CHAPTER 5: CIRCULAR MOTION

CHAPTER 5

CIRCULAR MOTION

KMM

Physics SP015 Topic 5

TOPIC 5:
CIRCULAR MOTION

1

Physics SP015 Topic 5

Overview: Circular Motion

Parameters in Uniform Centripetal force
circular motion circular motion

Define & Use Describe = 2 Explain

Angular Uniform Centripetal #! r %
displacement θ = circular acceleration " = $ = =
motion
= 2

Period = 1 = ω Centripetal
force
Frequency
ω = 2 = 2 Convert unit Solve " = % = % =
Angular 1 rev = 2π rad = 360∘ problems
velocity

2

Physics SP015 Topic 5

Learning Outcome:
5.1 Parameters in circular motion

At the end of this topic, students should be able to:
a) Define and use :

i. angular displacement, q
ii. period, T
iii. frequency, f
iv. angular velocity, ω

3

Physics SP015 Topic 5

a) Define and use :

5.1 Parameters in circular motion

5.1.1 Angular displacement, q
• 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 : radian (rad)

• From Figure 5.1, thus

θ = OR = θ

r s
where θ ∶ angle (angular displacement) in radian
θ ∶ length of circular arc

O

: radius of the circular path

Figure 5.1 4

Physics SP015 Topic 5

5.1.2 Period, T
• is a time taken for one complete revolution (cycle/rotation).
• Unit of the period: second (s).

5.1.3 Frequency, f

• is a number of revolutions (cycles/rotations) completed in one
second.

• Unit of the frequency: hertz (Hz) or s-1.

• Equation : 1

=

• For an object makes one complete circular motion:

• distance travelled = 2pr (circumference of the circle),

• time interval/taken = one period T

5

Physics SP015 Topic 5

5.1.4 Angular velocity, w
• is defined as the rate of change of angular displacement.

• Equation

ωav = θ2 - θ1 = Dθ
t2 - t1 Dt

where θ2 : final angular displacement in radian
θ1 :initial angular displacement in radian

Dt : time interval

• It is a vector quantity.
• Unit of the angular velocity (angular frequency) is radian per second

(rad s-1)
• Others unit is revolution per minute (rev min-1 or rpm)

6

Physics SP015 Topic 5

Learning Outcome:
5.2 Uniform circular motion

At the end of this topic, students should be able to:
a) Describe uniform circular motion.
b) Convert units between degrees, radian and revolution or rotation.

7

Physics SP015 Topic 5

a) Describe uniform circular motion.

5.2 Uniform circular motion

• is a motion in a circle (circular arc) at a constant speed, v.

5.2.1 Linear (tangential) velocity, ⃗

• It is directed tangentially to the circular path and always perpendicular
to the radius of the circular path. v!
• In uniform circular motion, the magnitude of v!

the linear velocity (speed) of an object is rr
constant but the direction is continually

changing causes the change in velocity. O

• Unit of the tangential (linear) velocity: m s-1. r v!

Figure 5.2

8

Physics SP015 Topic 5

• For an object makes one complete circular motion:

• distance travelled = 2pr (circumference of the circle),

• time interval/taken = one period T

• The speed of the body v is

v = distance travelled thus v = 2pr OR v = 2prf
time interval T

If ω = 2p = 2pf therefore v = rω
T
where

ω: angular velocity (angular frequency)

r : radius of the circular path

9

Physics SP015 Topic 5

a) Convert units between degrees, radian and revolution or rotation.
5.2.2 Unit conversion

• Others unit for angular displacement,q is degree (°) and

revolution (rev) or rotation.

• conversion factor 1 rev = 360o = 2p rad

üRevolution to degrees 1 rev = 360o

üDegrees to radian 360o = 2p rad
180o = p rad

üRadian to revolution 2p rad = 1 revolution or 1 rotation

10

Physics SP015 Topic 5

EXAMPLE 1

An object undergoes circular motion with uniform angular speed 100 rpm. Calculate :
(a) the period, T
(b) the frequency of revolution, f.

