CHAPTER 6:

Circular motion
(4 Hours)

1
 LEARNING OUTCOME:
6.1 Uniform circular motion (1 hour)

At the end of this chapter, students should

be able to:
• Describe uniform circular motion.
– In terms of velocity with constant
magnitude (only the direction of the
velocity changes).

2
 6.1 UNIFORM CIRCULAR MOTION

Circular motion – motion which occurs when bodies rotate in circular path.

Circular motion

Uniform Non - Uniform

(Horizontal plane) (Vertical plane)

Examples: Examples:
- a ball is swung in horizontal circle. - a bucket of water is swung in vertical
- a car / motorcycle turning a circle.
corner(on a banked track). - roller coaster cars
- cone pendulum.
- merry go round

3
 6.1 Uniform Circular Motion

Uniform circular motion is the motion of an object in a circle with a

constant / uniform speed.

 magnitude of its velocity remains constant.

 direction of its velocity changes continually.

Comparison Of Linear And

Circular Motion

Linear motion Circular motion

Displacement, s (m) Angular displacement, Ө (rad)

Velocity, (m/s) Angular velocity, (rad/s)

4
 6.1 Uniform Circular Motion

MOTION CHARACTERICTICS FOR CIRCULAR MOTION

B Linear Distance ( s)

r
s The arc length between A and B


A

Angular Displacement (  )

The angle subtended by the arc length.

Unit : radian ( rad )

Relation between , r, s  s  r

5
 6.1 Uniform Circular Motion
Motion Characterictics For Circular Motion

ANGULAR VELOCITY ( ω )

 Rate of change of angular displacement.

 angular displacement (rad)


t
time taken (s)

- Unit ω: rad s-1

- Other units: rps and rpm

- Vector quantity and its direction is perpendicular to the plane of motion
(right hand rule)

6
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

THE RELATIONSHIP BETWEEN LINEAR VELOCITY, V AND ANGULAR

VELOCITY, ω

s  r
Divide both sides by , t
s r
 v  r
t t

- The direction of linear velocity at every point along the circular path is
tangent to the point.

-The direction of the angular velocity, ω depends on the rotation

of the object (clockwise or counterclockwise ).

7
• Period, T
– is defined as the time taken for one complete revolution (cycle/rotation).
– The unit of the period is second (s).

• Frequency, f
– is defined as the number of revolutions (cycles/rotations) completed in
one second.
– The unit of the frequency is hertz (Hz) or s1.
– Equation : 1
f 
T
• Let the object makes one complete revolution in circular motion, thus
– the distance travelled is 2 r (circumference of the circle),
– the time interval is one period, T.

8
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

LINEAR VELOCITY can be written in terms of period, T and frequency, f :

2 r
v  2 r f
T

ANGULAR VELOCITY can be written in terms of period, T and frequency, f :

2
  2f
T

9
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

EXAMPLE 6.1.1

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

SOLUTION 6.1.1
Given : ω = 100 rpm
Convert to rad s-1 :
100 rev 1002 
ω   10.47 rads 1
1 min 60
2 2
(a) T   0.60 s
ω 10.47
1 1
(b) f    1.67 Hz
T 0.60
10
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

EXAMPLE 6.1.2

An object travels around the circumference of a circle of radius 6 m at a rate of

30 rev/min. Calculate
(a) its angular speed in rad/s.
(b) its linear speed around the circle.

SOLUTION 6.1.2
r = 6m
ω = 30 rev/min

30  2
(a) 
60
(b) v  rω
 3.14rad / s v = 6 π m/s

 rad / s or v = 18.85 m/s

11
EXAMPLE 6.1.3

The diameter of a tire is 64.8 cm. A tack is embedded in the tread of the right
rear tire. What is the magnitude and direction of the tack's angular
velocity vector if the vehicle is traveling at 10.0 km/h?

Solution 6.1.3:

(10 km/h)(1000/3600) = 2.78 m/s

r = 0.648/2 = 0.324 m

ω = v/r = (2.78)/(0.324) = 8.58 rad/s

 Learning Outcome:

6.2 Centripetal force (3 hours)

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

• Define centripetal force and use its formulae,

mv 2
Fc 
r
• Identify forces such as tension, T, friction, f, weight, W and
reaction, N that enable a body to perform circular motion on a
horizontal and vertical plane.

