Note Chap 6 SP015 (Circular motion) Ema Azura

Chapter 06 Physics

CHAPTER 6:
Circular motion

(F2F:3 Hours)

1

Chapter 06 Physics

Overview:

Circular motion

Definition Uniform circular Centripetal force
motion

2

Chapter 06 Physics
Learning Outcome:

6.1 Uniform circular motion (0.5 hour)

At the end of this chapter, students should be able to:
 Describe uniform circular motion.
 Convert units between degrees, radian and

revolution or rotation.

3

Chapter 06 Physics

6.1 Uniform circular motion

At the end of this chapter, students should be able to:
o Describe uniform circular.
o Convert units between degrees, radian and revolution or

rotation.

 is a motion in a circle (circular arc) at a constant speed.

 The length of a circular arc, s is
given by

r s s = rθ

θ

O where

θ : angle which the arc subtends

to the centre of the circle in radian

r : radius of the circular path

Figure 6.1

4

Chapter 06 Physics
v6.1.1 Linear (tangential) velocity ,

 It is directed tangentially to the circular path and always

perpendicular to the radius of the circular path.

 
v v

rr

O

Figure 6.2 r
v

 In uniform circular motion, the magnitude of the linear velocity

(speed) of an object is constant but the direction is

continually changing causes the change in velocity.

 Unit of the tangential (linear) velocity: m s−1.

5

Chapter 06 Physics

 Period, T

⚫ is a time taken for one complete revolution
(cycle/rotation).

⚫ Unit of the period: second (s).

 Frequency, f

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

⚫ Unit of the frequency: hertz (Hz) or s−1.

⚫ Equation : f =1
T

 For an object makes one complete circular motion:

⚫ distance travelled = 2r (circumference of the circle),

⚫ time interval/taken = one period T

6

Chapter 06 Physics

 The speed of the body v is

v = distance travelled
time interval

v = 2r OR v = 2rf
T

If ω = 2 = 2f therefore v = rω

T

wωhe: raengular velocity (angular frequency)

r : radius of the circular path

 Unit of angular velocity (angular frequency):

rad s−1 (radian per second).
 rad = 180

2 rad = 360 7

Chapter 06 Physics
Learning Outcome:

6.2 Centripetal force (2.5 hours)

At the end of this chapter, students should be able to:
 Define centripetal acceleration.
 Solve problems related to centripetal force for

uniform circular motion cases: conical pendulum,
horizontal circular motion and vertical circular
motion (exclude banked curve).

8

Chapter 06 Physics

6.2 Centripetal force 

6.2.1 Centripetal (radial) acceleration, ac or ar

At the end of this chapter, students should be able to:
o Define centripetal acceleration.

 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 = r 2 = v
r

where ac : centripetal acceleration 9
v : linear(tangential) velocity
r : radius of circular path

ω : angular velocity (angular frequency)

Chapter 06 Physics

 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 6.3.

  
ac ac ac


 ac
ac 

ac

Figure 6.3

 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.

10

Chapter 06 Physics

6.2.2 Centripetal force

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

Fc = mv2 = mr 2 = mv
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.

11

Chapter 06 Physics

 Its direction is in the same direction of the centripetal

acceleration.   
Fc ac v

v  
ac 
Fc
Fc

ac

v

Figure 6.4

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Chapter 06 Physics

 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). 
v
 
 Fc ac  Simulation 6.1
v Fc
  
Figure 6.5 ac 
ac Fc = 0 Fc = 0
 v ac = 0 ac = 0
Fc
v v
 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 mr22 13

22

Chapter 06 Physics

6.2.1 Examples of uniform circular motion
At the end of this chapter, students should be able to:
o Solve problems related to centripetal force for uniform

circular motion cases: conical pendulum, horizontal

circular motion and vertical circular motion (exclude

banked curve).

