2 x 3  9 x 2  12 x , this gives 6 x 2  18 x  12  0 

x 2  3x  2  0

The roots of this equation are at x = 2 and x = 1. The second derivative of the potential at these points is

  6  x  1 d 2V (x)  12x  18   2 dx   6  x  2

Thus the potential is minimum at x = 2. k  V  x  2  6 . This gives  



V ( x)  C 1  e

14.4



 a ( x  x0 ) 2

k m 



The corresponding spring constant is

6  2.45rad s 1 .





dV  2C 1  e  a ( x  x0 )  ae  a ( x  x0 ) dx

Equating the derivative to zero gives x = x0. At this point the second derivative





d 2V  2C 2a 2 e  2 a ( x  x0 )  a 2 e  a ( x  x0 )  2Ca 2 2 dx

is positive and therefore at this point the potential is minimum. (i)

The corresponding spring constant k  2Ca 2 .

(ii)

To plot the potential and harmonic approximation to it, we have taken C=2, x0=1.0 and two values of a: a=0.4 and a=1.2

224

a = 0.4

a = 1.2

k 2C a m m

(iii)



(iv)

As is clear from the figure, depending on the value of a the potential becomes softer or harder for larger displacements from x = x0. Thus the frequency will become smaller and time-period larger for a=1.2 and the frequency will become larger and time-period smaller for a=0.4.

14.5

  V ( x)        x 

6       x  

12





 dV  12 6     12 13  6 7  dx x x  

Equating the derivative to zero gives x  21 6  . At this point the second derivative    d 2V  12 6  12 6 18   156  42   156  42  13 2   2 14 8  14 13 8 13  dx x x    4 2   2 2  2   

Thus the corresponding spring constant is

is  

k 1  m 

18 and the frequency of small oscillations 21 3  2 18 9  21 3   21 3 m

225

 m

14.6 The surface charge density    x gives a electric field E 

x that pulls the 0

displaced electrons back, i.e. it applies a restoring force. We consider volume V of the displaced block, the force on it is V 

x V 2 x  . 0 0

The total mass of the block is = number of electrons in volume V electron mass V

= e me So the equation of motion for the displaced block of electrons is (writing x  y ) V V 2 me y   y e 0



y 

e y0  0 me

This gives the plasma frequency (here n is the number density of electrons) p 

14.7 In this problem k  2000 Nm 1 (i)  

k  m

e ne 2   0 me  0 me

m  75kg

2000  5.16 rad s 1 75

(ii) Velocity of platform and person on it is calculated by principle of linear momentum conservation and gives v

(iv)

55 2 gh

 55  20 

 4.59ms 1

If we take the equilibrium point of (person+platform) as y=0, then the displacement without the person on the platform is =

55  9.8  0.27 m . Thus this is a problem 2000

with the initial conditions y (t  0)  0.27

y (t  0)  4.6ms 1

Representin the motion as y (t )  A sin( t   ) , we have

Asin   0.27    A  0.93 and tan   0.30 A cos  4.6  A cos  0.89 226

sin   0    cos   0      which gives   2.85rad 2 tan   0  Thus the displacement is given by x(t )  0.93 sin  5.16t  2.85 14.8 In this case, it is the motion of a rigid body about a pivot point. Thus the equation of motion is written in terms of the moment of inertia, angular acceleration and the restoring torque. If the square is displaced by an angle  about the vertical (see figure) the torque is   mg

a 2

mga

sin  

2



 a

mg The moment of inertia about the pivot is calculated by the parallel axis theorem and is  I  m 

2

a  

2

m

a2 2  ma 2 6 3

The quation of motion is I  

mga  2

or

2 mga ma 2   0 3 2

This can be rewritten as 3g    0 2a 2

This gives the frequency of oscillation to be  

227

3g 2a 2

14.9 These two pendulums with their displacements are shown in the figure below.

2

1 L K

M

For visualizing the force applied by the spring, let us take  2  1 . Then the bob on the left feels a force to the right due to the spring and to the left due to the weight of the bob. The spring force is in the opposite direction on the pendulum on the right. (i) Since the rods are rigid, and therefre generate internal forces, we use the angularmomentum torque equation for describing the motion of the pendulums.

Thus the

equations of motion for the two pendulums are ML21   Mgl1  ML22   Mgl 2 

(ii)

L L  k  2  1  2 2 L L  k  2   1  2 2

 

g k 1  1   2  1  L 4M g k 2   2    2  1  L 4M

The equations above are solved by adding them to get the equation for  (t )  1 (t )   2 (t ) and for  (t )  1 (t )   2 (t ) . These are

g     0 L

 g k        0  L 2M 

These give  (t )  A cos 1t  B sin 1t

 

 1  

g  L 

 (t )  C cos  2 t  D sin  2 t

 

 2  

g k  L 2M

   

Combination of these two gives  1 (t ) and  2 (t ) . There are four unknowns that will be given in terms of four initial conditions. 228

Angular displacement of the rod by and angle  and the

14.10

corresponding angle  that the strings make with the vertical are shown in the figure below. Also shown are the tensions in the strings; these are equal by symmetry.



T T

l



L Since we are dealing with a rigid rod, the correct equation of motion to be used is the angular momentum-torque equation. There are three unknowns in the problem: ,  and T. We thus need three equations. These are 2T cos   mg

mL2     L  T sin  12

and

l sin  

L sin  2

In the small angle approximation we have cos   1

sin   

and

sin   

Then T

mg 2



L  2l

And the equation of motion becomes mL2  mg L   L    12 2 2l

Thus the frequency of oscillation is  



3g     0 l

3g . l

14.11We show in the figure below a displacement of the rod when the left spring is stretched by y1 and the right one by y2.

