Engineering Mechanics

ENG2016: ENGINEERING MECHANICS By: Engr. Kevin Lester B. Lobo

Mechanics

Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies that are subjected to the action of forces.

Mechanics Statics of Rigid-bodies

Engineering Mechanics

Dynamics of Rigid-bodies

• Kinematics • Kinetics

Mechanics of Deformable-bodies

Fluid Mechanics

Fundamental Concepts Basic Quantities Length Is used to locate the position of a point in space and thereby describe the size of a physical system. Once a standard unit of length is defined, one can then use it to define distances and geometric properties of a body as multiples of this unit.

Time Is conceived as a succession of events. Although the principles of statics are time independent, this quantity plays an important role in the study of dynamics.

Mass Is a measure of a quantity of matter that is used to compare the action of one body with that of another. This property manifests itself as a gravitational attraction between two bodies and provides a measure of the resistance of matter to a change in velocity.

Force In general, it is considered as the “push” or “pull” exerted by one body on another. This interaction can occur when there is direct contact between the bodies, such as a person pushing on a wall, or it can occur through a distance when the bodies are physically separated. Examples of the latter type include gravitational, electrical, and magnetic forces. In any case, a force is completely characterized by its magnitude, direction, and point of application.

Units of Measurement

Units of Measurement Conversion of Units

Length:

Mass:

Force:

1 𝑖𝑛 = 2.54 𝑐𝑚 1 𝑓𝑡 = 12 𝑖𝑛 1 𝑦𝑎𝑟𝑑 = 3 𝑓𝑡 1 𝑓𝑢𝑟𝑙𝑜𝑛𝑔 = 660 𝑓𝑡 1 𝑚𝑖𝑙𝑒 = 5,280 𝑓𝑡 1 𝑚𝑖𝑙𝑒 = 1,609.344 𝑚 1 𝑛𝑎𝑢𝑡𝑖𝑐𝑎𝑙 𝑚𝑖𝑙𝑒 = 1,852 𝑚

1 𝑙𝑏𝑚 = 0.45359237 𝑘𝑔 1 𝑡𝑜𝑛 = 2,000 𝑙𝑏𝑚 1 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑛𝑒 = 1,000 𝑘𝑔

1 𝑙𝑏 = 16 𝑜𝑢𝑛𝑐𝑒 𝑜𝑧 1 𝑁 = 100,000 𝑑𝑦𝑛𝑒 1 𝑘𝑖𝑝 = 1,000 𝑙𝑏

The International System of Units Prefixes

The International System of Units Rules for Use • Quantities defined by several units which are multiples of one another are separated by a dot to avoid confusion with prefix notation, as indicated by 𝑁 = 𝑘𝑔 · 𝑚Τ𝑠 2 . Also, 𝑚 · 𝑠 (meter-second), whereas 𝑚𝑠 (milli-second). • The exponential power on a unit having a prefix refers to both the unit and its prefix. For example, μ𝑁 2 = μ𝑁 2 = μ𝑁 · μ𝑁. Likewise, 𝑚𝑚2 represents 𝑚𝑚 2 = 𝑚𝑚 · 𝑚𝑚. • With the exception of the base unit the kilogram, in general avoid the use of a prefix in the denominator of composite units. For example, do not write 𝑁Τ𝑚𝑚, but rather 𝑘𝑁Τ𝑚; also, 𝑚Τ𝑚𝑔 should be written as 𝑀𝑚Τ𝑘𝑔. • When performing calculations, represent the numbers in terms of their base or derived units by converting all prefixes to powers of 10. The final result should then be expressed using a single prefix. Also, after calculation, it is best to keep numerical values between 0.1 and 1000; otherwise, a suitable prefix should be chosen. For example, 50 𝑘𝑁 60 𝑛𝑚 = 50 103 𝑁 60 10−9 𝑚 = 3000 10−6 𝑁 · 𝑚 = 3 10−3 𝑁 · 𝑚 = 3 𝑚𝑁 · 𝑚

Numerical Calculations Dimensional Homogeneity Each term must be expressed in the same units. Example: 1 𝑆 = 𝑉0 𝑡 + 𝑎 𝑡2 2 𝐿 1 𝐿 𝐿 = 𝑇 + 𝑇2 2 𝑇 2 𝑇

Significant Figures The number of significant figures contained in any number determines the accuracy of the number.

Rounding Off Numbers As a general rule, any numerical figure ending in five or greater is rounded up and a number less than five is rounded down.

Calculations Do not round off calculations until expressing the final result. This procedure maintains precision throughout the series of steps to the final solution.

General Procedure for Analysis The most effective way of learning the principles of engineering mechanics is to solve problems. To be successful at this, it is important to always present the work in a logical and orderly manner, as suggested by the following sequence of steps: • Read the problem carefully and try to correlate the actual physical situation with the theory studied. • Tabulate the problem data and draw any necessary diagrams. • Apply the relevant principles, generally in mathematical form. When writing any equations, be sure they are dimensionally homogeneous. • Solve the necessary equations, and report the answer with no more than four significant figures. • Study the answer with technical judgment and common sense to determine whether or not it seems reasonable.

Example 1.1 Convert 2 𝑘𝑚Τℎ to (a) 𝑚Τ𝑠, (b) 𝑓𝑡Τ𝑠 SOLUTION (a) Since 1 𝑘𝑚 = 1000 𝑚 and 1 ℎ = 3600 𝑠, the factors of conversion are arranged in the following order, so that a cancellation of the units can be applied: 2 𝑘𝑚 1000 𝑚 1ℎ 2 𝑘𝑚Τℎ = ℎ 𝑘𝑚 3600 𝑠 2000 𝑚 = 3600 𝑠 = 𝟎. 𝟓𝟓𝟔 𝒎Τ𝒔 𝑨𝒏𝒔. (b) Recall, 1 𝑓𝑡 = 0.3048 𝑚

0.556 𝑚Τ𝑠

= =

0.556 𝑚 1 𝑓𝑡 𝑠 0.3048 𝑚 𝟏. 𝟖𝟐 𝒇𝒕Τ𝒔

𝑨𝒏𝒔.

Example 1.2 Convert the quantities 300 𝑙𝑏 · 𝑠 and 52 𝑠𝑙𝑢𝑔Τ𝑓𝑡 3 to appropriate SI units.

