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## Chapter 7:Rotation of a Rigid Body

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Title: Chapter 7:Rotation of a Rigid Body

1
CHAPTER 8
Rotational of rigid body(8 Hours)
2
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.

3
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,

4
Learning Outcome
• Use equations for rotational motion with constant
angular acceleration,

5
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
• The S.I. unit of the angular displacement is
• Figure 8.1 shows a point P on a rotating compact
disc (CD) moves through an arc length s on a
through point O.

Figure 8.1
6
• 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
7
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
8
• It is a vector quantity.
• The unit of angular velocity is radian per second
• 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
9
• 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
10
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.
11
Solution a. At time, t1 2.2 s At
time, t2 4.8 s
12
Solution b. By applying the equation of arc
length, c. The average angular velocity in
13
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
14
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
15
• 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
16
• 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
17
Solution
18
Solution b. The instantaneous angular
acceleration as a function of time is
At time, t 3.0 s
19
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
20
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
21
• 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
22
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
23
• 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
24
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
25
• Table 8.2 shows the comparison of linear and
rotational motion with constant acceleration.

Linear motion Rotational motion

Table 8.2
26
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
27
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
28
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
29
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
30
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
31
Solution b. The number of revolutions of the
blade is c. The tangential speed of a
point is given by
32
Solution d. The magnitude of the resultant
acceleration is
33
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
34
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
35
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
36
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
37
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,

38
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
39
• 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.

40
• 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)
41
• 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
42
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
43
• 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.

44
• 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
45
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

46
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)
47
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?

48
Solution Since the beam remains at rest thus
the system in equilibrium. a. Hence b.
49
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
floor. (Given g 9.81 m s?2)
50
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?

51
Solution Since the ladder in equilibrium thus
52
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?

53
Solution Consider the ladder stills in
equilibrium thus
54
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)
55
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?

56
Solution b. The floodlight and beam remain at
rest thus i.
57
Solution b. ii. Therefore the magnitude of the
force is and its direction is given
by
58
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
59
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
60
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
61
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
62
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
• State and use torque,

63
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
64
• 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
65
• 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
66
Shape Diagram Equation

Uniform rod or long thin rod with rotation axis
through the centre of mass.
Solid Sphere
67
Shape Diagram Equation

Hollow Sphere or thin spherical shell
68
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.
69
Solution a. rotation axis about point
O, Since r1 r2 r3 r4 r thus
and the connecting rods are light therefore
70
Solution b. rotation axis along the line
AB, r1 r2 r3 r4 r0.6 m therefore
71
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.

72
• 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
73
• From the equation, the nett torque acting on the
rigid body is proportional to the bodys angular
acceleration.
• Note

is analogous to the
74
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, )
75
Solution a. The nett torque on the disc is
b. By applying the relationship between
torque and angular acceleration,
76
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.
77
Solution a. Free body diagram for
wheel, for block,
78
Solution b. For wheel, For block, By
substituting eq. (1) into eq. (2), thus
79
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)
80
Solution a. Free body diagram for
pulley, for block,
81
Solution a. By substituting eq. (1) into eq.
(2), thus b. By using the relationship
between a and ?, hence
82
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
83
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

84
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
85
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,

86
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

87
• 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
88
• 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
89
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)
90
Solution a. From the principle of
conservation of energy,
91
Solution b. The linear speed of the sphere is
given by c. By using the relationship
between v and ?, thus
92
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
93
• If the torque is constant thus
• Work-rotational kinetic energy theorem states

is analogous to the
where
94
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
95
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
96
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
97
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
98
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

99
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
100
• 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
101
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
102
Example 8.16
A 200 kg wooden disc of radius 3.00 m is rotating
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 )
103
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,
104
Solution b. The initial rotational kinetic
energy, The final rotational kinetic
energy, thus It is because
105
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
106
Solution
107
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
108
Solution b. The initial rotational kinetic
energy is given by and the
final rotational kinetic energy is
109
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)
110
Solution The moment of inertia of the pulley,
Final
Initial
111
Solution By using the principle of
conservation of energy, thus
112
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

113
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
114
Summary
Linear Motion Relationship Rotational Motion