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CHAPTER 8

Rotational of rigid body(8 Hours)

WHAT IS ROTATIONAL OF RIGID BODY ? The study of

the motions of a rigid body under the influence

of forces and torques.

WHAT IS RIGID BODY ? an object or system of

particles in which the distances between

particles are fixed and remain constant.

Learning Outcome

8.1 Rotational Kinematics (2 hour)

- At the end of this chapter, students should be

able to - Define and use
- angular displacement (?)
- average angular velocity (?av)
- instantaneous angular velocity (?)
- average angular acceleration (?av)
- instantaneous angular acceleration (?).

- Relate parameters in rotational motion with their

corresponding quantities in linear motion. Write

and use,

Learning Outcome

- Use equations for rotational motion with constant

angular acceleration,

8.1.1 Parameters in rotational motion

- 1) Angular displacement,?
- is defined as an angle through which a point or

line has been rotated in a specified direction

about a specified axis. - The S.I. unit of the angular displacement is

radian (rad). - Figure 8.1 shows a point P on a rotating compact

disc (CD) moves through an arc length s on a

circular path of radius r about a fixed axis

through point O.

Figure 8.1

- From Figure 8.1, thus
- Others unit for angular displacement is degree

(?) and revolution (rev). - Conversion factor
- Sign convention of angular displacement
- Positive if the rotational motion is

anticlockwise. - Negative if the rotational motion is clockwise.

OR

where

2) Angular velocity

- Average angular velocity, ?av
- is defined as the rate of change of angular

displacement. - Equation
- Instantaneous angular velocity, ?
- is defined as the instantaneous rate of change of

angular displacement. - Equation

where

- It is a vector quantity.
- The unit of angular velocity is radian per second

(rad s-1) - Others unit is revolution per minute (rev min?1

or rpm) - Conversion factor
- Note
- Every part of a rotating rigid body has the same

angular velocity. - Direction of the angular velocity
- Its direction can be determine by using right

hand grip rule where

Thumb direction of angular velocity Curl

fingers direction of rotation

- Figures 8.2 and 8.3 show the right hand grip rule

for determining the direction of the angular

velocity.

Figure 8.2

Figure 8.3

Example 8.1

The angular displacement,? of the wheel is given

by where ? in radians and t in seconds. The

diameter of the wheel is 0.56 m. Determine a. the

angle, ? in degree, at time 2.2 s and 4.8 s, b.

the distance that a particle on the rim moves

during that time interval, c. the

average angular velocity, in rad s?1 and in rev

min?1 (rpm), between 2.2 s and 4.8 s, d.

the instantaneous angular velocity at time 3.0 s.

Solution a. At time, t1 2.2 s At

time, t2 4.8 s

Solution b. By applying the equation of arc

length, c. The average angular velocity in

rad s?1 is given by

Solution c. and the average angular velocity in

rev min?1 is d. The instantaneous angular

velocity as a function of time is At

time, t 3.0 s

3) Angular acceleration

- Average angular acceleration, ?av
- is defined as the rate of change of angular

velocity. - Equation
- Instantaneous angular acceleration, ?
- is defined as the instantaneous rate of change of

angular velocity. - Equation

where

- It is a vector quantity.
- The unit of angular acceleration is rad s?2.
- Note
- If the angular acceleration, ? is positive, then

the angular velocity, ? is increasing. - If the angular acceleration, ? is negative, then

the angular velocity, ? is decreasing. - Direction of the angular acceleration
- If the rotation is speeding up, ? and ? in the

same direction as shown in Figure 8.4.

Figure 8.4

- If the rotation is slowing down, ? and ? have the

opposite direction as shown in Figure 8.5. - Example 8.3
- The instantaneous angular velocity, ? of the

flywheel is given - by
- where ? in radian per second and t in seconds.
- Determine
- a. the average angular acceleration between 2.2 s

and 4.8 s, - b. the instantaneous angular acceleration at

time, 3.0 s.