SOLUTION Recall, 1 𝑙𝑏 = 4.448 𝑁 300 𝑙𝑏 · 𝑠

=

= =

4.448 𝑁 300 𝑙𝑏 · 𝑠 𝑙𝑏 1334.5 𝑁 · 𝑠 𝟏. 𝟑𝟑 𝒌𝑵 · 𝒔

𝑨𝒏𝒔.

Since 1 𝑠𝑙𝑢𝑔 = 14.59388 𝑘𝑔 and 1 𝑓𝑡 = 0.3048 𝑚, then 52 𝑠𝑙𝑢𝑔 14.59 𝑘𝑔 1 𝑓𝑡 3 Τ 52 𝑠𝑙𝑢𝑔 𝑓𝑡 = 𝑓𝑡 3 1 𝑠𝑙𝑢𝑔 0.3048 𝑚 = 26,800 𝑘𝑔Τ𝑚3 = 𝟐𝟔. 𝟖 𝑴𝒈Τ𝒎𝟑

3

𝑨𝒏𝒔.

Example 1.3 Evaluate each of the following and express with SI units having an appropriate prefix: a 50 𝑚𝑁 ∗ 6 𝐺𝑁 , b 400 𝑚𝑚 ∗ 0.6 𝑀𝑁 2 , c 45𝑀𝑁 3 Τ900𝐺𝑔. SOLUTION 𝑷𝒂𝒓𝒕 𝒂

50 𝑚𝑁 6 𝐺𝑁

= = = =

50 ∗ 10−3 𝑁 6 ∗ 109 𝑁 300 ∗ 106 𝑁 2 1 𝑘𝑁 6 2 300 ∗ 10 𝑁 ∗ 103 𝑁 𝟑𝟎𝟎 𝒌𝑵𝟐

2

𝑨𝒏𝒔.

Example 1.3 SOLUTION 𝑷𝒂𝒓𝒕 𝒃 400 𝑚𝑚 0.6 𝑀𝑁

2

We can also write 144 109 𝑚 · 𝑁 2

= = = = = =

400 ∗ 10−3 𝑚 0.6 ∗ 106 𝑁 2 400 ∗ 10−3 𝑚 0.36 ∗ 1012 𝑁 2 144 ∗ 109 𝑚 · 𝑁 2 𝟏𝟒𝟒 𝑮𝒎 · 𝑵𝟐 1 𝑀𝑁 9 2 144 ∗ 10 𝑚 · 𝑁 ∗ 106 𝑁 𝟎. 𝟏𝟒𝟒 𝒎 · 𝑴𝑵𝟐

𝑨𝒏𝒔.

2

𝑨𝒏𝒔.

Example 1.3 SOLUTION 𝑷𝒂𝒓𝒕 𝒄 45 𝑀𝑁 3 900 𝐺𝑔

=

= = =

45 106 𝑁 3 900 106 𝑘𝑔 50 109 𝑁 3 Τ𝑘𝑔 1 𝑘𝑁 50 109 𝑁 3 103 𝑁 𝟓𝟎 𝒌𝑵𝟑 Τ𝒌𝒈

3

1 𝑘𝑔 𝑨𝒏𝒔.

Scalars and Vectors Scalar

A scalar is any positive or negative physical quantity that can be completely specified by its magnitude. Examples of scalar quantities: length, mass, time, temperature.

Scalars and Vectors Vector A vector is any physical quantity that requires both a magnitude and a direction for its complete description. Some examples of vectors are: force, position, velocity, moment. A vector is shown graphically by an arrow.

Scalars and Vectors Vector

The length of the arrow represents the magnitude of the vector, and the angle between the vector and a fixed axis defines the direction of its line of action. The head or tip of the arrow indicates the sense of direction of the vector. In print, vector quantities are represented by bold face letters such as 𝐀 and the magnitude of the vector is italicized, 𝐴. For handwritten work, it is often convenient to denote a vector quantity by simply drawing an arrow on top of it, 𝐴Ԧ

Resultant of Vector Parallelogram Law • Two “component” forces 𝐅1 and 𝐅2 add according to the parallelogram law, yielding a resultant force 𝐅𝑅 that forms the diagonal of the parallelogram. • If a force 𝐅 is to be resolved into components along two axes 𝑢 and 𝑣, then start at the head of force 𝐅 and construct lines parallel to the axes, thereby forming the parallelogram. The sides of the parallelogram represent the components, 𝐅𝑢 and 𝐅𝑣 . • Label all the known and unknown force magnitudes and the angles on the sketch and identify the two unknowns as the magnitude and direction of 𝐅𝑅 , or the magnitudes of its components.

Resultant of Vector Trigonometry • Redraw a half portion of the parallelogram to illustrate the triangular head-to-tail addition of the components. • From this triangle, the magnitude of the resultant force can be determined using the Law of cosines, and its direction is determined from the Law of sines. The magnitudes of the other two force components are determined from the Law of sines.

Resultant of Multiple Vector Component Method If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force.

For example, if three forces 𝐅1 , 𝐅2 , 𝐅3 act at a point O, the resultant of any two of the forces is found, say, 𝐅1 + 𝐅2 —and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., 𝐅𝑅 = 𝐅1 + 𝐅2 + 𝐅3 .

--OR--

Resultant of Multiple Vector Component Method If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force.

Problems of this type are easily solved by using the “Rectangular Component Method”.

𝐅𝑹 =

𝚺𝑭𝒙

𝟐

+ 𝚺𝑭𝒚

𝟐

Example 1 The screw eye is subjected to two forces, 𝐅1 and 𝐅2 . Determine the magnitude and direction of the resultant force.

Example 1 SOLUTION: Apply Parallelogram Law: The parallelogram is formed by drawing a line from the head of 𝐅1 that is parallel to 𝐅2 , and another line from the head of 𝐅2 that is parallel to 𝐅1 . The resultant force 𝐅𝑅 extends to where these lines intersect at point A. The two unknowns are the magnitude of 𝐅𝑅 and the angle θ (theta).

Example 1 SOLUTION: Then apply Trigonometry: From the parallelogram, the vector triangle is constructed. Using the law of cosines 𝐹𝑅 = 100 𝑁 2 + 150 𝑁 2 − 2 100 𝑁 150 𝑁 cos 115° 𝐹𝑅

= =

10,000 𝑁 2 + 22,500 𝑁 2 − 30,000 𝑁 2 −0.4226 𝟐𝟏𝟑 𝑵

𝑨𝒏𝒔.