229

L y1

y2

yCM



The equation of motion for the CM is M y CM  Fext  k ( y1  y 2 )

with

y CM 

y1  y 2 . 2

And the equation for rotation about the CM is I   ext

where we mesure counterclockwise  as positive. Then 

y1  y 2 L

L  ext   k ( y1  y 2 ) 2

Thus we have for the CM y CM 

2k y CM  0   CM  M

2k M

And for rotation about the CM ML2  kL2    12 2

6k or     0   rot  M

6k M

With the initial conditions y1  A1 and y 2  A2 and the rod held stationary in the beginning, we have A1  A2 , 2

y CM (t  0) 

A1  A2 ,  (t  0)  0 L

y CM (t  0)  0,  (t  0) 

This then gives the answer  A1  A2   cos  CM t 2  

y CM (t )  

and

230

 A1  A2   cos  rot t L  

 (t )  

14.12 Since we are writing the displacement as y (t )  A sin  t    , the projection of the phasor on the y-axis gives the displacement. The phase angle is measured from the x-axis. In the following we show the position of he phasor at t = 0.

After that it rotates

counterclockwise.

(i)

(ii)

(iv)

14.13

(iii)

(v)

(vi)

Assuming that the swing is performing simple harmonic motion, the period of the swing is 

g  l

9 .8  1.28s 1 6

Therefore the speed of the person on the swing as the swing passes through its equilibrium point is = A  1.28  2  2.56ms 1 (i)

When the swing is at the extremes and the child is handed over, the amplitude will not change. It is like strating a swing, ireespective of its mass, from a distance A from the equilibrium with zero initial speed.

Energetically it can be seen as

follows: When only the man is on the swing the total energy is =

M gA 2 1 M man 2 A 2  man 2 2l

When the child is handed over to the person at the extreme, the child brings in additional potential energy (with respect to the equilibrium point) = Mchildgthe height of the extreme point with respect to the equilibrium point 231



 M child g l  l  A 2

2



M child gA 2  2l

Thus the total energy that the swing posseses after the child is handed over is

 M man  M child  gA 2 2l If the new amplitude is Anew, this should equal (keep in mind that  remains unchanged) 2  M man  M child  gAnew 1 2  M man  M child  2 Anew  2 2l

A comparison in the total energies calculated in two ways immediately gives Anew  A . (ii) When the child is handed over at the equilibrium point, the speed os the swing decreses by the conservation of linear momentum. It is V 

M manVinitial 50  2.56   2.13ms 1 M man  M child 60

And the total energy of the system is now 1  M man  M child V 2  136.53J 2

If the new amplitude is Anew, then (keep in mind that  remains unchanged) 1 2 2  M man  M child  2 Anew  30  1.64  Anew  136.53 2

This gives Anew  1.67 m . (iii)

As is clear from the calculations above, energy is conserved in case (i) only.

14.14 The equation of motion now will be mx   kx  f

Its general solution is x (t )  C cos  o t  D sin  o t 

f k

With the initial conditions x (0)   A and x (0)  0 , the solution is f  f  x(t )   A   cos  o t  k k 

Since the velocity x of the block is

232

f   x   A    o sin  o t k 

It becomes zero at  0 t   , At this time f  f 2f  x (t )   A     A  k k k 

14.15

(i) Since this is the motion of a rigid body pivoted at a point under an external force, its motion is described by the angular momentum-torque equation. The torque on the body is applied by the weight acting at its CG, and it is a restoring torque (see figure). The value of the torque about the pivot point is  

l mg sin  2

The moment of inertia of the rod about the pivot point is I 

ml 2 3

mg Thus the equation of motion for the rod is ml 2  l    mg sin  3 2

For small angles this equation bocomes ml 2  l    mg 3 2

(ii)

3g or    0   2l

3g 2l

As the pendulum swings about, the frictional torque causes it to lose energy and therefore the amplitude of the pendulum decreases slowly. The equation of motion

233

can be written and solved exactly as in the problem above. When the pendulum in the figure above is moving clockwise, the equation of motion is 3g 3 or     2 2l ml

ml 2  l    mg   3 2

With the initial conditions  (t  0)   0 ,  (t  0)  0 , the solution of the equation above is (just like in the problem of spring-mass system with friction)  2  2  cos t   (t )    0  mgl mgl   4

Like in the problem above, this then leads to a reduction in the amplitude by mgl by the time the pendulum reaches the left extreme. Thus in one cycle the amplitude reduces by 8 . We will now derive this result from energy considerations also. mgl

Suppose at a certain instant the pendulum starts with the maximum displacement of 1 and goes to the maximum angle  2 on the other side. Then the total angle covered by it is

 1   2 

and the work done against the frictional torque is therefore  1   2  . The energy

loss during the motion is

1 1 I 212  I 2 22 . Equating this to the work done against 2 2

friction gives 1 2 2 1 2 2 I 1  I  2   1   2   2 2

1   2  

2 2 4   2 2 I ml 3 g mgl  3 2l

8

Thus in one cycle the amplitude reduces by mgl . Now if the pendulum starts with an angle  0 and completes N cycles, it will stop swinging 2

mgl

when it is at angle  such that 2       mgl . Thus we have 0  N

8 2  mgl mgl



234

N

mgl 0 1 mgl 0   8 4 8

14.16

As the mass leaks out, it carries with it the energy at the rate 1 dm 2 1 v  Cv 2 . 2 dt 2