SOLUTION

Given : ω = 100 rpm 100 rev 100(2p )
1 min
Convert to rad s-1 : ω = = 60 = 10.47 rad s -1

(a) 2p 2p (b) f = 1 = 1 = 1.67 Hz
T 0.60
T = ω = 10.47 = 0.60 s

11

Physics SP015 Topic 5

Exercise 12

1. Convert 50o to radians.
2. Convert 11 rad to revolutions.
3. Convert 71.5 rpm to rad/s.
4. Prove that 415o is equal to 7.25 rad.

Physics SP015 Topic 5

Learning Outcome:
5.3 Centripetal force

At the end of this topic, students should be able to:
a) Explain centripetal acceleration and centripetal force.

! = "! = r $ = and ! = " = " =
#

b) Solve problems related to centripetal force for uniform circular
motion cases: horizontal circular motion, vertical circular motion
and conical pendulum (exclude banked curve).

13

Physics SP015 Topic 5

a) Explain centripetal acceleration and centripetal force.

5.3 Centripetal force

5.3.1 Centripetal (radial) acceleration, ⃗ 9 or ⃗ :

¡ is an acceleration of an object moving in circular path whose
direction is towards the centre of the circular path and whose

magnitude is equal to the square of the speed divided by the
radius.

ac = v2 OR ac = rw 2 = vw
r

where avc : centripetal acceleration
: linear(tangential) velocity
r : radius of circular path

ω : angular velocity (angular frequency)

14

Physics SP015 Topic 5

• The direction of centripetal (radial) acceleration is always directed toward
the centre of the circle and perpendicular to the linear (tangential)
velocity as shown in Figure 5.3.

a!c a!c a!c

a!c a!c
a!c

Figure 5.3

• For uniform circular motion, the magnitude of the centripetal acceleration
always constant but its direction continuously changes as the object
moves around the circular path.
15

Physics SP015 Topic 5

5.3.2 Centripetal force, !

• is a nett force acting on a body causing it to move in a circular
path of magnitude

Fc = mv2 = mrw 2 = mvw
r

where Fc : centripetal force

v : linear(tangential) velocity

r : radius of circular path

ω : angular velocity (angular frequency)
m : mass of a body

and it always directed towards the centre of the circular path.

16

Physics SP015 Topic 5

• Its direction is in the same direction of the centripetal acceleration.

v! ! a!c v!
Fc
!
a!c ! Fc

Fc a!c
v!

Figure 5.4

17

Physics SP015 Topic 5

• If the centripetal force suddenly stops to act on a body in the circular
motion, the body flies off in a straight line with the constant
tangential (linear).
a!c v!
v! ! !
Fc Fc
! !
a!c a!c aF!cc = 0 aF!cc =0
v!
! v!= 0 =v! 0
Fc

Figure 5.5

• In uniform circular motion, the nett force on the system is centripetal
force.

• The work done by the centripetal force is zero but the kinetic energy
of the body is not zero and given by
K 1 mv2 1 mr 2w2
= 2 = 2

18

Physics SP015 Topic 5

b) Solve problems related to centripetal force for uniform circular
motion cases: horizontal circular motion, vertical circular motion
and conical pendulum (exclude banked curve).

5.3.3 Examples of uniform circular motion

• Motion on a horizontal circle.
• Motion rounds a curve on a flat (unbanked) track (for car,

motorcycle, bicycle, etc…).
• Motion in a vertical circle.
• Conical pendulum.

19

Physics SP015 Topic 5

Motion in a horizontal circle

Example 5.1 :

A ball of mass 150 g is attached to one end of a string 1.10 m long. The
ball makes 2.00 revolution per second in a horizontal circle on a smooth
table.

a. Sketch the free body diagram for the ball.

b. Determine

i. the centripetal acceleration of the ball,

ii. the magnitude of the tension in the string.