Example 1 SOLUTION: Applying the law of sine to determine θ, 150 𝑁 212.6 𝑁 = sin θ sin 115°

sin θ =

150 𝑁 sin 115° 212.6 𝑁 𝜽 = 𝟑𝟗. 𝟖°

Thus the direction φ (phi) of 𝐅𝑅 , measured from the horizontal, is 𝝋 = 39.8° + 15.0° = 𝟓𝟒. 𝟖° 𝑨𝒏𝒔.

𝑨𝒏𝒔.

Example 2 Resolve the horizontal 600-lb force into components acting along the 𝑢 and 𝑣 axes and determine the magnitudes of these components.

Example 2 SOLUTION: The vector addition using the triangle rule is shown. The two unknowns are the magnitudes of 𝐹𝑢 and 𝐹𝑣 . Applying the law of sine,

𝐹𝑢 sin 120°

=

600 𝑙𝑏 sin 30°

𝐹𝑢

=

1,039 𝑙𝑏

𝐹𝑣 sin 30°

=

600 𝑙𝑏 sin 30°

𝐹𝑣

=

600 𝑙𝑏

𝐴𝑛𝑠.

𝐴𝑛𝑠.

Example 3 Determine the magnitude of the component force 𝐅 in and the magnitude of the resultant force, 𝐅𝑅 if it is directed along the positive y-axis.

SOLUTION • The parallelogram law of addition. • The triangle rule. The magnitudes of 𝐅𝑅 and 𝐅 are the two unknowns. They can be determined by applying the law of sine. 𝐹 sin 60°

=

200 𝑙𝑏 sin 45°

𝐹

=

245 𝑙𝑏

𝐹𝑅 sin 75°

=

200 𝑙𝑏 sin 45°

𝐹𝑅

=

273 𝑙𝑏

𝐴𝑛𝑠.

𝐴𝑛𝑠.

Example 5 It is required that the resultant force acting on the eyebolt be directed along the positive x axis and that 𝐅2 have a minimum magnitude. Determine this magnitude, the angle θ, and the corresponding resultant force.

SOLUTION The triangle rule for is shown. Since the magnitudes (lengths) of 𝐅𝑅 and 𝐅2 are not specified, then 𝐅2 can actually be any vector that has its head touching the line of action of 𝐅𝑅 . However, as shown, the magnitude of 𝐅2 is a minimum or the shortest length when its line of action is perpendicular to the line of action of 𝐅𝑅 , that is, when 𝜽 = 𝟗𝟎° 𝑨𝒏𝒔. Since the vector addition forms a right triangle, the two unknown magnitudes can be obtained by trigonometry. 𝐹𝑅 = 800 𝑁 cos 60° = 𝟒𝟎𝟎 𝑵 𝑨𝒏𝒔. 𝐹2 = 800 𝑁 sin 60° = 𝟔𝟗𝟑 𝑵 𝑨𝒏𝒔.

Cartesian Vectors Rectangular Components of a Vector 𝐀 = 𝐀𝑥 + 𝐀𝑦 + 𝐀𝑧

Cartesian Vectors Cartesian Unit Vectors In three dimensions, the set of Cartesian unit vectors, i, j, k, is used to designate the directions of the x, y and z axes, respectively.

Cartesian Vectors Rectangular Components of a Vector 𝐀 = 𝐴𝑥 𝐢 + 𝐴 𝑦 𝐣 + 𝐴𝑧 𝐤

Cartesian Vectors Magnitude of a Cartesian Vector 𝐴=

𝐴2𝑥 + 𝐴2𝑦 + 𝐴𝑧2

Cartesian Vectors Magnitude of a Cartesian Vector cos α =

𝐴𝑥 |𝐴|

cos β =

𝐴𝑦

|𝐴|

cos γ =

𝐴𝑧 |𝐴|

Cartesian Vectors Magnitude of a Cartesian Vector cos α =

𝐴𝑥 |𝐴|

cos β =

𝐴𝑦

|𝐴|

cos γ =

𝐴𝑧 |𝐴|

Cartesian Vectors Magnitude of a Cartesian Vector cos α =

𝐴𝑥 |𝐴|

cos β =

𝐴𝑦

|𝐴|

cos γ =

𝐴𝑧 |𝐴|

Cartesian Vectors Magnitude of a Cartesian Vector cos α =

𝐴𝑥 |𝐴|

cos β =

𝐴𝑦

|𝐴|

cos γ =

𝐴𝑧 |𝐴|

Cartesian Vectors Unit Vectors An easy way of obtaining these direction cosines is to form a unit vector 𝐮𝐴 in the direction of A

𝐴𝑦 𝐀 𝐴𝑥 𝐴𝑧 𝐮𝐴 = = 𝐢+ 𝐣+ 𝐤 |𝐴| 𝐴 𝐴 𝐴

Cartesian Vectors Unit Vectors cos 2 α + cos2 β + cos2 γ = 1

Problem 1 The cable attached to the eye bolt is pulled with the force ‘F’ of magnitude 500-lb. Determine (a) the rectangular representation of this force and (b) the force acting along the x, y and z axes

Problem 2 Referring to Figure show, determine (a) the rectangular representation of the position vector A and (b) the angle between vector A and each of the positive coordinate axes.

Motion Concepts Position and Displacement

∆𝒔 = 𝒔′ − 𝒔

Rectilinear Kinematics: Continuous Motion Velocity

𝒗𝒂𝒗𝒈

∆𝒔 = ∆𝒕

the Instantaneous Velocity is a vector defined as: ∆𝑠 𝑣 = lim ∆𝑡→0 ∆𝑡

𝒅𝒔 𝒗= 𝒅𝒕

Rectilinear Kinematics: Continuous Motion Velocity

Velocity can either be positive or negative. For example, if the particle is moving to the right, the velocity is positive; whereas if it is moving to the left, the velocity is negative.