Figure 8.5

Solution

Solution b. The instantaneous angular

acceleration as a function of time is

At time, t 3.0 s

Exercise 8.1A

1. If a disc 30 cm in diameter rolls 65 m along a

straight line without slipping, calculate a. the

number of revolutions would it makes in the

process, b. the angular displacement would be

through by a speck of gum on its

rim. ANS. 69 rev 138? rad 2. During a certain

period of time, the angular displacement of a

swinging door is described by where ? is in

radians and t is in seconds. Determine the

angular displacement, angular speed and angular

acceleration a. at time, t 0, b. at time, t

3.00 s. ANS. 5.00 rad, 10.0 rad s?1, 4.00 rad

s?2 53.0 rad, 22.0 rad s?1, 4.00 rad s?2

8.1.2 Relationship between linear and

rotational motion

- Relationship between linear velocity, v and

angular velocity, ? - When a rigid body is rotates about rotation axis

O , every particle in the body moves in a circle

as shown in the Figure 8.6.

Figure 8.6

- Point P moves in a circle of radius r with the

tangential velocity v where its magnitude is

given by - The direction of the linear (tangential) velocity

always tangent to the circular path. - Every particle on the rigid body has the same

angular speed (magnitude of angular velocity) but

the tangential speed is not the same because the

radius of the circle, r is changing depend on the

position of the particle.

and

Relationship between tangential acceleration, at

and angular acceleration, ?

- If the rigid body is gaining the angular speed

then the tangential velocity of a particle also

increasing thus two component of acceleration

are occurred as shown in Figure 8.7.

Figure 8.7

- The components are tangential acceleration, at

and centripetal acceleration, ac given by - but
- The vector sum of centripetal and tangential

acceleration of a particle in a rotating body is

resultant (linear) acceleration, a given by - and its magnitude,

and

Vector form

8.1.3 Rotational motion with uniform

angular acceleration

- Table 8.1 shows the symbols used in linear and

rotational kinematics.

Linear motion Quantity Rotational motion Quantity

Angular displacement

Angular velocity (initial)

Angular velocity (final)

Angular acceleration

time

Displacement

Initial velocity

Final velocity

Acceleration

Time

Table 8.1

- Table 8.2 shows the comparison of linear and

rotational motion with constant acceleration.

Linear motion Rotational motion

where ? in radian.

Table 8.2

Example 8.3

A car is travelling with a velocity of 17.0 m s?1

on a straight horizontal highway. The wheels of

the car has a radius of 48.0 cm. If the car then

speeds up with an acceleration of 2.00 m s?2 for

5.00 s, calculate a. the number of revolutions of

the wheels during this period, b. the angular

speed of the wheels after 5.00 s. Solution a.

The initial angular velocity is and the

angular acceleration of the wheels is given by

Solution a. By applying the equation of

rotational motion with constant angular

acceleration, thus b. The angular speed of

the wheels after 5.00 s is

Example 8.5

The wheels of a bicycle make 30 revolutions as

the bicycle reduces its speed uniformly from 50.0

km h-1 to 35.0 km h-1. The wheels have a diameter

of 70 cm. a. Calculate the angular

acceleration. b. If the bicycle continues to

decelerate at this rate, determine the

time taken for the bicycle to stop. Solution

Solution a. The initial angular speed of the

wheels is and the final angular speed of

the wheels is therefore b. The car stops

thus Hence

Example 8.5

A blade of a ceiling fan has a radius of 0.400 m

is rotating about a fixed axis with an initial

angular velocity of 0.150 rev s-1. The angular

acceleration of the blade is 0.750 rev s-2.

Determine a. the angular velocity after 4.00

s, b. the number of revolutions for the blade

turns in this time interval, c. the tangential

speed of a point on the tip of the blade at

time, t 4.00 s, d. the magnitude of

the resultant acceleration of a point on the tip

of the blade at t 4.00 s. Solution a.