Conical Pendulum

Example 6.1 :

Figure 6.6 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, Figure 6.6
iii. the radial acceleration of the bob.
14

Chapter 06 Physics

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

a. The free body diagram of the bob :

  θ T cosθ
ac T

T sin θ


mg

b. i. Fy = 0 T = 788 N
T cosθ = mg

T cos5.00 = (80.0)9.81

15

Chapter 06 Physics

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

sin θ = r tension. Fx = Fmc v 2
l
T sin θ = r
r = l sin θ mv
l 2

T sin θ = l 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

16

Chapter 06 Physics

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

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

v = 2r ( )0.865 = 2 10.0 sin 5.00
T T
T = 6.33 s
v = 2(l sin θ )

T

iii. ar = v2 ar = v2 θ
r l sin

ar = (0.865)2

10.0 sin 5.00

ar = 0.859 m s−2

(towards the centre of the horizontal circle)

17

Chapter 06 Physics

Motion rounds a curve on a flat (unbanked) track (for car,

motorcycle, bicycle, etc…)

Example 6.2 : Picture 6.1

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 
N
a. The free body diagram of the car : 
ac



Centre of f
circle

mg

18

Chapter 06 Physics

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

b. From the diagram in (a),

y-component : Fy = 0 N = mg

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

force between the wheel (4 tyres) and the road.

 Fx = mv 2
=
f r 2
mv

r 2
mv
μmg =
r
v = μrg

v = (0.70)(20)(9.81)

v = 11.7 m s−1

19

Chapter 06 Physics

Motion in a horizontal circle

Example 6.3 :

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

a. The free body diagram for the ball :
N ac
v T

r
mg 20

Chapter 06 Physics

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

b. i. v = 2r = 2rf v = 13.8 m s−1
T

v = 2(1.10)(2.00)

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.

 Fx = Fc = mac

T = mac

T = (0.150)(173)

T = 26.0 N

21

Chapter 06 Physics

Motion in a vertical circle B
Example 6.4 :

v

3.00 m


v

Figure 6.7

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 6.7. Determine the

magnitude of the reaction force exerted on the car by the track

at

a. point A,

b. point B. 22

Chapter 06 Physics

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

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

 
ac NA


mg

 F = 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

23

Chapter 06 Physics

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

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

 
NB mg
ac

 F = mv 2 NB + mg = mv 2
r r
(1.20)(15.0)2
NB + (1.20)(9.81) =
3.00

N B = 78.2 N

24

Example 6.5 : Chapter 06 Physics

v

v

Figure 6.8

A rider on a Ferris wheel moves in a vertical circle of radius,
r = 8 m at constant speed, v as shown in Figure 6.8. 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.

25

Chapter 06 Physics

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

a. The constant speed of the rider is

v = 2r 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
 F = mv 2
  r
ac mg mv 2
mg − Nt = r

(60)(9.81) − Nt = (60)(5.03)2

8

Nt = 399 N

26

Chapter 06 Physics

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

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

 F = mv 2
r
mv 2
Nb − mg = r

 Nb − (60)(9.81) = (60)(5.03)2
ac Nb
8

mg Nb = 778 N

27

Example 6.6 : Chapter 06 Physics

3.0 m s−1 A

3.0 m s−1
D

Figure 6.9 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 6.9. By the aid of the free body diagram,
determine the tension in the string at points A, D and E.

28

Chapter 06 Physics

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  F = mv 2 TA + mg = mv2
Point D, r r

TA  TA + (5.0)(9.81) = (5.0)(3.0)2

 mg 0.55
ac

TA = 32.8 N

 TD = mv2
r
ac  (5.0)(3.0)2
TD D
0.55
 TD = TD = 81.8 N
mg

29

Chapter 06 Physics

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,

  TE − mg = mv2
ac 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,

⚫ the normal force or tension is maximum at the bottom of
the circle.

⚫ the normal force or tension is minimum at the top of the

circle. 30

Chapter 06 Physics

Exercise 6.1 :

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

2. A small mass, m is set on the surface m
θ
of a sphere as shown in Figure 6.10.
O
If the coefficient of static friction is s

= 0.60, calculate the angle  would

the mass start sliding.

ANS. : 31

Figure 6.10

31

Chapter 06 Physics

3. A ball of mass 1.34 kg is connected Figure 6.11
by means of two massless string to
a vertical rotating rod as shown in
Figure 6.11. 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

32

Chapter 06 Physics

THE END.

Next Chapter…

CHAPTER 7 :
Gravitation

33