Rectilinear Kinematics: Continuous Motion Average Speed and Velocity

𝒔𝒑𝒂𝒗𝒈

𝑺𝑻 = ∆𝒕

𝒗𝒂𝒗𝒈

∆𝒔 = ∆𝒕

Rectilinear Kinematics: Continuous Motion Acceleration

𝒂𝒂𝒗𝒈

∆𝒗 = ∆𝒕

the Instantaneous Acceleration is a vector defined as ∆𝑣 𝑎 = lim ∆𝑡→0 ∆𝑡

𝒅𝒗 𝒂= 𝒅𝒕 𝒅𝟐 𝒔 𝒂= 𝟐 𝒅𝒕

Rectilinear Kinematics: Continuous Motion Deceleration

Acceleration can either be positive or negative. In particular, when the particle is slowing down, or its speed is decreasing, the particle is said to be decelerating. Thus, acceleration is negative

Rectilinear Kinematics: Continuous Motion Differential relation involving ‘s’, ‘v’ and ‘a’ along the path Finally, an important differential relation involving the displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential, dt between the instantaneous velocity and instantaneous acceleration which gives: Instantaneous Velocity:

𝒅𝒔 𝒗= 𝒅𝒕 Instantaneous Acceleration:

𝒅𝒗 𝒂= 𝒅𝒕 Derived Relationship:

𝒂 𝒅𝒔 = 𝒗 𝒅𝒗

Rectilinear Kinematics: Continuous Motion Velocity as a Function of Time 𝑣

𝑡

න 𝑑𝑣 = න 𝑎𝑐 𝑑𝑡 𝑣0

0

𝑣 − 𝑣0 = 𝑎𝑐 𝑡

𝒗 = 𝒗𝟎 + 𝒂 𝒄 𝒕

Rectilinear Kinematics: Continuous Motion Position as a Function of Time 𝑠

𝑡

න 𝑑𝑠 = න 𝑣0 + 𝑎𝑐 𝑡 𝑑𝑡 𝑠0

0

𝑠 − 𝑠0 = 𝑣0 𝑡 + 𝑎𝑐

𝑡2 2

𝟏 𝒔 = 𝒔𝟎 + 𝒗 𝟎 𝒕 + 𝒂 𝒄 𝒕 𝟐 𝟐

Rectilinear Kinematics: Continuous Motion Velocity as a Function of Position 𝑣

𝑠

න 𝑣 𝑑𝑣 = න 𝑎𝑐 𝑑𝑠 𝑣0

𝑠0

𝑣 2 − 𝑣02 = 𝑎𝑐 𝑠 − 𝑠0 2

𝒗𝟐 = 𝒗𝟐𝟎 + 𝟐𝒂𝒄 𝒔 − 𝒔𝟎

Rectilinear Kinematics: Continuous Motion Formulas Differential Relations

𝑑𝑠 𝑣= 𝑑𝑡 𝑑𝑣 𝑎= 𝑑𝑡 𝑎 𝑑𝑠 = 𝑣 𝑑𝑣

(12-1)

(12-2)

(12-3)

Velocity as a Function of Time

𝑣 = 𝑣0 + 𝑎𝑐 𝑡

(12-4)

Position as a Function of Time

1 𝑠 = 𝑠0 + 𝑣0 𝑡 + 𝑎𝑐 𝑡 2 2

(12-5)

Velocity as a Function of Position

𝑣 2 = 𝑣02 + 2𝑎𝑐 𝑠 − 𝑠0

(12-6)

Motion of a Projectile Vertical Motion Only 𝟏 𝟐 𝒉 = 𝒗𝟎 𝒕 + 𝒈𝒕 𝟐 𝒗 = 𝒗𝟎 + 𝒈𝒕 𝒗𝟐 = 𝒗𝟎 𝟐 + 𝟐𝒈𝒉 Acceleration due to gravity:

𝒈 = 𝟗. 𝟖𝟏 𝒎Τ𝒔𝟐 = 𝟑𝟐.2 𝒇𝒕Τ𝒔𝟐

• Acceleration due to gravity, ‘g’ is positive when the object is falling or going in a downwards direction; otherwise it is negative. • When the object is “dropped” and the one that drops the object is not in motion, the object will have zero initial velocity; otherwise, if the one that drops the object is in motion, the object would have an initial velocity and direction equal to the velocity and direction that drops it. • Air resistance is neglected. • Vertical velocity at the peak is zero.

Motion of a Projectile Horizontal Motion Displacement : 𝒙 = 𝒗𝟎 𝒙 𝒕 Velocity

:

𝒗𝟎

𝒙

= 𝒗𝟎 ∗ 𝐜𝐨𝐬𝜽

Acceleration : 𝟎

Vertical Motion 𝒗𝒚 = 𝒗 𝟎

𝒗𝟎

𝒚

𝒚

− 𝒈𝒕

= 𝒗𝟎 ∗ 𝒔𝒊𝒏𝜽

𝟏 𝟐 ±𝒚 = 𝒗𝟎 𝒚 𝒕 − 𝒈𝒕 𝟐 𝒗𝟐𝒚 = 𝒗𝟎

𝟐 𝒚

− 𝟐𝒈𝒚

Motion of a Projectile General Equation of Projectile

𝒈𝒙𝟐 ±𝒚 = 𝒙𝒕𝒂𝒏𝜽 − 𝟐 𝒗𝒐 𝟐 𝒄𝒐𝒔𝟐 𝜽 • the height, ‘y’ is positive for the projectile that hits above the firing point and it is negative when the projectile hits below the firing point. • when ‘y’ is zero; then x = R (Max Range) • at max height, h; the y-component velocity, (Vo)y is zero • when Θ = 0°, then y = h (Max Height)