Solution : m = 0.150 kg; l = r = 1.10 m; f = 2.00 Hz N! !a!c
T
a. The free body diagram for the ball :

v!

r mg! 20

Physics SP015 Topic 5

Solution : m = 0.150 kg; l = r = 1.10 m; f = 2.00 Hz

b. i. v = 2pr = 2prf
T
v = 2p(1.10)(2.00) v = 13.8 m s-1

Therefore the centripetal acceleration is

ac = v2 ac = (13.8)2
r
ac = 1.10 s -2

173 m

(towards the centre of the horizontal circle)

ii. The tension in the string enables the ball to move in a circle. 21

å Fx = Fc = mac
T = mac
T = (0.150)(173)
T = 26.0 N

Physics SP015 Topic 5

Motion rounds a curve on a flat (unbanked) track (for car, motorcycle,
bicycle, etc…)

Example 5.2 :

A car of mass 2000 kg rounds a circular turn of radius 20 m. The road is flat
and the coefficient of friction between tires and the road is 0.70.

a. Sketch a free body diagram of the car.

b. Determine the maximum car’s speed without skidding.

Solution : m = 2000 kg; r = 20 m; μ = 0.70

a. The free body diagram of the car : !
a!c N

f!

Centre of mg! 22
circle

Physics SP015 Topic 5

Solution : m = 2000 kg; r = 20 m; μ = 0.70

b. From the diagram in (a),

åy-component : Fy = 0 N = mg

x-component : The centripetal force is provided by the frictional force

between the wheel (4 tyres) and the road.
mv 2
åμ Fx = mrv 2
f = mrv 2
mg =
r
v = μrg

v = (0.70)(20)(9.81)

v = 11.7 m s-1

23

Physics SP015 Topic 5

Motion in a vertical circle B
Example 5.3 :
v!

3.00 m

v!

Figure 5.6 A

A small remote control car with mass 1.20 kg moves at a constant speed
of v = 15.0 m s-1 in a vertical circle track of radius 3.00 m as shown in
Figure 5.6. Determine the magnitude of the reaction force exerted on the
car by the track at

a. point A,
b. point B.

24

Physics SP015 Topic 5

Solution : m = 1.20 kg; r = 3.00 m; v = 15.0 m s-1 25

a. The free body diagram of the car at point A :

!
a!c N A

mg!

åF = mv 2 N A - mg = mv2
r r
(1.20)(15.0)2
N A - (1.20)(9.81) =
3.00

N A = 102 N

Physics SP015 Topic 5

Solution : m = 1.20 kg; r = 3.00 m; v = 15.0 m s-1 26

b. The free body diagram of the car at point B :

!
a!cN B mg!

åF = mv 2 NB + mg = mv 2
r r
(1.20)(15.0)2
N B + (1.20)(9.81) =
3.00

N B = 78.2 N

Physics SP015 Topic 5

Example 5.4 :

v

v

Figure 5.7

A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m at
constant speed, v as shown in Figure 5.7. If the time taken to makes one
rotation is 10 s and the mass of the rider is 60 kg, Calculate the normal force
exerted on the rider

a. at the top of the circle,

b. at the bottom of the circle.

27

Physics SP015 Topic 5

Solution : m = 60 kg; r = 8 m; T = 10 s 28

a. The constant speed of the rider is

v = 2pr v = 2π(8)
T
10
v = 5.03 m s-1

The free body dia!gram of the rider at the top of the circle :
Nt mv 2
a!c mg! åF = r

mg - Nt = mv 2
r
(60)(5.03)2
(60)(9.81) - Nt =
8

Nt = 399 N

Physics SP015 Topic 5

Solution : m = 60 kg; r = 8 m; T = 10 s

b. The free body diagram of the rider at the bottom of the circle :

åF = mv 2
r
mv 2
Nb - mg = r

a!c ! Nb - (60)(9.81) = (60)(5.03)2
Nb
8

mg! Nb = 778 N

29

Physics SP015 Topic 5

Example 5.5 : 3.0 m s-1 A

3.0 m s-1
D

Figure 5.8 E 3.0 m s-1

A sphere of mass 5.0 kg is tied to an inelastic string. It moves in a vertical
circle of radius 55 cm at a constant speed of 3.0 m s-1 as shown in Figure
5.8. By the aid of the free body diagram, determine the tension in the
string at points A, D and E.