If at time t, the amplitude is A(t) and the frequency (t), then

v(t )   (t ) A(t ) sin  t    and the average rate of loss

dE (t ) of energy E(t) of the dt

oscillator is dE (t ) 1 1   C 2 (t ) A 2 (t ) sin 2  t      C 2 (t ) A 2 (t ) dt 2 4

In time averaging A(t) and (t) come out of the integral because they vary slowly over a few oscillations (that is precisely why they can be defined). The negative sign shows that the energy is being lost with time.  2 (t ) 

k k  m(t ) m0  Ct

And if amplitude is A(t), the energy E (t ) 

1 kA 2 (t ) and the rate of change of energy 2

dE (t ) dA(t )  kA(t ) . Combining all this gives dt dt

kA(t )

dA(t ) 1 k  C A 2 (t )  dt 4 m0  Ct

dA(t ) C  A(t ) dt 4 m0  Ct 

If the amplitude at t = 0 is A0, this equation is integrated as dA dt  A A   0 4 m0  Ct  0 A

t

This gives Ct A(t )  A0  1  m0 

14.17



14



 

The energy loss is given by the formula dE  4e 2 A2  cos 2 t dt 6 0 c 3

Thus the average energy loss is

235

dE  4e 2 A2 1  4 e 2 A2 2  cos t   dt 2 6 0 c 3 6 0 c 3

If the damping factor of the system is , then E

dE dt

 E

where the energy E is

1 me 2 A 2 . Thus for the current problem we get 2

 

1 dE 2 1  4 e 2 A2  2e 2    E dt me 2 A 2 2 6 0 c 3 6 0 me c 3

14.18

We consider an oscillator with initial conditions so that its motion is given by

x (t ) 

 t Ae 2 sin  0 t

For undamped oscillatot the siplacement is maximum when  0 t 

 . For the damped 2

oscillator, it will be given at time when 

2 0 dx     t  A  sin  0 t   0 cos  0 t  e 2  0  tan  0 t   2Q dt   2 

Since Q is very high, the maximum occurs when  0 t  the maximum occurs before

 2

   where  is very small. Since 2

, it is advanced by phase angle . Thus        2Q  cot   2Q  2 

tan

For small , we have cot  

cos  1  sin  

1

This gives   2Q . 14.19

If the person gives an impulse J, the energy gained by him is Jv, where v is the speed at the equilibrium point. Since v  A , we have Energy gained per cycle = JA 236

This will equal the energy loss per cycle due to damping. E

Energy loss per cycle = 2πenergy loss per radian = 2 Q Here E is the energy of the swing. This follows from the definition of the quality factor. E

1 mgA 2 m 2 A 2  2 2l

Thus we get JA  2 

1 m 2 A 2  Q 2



mA  mA  Q Q

J 

g l

Putting in all the numbers gives J 

  45  2 50

9.8  7.9 Ns 5

It is given that k  5000 Nm 1 and the mass of the platform

14.20

is 50kg. Since the system is critically damped for 500kg put on the platform, we have  2  4 2  4 

k 5000  4 m 550



  6.03

So the coefficient of drag force b  m  550  6.03  3316.5N/ms 1 Thus when a 200kg weight is on the platform, the value of   And the angular frequency  0 

k  m

b 3316.5   13.27s 1 m 250

5000  4.47 rad s 1 250

Thus  2  176.1  4 02  80 , which makes the system hevily damped. The solution then is y (t )  Ae

 1t

 Be

 2t

with 1 

 2    02  1.735 and 2 4

2 

 2    02  11.535 2 4

So y (t )  Ae 1.735t  Be 11.535t

Here A and B are to be determined by the initial conditions. Initial conditions:

237

Since we are taking the equilibrium position of the (load+platform) as y = 0, the initial displacement from that position is y (t  0)  (i)

200 g  0.4m . k

The speed of the load when it lands on the platform is

2 gh 

20  4.47 ms 1 .

Then by momentum conservation, the load+platform will move with the speed 200  4.47  3.58ms 1 . 250

(ii)

From the above, the initial conditions are y (t  0)  0.4

y (t  0)  3.58ms 1 .

Thus we have from the solution above A  B  0 .4

1.735 A  11.535 B  3.58

Thses give A  0.106

and

B  0.294

The solution therefore is y (t )  0.106e 1.735t  0.294e 11.535t

For a forces oscillation the angle  by which the displacement

14.21

lags the applied force is given by equations

sin   and

2

 2

( o   2 ) 2   2 2 tan  

cos  

( o   2 ) 2

( o   2 ) 2   2 2

 2 ( o   2 )

For a very heavily damped oscillator    . Thus we have sin   1

This gives  

cos   0

 2

238

and

tan   

Chapter 15 15.1 We perform our calculations in a noninertial frame attached with the box and take the origin (x = 0) at the point where the spring is connected to the box. In this frame, there is a pseudo force ma acting in the negative x direction. Thus the system in the noninertial frame looks as shown below.

x=0

ma

l The equilibrium point x0 is given by  k  x0  (l )  ma  0 

x 0  l 

ma k

Thus the equation of motion for the mass in this frame is  ma    mx  k  x    l   k    

 x 

k ma   xl   0 m k 

At t  0 , the initial conditions are x(t  0)  l , x (t  0)  0. 