Motion of a Projectile Range at an Inclined or Declined Angle Inclined

Declined

𝟐 𝒗𝒐 𝟐 𝒄𝒐𝒔𝜽 ∗ 𝒔𝒊𝒏(𝜽 − 𝜷) 𝑹= 𝒈 ∗ 𝒄𝒐𝒔𝟐 𝜷

𝟐 𝒗𝒐 𝟐 𝒄𝒐𝒔𝜽 ∗ 𝒔𝒊𝒏(𝜽 + 𝜷) 𝑹= 𝒈 ∗ 𝒄𝒐𝒔𝟐 𝜷

Rectilinear Kinematics: Erratic Motion Given s-t graph, find v-t graph

𝑣=

𝑑𝑠 𝑑𝑡

𝑣 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑠 − 𝑡 𝑔𝑟𝑎𝑝ℎ

Rectilinear Kinematics: Erratic Motion Given v-t graph, find a-t graph

𝑎=

𝑑𝑣 𝑑𝑡

𝑎 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ

Rectilinear Kinematics: Erratic Motion Given a-t graph, find v-t graph

Δ𝑣 = න 𝑎 𝑑𝑡

∆𝑣 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑎 − 𝑡 𝑔𝑟𝑎𝑝ℎ

Rectilinear Kinematics: Erratic Motion Given v-t graph, find s-t graph

Δ𝑠 = න 𝑣 𝑑𝑡

Δ𝑠 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ

Rectilinear Kinematics: Erratic Motion Given a-s graph, find v-s graph

𝑠1 1 2 2 𝑣 − 𝑣0 = න 𝑎 𝑑𝑠 2 1 𝑠0

𝑠1

න 𝑎 𝑑𝑠 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑎 − 𝑠 𝑔𝑟𝑎𝑝ℎ 𝑠0

Rectilinear Kinematics: Erratic Motion Given v-s graph, find a-s graph

𝑑𝑣 𝑎=𝑣 𝑑𝑠

𝑣 ∗ 𝑎 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑣 − 𝑠 𝑔𝑟𝑎𝑝ℎ

Problem 1 The car in starts from rest and travels along a straight track such that it accelerates at 10 𝑚Τ𝑠 2 for 10 s, and then decelerates at 2 𝑚Τ𝑠 2 . Draw the v-t and s-t graphs and determine the time, ‘t‘ needed to stop the car. How far has the car travelled?

Problem 2 A bicycle moves along a straight road such that its position is described by the graph shown. Construct the v-t and a-t graphs for 0 ≤ 𝑡 ≤ 𝑠.

Problem 3 A two-stage missile is fired vertically from rest with the acceleration shown. In 15-s the first stage A burns out and the second stage B ignites. Plot the v-t and s-t graphs which describe the two-stage motion of the missile for 0 ≤ t ≤ 20 seconds

Activity 1 The velocity of a particle is shown in the graph below. Take note that at t=0, s=0. Draw the a-t and s-t graph of the particle

v(t), m/s 6

3

-4

6

8

t, seconds

General Curvilinear Motion

Curvilinear Motion occurs when a particle moves along a curved path.

General Curvilinear Motion Position Consider a particle located at a point on a space curve defined by the path function, s(t). The position of the particle, measured from a fixed point 0, will be designated by the position vector: 𝐫 = 𝐫 𝑡

General Curvilinear Motion Displacement Suppose that during a small time interval, Δt the particle moves a distance, Δs along the curve to a new position, defined by: 𝐫 ′ = 𝐫Ԧ + 𝚫𝐫 The displacement, 𝚫𝐫 represents the change in the particle's position and is determined by vector subtraction; 𝚫𝐫 = 𝐫 ′ − 𝐫Ԧ

General Curvilinear Motion Velocity Since 𝚫𝐫 will be tangent to the curve, the direction of the velocity, 𝐯 is also tangent to the curve. The magnitude of |v|, which is called the speed, is obtained by realizing that the length of the straight line segment 𝚫𝐫 in approaches the arc length, Δs as 𝛥𝑡 → 0,

𝑑𝒔 𝑣= 𝑑𝒕

General Curvilinear Motion Acceleration

The Average Acceleration, aavg of the particle during the time interval, 𝚫𝐭 is:

𝒂𝑎𝑣𝑔

∆𝑣 = ∆𝑡

General Curvilinear Motion Acceleration

A Hodograph describes the locus of points for the arrowhead of the velocity vector in the same manner as the path, ‘s’ describes the locus of points for the arrowhead of the position vector The Instantaneous Acceleration, a as 𝚫𝐭 → 0

𝑑𝐯 𝐚= 𝑑𝒕 𝑑2 𝐫 𝐚= 2 𝑑𝒕

General Curvilinear Motion Acceleration

Thus by definition of the derivative, Acceleration, ′𝐚′ acts tangent to the hodograph; and in general it is not tangent to the path of motion.

Curvilinear Motion: Rectangular Components Position

𝐫 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 Magnitude of ′𝐫′:

𝑟=

𝑥2 + 𝑦2 + 𝑧2

Direction of ′𝐫′:

𝐫 𝐮𝑟 = 𝑟

Curvilinear Motion: Rectangular Components Velocity

𝑑𝐫 𝐯= = 𝑣𝑥 𝐢 + 𝑣𝑦 𝐣 + 𝑣𝑧 𝐤 𝑑𝑡 Magnitude of ′𝐯′:

𝑣=

𝑣𝑥2

+

𝑣𝑦2

+ 𝑣𝑧2

Direction of ′𝐯′:

𝐯 𝐮𝑣 = 𝑣

Curvilinear Motion: Rectangular Components Velocity

𝑣𝑥 = 𝑥ሶ 𝑣𝑦 = 𝑦ሶ 𝑣𝑧 = 𝑧ሶ

Curvilinear Motion: Rectangular Components Acceleration

𝑑𝐯 𝐚= = 𝑎𝑥 𝐢 + 𝑎𝑦 𝐣 + 𝑎𝑧 𝐤 𝑑𝑡 Magnitude of ′𝐚′:

𝑎=

𝑎𝑥2

+

𝑎𝑦2

+

Direction of ′𝐚′:

𝑎𝑧2

𝐚 𝐮𝑎 = 𝑎

Curvilinear Motion: Rectangular Components Acceleration

𝑎𝑥 = 𝑣𝑥ሶ = 𝑥ሷ 𝑎𝑦 = 𝑣𝑦ሶ = 𝑦ሷ 𝑎𝑧 = 𝑣𝑧ሶ = 𝑧ሷ

Problem 1 At any instant, the horizontal position of the weather balloon in the figure below is defined by x = (8t) ft, where ‘t’ is time in seconds. If the equation of the path is y = x2/10, determine the magnitude and direction of the velocity and the acceleration when t = 2s.

Problem 2 The path of the plane is described by 𝑦 = (0.001𝑥 2 ) 𝑚. If the plane is rising with a constant velocity of 10 𝑚/𝑠, determine the magnitudes of the velocity and acceleration of the plane when it is 100-m above the ground.

