CHAPTER 6: ROTATION OF RIGID BODY

CHAPTER 6

ROTATION OF RIGID
BODY

KMM

TOPIC 6:
ROTATION
OF RIGID

BODY

1

TOPIC 6:

Rotation of rigid body

(8.5 Hours)

6.1 Rotational kinematics
6.2 Equilibrium of a uniform rigid body
6.3 Rotational dynamics
6.4 Conservation of angular momentum

Learning Outcome:

6.1 Rotational kinematics

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

a) Define and use:

i) angular displacement,  ;

ii) average angular velocity, av ;
iii) instantaneous angular velocity,  ;

iv) average angular acceleration, av and
v) instantaneous angular acceleration,  .

b) Analyse parameters in rotational motion with their

corresponding quantities in linear motion:

s  rθ, v  r, at  r , ac  r 2  v2
r

Learning Outcome:

6.1 Rotational kinematics

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

c) Solve problem related to rotational motion with constant angular
acceleration:

ω  ω0  αt θ  ω0t  1 t 2 ω2  ω02  2αθ   1 o   t
2 2

4

OVERVIEW

Rotation of a rigid body

Rotational Equilibrium of Rotational Conservation
kinematics a uniform dynamics of angular
rigid body momentum
Rotational Moment of
linear Torque inertia Angular
momentum
relationship  rF   I
Principle of
Rotation Conditions for conservation
motion with Equilibrium
of angular
uniform F 0 momentum
angular   0
acceleration 5

6.1 ROTATIONAL KINEMATICS Figure 6.1 (a)

i. 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 6.1(a) 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.

From Figure 6.1 (a) Figure 6.1 (a)

θ  s OR s  rθ

r

where θ : angle (angular displacement) in radian
s : arc length
r : radius of the circle

Others unit for angular displacement is degree () and revolution (rev).

Conversion factor : 1rev  2π rad  360

Sign convention of angular displacement :

Positive – if the rotational motion is anticlockwise.

Negative – if the rotational motion is clockwise.

ii. Average angular velocity, av

• is defined as the rate of change of angular displacement.

• Equation : θ2  θ1 θ
t2  t1 t
ωav  

where θθ21 :final angular displacement in radian
:initial angular displacement in radian

t : time interval

iii. Instantaneous angular velocity, 

is defined as the instantaneous rate of change of angular

displacement.   limit θ  dθ
Equation : t0 t dt

• 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:
1 rpm  2 rad s1   rad s1
60 30

• 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 6.1 (b) and 6.1 (c) show the right hand grip rule for
determining the direction of the angular velocity.

Figure 6.1 (b) Figure 6.1 (c)

iv. Average angular acceleration, av

• is defined as the rate of change of angular velocity.

• Equation : ω2  ω1 ω
t2  t1 t
 av  

where ωω21 : final angular ve locity
: initial angular ve locity

t : time interval

v. Instantaneous angular acceleration, 

• is defined as the instantaneous rate of change of angular velocity.

• Equation : α  limit ω  dω
t0 t dt

• 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 6.1 (d).

Figure 6.1 (d)

Direction of the angular acceleration

• If the rotation is speeding up,  • If the rotation is slowing down,
and  in the same direction as
shown in Figure 6.1 (d).  and  have the opposite
direction as shown in Figure

6.1 (e).

റ



Figure 6.1 (d) Figure 6.1 (e)

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 6.1 (f).

y

റ Figure 6.1 (f)

P x

rs


O

• Point P moves in a circle of radius r with the tangential velocity v where

its magnitude is given by y

v  ds and s  rθ റ
dt
P
v  r d
r s
dt
 x
v  r
O

• 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.

Relationship between tangential acceleration, at and angular

acceleration, 

• If the rigid body is gaining the angular y
speed then the tangential velocity of a
particle also increasing thus two റ
component of acceleration are റ P
occurred as shown in Figure 6.1 (g).
റ

x

O

Figure 6.1 (g)

• The components are tangential acceleration, at and centripetal
acceleration, ac given by
dv
at  dt and v  rω

at  r d at  r

dt

but ac  v2  r 2  v
r

• The vector sum of centripetal and tangential acceleration of a

particle in a rotating body is resultant (linear) acceleration, a given by

റ = റ + റ Vector form

and its magnitude, റ = 2 + 2

Rotational motion with uniform angular acceleration

• Table 6.1 shows the symbols used in linear and rotational kinematics.

Linear Motion Rotational Motion

Symbol Quantity Relationship Symbol Quantity

s Displacement srθ θ Angular displacement
u Initial velocity ω0 Initial angular velocity
v Final velocity u  r o
a Acceleration vr ω Final angular velocity
t Time ar
α Angular acceleraton
- t Time
Table 6.1

• Table 6.2 shows the analogy of linear motion with rotational
motion parameters.