We change the variables to y   x  l  

y 

ma   and write the above equation as k 

k ma, y  0 with the initial conditions y (t  0)  m k

y (t  0)  0.

This gives the

soulution y (t ) 

ma cos k

k t m



x(t )  l 

ma  1  cos k 

k  t m 

15.2 This problem is very easy to solve in the accelerating frame attached with the car. The free body diagram of the door in the car frame with pseudo force included is shown below.

239

car Ma

w

In the car frame therefore the door is being pulled at its CM by a force Ma and rotates about the hinges. Thus the problem becomes like examples 12.7 and 15.3 where a ruler/rod, pivoted at one of its ends’ rotates about an axis at that point when the pivot point starts accelerating with an acceleration. Then by energy conservation this gives the answer for the angular speed of the door when it is about to close 

3a w

15.3 This problem is similar to problem 12.7 and the problem above and is solved in exactly the same manner. 15.4 We substitute

  A  v rotating

in

 dA dt

 inertial

 dA dt

  A rotating

to get  dv rotational dt



 dv rotating dt

inertial

     v rotating rotating



dv rotating      Now from equation 15.8 v rotating  vinertial    r and a rotating  dt

 dvinertial dt 

Since vinertilal

 dr  , dt inertial

 inertial

    a rotating    v rotatingl inertilal

the above equation is

 dvinertial dt

Again substituting

 d   r  dt

so we have rotating

        v inertial  a rotating    v rotating inertial

    vinertial  v rotating    r

 dv inertial  from equation 15.8 and ainertial  dt

we get

240

, inertial

        ainertial    v rotational       r   a rotating    v rotational

which gives        ainertial  a rotating  2  v rotating    (  r )

15.5 In the rotating frame the rod is stationary and there is a pseudo force – the centrifugal force – acting on it. Thus the free-body diagram of the rod in the rotating frame is as shown below. Notice that we are not taking the centrifugal force to be acting at the centre of mass because it is different for different portions of the rod; it increases with the distance of fom the axis.

Centrifugal force

mg

(i)

It is clear from the figure above that the rod is pulled out horizontally by the centrifugal force (in the rotating frame) and would therefore move out when displaced from the vertical position. Since we are dealing with a rigid body, we should be working in terms of totques. Thus at the given position, the centrifugal force applies a counterclockwise torque on the rod but at the vertical position, neither its weight nor the force by the pivot apply any torque. Thus the rod will tend to swing away from the axis. It will be clearer mathematically in part (ii).

(ii)

When the rod is at an angle  from the vertical, the component of force on a portion of length ds at distance s from the pivot is

241

m ds ( s sin  ) 2 (see figure). l

s

m ds ( s sin  ) 2 l mg

Thus the torque due to the centrifugal force is l

 centrifugal   s cos   0

m ml 2 2 ds( s sin  ) 2  sin  cos  l 3

At the same time there is an opposing torque due to gravity and its value is  gravity 

mgl sin  2

Balancing these two gives cos  

Thus as long as cos   1, i.e. l 2 

3g 2l 2

3g , the rod will be at equilibrium at the angle whose 2

cosine is given above. Thus when disturbed from the vertical position it will move out to this equilibrium position (with very small friction damping the motion but not affecting the 2 equilibrium position). For angular speeds such that cos   1, i.e. l 

3g , the rod will 2

not move out because the torque due to the centrifugal force will not be able to overcome the torque due to the rod’s weight. 15.6

This problem is similar to the problem above except that there is no gravitational force involved here. If looked at from the top along CD, the free body diagram of the rod in the rotating frame, when it is at an angle  from the frame, looks as shown below.

242

m ds ( s sin  ) 2 l

s

A



B

If we take a strip of width ds at a distance s from the axis CD, the mass of the strip will be m m m wds  ds and the centrifugal force on it = ds ( s sin  ) 2 . The net torque on it is lw l l

therefore l 2

 centrifugal 



s cos  

l 2

This shows that both   0 and   (i)

m ml 2 2 ds ( s sin  ) 2  sin  cos  l 12

 are equilibrium positions of the sheet. 2

ml 2 2  Thus when disturbed slightly  ~ 0  , the torque on the sheet is  centrifugal  12

and tends to take it away from that position. The equation of motion for slight disturbance from the position parallel to the frame is therefore I 

ml 2  ml 2 2    12 12

or    2  0

This gives  (t )  A exp(t )  B exp(t )

Therefore the equilibrium position at   0 is an unstable equilibrium position. (ii)

If we call the angle from the perpendicular position of the sheet  then   The expression for torque in terms of  remains  centrifugal 

ml 2 2 sin  cos  12

243

  . 2

However, as the sheet is disturbed from this position, the torque has restoring nature. For small  therefore the equation of motion is I 

ml 2  ml 2 2    12 12

or    2  0

Thus the motion about this position is oscillatory and is given by  (t )  A cos(t )  B sin(t )

or equivalently by  (t )  A cos(t )  B sin(t )