Answer: m Velocity: |v| = 18.7083 s Acceleration: |a| = 0.7906 m/s 2

Curvilinear Motion: Normal & Tangential Components Planar Motion

When the path along which a particle travels is ‘known’, then it is often convenient to describe the motion using ‘n’ and ‘t’ coordinate axes which act normal and tangent to the path, respectively and having a fixed origin that is coincident with the particle at the instant considered

Curvilinear Motion: Normal & Tangential Components Planar Motion

Consider the particle shown in the figure, which moves in a plane along a fixed curve, such that at a given instant it is at position, s, measured from point, O Each segment, ds is formed from the arc of an associated circle having a radius of curvature, p (rho) and center of curvature, O’

Curvilinear Motion: Normal & Tangential Components Velocity

Since the particle moves, ‘s’ is a function of time. The particle's velocity, v has a direction that is always tangent to the path, and a magnitude that is determined by taking the time derivative of the path function, s = s(t) i.e., v = ds/dt. Hence:

𝐯 = 𝑣𝐮𝑡 𝑣 = 𝑠ሶ

Curvilinear Motion: Normal & Tangential Components Acceleration

The acceleration of the particle is the time rate of change of the velocity. Thus

𝐚 = 𝐯ሶ = 𝑣𝐮 ሶ 𝑡 + 𝑣 𝐮𝑡ሶ

Curvilinear Motion: Normal & Tangential Components Acceleration, ′𝐚′ can be written as the sum of its two components

𝐚 = 𝑎 𝑡 𝐮 𝑡 + 𝑎𝑛 𝐮 𝑛 Where:

𝑑𝐯 𝑎𝑡 = 𝑣ሶ = 𝑑𝑡 or

𝑎𝑡 𝑑𝑠 = 𝑣𝑑𝑣 and

𝑣2 𝑎𝑛 = 𝜌 The magnitude of acceleration

𝑎=

𝑎𝑡2 + 𝑎𝑛2

Curvilinear Motion: Normal & Tangential Components Acceleration

To better understand, consider the following two special cases of motion. 1. If the particle moves along a straight line, then 𝜌 → ∞ and 𝑎𝑛 = 0 . Thus 𝑎 = 𝑎𝑡 = 𝑣,ሶ and we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of the velocity. 2. If the particle moves along a curve with a constant speed, then 𝑎𝑡 = 𝑣ሶ = 0 and 𝑎 = 𝑎𝑛 = 𝑣 2 /𝜌. Therefore, the normal component of acceleration represents the time rate of change in the direction of velocity. Since 𝐚𝑛 always acts towards the center of curvature, this component is sometimes referred to as the centripetal (or center seeking) acceleration.

Problem 3 A race car, ‘C’ starts from rest and travels around the horizontal circular track that has a radius of 300 𝑓𝑡. If the car increases its speed at a constant rate of 7 𝑓𝑡/𝑠 2 , starting from rest, determine the time needed for it to reach an acceleration of 8 𝑓𝑡/𝑠 2 . What is its speed at this instant?

Answer: Acceleration: 𝑡 = 4.87 𝑠 Velocity: 𝑣 = 34.1 𝑓𝑡/𝑠

Problem 4 The boxes travel along the industrial conveyor. If a box starts from rest at ‘A’ and increases its speed such that 𝑎𝑡 = (0.2𝑡) 𝑚/𝑠 2 , where ‘t’ is in seconds, determine the magnitude of its acceleration when it arrives at point ‘B’.

Answer: Acceleration: 𝑎𝐵 = 5.36 𝑚/𝑠 2

Curvilinear Motion: Normal & Tangential Components Equation of Motion When a particle moves along a curved path which is known, the equation of motion for the particle may be written in the tangential, normal, and binormal directions Note that there is NO motion of the particle in the binormal direction, since the particle is constrained to move along the path.

Curvilinear Motion: Normal & Tangential Components Equation of Motion Recall that Tangential Acceleration, at (= dv/dt) represents the time rate of change in the magnitude of velocity. So if ΣFt acts in the direction of motion, the particle's speed will increase, whereas if it acts in the opposite direction, the particle will slow down. Likewise, Normal Acceleration, an (= v2 /ρ) represents the time rate of change in the velocity's direction. It is caused by ΣFn, which always acts in the positive ‘n’ direction, i.e., toward the path's center of curvature. From this reason it is often referred to as the centripetal force.

Curvilinear Motion: Normal & Tangential Components Banking of Curve Case 1: Circular path on a Flat Horizontal Road • When a vehicle is moving in an arc, there is a centripetal acceleration and thus force directed towards the center of the circle. • The force causing the centripetal acceleration is the “sideways” friction force on the vehicle by the road. • Without that force, the vehicle will tend to continue in a straight line rather than turning the corner • There is a limit to the frictional force provided by the road thus limiting the cornering speed of an arc given radius.

෍ 𝐹𝑥 = 𝑚𝑎𝑛

𝑣2 𝐹𝑓𝑟 = 𝑚 ∗ 𝑟 𝑣2 𝜇𝑠 ∗ 𝑁 = 𝑚 ∗ 𝑟

Curvilinear Motion: Normal & Tangential Components Banking of Curve Case 2: Banked Curved without Friction (Ideal Velocity) • This expression can be understood by considering how the banking angle, θ depends on the ideal velocity, v of the vehicle and the radius of curvature, r of the path.

෍ 𝐹𝑛 = 𝑚𝑎𝑛

෍ 𝐹𝑏 = 0

𝒗𝟐 𝑵 𝐬𝐢𝐧 𝜽 = 𝒎 ∗ 𝒓

𝑵 𝐜𝐨𝐬 𝜽 − 𝒎𝒈 = 𝟎

Taking Eq.1/Eq.2 we have:

𝑁 sin 𝜃 𝑚𝑣 2 = 𝑁 cos 𝜃 𝑚𝑔𝑟 𝒗𝟐 𝐭𝐚𝐧 𝜽 = 𝒈𝒓

Curvilinear Motion: Normal & Tangential Components Banking of Curve Case 2A: Banked Curved with Friction (Actual Velocity > Ideal Velocity) • The centripetal force, N*sinΘ will not be enough thus friction will now be a factor and will act at a downward direction in the incline.

෍ 𝐹𝑛 = 𝑚𝑎𝑛 (𝒗𝒂𝒄𝒕𝒖𝒂𝒍 )𝟐 𝑵 𝐬𝐢𝐧 𝜽 + 𝒇 cos 𝜽 = 𝒎 ∗ 𝒓 (𝒗𝒂𝒄𝒕𝒖𝒂𝒍 )𝟐 𝑵 𝐬𝐢𝐧 𝜽 + 𝝁𝒔 𝑵 cos 𝜽 = 𝒎 ∗ 𝒓 ෍ 𝐹𝑏 = 0 𝑵 𝐜𝐨𝐬 𝜽 − 𝒎𝒈 − 𝒇 sin 𝜽 = 𝟎

Curvilinear Motion: Normal & Tangential Components Banking of Curve Case 2B: Banked Curved with Friction (Actual Velocity < Ideal Velocity) • The centripetal force, N*sinΘ will not be enough thus friction will now be a factor and will act at an upward direction in the incline.

