CHAPTER 7 ROTATION OF RIGID BODY

CHAPTER 7:
Rotation of Rigid Body

1

• Motion in which an
entire object moves is
called translation

• Motion in which an
object spins is called
rotation

• An object can both
translate and rotate

2

What is meant by rigid body ?

Ö Rigid body is any real body which
has a definite size and shape and can move.

Ö In ideal cases, rigid bodies do not change shapes
& sizes.

Ö They do not bend, dent or stretch when ever any

force or forces are applied on them. 3

As the rigid body rotate, all points on the body
move in circular path.

4

Learning Outcome: 5

7.1 Rotational kinematics

a) Define:

C1

 angular displacement ()
 average angular velocity (av)
 instantaneous angular velocity ()
 average angular acceleration (av)
 instantaneous angular acceleration ().

b) Use:

C3  angular displacement ()
 average angular velocity (av)
 instantaneous angular velocity ()
 average angular acceleration (av)
 instantaneous angular acceleration ().

Learning Outcome:

7.1 Rotational kinematics

C1 c) State parameters in rotational motion with
their corresponding quantities in linear motion

d) Use parameters in rotational motion with

C3 their corresponding quantities in linear motion

= θ; = ; = ; = 2 =
2

e) Solve problem related to rotational motion
C3 with constant angular acceleration.

ω  ω0  αt θ  ω0t  1 t 2 ω2  ω02  2αθ
2

6

 an angle through Angular velocity,ω Angular acceleration,α
which a point or
line has been Average Average angular
rotated in a angular
specified velocity, av acceleration, av:
direction about a the rate of
specified axis change of the rate of change
angular of
displacement
angular velocity
Instantaneous
angular Instantaneous
angular
velocity, 
the acceleration, :

instantaneous the instantaneous
rate of change rate of change of
angular velocity
of angular
displacement 7

θ  s OR s  rθ
r

θ : angular displaceme nt
s : arc length

r :radiusof thecircle

 The S.I. unit of the angular
displacement is radian (rad).

1 revolution = 360o = 2 rad  Odreitsvhpoellrauscteuiomnniet (fnorterivsa)ndgeuglraere () and

 Figure shows a point P on a  Sign convention of angular
rotating compact disc (CD) displacement :
moves through an arc length  Positive – if the rotational
s on a circular path of radius motion is anticlockwise.
r about a fixed axis through  Negative – if the rotational
point O motion is clockwise.

8

Angular • It is a vector quantity
velocity,ω • Unit: rad s-1/ rev min-1/ rpm

ii. Average 1 rpm  2 rad s1   rad s1
angular
60 30
velocity, av
the rate of ωav  θ2  θ1  θ – Every
change of t2  t1 t part of a
rotating
angular θ2 : final angular displacement (rad) rigid
displacement θ1 : initial angular displacement (rad) body has
t : time interval the same
iii. angular
Instantaneous   limit θ  dθ velocity
t0 t dt
angular 9
velocity, 

the
instantaneous
rate of change

of angular
displacement

Angular • It is a vector quantity
acceleration,α • Unit: rad s2
• Note, if:
Average angular
-  is positive   is increasing
acceleration, av: -  is negative   is decreasing

the rate of  av  ω2  ω1  ω
change of t2  t1 t
angular velocity
ω : final angular velocity
Instantaneous 2
angular ω
1 : initial angular velocity
acceleration, :
t : timeinterval
the
instantaneous α  limit ω  dω
rate of change of t0 t dt
angular velocity

10

Linear motion Rotational motion

11

Rotational motion with uniform angular
acceleration

 Table 8.1 shows the symbols used in linear and rotational
kinematics.