Therefore the equilibrium position at  

15.7

 is a stable equilibrium position. 2

We follow the convention for the directions as in the main text. There is no force in the horizontal plane. The equations of motion in the northern hemisphere are therefore dv x  2v y  sin   2v z  cos  dt dv y  2v x  sin  dt

dv z   g  2v x  cos  dt

If   0 , vx and vy would remain zero for a particle thrown up. Thus correct to the linear order in  , the equations above are dv x  2v z  cos  dt

and

dv z  g dt

This gives v z (t )  v z 0  gt ; z (t )  v z 0 t 

gt 2 2

and

dv x d 2 x  2  2 v z 0  gt   cos  dt dt

Upon integration, the second equation gives





v x (t )   2v zo t  gt 2  cos 

and

 gt 3    cos  x(t )    v zo t 2  3  

Straightforward approach to solve for the deflection by the time the particle reaches the earth again would be to find the time for it to come back to the ground and substitute it in the equation for x(t). We have 244

vz0 

2 gh 

2  9.8  100  44.27 ms 1 ,

Time taken to reach the highest point T 

vz0  4.52s g

And the total time of flight = 2T = 9.04s. Thus the total deflection by the time the stone comes back to the ground is  9.8  9.04 3 x(2T )    44.27  9.04 2  3 



  cos   1206.65 cos   8.8 cos  cm 

The negative sing in front indicates that the deflection is towards the west. This is an interesting result. One’s immediate answer would normally have been that the stone deflects one way while going up and the other way while coming down, since the sign of vz changes, the net deflection would have been zero. This does not happen because by the time the stone reaches its highest point, it already has a westward velocity. To see its effect, let us calculate the distances in two steps: first when the stone goes up and the second when it comes down. While going up, the deflection of the particle by the time it reaches the highest point is  9.8  4.52 3 x (T )    44.27  4.52 2  3 



  cos   603.33 cos   4.4 cos  

At this point it has the horizontal velocity





v x (T )   2  44.27  4.52  9.8  4.52 2  cos   200.22 cos   1.46 cos  cm s1

The negative sing in front indicates that the velocity is towards the west. Now, as shown in example 15.9, if a stone is dropped from a height of 100m, it deflects to the east by 2.2 cos  cm. However, now the stone has an initial westward speed of 1.46 cos  cm s1. Thus the nest deflection fo th stone while it is coming down will be  1.46 cos   4.52  2.2 cos   4.4 cos 

This gives a total deflection of  8.8 cos  during the entire flight. 15.8

The initial components of the velocity in the east, north and vertical direction are v x 0  v0 cos  cos 

v y 0  v 0 cos  sin 

Equations of motion in are

245

v z 0  v 0 sin 

dv x  2v y  sin   2v z  cos  dt dv y  2v x  sin  dt dv z   g  2v x  cos  dt

Since  is very small, we will do our calculations correct to the first order in . This means that for vx and vy, we substitute their initial values and for vz we substitute v z (t )  v 0 sin   gt in the equations above. This gives dv x d 2 x  2  2v0 cos  sin   sin   2 v 0 sin   gt   cos  dt dt dv y d 2 y  2  2v 0 cos  cos   sin  dt dt dv z d 2 z  2   g  2v0 cos  cos   cos  dt dt

Integration of the last equation, with proper initial conditions v z 0 (t  0)  0, z (t  0)  0 , gives v z  v0 sin   gt  2v 0 t cos  cos  cos  z  v 0 sin  t 

1 2 gt  v0 t 2 cos  cos  cos  2

For time of flight T calculation, we substitute z = 0 and get T

2v0 sin  2v sin   0 g  2v0  cos  cos  cos  g  

 1 

2v 0  cos  cos  cos    g 

Thus the time of flight of the projectile increases. Now we integrate the equation for motion in the x direction, with the initial conditions v x 0 (t  0)  v 0 cos  cos  , x (t  0)  0 to get 1   v x (t )  v0 cos  cos   2v 0  t cos  sin  sin   2 v0 t sin   gt 2   cos  2    gt 3    cos  x(t )  v0 t cos  cos   v 0  t 2 cos  sin  sin    v0 t 2 sin   3  

Similarly, integration of the equation for the motion in the y direction gives

246

v y (t )  v0 cos  sin   2v 0  t cos  cos  sin  y (t )  v 0 t cos  sin   v 0  t 2 cos  cos  sin 

Thus the deflection of the projectile in the east and north direction are known. As a check, we take a projectile thrown staright up   90 and find that the answer matches with that obtained in problem 15.7.

To know the deflection x (t ) in the direction of launch and y (t )

perpendicular to it, we transform these as (see figure) Y’

Y X’

α O

X

x '  x cos   y sin  y '   x sin   y cos 

This transformation gives  gt 3    cos  cos  x (t )  v0 cos  t   v0 sin  t 2  3    gt 3    sin  cos  y (t )  v0  t 2 cos  sin    v0 sin  t 2  3  

To calculate the range and deflection from the path during the flight, we substitute the time of flight T for t. To the first order in , we then have T 

T2 

2v 0 sin  g 

 1  

4v02 sin 2  g2 

2v 0  cos  cos  cos    g 



 1 

4v0  cos  cos  cos    g 

247

8v03 sin 3  T  g3

6v 0  cos  cos  cos    g 



 1 

3



Therefor the range, up to the first order in  is x (T )  v0 cos  

2v 0 sin  g 

 1  

2v 0  cos  cos  cos    g 

 4v sin  g 8v 03 sin 3    v0 sin     3 g2 g3  2 0





2

2v 02 sin  cos  4v 03  sin   g g2

  cos  cos  



1 2  2  cos   sin   cos  cos  3  

4v03  sin  So the change in the range due to the coriolis force is g2 

1 2  2  cos   sin   cos  cos  . 3  

The sideways deflection is given by y (t )  v 0   

 4v 02 sin 2  4v 02 sin 2  g 8v 03 sin 3      sin  cos  cos  sin   v sin      0 3 g2 g2 g3  