Given t 4.00 s, thus

Solution b. The number of revolutions of the

blade is c. The tangential speed of a

point is given by

Solution d. The magnitude of the resultant

acceleration is

Example 8.6

A coin with a diameter of 2.40 cm is dropped on

edge on a horizontal surface. The coin starts out

with an initial angular speed of 18 rad s?1 and

rolls in a straight line without slipping. If the

rotation slows down with an angular acceleration

of magnitude 1.90 rad s?2, calculate the distance

travelled by the coin before coming to

rest. Solution The radius of the coin is

Solution The initial speed of the point at the

edge the coin is and the final speed is The

linear acceleration of the point at the edge the

coin is given by Therefore the distance

travelled by the coin is

Exercise 8.1B

1. A disk 8.00 cm in radius rotates at a

constant rate of 1200 rev min-1 about its central

axis. Determine a. its angular speed, b. the

tangential speed at a point 3.00 cm from its

centre, c. the radial acceleration of a point on

the rim, d. the total distance a point on the

rim moves in 2.00 s. ANS. 126 rad s?1 3.77 m

s?1 1.26 ? 103 m s?2 20.1 m 2. A 0.35 m

diameter grinding wheel rotates at 2500 rpm.

Calculate a. its angular velocity in rad

s?1, b. the linear speed and the radial

acceleration of a point on the edge of

the grinding wheel. ANS. 262 rad s?1 46 m

s?1, 1.2 ? 104 m s?2

3. A rotating wheel required 3.00 s to rotate

through 37.0 revolution. Its angular speed at the

end of the 3.00 s interval is 98.0 rad s-1.

Calculate the constant angular acceleration of

the wheel. ANS. 13.6 rad s?2 4. A wheel

rotates with a constant angular acceleration

of 3.50 rad s?2. a. If the angular speed of the

wheel is 2.00 rad s?1 at t 0, through what

angular displacement does the wheel rotate in

2.00 s. b. Through how many revolutions has

the wheel turned during this time

interval? c. What is the angular speed of the

wheel at t 2.00 s? ANS. 11.0 rad 1.75 rev

9.00 rad s?1

8.2 Equilibrium of a uniform rigid body (2 hours)

Learning Outcome

- At the end of this chapter, students should be

able to - Define and use torque,
- State and use conditions for equilibrium of rigid

body,

8.2.1 Torque (moment of a force),

- The magnitude of the torque is defined as the

product of a force and its perpendicular distance

from the line of action of the force to the point

(rotation axis). - OR
- Because of
- where r distance between the pivot point

(rotation axis) and the point of

application of force. - Thus

where

OR

where

- It is a vector quantity.
- The dimension of torque is
- The unit of torque is N m (newton metre), a

vector product unlike the joule (unit of work),

also equal to a newton metre, which is scalar

product. - Torque is occurred because of turning (twisting)

effects of the forces on a body. - Sign convention of torque
- Positive - turning tendency of the force is

anticlockwise. - Negative - turning tendency of the force is

clockwise. - The value of torque depends on the rotation axis

and the magnitude of applied force.

- Case 1
- Consider a force is applied to a metre rule which

is pivoted at one end as shown in Figures 8.12a

and 8.12b.

(anticlockwise)

(anticlockwise)

- Case 2
- Consider three forces are applied to the metre

rule which is pivoted at one end (point O) as

shown in Figures 8.13. - Caution
- If the line of action of a force is through the

rotation axis then

Therefore the resultant (nett) torque is

and

8.2.2 Equilibrium of a rigid body

- Rigid body is defined as a body with definite

shape that doesnt change, so that the particles

that compose it stay in fixed position relative

to one another even though a force is exerted on

it. - If the rigid body is in equilibrium, means the

body is translational and rotational equilibrium. - There are two conditions for the equilibrium of

forces acting on a rigid body. - The vector sum of all forces acting on a rigid

body must be zero.

OR

- The vector sum of all external torques acting on

a rigid body must be zero about any rotation

axis. - This ensures rotational equilibrium.
- This is equivalent to the three independent

scalar equations along the direction of the

coordinate axes, - Centre of gravity, CG
- is defined as the point at which the whole weight

of a body may be considered to act. - A force that exerts on the centre of gravity of

an object will cause a translational motion.

- Figures 8.14 and 8.15 show the centre of gravity

for uniform (symmetric) object i.e. rod and

sphere - rod refer to the midway point between its end.
- sphere refer to geometric centre.