30

Physics SP015 Topic 5

Solution : m = 5.0 kg; r = 0.55 m; v = 3.0 m s-1

The free body diagram of the sphere at :

Point A, A mv 2 mv2
r r
! åF = TA + mg =
TA
mg! TA + (5.0)(9.81) = (5.0)(3.0)2
a!c
0.55

TA = 32.8 N

Point D,

a!c ! TD = mv2
TD r
(5.0)(3.0)2
D TD = TD = 81.8 N
0.55
mg!
31

Physics SP015 Topic 5

Solution : m = 5.0 kg; r = 0.55 m; v = 3.0 m s-1

The free body diagram of the sphere at :
Point E,

a!c ! TE - mg = mv2
TE r
(5.0)(3.0)2
TE - (5.0)(9.81) =
0.55
E
TE = 131 N
mg!

Caution :

¡ For vertical uniform circular motion only,
l the normal force or tension is maximum at the bottom of the circle.
l the normal force or tension is minimum at the top of the circle.

32

Physics SP015 Topic 5

Conical Pendulum
Example 5.6 :
Figure 5.9 shows a conical pendulum with a bob of mass
80.0 kg on a 10.0 m long string making an angle of 5.00° to
the vertical.
a. Sketch a free body diagram of the bob.
b. Determine

i. the tension in the string,
ii. the speed and the period of the bob,
iii. the radial acceleration of the bob.

Figure 5.9

33

Physics SP015 Topic 5

Solution : m = 80.0 kg; l = 10.0 m; θ = 5.00! 34

a. The free body diagram of the bob :

a!c ! θ T cosθ
T

T sin θ
mg!

å Fy = 0 T = 788 N

b. i. T cosθ = mg

T cos5.00! = (80.0)9.81

Physics SP015 Topic 5

Solution : m = 80.0 kg; l = 10.0 m; θ = 5.00!

b. ii.
The centripetal force is contributed by the
horizontal component of the tension.

sin θ = r l åT Fx = Fmc v 2
l mrv 2
T sin θ = l sin θ
r = l sin θ
sin θ =

r v= Tl sin 2 θ
m

( )v =
(788)(10.0) sin 5.00! 2

80.0

v = 0.865 m s-1 35

Physics SP015 Topic 5

Solution : m = 80.0 kg; l = 10.0 m; θ = 5.00!

b. ii. and the period of the bob is given by

v = 2pr ( )0.865
T 2p 10.0 sin 5.00!
2p(l sin θ ) = T
v =
T T = 6.33 s

iii. ar = v2 ar = v2
r l sin θ

ar = (0.865)2

10.0 sin 5.00!

ar = 0.859 m s-2

(towards the centre of the horizontal circle)

36

Physics SP015 Topic 5

Exercise 5.1 :

1. A ball of mass 0.35 kg is attached to the end of a horizontal cord
and is rotated in a circle of radius 1.0 m on a frictionless
horizontal surface. If the cord will break when the tension in it
exceeds 80 N, determine

a. the maximum speed of the ball,

b. the minimum period of the ball.

ANS. : 15.1 m s-1; 0.416 s m

2. A small mass, m is set on the surface θ
of a sphere as shown in Figure 5.10. If
O
the coefficient of static friction is µs =
0.60, calculate the angle q would the

mass start sliding.

ANS. : 31° Figure 5.10

37

Physics SP015 Topic 5

3. A ball of mass 1.34 kg is connected by means of two Figure 5.11
massless string to a vertical rotating rod as shown in
Figure 5.11. The strings are tied to the rod and are
taut. The tension in the upper string is 35 N.
a. Sketch a free body diagram for the ball.
b. Calculate
i. the magnitude of the tension in the lower string,
ii. the nett force on the ball,
iii. the speed of the ball.

ANS. : 8.74 N; 37.9 N (radially inward); 6.45 m s-1

38

Physics SP015 Topic 5

THE END. 39

Next Topic…

TOPIC 6 :
Rotation of Rigid Body