4v03  sin 2  g2

 1   sin  sin  cos   cos  sin    3 

4v 03  sin 2  Thus the sideways deflection is g2

 1   sin  sin  cos   cos  sin   . 3  

This shows that a projectile fired eastward    0  will deflect towards the south and that fired westward    180 will deflect to the north. If we look at the equations of motion in the northern and the southern hemisphere, they differ by the sign of the terms containing sin  , which is equivalent to changing the sign of λ to λ. From the expression derived above, it is then clear that the change in the range will not change but the sideways deflection will change to

4v03  sin 2   1   sin  sin  cos   cos  sin   in the southern hemisphere. 2 g  3 

248

15.9 Exact solution of the equations of motion on the surface of the earth. Thse are dv x  2v y  sin   2v z  cos  dt dv y  2v x  sin  dt dv z   g  2v x  cos  dt

To get the exact solutions, we differentiate the first equation once and get

dv y d 2vx dv  2  sin   2 z  cos  2 dt dt dt Substituting for the derivatives of the velocity components appearing on the right, we get

d 2vx d 2vx 2  4 v x  2 g cos    4 2 v x  2 g cos  2 2 dt dt The solution of this differential equation is given by the sum of the solution for its homogeneous part and the particular solution. Thus it is v x (t )  A cos 2t   B sin  2t  

g cos  2

Here A and B are determined by the initial conditions. Subztituting this solution in the equation for vy gives v y (t )    A sin  2t   B cos 2t   gt cos   sin   C

Here C is another constant. Similarly from the last equation we get

v z (t )   gt   A sin  2t   B cos 2t   gt cos   cos   D Here D is another constant.

249

No wlet us determine the constants appearing in the equations above for a stone dropped from height h. Since we have solved a second-order differential equation for vx , we need two initial

condtions for it. These are v x (t  0)  0,

dv x (t  0)  0 . Then we have dt

A

g cos  , B0 2

This gives v x (t ) 

 sin  2t    t  cos  sin   C 2 

g cos  1  cos 2t   2

v y (t )  g  

 sin  2t   v z (t )   gt  g    t  cos 2   D 2  

For the y and the z components of the velocity we have v y (t  0)  0, v z (t  0)  0 . These give C = 0 and D = 0. Thus the complete solution for the velocity is v x (t ) 

 sin  2t    t  cos  sin  2 

g cos  1  cos 2t   2

v y (t )  g  

 sin  2t   v z (t )   gt  g    t  cos 2  2   

In the limit of   0 this gives, to the first order in , v x (t )  gt 2 cos 

v y (t )  0

v z (t )   gt

This leads to the same answers as in example 15.9.

15.10 Cyclone building up in the Bay of Bengal. If only the motion in the horizontal plane is considered, the equations of motion are

250

dv x dy  2v y  sin   2  sin  dt dt dv y dx  2v x  sin   2 sin dt dt

We choose the coordinate system such that its origin is where the storm starts from. Thus the initial conditions are x(t  0)  0, v x (t  0)  50kmph, y (t  0)  0, v y (t  0)  0 . The second equation above can be immediately integrated to obtain v y (t )  2x(t ) sin   C

where C is a constant. Since x(t  0)  0, v y (t  0)  0 , we get C = 0. Thus v y (t )  2x (t ) sin 

We substitute this in the first equation above to get dv x d 2 x 2  2   2 sin   x or dt dt

d 2x 2   2 sin   x  0 dt 2

This gives x(t )  A sin  2t sin    B cos 2t sin   1 With the initial conditions x(t  0)  0, v x (t  0)  50kmph  13.9ms , this gives

A

13.9 , B0 2 sin 



x (t )  

13.9 sin  2t sin   2 sin 

Thus v y (t ) 

dy  13.9 sin  2t sin   dt



y (t )  

13.9 cos 2t sin    C 2 sin 

With the initial conditions y (t  0)  0, v y (t  0)  0 we get C 

13.9 . Thus finally we have 2 sin 

for the coordinates of the eye of the cyclone x(t )  

13.9 sin  2t sin   2 sin 

and y (t ) 

251

13.9 1  cos 2t sin    2 sin 

(i) This gives the equation for the trajectory as 13.9   x2   y   2  sin   

2



13.9   2  sin   

2



13.9 13.9  10 -3 m  278km , the above equation can also be written in 2 sin  2  7.3  10 5  sin 20

Since

kilometers as x 2   y  278  278 2 2

Thus the trajectory is a circlular one with the centre at  0, 278 km. (ii)

Thus the sense of the trajectory is clockwise as seen from above. On the other hand, since

we are in the northern hemisphere, the winds around the eye move in counterclockwise direction. (iii) To find out where the cyclone hits the coast, we substitute in the equation above x  10 km and

get y  18.6 km and y  537.4 km . The second coordinate is for the return

path. Thus the cyclone hits the coast 18km north of where it started from. 15.11 The solution here is similar to the solution above except for the initial conditions. We take the origin to be where the striker is played from. The x and the y axes are shown in the figure below.