Linear motion Rotational motion

a  constant α  constant

v  u  at ω  ω0  αt

s  ut  1 at2 θ  ω0t  1 αt 2
2 2

v2  u2  2as ω2  ω02  2αθ

s  1 v  ut θ  1 ω  ω0 t
2
2

where  in radian. Table 6.2

SUMMARY
6.1 ROTATIONAL KINEMATICS

PARAMETERS IN ROTATIONAL MOTION

Physical Definition Equation Unit
Quantity

Angular The angle turned by the θ rad
displacement, line joining a point on
θ an object from the axis
of rotation of the object.

Average The rate of change of θ2  θ1 θ
t2  t1 t
angular angular displacement ωav   rads-1

velocity,av

Instantaneous The instantaneous rate   limit θ  dθ rads-1
t0 t dt
angular of change of angular

velocity, displacement.

Average ω2  ω1 ω
t2  t1 t
angular The rate of change of  av   rads-2
rads-2
acceleration, angular velocity α  limit ω  dω
av t0 t dt

Instantaneous The instantaneous rate

angular of change of angular

acceleration, velocity

ANALOGY OF LINEAR MOTION WITH ROTATIONAL MOTION
PARAMETERS.

Quantities Equations
Relationship
Linear Motion Rotational Linear Motion Rotational
Displacement, s  r Motion
Angular s  1 v ut Motion
s v  rω
Displacement, 2 θ  1  ω  ω0  t
Velocity, at  r 2
θ s  ut  1 at2 1
v 2 θ  ω0t  2 αt 2

Acceleration, Angular v  u  at ω  ω0  αt
v2  u2  2as ω2  ω02  2αθ
a velocity,

ω

Angular a  vu α  ω  ω0
t t
accelαeration,

PROBLEM SOLVING
EXAMPLE

Example 6.1 :

The angular displacement, of the wheel is given by

θ  5t 2  t

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 : r  d  0.56  0.28 m
22

a. At time, t1 =2.2 s :

θ1  52.22  2.2

θ1  22 rad 180∘ = 1260.51∘
π rad
θ1 = 22 rad

At time, t2 =4.8 s : θ2  54.82  4.8 = 6302.54∘

θ2  110 rad 180∘

θ2 = 110 rad π rad

Solution : r  d  0.56  0.28 m
22

b. By applying the equation of arc length,

s  rθ

s  r  r2 1

 0.28110  22  24.64 m

c. The average angular velocity in rad s1 is given by

ωav  θ  2  1 
t t2  t1 

 110  22  33.84 rad s-1
4.8  2.2

Solution :
c. and the average angular velocity in rev min1 is

ωav = 33.84 rad 1 rev 60 s

1s 2 rad 1 min

ωav = 323.15 rev min−1 = 323.15 rpm

d. The instantaneous angular velocity as a function of time is

 ω  dθ  d 5t 2  t
dt dt

At time, t =3.0 s : ω  10t 1

ω  103.0 1  29 rad s1

Example 6.2 :

A diver makes 2.5 revolutions on the way down from a 10 m high
platform to the water. Assuming zero initial vertical velocity, calculate
the diver’s average angular (rotational) velocity during a dive.
(Given g = 9.81 m s2)
Solution :

u0

θ0  0

10 m s  10 m

θ1  2.5 rev

water

Solution :

From the diagram, θ1  2.5  2π  5π rad
s  10 m

Thus s  ut  1 gt 2

10  0  129.81t 2 t  1.43 s

2

Therefore the diver’s average angular velocity is

ωav  θ1  θ0
t
 5π  0
1.43 ωav  11.0 rad s1

Example 6.3 :

The instantaneous angular velocity,  of the flywheel is given by

ω  8t3  t 2

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.

Solution : ω1 = 80.34 rad s−1
a. At time, t1 =2.2 s :

ω1 = 8 2.2 3 − 2.2 2

At time, t2 =4.8 s : ω2 = 861.70 rad s−1

ω2  84.83  4.82

Therefore the average angular acceleration is

αav  ω2  ω1 861.70 − 80.34 αav = 300.52 rad s−
t2  t1 = 4.8 − 2.2

Solution :
b. The instantaneous angular acceleration as a function of time is

 α  dω  d 8t3  t 2 α  24t 2  2t
dt dt

At time, t =3.0 s : α  210 rad s2

α  243.02  23.0

Exercise

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 θ  5.00 10.0t  2.00t 2
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