Figure 8.14

Figure 8.15

Problem solving strategies for equilibrium of a

rigid body

- The following procedure is recommended when

dealing with problems involving the equilibrium

of a rigid body - Sketch a simple diagram of the system to help

conceptualize the problem. - Sketch a separate free body diagram for each

body. - Choose a convenient coordinate axes for each body

and construct a table to resolve the forces into

their components and to determine the torque by

each force. - Apply the condition for equilibrium of a rigid

body - Solve the equations for the unknowns.

and

Example 8.7

A hanging flower basket having weight, W2

23 N is hung out over the edge of a balcony

railing on a uniform horizontal beam AB of length

110 cm that rests on the balcony railing. The

basket is counterbalanced by a body of weight, W1

as shown in Figure 8.9. If the mass of the beam

is 3.0 kg, calculate a. the weight, W1 needed, b.

the force exerted on the beam at point O. (Given

g 9.81 m s?2)

Solution The free body diagram of the beam

Let point O as the rotation axis.

0.20 m

Force y-comp. (N) Torque (N m), ?oFdFrsin?

Solution Since the beam remains at rest thus

the system in equilibrium. a. Hence b.

Example 8.9

A uniform ladder AB of length 10 m and mass 5.0

kg leans against a smooth wall as shown in Figure

8.10. The height of the end A of the ladder is

8.0 m from the rough floor. a. Determine the

horizontal and vertical forces the floor

exerts on the end B of the ladder when a

firefighter of mass 60 kg is 3.0 m from

B. b. If the ladder is just on the verge of

slipping when the firefighter is 7.0 m up

the ladder , Calculate the coefficient of

static friction between ladder and

floor. (Given g 9.81 m s?2)

Solution a. The free body diagram of the ladder

Let point B as the rotation axis.

Force x-comp. (N) y-comp. (N) Torque (N m), ?BFdFrsin?

Solution Since the ladder in equilibrium thus

Solution b. The free body diagram of the ladder

Let point B as the rotation axis.

Force x-comp. (N) y-comp. (N) Torque (N m), ?BFdFrsin?

Solution Consider the ladder stills in

equilibrium thus

Example 8.9

A floodlight of mass 20.0 kg in a park is

supported at the end of a 10.0 kg uniform

horizontal beam that is hinged to a pole as shown

in Figure 8.11. A cable at an angle 30? with the

beam helps to support the light. a. Sketch a free

body diagram of the beam. b. Determine i. the

tension in the cable, ii. the force exerted on

the beam by the pole. (Given g 9.81 m s?2)

Solution a. The free body diagram of the beam

b. Let point O as the rotation axis.

Force x-comp. (N) y-comp. (N) Torque (N m), ?oFdFrsin?

Solution b. The floodlight and beam remain at

rest thus i.

Solution b. ii. Therefore the magnitude of the

force is and its direction is given

by

Exercise 8.2

Use gravitational acceleration, g 9.81 m

s?2 1. Figure 8.12 shows the forces,

F1 10 N, F2 50 N and F3 60 N are applied to a

rectangle with side lengths, a 4.0 cm and b

5.0 cm. The angle ? is 30?. Calculate the

resultant torque about point D. ANS. -3.7 N m

2. A see-saw consists of a uniform board

of mass 10 kg and length 3.50 m supports a father

and daughter with masses 60 kg and 45 kg,

respectively as shown in Figure 8.13. The fulcrum

is under the centre of gravity of the board.

Determine a. the magnitude of the force exerted

by the fulcrum on the board, b.

where the father should sit from the fulcrum to

balance the system. ANS. 1128 N 1.31 m

3. A traffic light hangs from a

structure as show in Figure 8.14. The uniform

aluminum pole AB is 7.5 m long has a mass of 8.0

kg. The mass of the traffic light is 12.0 kg.

Determine a. the tension in the horizontal

massless cable CD, b. the vertical and

horizontal components of the force exerted

by the pivot A on the aluminum pole. ANS. 248

N 197 N, 248 N

Exercise 5.2

4. A uniform 10.0 N picture frame is

supported by two light string as shown in Figure

8.15. The horizontal force, F is applied for

holding the frame in the position shown. a.