• Use the relationship of the forces above and centripetal force.

13
6.2 Centripetal Acceleration, ac

 When an object moving in a circle of radius, r at a constant speed, v, the

direction of the object changes. Thus it has an acceleration called the centripetal
acceleration.

 ac is defined as the acceleration of an object moving in circular path and it

directed towards the center of the circle.

 Direction of ac – graphically.

Δv
a
Δt
FIGURE 6.2.1

14
6.2 Centripetal Acceleration, ac

v2 speed of the object

Magnitude of ac : ac 
r
radius

 ac  rω 2 v : linear tangential velocity

since v = r ω, ω: angular velocity (angular
ac  vω frequency)
r : radius of circular path

2πr 4π 2 r
If v  then, ac 
T T2

15
6.2 Centripetal Acceleration, ac
Example 6.2.1
Calculate the centripetal acceleration of a car traveling on a circular racetrack of
1000 m radius at a speed of 180 km h-1.

Solution

Given : r = 1000 m
v = 180 km h-1

180(1000)
v  50 ms 1
60(60)

2 2
v 50
Hence , ac    2.5 ms 2
r 1000

16
Centripetal Force, Fc

Fc is defined as the net force required to keep an object of mass, m

moving at a speed v on a circular path of radius, r.

Examples :

As the moon orbits the Earth, the force of gravity acting

upon the moon provides the centripetal force required for
circular motion.

FIGURE 6.2.2

As a car makes a turn, the force of friction acting upon

the turned wheels of the car provides the centripetal force
required for circular motion.

FIGURE 6.2.3
17
Centripetal Force, Fc

Magnitude of Fc :

since v = r ω , thus

Fc  mrω  mvω 2

Direction of Fc - towards the centre of the circle and same direction of the
centripetal acceleration.

Fc is perpendicular to the direction v, so it does

no work on the object.

FIGURE 6.2.4
18
Circular motion in horizontal plane (uniform circular motion)

Case 1 : object moves in a horizontal circle with steady speed.

FIGURE 6.2.5 mg

Two forces acting on the object :

1.The force of gravity ( weight i.e. mg )
2.The tension in the string - is the only component in the radial direction that
provided the centripetal force.

Applying Newton 2nd Law :

2 mv 2
 Fnet  mac 
mv T
r r
19
 Circular motion in horizontal plane
(uniform circular motion)
Example 6.2.2

A 0.25 kg rock attached to a string is whirled in a horizontal circle at a constant

speed of 10.0 ms-1. The length of the string is 1.0m. Neglecting the effects of
gravity, find the tension in the string.

Solutio 6.2.2

T  Fc
mv 2 0.25(10) 2
T   25 N
r 1.0

20
 Case 2 : Motion of car round a curve

(a) Flat curve road

If the coefficient of static friction between the tires & the road is μ
then,
f s= µR R
Vertical comp. : Horizontal comp.:

Σ Fy  0 mv 2 fs
fs  Fc 
r mg
R  mg  0
R  mg … ( 1 ) mv 2 …( 2 )
μR 
r

mv 2
(1) into (2)  mg 
: r
∴ The maximum velocity without slipping on the road is : v  μr g
21
Motion of car round a curve

Example 6.2.3

A car travels around a flat curve of radius r = 50m. The coefficient of the
static friction between the tires & the road is μs = 0.75. Calculate the
maximum speed at which the car can travel without skidding.

Solution 6.2.3
R
Given: μs = 0.75 , r = 50m
fs supplies the centripetal force
2 fs
mv
fs  But fs= µsR mg
r
mv 2
mg  v rg
r
 (0.75) (50) (9.81)
 19.18 m s 1 22
Motion of car round a curve

Example 6.2.4

A 1200 kg car with a velocity of 8.0 m/s travels around a flat curve of radius r =
9.0m.
a) Calculate the horizontal force must the
pavement exert on the tires to hold the car
in the circular path ? R
b) What coefficient of friction must exist for
the car not to slip ?
Solution 6.2.4 fs
12008.0 
2
mv 2 mg
a) f s  Fc    8533 N
r 9.0
b) mv2 mg 
mv2
fs  R = mg
r
r

v2

8.0 
2
 0.72
rg 9.0 9.81

23
Case 2 : Motion of car round a curve

(b) Frictionless Banked Road:

With no friction : R cos Ө

R
comp.  x :
mv 2
R sin    (1)
r
R sin Ө

comp.  y :
R cos   mg  (2)

2 FIGURE 6.2.6
mv
( ) v2
(1) R sin  r
:  tan  
(2) R cos  mg rg

v  r g tan
24
Motion of car round a curve

Example 6.2.5

A curve of radius 30 m is to be banked so that a car may make the turn at a

speed of 13 m/s without depending on friction. What must be the slope of the
curve (the banking angle) ?

Solution 6.2.5

tan  
v2
   tan 1 v
2
 tan 1 13
 300
2

rg rg 309.81

25
Case 3 : Conical Pendulum

A conical pendulum moving in uniform circular motion with speed v:

T sin θ supplies centripetal force.

r = L sin θ

Component – x
: r

mv 2
Fx 
r FIGURE 6.2.7

2 Component – y :
mv … (1)
Tsin  
r Fy  0
Tcos   mg … (2)
26
 Case 3 : Conical Pendulum

r = L sin θ

mv 2 … (1)
Tsin  
r

Tcos   mg … (2)

mv 2 r
( )
(1) Tsin 
:  r
(2) Tcos  mg
FIGURE 6.2.8

v2
tan  
rg

v  r g tan   Lg sin  tan 

27
Conical Pendulum
Example 6.2.6

A 0.15 kg ball attached to a string which is 1.2

m in length moves in a horizontal circle. The
string makes an angle of 30° with the vertical.
Find the tension in the string & the speed of the
ball. r = L sin 30˚
Solution 6.2.6
Component - y : Component - x :

T cos 30° = mg mv 2
r
Tsin θ 
r
0.15(9.81)
T r Tsin θ
cos 30 v
T  1.70 N m
(Lsin θ) Tsin θ
v
m
 1.84 m s 1

28
 CIRCULAR MOTION IN VERTICAL PLANE
(NON-UNIFORM CIRCULAR MOTION)

Case 1 : A ball is attached to a string & moves in a vertical circle.

FIGURE 6.2.9 FIGURE 6.2.10

At the top of the circle ( point A ) : At the bottom of the circle ( point B ) :
both T & mg are directed downwards T & mg point in opposite direction

 Fnet  mac  Fnet  mac

2
mv mv2
T  mg  T  mg 
r r
mv2 mv 2
T  mg (T is minimum) T  mg (T is maximum)
r r
29
 Circular motion in vertical plane
(non-uniform circular motion)

Circular motion is possible as long as the cord remain taut, thus there is a critical
(minimum) speed to be maintained.

mv 2
T  mg 
r

If the rope is sagging, T = 0 , thus :

mv 2
mg 
r
T=0
v min  r g

30
 Circular motion in vertical plane
(non-uniform circular motion)
Example 6.2.7 constant speed
A 1.2 kg rock is tied to the end of a 90 cm length of string. The rock is then is
whirled in a vertical circle at a constant speed of 8 m/s. What are the tensions
in the string at the top and bottom of the circle ?
Solution 6.2.7
m = 1.2 kg , r = 90 cm , v = 8 m/s

Top Bottom

mv2 mv2
T  mg  T  mg 
r r
mv2 mv 2
T  mg T  mg
r r
T = 73.6 N T = 97.1 N

31
 Circular motion in vertical plane
(non-uniform circular motion)
Example 6.2.8 not constant speed

A 2 kg ball is tied to the end of a 80 cm length of string. The ball is then is whirled
in a vertical circle and has a velocity of 5 m/s at the top of the circle.
a) What is the tension in the string at that
instant ?
b) What is the minimum speed at the top necessary to maintain circular motion ?

Solution 6.2.8
m = 2 kg , r = 80 cm , v = 5 m/s at the top.
mv2
a) b) T  mg
Top r
For vmin , T = 0
2
mv
T  mg mg 
mv 2
r r
T = 42.9 N vmin  r g
 0.89.81  2.8 m/s
32
 Circular motion in vertical plane
(non-uniform circular motion)

Example 6.2.9

A rope is attached to a bucket of water and the bucket is then rotated in a

vertical circle of 0.70 m radius. Calculate the minimum speed of the bucket of
water such that the water will not spill out.