Example 6.4 :

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 : u  17.0 m s1, r  0.48 m, a  2.00 m s2 , t  5.00 s

a. The initial angular velocity is u  0r.ω480ω0

17.0  ω0 = 35.42 rad s−1

and the angular acceleration of the wheels is given by

a  rα

2.00  0.48α α  4.17 rad s2

Solution : u  17.0 m s1, r  0.48 m, a  2.00 m s2 , t  5.00 s

a. By applying the equation of rotational motion with constant

angular acceleration, thus

θ  ω0t  1 αt 2 = 35.42 5.00 1 5.00 2
2 + 2 4.17

therefore θ = 229.23 rad
1 rev

θ = 229.23 rad 2π rad = 36.48 rev

b. The angular speed of the wheels after 5.00 s is

ω  ω0  αt = 35.42 + 4.17 5.00

ω = 56.27 rad s−

Example 6.5 :

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 : θ  30  2π  60π rad, r  0.70  0.35 m,
2

u  50.0 km  103 m  1h s   13.9 m s1,
1h 1 km 3600 

v  35.0 km  103 m  1h s   9.72 m s 1
1h 1 km 3600 

Solution :

a. The initial angular sp1e3e.d9uof t0hr.eω3w05hωe0els is ω0 = 39.71rad s−1

and the final angularvspereωd of the wheels is

9.72  0.35ω ω = 27.77 rad s−1

ω2  ω02  2αθ

27.77 2 = 39.71 2 + 2α 60π

= −2.14 rad s−

b. The car stops thus ω = 0 and ω0 = 27.77 rad s−1

Hence ω  ω0  αt

0 = 27.77 + −2.14
= 12.98 s

Example 6.6 :

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 : r  0.400 m, ω0  0.150  2π  0.300π rad s1,
a. Given t =4.00 s, thus α  0.750  2π  1.50π rad s2

ω  ω0  αt ω  0.300π  1.50π4.00 ω = 19.79 rad s−1

Solution :
b. The number of revolutions of the blade is

θ  ω0t  1 αt 2  0.300 4.00 1 1.50 4.002
2
2

θ = 41.47rad

1 rev
θ = 41.47 rad 2π rad = 6.60 rev

c. The tangential speed of a point is given by

v  rω = 0.400 19.79

v  7.92 m s1

Solution :

d. The magnitude of the resultant acceleration is

a  ac2  at 2

  v2 2  rα 2
r

  7.922 2  0.400 1.50π2

0.400

= 156.83 m s−2

Example 8.7 :

Calculate the angular velocity of
a. the second-hand,
b. the minute-hand and
c. the hour-hand,

of a clock. State in rad s-1.
d. What is the angular acceleration in each case?

Solution :

a. The period of second-hand of the clock is T = 60 s, hence

ω  2π ω  2π ω = 0.105 rad s−1
T 60

Solution :

b. The period of minute-hand of the clock is T = 60 min = 3600 s,

hence ω  2π ω  1.74 103 rad s1
3600

c. The period of hour-hand of the clock is T = 12 h = 4.32 104 s,

hence ω  1.45 104 rad s1

ω  2π
4.32 104

d. The angular acceleration in each cases is zero.

Example 6.8 :

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 : ω0  18 rad s1 ω  0 rad s1

d  2.40 102 m α  1.90 rad s2

s

The radius of the coin is r  d  1.20 102 m
2

Solution :

The initial speed of the point at the edge the coin is

u  rω0  u  1.20 102 18 u  0.216 m s1

and the final speed is v  0 m s1

The linear acceleration of the point at the edge the coin is given by

a  rα  a  1.20 102 1.90 a  2.28 102 m s2

Therefore the distance travelled by the coin is s  1.02 m

 v2  u2  2as 0  0.2162  2  2.28 102 s

Exercise

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

5. A bicycle wheel is being tested at a repair shop. The angular velocity
of the wheel is 4.00 rad s-1 at time t = 0 , and its angular acceleration
is constant and equal 1.20 rad s-2. A spoke OP on the wheel
coincides with the +x-axis at time t = 0 as shown in below.

y

Px

O

a. What is the wheel’s angular velocity at t = 3.00 s?
b. What angle in degree does the spoke OP make with the positive

x-axis at this time?
ANS. : 0.40 rad s1; 18

Learning Outcome:

6.2 Equlibirium of a uniform rigid body

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

a) Stater the physical meaning of cross (vector) product for torque,

F sin θ

b) Define and apply torque 

c) State conditions for equilibrium of rigid body, Fx  0, Fy  0

d) Solve problem related to equilibirium of a uniform rigid body

*limit to 5 forces