Sketch the free body diagram of the picture

frame. b. Calculate i. the tension in the

ropes, ii. the magnitude of the horizontal

force, F . ANS. 1.42 N, 11.2 N 7.20 N

8.3 Rotational dynamics (1 hours)

Learning Outcome

- At the end of this chapter, students should be

able to - Define the moment of inertia of a rigid body

about an axis, - State and use torque,

8.3.1 Moment of inertia, I

- Figure 8.16 shows a rigid body about a fixed axis

O with angular velocity ??. - is defined as the sum of the products of the mass

of each particle and the square of its respective

distance from the rotation axis.

Figure 8.16

- OR
- It is a scalar quantity.
- Moment of inertia, I in the rotational

kinematics is analogous to the mass, m in linear

kinematics. - The S.I. unit of moment of inertia is kg m2.
- The factors which affect the moment of inertia, I

of a rigid body - a. the mass of the body,
- b. the shape of the body,
- c. the position of the rotation axis.

where

- Moments of inertia of various bodies
- Table below shows the moments of inertia for a

number of objects about axes through the centre

of mass.

Shape Diagram Equation

Hoop or ring or thin cylindrical shell

Solid cylinder or disk

Shape Diagram Equation

Uniform rod or long thin rod with rotation axis

through the centre of mass.

Solid Sphere

Shape Diagram Equation

Hollow Sphere or thin spherical shell

Example 8.10

Four spheres are arranged in a rectangular shape

of sides 250 cm and 120 cm as shown in Figure.

The spheres are connected by light rods

. Determine the moment of inertia of the system

about an axis a. through point O, b. along the

line AB.

Solution a. rotation axis about point

O, Since r1 r2 r3 r4 r thus

and the connecting rods are light therefore

Solution b. rotation axis along the line

AB, r1 r2 r3 r4 r0.6 m therefore

8.3.2 Torque,?

- Relationship between torque,? and angular

acceleration, ? - Consider a force, F acts on a rigid body freely

pivoted on an axis through point O as shown in

Figure. - The body rotates in the anticlockwise direction

and a nett torque is produced.

- A particle of mass, m1 of distance r1 from the

rotation axis O will experience a nett force F1 .

The nett force on this particle is - The torque on the mass m1 is
- The total (nett) torque on the rigid body is

given by

and

and

- From the equation, the nett torque acting on the

rigid body is proportional to the bodys angular

acceleration. - Note

is analogous to the

Example 8.11

Forces, F1 5.60 N and F2 10.3 N are applied

tangentially to a disc with radius 30.0 cm and

the mass 5.00 kg as shown in Figure. Calcula

te, a. the nett torque on the disc. b. the

magnitude of angular acceleration influence by

the disc. ( Use the moment of inertia, )

Solution a. The nett torque on the disc is

b. By applying the relationship between

torque and angular acceleration,

Example 8.12

A wheel of radius 0.20 m is mounted on a

frictionless horizontal axis. The moment of

inertia of the wheel about the axis

is 0.050 kg m2. A light string wrapped around the

wheel is attached to a 2.0 kg block that slides

on a horizontal frictionless surface. A

horizontal force of magnitude P 3.0 N is

applied to the block as shown in Figure. Assume

the string does not slip on the wheel. a.

Sketch a free body diagram of the wheel and the

block. b. Calculate the magnitude of the angular

acceleration of the wheel.

Solution a. Free body diagram for

wheel, for block,

Solution b. For wheel, For block, By

substituting eq. (1) into eq. (2), thus

Example 8.13

An object of mass 1.50 kg is suspended from a

rough pulley of radius 20.0 cm by light string as

shown in Figure. The pulley has a moment of

inertia 0.020 kg m2 about the axis of the pulley.

The object is released from rest and the pulley

rotates without encountering frictional force.

Assume that the string does not slip on the

pulley. After 0.3 s, determine a. the linear

acceleration of the object, b. the angular

acceleration of the pulley, c. the tension in the

string, d. the liner velocity of the object, e.

the distance travelled by the object. (Given g

9.81 m s-2)

Solution a. Free body diagram for

pulley, for block,

Solution a. By substituting eq. (1) into eq.