Solution 6.2.9

 Fnet  mac
mv 2
T  mg 
r

The water will not spill out if the T=0, thus :

mv 2
mg  v min  r g
r
v min  (0.7) 9.81  2.62 m s 1
33
Circular motion in vertical plane
(non-uniform circular motion)

Case 2 : Roller coaster on a circular track / Ferris wheel

Top of the circle : Bottom of the circle :

R
 Fnet  mac  Fnet  mac
2 mv 2
mv R R  mg 
R  mg  mg r
r mg

mv 2 mv 2
R  mg R  mg
r r

A minimum velocity (when R = 0) is required in order to keep a roller coaster car

on a circular track.

mv 2 v min  r g
0  mg 
r
34
 Circular motion in vertical plane
(non-uniform circular motion)
Example 6.2.10
What minimum speed must a roller coaster be traveling when
upside down at the top of a circle (refer to the figure) if the
passengers are not to fall out ? Assume R = 8.0 m.

Solution 6.2.10
r = 8.0 m

Top  Fnet  mac

mv 2
R  mg 
r
For vmin , R = 0

v min  r g

vmin  8.09.81  8.9 m/s

35
Example 6.2.11 :

v
Figure 6.13

A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m

at constant speed, v as shown in Figure 6.13. 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.
(Given g = 9.81 m s-2)
36
Solution 6.2.11:
m  60 kg; r  8 m; T  10 s
a. The constant speed of the rider is

2r 2π8
v v
T 10
v  5.03 m s 1
The free body diagram of the rider at the top of the circle :


Nt mv 2
 F
r

ac  mv 2
mg mg  N t 
r

60 9.81  N t 
60 5.03
2

8
N t  399 N
37
Solution 6.2.11:
m  60 kg; r  8 m; T  10 s
b. The free body diagram of the rider at the bottom of the circle :

mv 2
 F
r
mv 2
N b  mg 
r
 
N b  60 9.81 
60 5.03
2
ac Nb
8
N b  778 N

mg

38
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

EXERCISES 6.1
1. A particle is moving on a circular path of radius 0.5 m at a constant
speed of 10 m/s. Calculate the time taken to complete 20 revolutions.
(t = 6.28 s)

2. Two wheels of a machine are connected by a transmission belt. The radius

of the first wheel r1 = 0.5 m, the radius of the second wheel r2 = 0.125 m.
The frequency of the bigger wheel equals 3.5 Hz. What is the frequency of
the smaller wheel ? (f2 = 14 Hz)

39
Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion

3. The astronaut orbiting the Earth is preparing to dock with Westar VI satellite.
The satellite is in a circular orbit 600 km above the Earth’s surface, where the
free fall acceleration is 8.21 m s2. Take the radius of the Earth as 6400 km.
Determine :
a. the speed of the satellite,
b. the time interval required to complete one orbit around the Earth.
ANS. : 7581 m s1; 5802 s

4. A pendulum bob of mass 1 kg is attached to a string 1 m long and made to

revolve in a horizontal circle of the radius 60 cm. Calculate the period of the
motion and the tension of the string.
2r
T  1.8 s Tension , T = 12.25 N
v
Exercise 6.2 :

Use gravitational acceleration, g = 9.81 m s2

1. A cyclist goes around a curve of 50 m radius at a speed of 15 m s1.
The road is banked at an angle  to the horizontal and the cyclist travels
at the right angle with the surface of the road. The mass of the bicycle
and the cyclist together equals 95 kg. Calculate
a. the magnitude of the centripetal acceleration of the cyclist,
b. the magnitude of the normal force which the road exerts on
the bicycle and the cyclist,
c. the angle .
ANS. : 4.5 m s2; 1.02 kN; 24.6

41
 Exercise 6.2 :

2. 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

3. A small mass, m is set on the surface of a m

sphere as shown in Figure 6.14. If the
coefficient of static friction is s = 0.60, θ
calculate the angle  would the mass start
sliding. O
ANS. : 31

42
 Exercise 6.2 :

4. A ball of mass 1.34 kg is connected by means of

two massless string to a vertical rotating rod as
shown in Figure 6.15. 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

43