Linear Quantity Rotational
motion motion
Displacement
s Initial velocity θ
Final velocity ω0
u Acceleration ω

v Time α
a
t t

Table 7.1

12

Linear motion Rotational motion

 s a  constant α  constant
o  u
 v v  u  at ω  ω0  αt
 a
s  ut  1 at2 θ  ω0t  1 αt 2
s  r 2 2
v  r
a  r v2  u2  2as ω2  ω02  2αθ

s  1 v  ut θ  1 ω  ω0 t
2
2

where  in radian13

• When a rigid body is  Point P moves in a circle of
rotates about rotation axis radius r with the tangential
O, every particle in the velocity v where its magnitude
body moves in a circle as is given by
shown
v  ds and s  rθ
dt

v  r d v  r
dt
 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 direction of the linear the radius of the circle, r is
(tangential) velocity changing depend on the
always tangent to the
circular path position of the particle. 14

• Point P on the rigid body at  dv and v  rω
dt
experiences 2 types of

acceleration: at  r d at  r

y dt v2
at r
  P but ac   r 2  v
a
ac
• resultant (linear)

x acceleration, a given by
a  at  ac
O

• The components are and its magnitude,
tangential acceleration, at 
and centripetal a at2  ac2
acceleration, ac
15

Example 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.

16

Solution : r  d  0.56  0.28 m
22

a. At time, t1 =2.2 s : 52.22 2.2

θ1  

θ1  22 rad 180   1261
π rad
θ1  22 rad 

At time, t2 =4.8 s : 54.82  4.8

θ2 

θ2  110 rad 180   6303
π rad
θ2  110 rad 

17

Solution : r  d  0.56  0.28 m
22

b. By applying thesequraθtion of arc length,

Therefore s  r  r2 1 

 0.28110  22

s  24.6 m

c. The average angular velocity in rad s1 is given by
 ωavθ
 t  2  1
t2  t1
  110  22
 4.8  2.2

ωav  33.9 rad s1 18

Solution :

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

ωav   33.9 rad  1 rev  60 s 
 1s  2 rad 1 min 

ωav  324 rev min1 OR 324 rpm

d. The instantaneous aωnguldarθvelocity as a function of time is
dt

 d 5t2  t
dt

ω  10t 1

At time, t =3.0 s :ω  103.0 1

ω  29 rad s1

19

Example 2 :

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 initial1a7ng.u0ular v0reωl.o4c08ityωis0 ω0  35.4 rad s1

a  rαand the angular acceleration of the wheels is given by
2.00  0.48α α  4.17 rad s2

20

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 1
2
θ  ω0t  αt 2

 35.45.00  1 4.175.002

θ  229 rad 2

therefore θ  229 rad  1 rev   36.5 rev
 2π rad 

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

ω  ω0  αt

 35.4  4.175.00

ω  56.3 rad s1

21

Example 3 :

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.  0.70  0.35 m,

Solution : θ  30  2π  60π rad, r 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 

22

Solution :

a. The ini1tia3l .au9ngular0rω.s3p05eeωd 0of the wheels is

ω0  39.7 rad s 1

and the final vangular sp0ereω.d3o5fωthe wheels is
9.72
ω 27.8 rad s 1

therefore ω2  ω3029.722αθ 2α60π

27.82
α  2.13 rad s2

b. The car stops thus ω  0 andω0  27.8 rad s1

Hence ω  ω0  αt

0  27.8  2.13t

t  13.1s 23

Example 4 :

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,

α  0.750  2π  1.50π rad s2

a. Given ωt =4.0ω00s,thαust ω  0.300π 1.50π4.00

ω  19.8 rad s1

24

Solution :

b. The number of revolutions of the blade is
1
θ  ω0t  2 αt 2

0.300 4.00  1 1.50 4.002

θ  41.5 rad 2

θ  41.5 rad  1 rev   6.61 rev
 2π rad 

c. The tangential speed of a point is given by

v  rω

 0.40019.8

v  7.92 m s1

25

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

a  157 m s2

26

Example 5 :

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
T 60
ω  0.11 rad s1

27

Solution :

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

hence ω  2π
3600

ω  1.74 103 rad s1

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

hence ω  2π
4.32 104

ω  1.45 104 rad s1

d. The angular acceleration in each cases is zero.

28

Example 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 : 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

29

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 trhαe point at the edge the coin is given by

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

Therefore the distance travelled by the coin is

v2  u 2  2as

 0  0.2162  2  2.28102 s

s  1.02 m

30

Exercise 1 :

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

31

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

32