CHAPTER 6: ROTATION OF RIGID BODY

Example 6.15 :

Four spheres are arranged in a rectangular shape of sides 250 cm and
120 cm as shown in Figure.

2 kg 3 kg

60 cm

A O B

60 cm

5 kg 250 cm 4 kg

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 : m2

m1  2 kg; m2  3 kg; m3  4 kg; m4  5 kg
a. rotation axis about point O, m1

r1 r2 0.6 m

O 1.25 m

r4 r3

m4 m3

Since r1= r2= r3= r4= r thus r  0.62  1.252  1.39 m

and the connecting rods are light therefore

IO  m1r12  m2r22  m3r32  m4r42

IO  r 2 m1  m2  m3  m4   1.392 2  3  4  5

O = 27.05 kg m2

Solution : m1  2 kg; m2  3 kg; m3  4 kg; m4  5 kg m2
b. rotation axis along the line AB, m1

r1 r2

A B

r4 r3

m4 m3
r1= r2= r3= r4= r=0.6 m therefore

IAB  m1r12  m2r22  m3r32  m4r42

IAB  r 2 m1  m2  m3  m4 
IAB  0.62 2  3  4  5

IAB  5.04 kg m2

Example 6.16 :

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.



O

30.0 cm

Calculate,

a. the nett torque on the disc.

b. the magnitude of angular acceleration influence by the disc.

( Use the moment of inertia, I CM  1 MR 2 )
2

Solution : R  0.30 m; M  5.00 kg

a. The nett torque on the disc is

  1  2

  RF1  RF2  R F1  F2 

  0.30 5.60 10.3

  1.41 N m

b. By applying the relationship between torque and angular

acceleration,   I    1 MR 2 
2


1.41   1 5.000.302 

2 
  6.27 rad s2

Example 6.17 :

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 : R  0.20 m; I  0.050 kg m2; P  3.0 N; m  2.0 kg
a. Free body diagram :

for wheel, for block,

റ

റ

 റ





Solution : R  0.20 m; I  0.050 kg m2; P  3.0 N; m  2.0 kg
b. For wheel,

For block, τ  Iα T  Iα (1)
R (2)
RT  Iα
P  T  ma
F  ma
By substituting eq. (1) into eq. (2), thus

P   Iα   ma and a  Rα α  4.62 rad s2
R

P   Iα   mRα
R

3.0   0.050α   2.00.20α

 0.20 

Example 6.18 : R

An object of mass 1.50 kg is suspended from a rough 1.50 kg
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 linear velocity of the object,
e. the distance travelled by the object.

Solution : റ  τ  Iα
a. Free body diagram :
 RT  Iα and α  a
for pulley, RT  I  a  R

R

for block, T  Ia (1)
R2

 റ F  ma

mg  T  ma (2)

m റ

Solution : R  0.20 m; I  0.020 kg m2; m  1.50 kg;
u  0; t  0.3 s

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

mg   Ia   ma
 R2 

1.509.81   0.020a   1.50a a  7.36 m s2

0.202

b. By using the relationship between a and , hence

a  Rα

7.36  0.20α
α  36.8 rad s2

Solution : R  0.20 m; I  0.020 kg m2; m  1.50 kg;
u  0; t  0.3 s

c. From eq. (1), thus T  Ia T  0.020 7.36  T  3.68 N
R2 0.202

d. By applying the equation of liner motion, thus

v  u  at v  0  7.360.3 v  2.21 m s1

(downwards)

e. The distance travelled by the object in 0.3 s is

s  ut  1 at2 s  0  1 7.360.32 s  0.331 m
2
2

Exercise : 70 g

1. 40 cm

80 cm

B

150 g A 150 g
80 cm

70 g

Figure 8.26 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 2.00 m s2
horizontal table is connected to a string that T1
passes over a pulley and then is fastened to a
hanging 9.00 kg object as in Figure 8.27. The T2
pulley has a radius of 0.250 m and moment of
inertia I. The block on the table is moving with a Figure 8.27
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

Learning Outcome:

6.4 Conservation of angular momentum

At the end of this chapter, students should be able to:

a) Explain and use angular momentum, L  I

b) State and use principle of conservation of angular momentum

6.4 Conservation of angular momentum

Angular momentum,
• is defined as the product of the angular velocity of a body and

its moment of inertia about the rotation axis.