(2), thus b. By using the relationship

between a and ?, hence

Solution c. From eq. (1), thus d. By

applying the equation of liner motion, thus e.

The distance travelled by the object in 0.3 s is

Exercise 8.3

- Use gravitational acceleration, g 9.81 m s?2
- 1. Figure 7.16 shows four masses that are held at

- the corners of a square by a very light
- frame. Calculate the moment of inertia
- of the system about an axis perpendicular
- to the plane
- a. through point A, and
- b. through point B.
- ANS. 0.141 kg m2 0.211 kg m2

2. A 5.00 kg object placed on a frictionless

horizontal table is connected to a string that

passes over a pulley and then is fastened to a

hanging 9.00 kg object as in Figure 8.17. The

pulley has a radius of 0.250 m and moment of

inertia I. The block on the table is moving with

a constant acceleration of 2.00 m s?2. a. Sketch

free body diagrams of both objects and

pulley. b. Calculate T1 and T2 the tensions

in the string. c. Determine I. ANS. 10.0 N,

70.3 N 1.88 kg m2

Figure 8.17

8.4 Work and energy of rotational motion(2 hours)

Learning Outcome

- At the end of this chapter, students should be

able to - Solve problems related to
- Rotational kinetic energy,
- Work,
- Power,

8.4 Work and energy of rotational motion.

- 8.4.1 Rotational kinetic energy, Kr
- Consider a rigid body rotating about the axis OZ

as shown in Figure. - Every particle in the body is in the circular

motion about point O.

- The rigid body has a rotational kinetic energy

which is the total of kinetic energy of all the

particles in the body is given by

and

- From the formula for translational kinetic

energy, Ktr - After comparing both equations thus
- For rolling body without slipping, the total

kinetic energy of the body, K is given by

? is analogous to v I is analogous to m

where

Example 8.14

A solid sphere of radius 15.0 cm and mass 10.0 kg

rolls down an inclined plane make an angle 25? to

the horizontal. If the sphere rolls without

slipping from rest to the distance of 75.0 cm and

the inclined surface is smooth, calculate a. the

total kinetic energy of the sphere, b. the linear

speed of the sphere, c. the angular speed about

the centre of mass. (Given the moment of inertia

of solid sphere is and the

gravitational acceleration, g 9.81 m s?2)

Solution a. From the principle of

conservation of energy,

Solution b. The linear speed of the sphere is

given by c. By using the relationship

between v and ?, thus

8.4.2 Work, W

- Consider a tangential force, F acts on the solid

disc of radius R freely pivoted on an axis

through O as shown in Figure 8.19. - The work done by the tangential force is given by

Figure 8.19

and

- If the torque is constant thus
- Work-rotational kinetic energy theorem states

is analogous to the

where

8.4.3 Power, P

- From the definition of instantaneous power,
- Caution
- The unit of kinetic energy, work and power in the

rotational kinematics is same as their unit in

translational kinematics.

and

and

is analogous to the

Example 8.15

A horizontal merry-go-round has a radius of 2.40

m and a moment of inertia 2100 kg m2 about a

vertical axle through its centre. A tangential

force of magnitude 18.0 N is applied to the edge

of the merry-go- round for 15.0 s. If the

merry-go-round is initially at rest and ignore

the frictional torque, determine a. the

rotational kinetic energy of the

merry-go-round, b. the work done by the force on

the merry-go-round, c. the average power supplied

by the force. (Given g 9.81 m s?2) Solution

Solution a. By applying the relationship

between nett torque and angular

acceleration, thus Use the equation of

rotational motion with uniform angular

acceleration, Therefore the rotational

kinetic energy for 15.0 s is

Solution b. The angular displacement, ? for

15.0 s is given by By applying the

formulae of work done in rotational motion,

thus c. The average power supplied by the

force is

Learning Outcome

8.5 Conservation of angular momentum (1 hour)

- At the end of this chapter, students should be

able to - Define and use the formulae of angular momentum,
- State and use the principle of conservation of

angular momentum

8.5 Conservation of angular momentum

- 8.5.1 Angular momentum,
- is defined as the product of the angular velocity

of a body and its moment of inertia about the

rotation axis. - OR
- It is a vector quantity.
- Its dimension is M L2 T?1
- The S.I. unit of the angular momentum is kg m2

s?1.