252

B

y



1

2 x A



For the problem    zˆ . Thus the equation of motion in the xy plane are dv x dy  2v y   2  dt dt dv y dx  2v x   2  dt dt

Note that the signs on th right hand side are opposite to what they are in the equations of motion in the orther hemisphere of the earth because the direction of the angular velocity is opposite. Assuming that the striker is played at an angle  from the x axis, the initial conditions are x (t  0)  0,

v x (t  0)  v cos  ,

y (t  0)  0,

v y (t  0)  v sin 

The second equation above is integrated to give v y (t )  2 x(t )  C

With the initial condition x(t  0)  0, v y (t  0)  v sin  , we get C  v sin  . Thus v y (t )  2 x(t )  v sin 

Substituting this in the first equation of motion, we get dv x d 2 x 2  2   2  x  2v sin  dt dt

or

d 2x 2   2  x  2v sin  2 dt

The solution for the equation is then obtained as the sum of the solution for the homogeneous part and the particular solution. Thus

253

v sin  2

x(t )  A sin  2t   B cos 2t  

With the initial condition x(t  0)  0, v x (t  0)  v cos  , we get A 

v cos  v sin  ,B . Thus 2 2

v cos  v sin  v sin  sin  2t   cos 2t   2 2 2

x (t ) 

When substituted in v y (t )  2 x(t )  v sin  , this gives dy  v cos  sin  2t   v sin  cos 2t  dt

Solved with the initial condition y(0) = 0, this gives y (t )  

v cos  v sin  v cos  cos 2t   sin  2t   2 2 2

Thus the trajectory is described by the coordinates v cos  v sin  v sin  sin  2t   cos 2t   2 2 2

x (t ) 

y (t )  

v cos  v sin  v cos  cos 2t   sin  2t   2 2 2

These can be combined together to give the equation of trajectory as 

 x 

v sin    2 

2

v cos     y  2  



Thus the trajectory is a circle with centre at   

2



v   2  

2



v sin  v cos   ,  and the sence of direction is 2 2 

therefore counterclockwise. This implies that the striker should be played in direction 2. Now we want the trajectory to be such that it pass through point B, i.e. through coordinates  0, L  . When substituted in the equation for the trajectory, this gives cos  

L v

If the angle with AB is α, then sin  

L  L     sin 1   v  v   L   from AB.  v 

1 The striker is played in direction 2 making an angle   sin 

254

(ii) We first solve this problem for v  L correct to first order in

L v

in an easy way. We

then solve the problem exactly. Easy approximate solution correct to first order in

L v

:

dv x  2v y  dt dv y  2v x  dt

It is given that initially v y  v and v x  0 . Thus as we integrate the equations above v x   and therefore

dv y dt

  2 . So up to order , vy can be taken to be a constant equal to v. Thus y (t )  vt

and the integration of the first equation above gives v x (t )  2y  C

Since v x ( y  0)  0 , we get C = 0. This gives v x (t )  2vt

and

x(t )  vt 2 because x(t=0)=0.

Therefore by the time the striker reaches the other end, time taken by it is T 

moved a distance x (T )  

L and its has v

L2 and has x-component of the velocity x (T )  2L . Now the v

striker starts its journey back after hitting then board. It again takes time T 

L to reach the v

side it started from. Now the initial x-componet of the velocity is x (T )  2L and the displacement x (T )  

L2 . Now if we integrate the equation with these conditions, then v vx

T t

 2L

T

 dvx  2  v y dt   2  y(T  t )  y(T )  2  y (T  t )  L   2  L  vt  L   2vt

This gives

255

v x (T  t )  2L  2vt

v x (T  t )  2L  2vt

or

Keep in mind that t is being measured from T onwatds. Thus t = 0 inplies when the striker starts its journey back. Integrating this with the initial condition x (T )   x (T  t )

t

 dx     2L  2vt dt   2Lt  vt



L2 v

For t  T 

L2 , we get v



x(T  t )  

0

L2  2Lt  vt v

L this gives v x( 2T )  

2L2 v

This is the final displacement when the striker comes back. One would have thought that the striker will come back to its original position; that does not happen because when it starts its journey back, it has an x-component of velocity that makes it drift. Had the x-component been zero when it started back, the striker would have reached its original position. Exact solution: If the striker is played along line AB then   90 . The trajectory of the striker is then 

 x 

v   2 

2



v   2  

2

 y2  

and with time x(t) and y(t) vary as x (t ) 

v v cos 2t   2 2

y (t ) 

v sin  2t  2

Thus for y = L, we have from the trajectory equation x2 

v x  L2  0 



x

v v 4 2 L2  1 2 2 v2

Thus the striker gets displaced in x direction as it reached point B. One of the two answers (smaller magnitude  

v v 4 2 L2 ) is when the particle is moving upward. The  1 2 2 v2

other one (larger magnitude  

v v 4 2 L2 ) is when the striker would have returned  1 2 2 v2

256

had it not hit the other side of the board. Further, when the striker reaches the other end, it would have taken time T such that y(T) = L. This gives L

v 2L 4 2 L2 sin  2T   sin  2T   , cos 2T   1  2 v v2

The x and y components of the velocity at the other end are therefore x (t ) t T  v sin  2T   2L