෍ 𝐹𝑛 = 𝑚𝑎𝑛 (𝒗𝒂𝒄𝒕𝒖𝒂𝒍 )𝟐 𝑵 𝐬𝐢𝐧 𝜽 − 𝒇 cos 𝜽 = 𝒎 ∗ 𝒓 (𝒗𝒂𝒄𝒕𝒖𝒂𝒍 )𝟐 𝑵 𝐬𝐢𝐧 𝜽 − 𝝁𝒔 𝑵 cos 𝜽 = 𝒎 ∗ 𝒓 ෍ 𝐹𝑏 = 0 𝑵 𝐜𝐨𝐬 𝜽 − 𝒎𝒈 + 𝒇 sin 𝜽 = 𝟎

Problem 5 Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed (in km/hr) at which a 100-m radius curve banked at 65.0° should be driven if the road is frictionless. Calculate the allowable velocity if the track was not banked with the standard coefficient of static friction for dry asphalt is 0.9 and an average weight for a Formula-One race car with its driver is 740-kg.

Answer: Velocity: 𝑣 = 165.12 𝑘𝑚/ℎ𝑟 Velocity: 𝑣 = 106.97 𝑘𝑚/ℎ𝑟

Problem 6 The sports car, having a mass of 1,700 kg travels horizontally along a 20° banked track which is circular and has a radius of curvature of 100-m. If the coefficient of static friction between the tires and the road is 0.2, determine (a) the maximum constant speed at which the car can travel without sliding up and (b) the minimum constant speed at which the car can travel without sliding down the slope.

Answer: (a) Velocity: 𝑣 = 24.4272 𝑚/𝑠 (b) Velocity: 𝑣 = 12.245 𝑚/𝑠

Problem 7 The device shown is used to produce the experience of weightlessness in a passenger when he reaches point A, ϴ=90°, along the path. If the passenger has a mass of 75-kg, determine the minimum speed he should have when he reaches ‘A’ so that he will experience the feeling of weightlessness. The chair is pin-connected to the frame ‘BC’ so that he is always seated in an upright position. During the motion his speed remains constant.

Answer: Velocity: 𝑣 = 9.9045 𝑚/𝑠

Problem 8 The 2-kg block ‘B’ and 15-kg cylinder ‘A’ are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5 m, determine the speed of the block.

Answer: Velocity: 𝑣 = 10.5054 𝑚/𝑠

Activity 2 1. A particle is traveling along the parabolic path y = 0.25x2. If x=(2t2) m, where ‘t’ is in seconds, determine the magnitude of the particle's velocity and acceleration when t=2s.

2. A particle is constrained to travel along the path. If x=(4t4) m, where ‘t’ is in seconds, determine the magnitude of the particle's velocity and acceleration when t=0.5s 3. The boat is traveling along the circular path with a speed of v=(0.0625t2) m/s, where t is in seconds. Determine the magnitude of its acceleration when t=10s.

Absolute Dependent Motion Analysis of Two Particles

When the motion of one particle depends on the corresponding motion of another particle. This dependency commonly occurs if the particles are interconnected by inextensible cords which are wrapped around pulleys

Absolute Dependent Motion Analysis of Two Particles

For this set-up: If the movement of block ‘A’ is downward along the inclined plane it will cause a corresponding movement of block ‘B’ up the other incline.

Note that each of the coordinate axes is: (1) measured from a fixed point, ‘O’ or fixed datum line, (2) measured along each inclined plane in the direction of motion of each block, and (3) has a positive sense from C to A and D to B If the total cord length is 𝑙 𝑇 , the two position coordinates are related by the equation:

𝒔𝑨 + 𝒍𝑪𝑫 + 𝒔𝑩 = 𝒍𝑻

Absolute Dependent Motion Analysis of Two Particles

Here 𝑙𝐶𝐷 is the length of the cord passing over arc, ‘CD’. Taking the time derivative of this expression, realizing that 𝑙𝐶𝐷 and 𝑙 𝑇 remain constant, while 𝑠𝐴 and 𝑠𝐵 measure the segments of the cord that change in length. We have:

𝒅𝒔𝑨 𝒅𝒔𝑩 + = 𝟎 𝐨𝐫 𝒗𝑩 = −𝒗𝑨 𝒅𝒕 𝒅𝒕 In a similar manner:

𝒂𝑩 = −𝒂𝑨

Absolute Dependent Motion Analysis of Two Particles For this set-up: The position of block ‘A’ is specified by SA, and the position of the end of the cord from which block B is suspended is defined by SB Note that the chosen position of coordinate axes: (1) have their origin at fixed points or datums, (2) are measured in the direction of motion of each block, and (3) are positive to the right for SA and positive downward for SB. The two position coordinates are related by the equation:

𝟐𝒔𝑩 + 𝒉 + 𝒔𝑨 = 𝒍𝑻 Taking the time derivative of this expression, we have:

𝟐

𝒅𝒔𝑩 𝒅𝒔𝑨 =− 𝒅𝒕 𝒅𝒕

𝒐𝒓 𝟐𝒗𝑩 = −𝒗𝑨

In a similar manner:

𝟐𝒂𝑩 = −𝒂𝑨

Absolute Dependent Motion Analysis of Two Particles The same set-up can also be worked by defining the position of block ‘B’ from the center of the bottom pulley (a fixed point), In this case: The two position coordinates are related by the equation:

𝟐 𝒉 − 𝒔𝑩 + 𝒉 + 𝒔𝑨 = 𝒍𝑻 Taking the time derivative of this expression, we have:

𝟐

𝒅𝒔𝑩 𝒅𝒔𝑨 = 𝒅𝒕 𝒅𝒕

𝒐𝒓 𝟐𝒗𝑩 = 𝒗𝑨

In a similar manner:

𝟐𝒂𝑩 = 𝒂𝑨 Here the signs are the same. Why?

Problem 1 Determine the speed of block A if block B has an upward speed of 6 ft/s.

Answer: Velocity: 𝑣𝐴 = 18 𝑓𝑡/𝑠 ↓

Problem 2 Determine the speed of A if B has an upward speed of 6 ft/s.