OR L  I is analogous to the p  mv

where L : angular momentum

I : moment of inertia of a body
ω : angular velocity of a body

• It is a vector quantity.
• The S.I. unit of the angular momentum is

L  I

  unit of L  kg m2 s1  kg m2 s1

• The relationship between angular momentum, L with linear
momentum, p is given by

vector notation : = റ × റ = റ × m റ

magnitude form : L  rpsin θ  mvr sin θ

where

r : distance from the particle to the rotation axis

= angle between റ and റ

• Newton’s second law of motion in term of linear momentum is

σ റ = =



• hence we can write the Newton’s second law in angular form as

σ = =



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.

Principle of conservation of angular momentum
• states that a total angular momentum of a closed system about an

rotation axis is constant.
OR

I  constant

If the ෍ = 0 σ = =0

  =0 and dL  Lf - Li

Therefore

σ =σ

Example 6.18 :

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.

R ω0 

R

Before After

Calculate the angular speed of the system after the bag of sand

falling onto the disc. (treat the bag of sand as a particle)

(Use the moment of inertia of disc is 1 MR2)

2

Solution : R  3.00 m; ω0  4.0 rad s1; mw  200 kg; mb  50 kg

The moment of inertia of the disc,

Iw  1 mw R 2  1 2003.002
2 2

Iw  900 kg m2

The moment of inertia of the bag of sand,

Ib  4m5b0Rk2gm5203.002
Ib 

By applying the principle of conservation of angular momentum,



 Li   Lf
Iwω0  Iw  Ib ω

9004.0  900  450ω

ω  2.67 rad s1

Example 6.19 :

A raw egg and a hard-boiled egg are rotating about the same axis
of rotation with the same initial angular velocity. Explain which egg
will rotate longer.
Solution :
The answer is hard-boiled egg.

Solution :
Reason
Raw egg :
When the egg spins, its yolk being denser moves away from the axis of
rotation and then the moment of inertia of the egg increases because of

I  mr2

From the principle of conservation of angular momentum,

I  constant

If the I is increases hence its angular velocity,  will decreases.

Hard-boiled egg :
The position of the yolk of a hard-boiled egg is fixed. When the egg is
rotated, its moment of inertia does not increase and then its angular velocity
is constant. Therefore the egg continues to spin.

Example 6.20 :

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. Determine the new distance of
each mass from the rotation axis.

Solution :

2.95 rev 2 rad = 18.54 rad s−1
0 = 1 s 1 rev

3.54 rev 2 rad = 22.24 rad s−1
= 1 s 1 rev

Solution : m  10..27559kgm;;ωω0  18.5 rad ss11;; I ss  5.43 kg m2;
rb   22.2 rad

0 

rb rb
m m

ra ra

Before After

Solution : m  10..27559kgm; ;ωω0  18.5 rad s1; I ss  5.43 kg m2;
rb   22.2 rad s1;

The moment of inertia of the system initially is

 Ii

Ii
Ii  Iss  Im  II5ssss.432mmr2rbb212.2m5 rb02.759 2

  6.87 kg m2

The moment of inertia of the system finally is
I5ss.432 m2ra12.25
If  ra 2
If 5.43  2.5ra2



By using the principle of conservation of angular momentum, thus

 Li  Lf

 Iiω0  I f ω

6.8718.5  5.43  2.5ra2 22.2

ra  0.344 m

Exercise :

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),

b. about its orbit around the Sun (treat the Earth as a particle

orbiting the Sun).
Given the Earth’s 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 1 ML2 )

ANS. : 4.71 x 106 kg1m2 2 s1

Summary:

Linear Motion Relationship Rotational Motion

v  ds v  r   d
dt
a  r dt
a  dv
dt n   d

m I  miri2 dt
i 1
 F  ma I
  rF sin
p  mv   I
L  rp sin
W  Fs L  I
129
P  Fv W  
P  

THE END.

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TOPIC 7 :
Simple harmonics motion & waves