From the above, one sees that the speed

y (t ) t T  v cos 2T   x 2  y 2

v 2  4 2 L2

remains unchanged during the motion. This is

expected since the coriolis force is perpendicular to the velocity and therefore does no work on the striker. As the striker hits the opposite side of the carom and returns back, its x velocity remains unchanged and the y component of the velocity changes direction and it traverses a new trajectory. Taking the time when it starts back to be t = 0, the new trajectory is thus determined by the initial conditions x ( 0)  

v v 4 2 L2  1 , 2 2 v2

x (0)  2L,

y ( 0)  L ,

y (0)   v 2  4 2 L2

One could solve this problem with these initial conditions. However, we can make use of the equations already derived for the trajectory if we describe the journey back with new coordinate system with the origin at the point where the striker starts back. This is shown in the figure below. In this coordinate system the initial conditions are x (0)  0,

x (0)  2L,

y (0)  0,

y (0) 

v 2  4 2 L2

So that we have initial speed v at an angle  such that tan  

v 2  4 2 L2 , sin   2L

257

v 2  4 2 L2 2L , cos   v v

x

B

 v

v 4 2 L2 v x   1 2  2 v 2 y A

Thus the x and y displacement of the striker in the new coordinate system is x(t ) 

v cos  v sin  v sin  sin  2t   cos 2t   2 2 2

 L sin  2t  

y (t )  

v 4 2 L2 v 4 2 L2   1 cos 2  t  1  2 2 v2 v2

v cos  v sin  v cos cos 2t   cos 2t   2 2 2

  L cos 2t  

v 4 2 L2 1 sin  2t   L 2 v2

These can be combined together to give the equation of trajectory as 2 2  x  v 1  4 L  2 v2 



2



  

  y  L

2



v     2 

2

For y = L this then gives x

v 4 2 L2 v 1  2 2 2 v

While cutting the y = L line for the first time, the value of x will be x

v 4 2 L2 v 1  2 2 2 v

This is in addition to the previous displacement of the origin in A  B journey. Thus the total displacement is

258

x  

v 4 2 L2 v 1  2   v

It is easily seen that for v  L this goes over to the approximate answer obtained earlier.

259

Appendix A A.1 In this problem there are four variables. So we make one dimensionless variable and equate it to a constant. The variables of the problem and the corresponding dimensions are p  ML1T 2

n  L3

mM

v  LT 1

Let the dimensionless variable be   p a m b n c v d . Since this is a dimensionless variable we have, on equating the dimensions of the right hand side in the equation above to zero ab  0

 a  3c  d  0

 2a  d  0

Since we want an expression for p, we take a = 1. b  1, d  2,

Then the equations above give

and c  1 . Therefore the dimensionless variable is



p mnv 2

Equating it to a constant k, we get p  kmnv 2

A.2 This problem has three variables. Therefore to apply Buckingham’s π theorem we form one dimensionless variable and equate it to a constant. The variables of the problem and the corresponding dimensions are p  ML1T 2

T  MT 2

RL

Let the dimensionless variable be   p a T b R c . On equating the dimensions of the right hand side in the equation above to zero we get ab  0

 2a  2b  0

ac 0

Taking a = 1, since we want an expression for p, we get b  1 and c  1 . Therefore the dimensionless variable is  

pR T

Equating it to a constant k, we get pk

260

T R

A.3 This problem has four variables. Therefore to apply Buckingham’s π theorem we form one dimensionless variable and equate it to a constant. The variables of the problem and the corresponding dimensions are E  ML2T 2

h  ML2T 1

mM

RL

If the dimensionless variable is   E a h b m c R d , on equating the dimensions of the right hand side in the equation above to zero we get abc  0

2a  2b  d  0

 2a  b  0

Since we want an expression for E, we take a = 1. Then the equations above give b  2, d  2, and c  1 . Therefore the dimensionless variable is



EmR 2 h2

Equating it to a constant k, we get pk

h2 mR 2

A.4 This problem has four variables. Therefore we form one dimensionless variable and equate it to a constant. The variables of the problem and the corresponding dimensions are E  ML1T 2 V

h  ML2T 1

mM

n  L3

If the dimensionless variable is   E a h b m c n d , on equating the dimensions of the right hand side in the equation above to zero we get abc  0

Since we want an expression for

 a  2b  3d  0

 2a  b  0

E , we take a = 1. Then the equations above give c  1 and V

5 b  2, d   , . Therefore the dimensionless variable is 3



 E V m h2n5 3

Equating it to a constant k, we get E h2 5 3 k n V m

261

A.5 This problem has five variables. Therefore to apply Buckingham’s π theorem we form twodimensionless variables and write one of them as a function of the other. The variables of the problem and the corresponding dimensions are dV  L3T 1 dt

P  ML1T  2

  ML1T 1

 If the dimensionless variables are expressed as   

dV    dt 

LL

RL

a

P b c Ld R e , on equating the

dimensions of the right hand side in the equation above to zero we get bc 0

3a  b  c  d  e  0

 a  2b  c  0

These give b  a, c  a and d  e  3a . Since we want an expression for

dV , we form one dt

of the dimensioless variables with a = 1. Then the equations above give b  1, c  1, and d  e  3 . Taking d = 0 gives e  3 giving one dimensionless variable as  dV    3  dt  PR

1  

For the other dimensionless variable, we take a = 0. This gives b = c = 0 and d  e . Taking e = 1, the second dimensionaless variable is 2 

R L

Thus the functional relationship between the variables can be expressed as

 1  f  2  Now we have several choices for the function f. If we take it to be a constant, we get PR 3  dV    k   dt 

But this is not consistent with observations. We next take f   2   k 2 where k is a constant. Then we get PR 4  dV    k L  dt 

This gives answer consistent with observations. For corrections to higher orders, we can take higher powers of  2 and construct higher order functional relationships.

262