Answer: Velocity: 𝑣𝐴 = 24 𝑓𝑡/𝑠 ↓

Problem 3 Determine the speed of block B if the end of the cord at A is pulled down with a speed of 2 m/s.

Answer: Velocity: 𝑣𝐵 = 0.5 𝑚/𝑠 ↑

Seatwork 1. Determine the speed of block ‘D’ if end ‘A’ of the rope is pulled down with a speed of 3m/s.

Seatwork 2. Determine the speed of block ‘A’ if end B of the rope is pulled down with a speed of 6 m/s.

Seatwork 3. Determine the speed of block ‘A’ if end B of the rope is pulled down with a speed of 1.5 m/s

Rotation about a Fixed Axis

When a body rotates about a fixed axis, any point where in the body travels moves along a circular path.

Rotation about a Fixed Axis Angular Position, ϴ  At the instant shown, the angular position of ‘r’ is defined by the angle ϴ, measured from a fixed reference line to ‘r’. Angular Displacement, dϴ  The change in the angular position, which can be measured as a differential dϴ  This vector has a magnitude of dϴ, measured in degrees, radians, or revolutions, where 1 rev = 2π rad.  Since motion is about a fixed axis, the direction of dϴ is always along this axis. Specifically, the direction is determined by the right-hand rule. Angular Velocity, ω  The time rate of change in the angular position.  This vector has a magnitude which is often measured in rad/s.

𝑑𝜃 𝜔= 𝑑𝑡

𝜔 = 2𝜋𝑓

1 T= 𝑓

Rotation about a Fixed Axis Angular Acceleration, α  measures the time rate of change of the angular velocity.

𝑑𝜔 𝛼= 𝑑𝑡

𝑑2𝜃 𝛼= 𝑑𝑡 Differential relation between the Angular acceleration, Angular velocity, and Angular displacement:

𝛼𝑑𝜃 = 𝜔𝑑𝜔

Rotation about a Fixed Axis Constant Angular Acceleration  If the angular acceleration of the body is constant, α=αc, then these equations, when integrated, yield a set of formulas which relate the body's angular velocity, angular position, and time. These equations are similar to equations used for rectilinear motion.

Relationship of Angular and Linear Motion Motion of Point, ‘P’  As the rigid body in the figure rotates, point ‘P’ travels along a circular path of radius, r with center at point O. This path is contained within the shaded plane shown in top view Position and Displacement of Point, ‘P’  The position of ‘P’ is defined by the position vector ‘r’, which extends from ‘O’ to ‘P’. If the body rotates dϴ then ‘P’ will displace ds = rdϴ

∆𝑠 = 𝑟 ∗ ∆𝜃 Velocity of Point, ‘P’  The velocity of P has a magnitude which can be found by dividing ds = rdϴ by ‘dt’ so that

𝑣 = 𝜔𝑟

Relationship of Angular and Linear Motion Acceleration of Point, ‘P’  The acceleration of ‘P’ can be expressed in terms of its normal and tangential components.

𝑎𝑡 = 𝛼 ∗ 𝑟

𝑎 2 = 𝛼𝑛 2 ∗ 𝛼 𝑡 2

𝑎𝑛 = 𝜔 2 ∗ 𝑟

Problem 1 Load ‘B’ is connected to a double pulley by one of the two inextensible cables shown. The motion of the pulley is controlled by cable ‘C’, which has a constant acceleration of 9in/s2 and an initial velocity of 12in/s, both directed to the right. Determine (a) the number of revolutions executed by the smaller pulley in 2-seconds, (b) the velocity and change in position of the load ‘B’ after 2-seconds, and (c) the acceleration of point ‘D’ on the rim of the inner pulley at t=0

Problem 2 A cylinder of radius 12-cm starts from rest and rotates about its axis with a constant angular acceleration of 5.0rad/s2. At t = 3.0-sec, what is its (a) angular velocity, (b) linear speed of the point on the rim and (c) radial and tangential components of acceleration of a point on the rim.

Problem 3 A record player is spinning at 33.3 rpm. (a) How far does it turn in 2-sec? (b) When the motor is shut off, the record player spins down for 20-sec before coming to rest. What is the angular acceleration assuming that it is constant? How many turn does it make during this coast down?

Problem 4 The motion of a cam is defined by the relation ϴ = t3 - 9t2 + 15t where ϴ is expressed in radians and ‘t’ in seconds. Determine the angular position, the angular velocity, and the angular acceleration of the cam when (a) t = 0s, (b) t = 3s.

Problem 5 The motion of an oscillating crank is defined by the relation ϴ=6sin(πt/4)-3sin (πt/2) where ‘ϴ’ is expressed in radians and ‘t’ in seconds. Determine the angular displacement, the angular velocity, and the angular acceleration of the crank when (a) t=0-s, (b) t=2-s

Problem 6 The angular acceleration of a shaft is defined by the relation α= -0.25ω, where ‘α’ is expressed in rad/s2 and ω in rad/s. Knowing that at t=0 the angular velocity of the shaft is 20 rad/s, determine (a) the number of revolutions the shaft will execute before coming to rest, (b) the time required for the shaft to come to rest, (c) the time required for the angular velocity of the shaft to be reduced to 1 percent of its initial value.

Problem 7 When studying whiplash resulting from rear end collisions, the rotation of the head is of primary interest. An impact test was performed, and it was found that the angular acceleration of the head is defined by the relation α=700cos(θ) + 70sin(θ) where ‘α’ is expressed in rad/s2 and ‘θ’ in radians. Knowing that the head is initially at rest, determine the angular velocity of the head when θ = 30°

Simple Pendulum

A simple pendulum is defined to have a point mass, which is suspended from a string of with negligible mass

Simple Pendulum • A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length ’L’ with negligible. • The only forces acting on the bob are the force of gravity acting through the weight of the bob and tension from the string. • The mass of the string is assumed to be negligible as compared to the mass of the bob.

Simple Pendulum Angular Frequency, ω

𝜔=

𝑔 𝐿

Where: g - acceleration due to gravity L - length of the cord

Period, T

𝐿 𝑇 = 2𝜋 𝑔 Where: g - acceleration due to gravity L - length of the cord

Problem 1 What is the acceleration due to gravity in a region where a simple pendulum having a length 75-cm has a period of 1.7357-s?