is analogous to the

where

- The relationship between angular momentum, L with

linear momentum, p is given by - vector notation
- magnitude form
- Newtons second law of motion in term of linear

momentum is - hence we can write the Newtons second law in

angular form as - and states that a vector sum of all the torques

acting on a rigid body is proportional to the

rate of change of angular momentum.

where

8.5.2 Principle of conservation of angular

momentum

- states that a total angular momentum of a system

about an rotation axis is constant if no external

torque acts on the system. - OR
- Therefore

If the

and

Example 8.16

A 200 kg wooden disc of radius 3.00 m is rotating

with angular speed 4.0 rad s-1 about the rotation

axis as shown in Figure. A 50 kg bag of sand

falls onto the disc at the edge of the wooden

disc. Calculate, a. the angular speed of

the system after the bag of sand falling

onto the disc. (treat the bag of sand as a

particle) b. the initial and final rotational

kinetic energy of the system. Why the

rotational kinetic energy is not the same? (Use

the moment of inertia of disc is )

Solution a. The moment of inertia of the disc,

The moment of inertia of the bag of

sand, By applying the principle of

conservation of angular momentum,

Solution b. The initial rotational kinetic

energy, The final rotational kinetic

energy, thus It is because

Example 8.17

A student on a stool rotates freely with an

angular speed of 2.95 rev s?1. The student holds

a 1.25 kg mass in each outstretched arm that is

0.759 m from the rotation axis. The moment of

inertia for the system of student-stool without

the masses is 5.43 kg m2. When the student pulls

his arms inward, the angular speed increases to

3.54 rev s?1. a. Determine the new distance of

each mass from the rotation axis. b. Calculate

the initial and the final rotational kinetic

energy of the system. Solution

Solution

Solution a. The moment of inertia of the

system initially is The moment of inertia

of the system finally is By using the

principle of conservation of angular momentum,

thus

Solution b. The initial rotational kinetic

energy is given by and the

final rotational kinetic energy is

Example 8.18

The pulley in the Figure has a radius of 0.120 m

and a moment of inertia 0.055 g cm2. The rope

does not slip on the pulley rim. Calculate the

speed of the 5.00 kg block just before it strikes

the floor. (Given g 9.81 m s?2)

Solution The moment of inertia of the pulley,

Final

Initial

Solution By using the principle of

conservation of energy, thus

Exercise 8.5

- Use gravitational acceleration, g 9.81 m s?2
- 1. A woman of mass 60 kg stands at the rim of a

horizontal turntable having a moment of inertia

of 500 kg m2 and a radius of 2.00 m. The

turntable is initially at rest and is free to

rotate about the frictionless vertical axle

through its centre. The woman then starts walking

around the rim clockwise (as viewed from above

the system) at a constant speed of 1.50 m s?1

relative to the Earth. - a. In the what direction and with what value of

angular speed - does the turntable rotate?
- b. How much work does the woman do to set

herself and the - turntable into motion?
- ANS. 0.360 rad s?1 ,U think 99.9 J

2. Determine the angular momentum of the

Earth a. about its rotation axis (assume the

Earth is a uniform solid sphere),

and b. about its orbit around the Sun (treat the

Earth as a particle orbiting the

Sun). Given the Earths mass 6.0 x 1024

kg, radius 6.4 x 106 m and is 1.5 x 108 km from

the Sun. ANS. 7.1 x 1033 kg m2 s?1 2.7 x

1040 kg m2 s?1 3. Calculate the magnitude of the

angular momentum of the second hand on a clock

about an axis through the centre of the clock

face. The clock hand has a length of 15.0 cm and

a mass of 6.00 g. Take the second hand to be a

thin rod rotating with angular velocity about one

end. (Given the moment of inertia of thin rod

about the axis through the CM is

) ANS. 4.71 x 10?6 kg m2 s?1

Summary

Linear Motion Relationship Rotational Motion