PHYSICS CHAPTER 6

1

CHAPTER 6:

Circular motion

(3 Hours)

PHYSICS CHAPTER 6

2

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

Describe graphically the uniform circular motion.

In terms of velocity with constant magnitude (only the

direction of the velocity changes).

Learning Outcome:

6.1 Uniform circular motion (1 hour)

PHYSICS CHAPTER 6

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6.1 Uniform circular motion is defined as a motion in a circle (circular arc) at a constant

speed.

Consider an object which does move with uniform circular

motion as shown in Figure 6.1.

Figure 6.1

r

θ

O

s

The length of a circular arc, s is given

by

rθs

pathcircular theof radius : r

subtends arc which theangle :θ where

radianin circle theof centre theto

PHYSICS CHAPTER 6

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It is directed tangentially to the circular path and always

perpendicular to the radius of the circular path as shown in

Figure 6.2.

In uniform circular motion, the magnitude of the linear velocity

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

continually changing.

The unit of the tangential (linear) velocity is m s1.

6.1.1 Linear (tangential) velocity ,

r

O

v

r

v

r

vFigure 6.2

v

PHYSICS CHAPTER 6

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The linear velocity, v is difficult to measure but we can measure

the period, T of an object in circular motion.

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 s1.

Equation :

Let the object makes one complete revolution in circular motion, thus

the distance travelled is (circumference of the circle),

the time interval is one period, T.

Tf

1

r2

PHYSICS CHAPTER 6

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From the definition of speed,

If therefore

Note:

The unit of angular velocity (angular frequency) is rad s1

(radian per second).

Unit conversion of angle, :

interval time

distance of changev

T

rv

2OR rfv 2

fT

ω

22

rωv

pathcircular theof radius : rfrequency)(angular locity angular ve :ω

where

360rad 2

180rad

PHYSICS CHAPTER 6

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

Define and use centripetal acceleration and use

centripetal acceleration,

Define and solve problem on centripetal force,

Learning Outcome:

6.2 Centripetal force (2 hours)

r

va

2

c

r

mvF

2

c

PHYSICS CHAPTER 6

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Figure 6.3 shows a particle moving with constant speed in a

circular path of radius, r with centre at O. The particle moves

from A to B in a time, t.

6.2.1 Centripetal (radial) acceleration,rc aa

or

Figure 6.3

1v

2v

The arc length AB is given by

The velocities of the particle at A

and B are v1 and v2 respectively

where

rΔΔs

r

ΔsΔ

vvv 21

(1)

PHYSICS CHAPTER 6

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Let PQ and PR represent the velocity vectors v1 and v2respectively, as shown in Figure 6.4.

Then QR represent the change in velocity vector v of the particle in time interval t. Since the angle between PQ and PR is small hence

By equating (1) and (2) then

12 vvvΔ

2v

1v

P Q

RFigure 6.4

PQQR vΔΔv

v

ΔvΔ (2)

v

Δv

r

Δs

PHYSICS CHAPTER 6

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Dividing by time, t, thus

Δt

Δv

v

1

Δt

Δs

r

1

v

a

r

v

r

va

2

c

pathcircular of radius : r

onaccelerati lcentripeta : cawhere

velocity gential)linear(tan : v

OR vra 2

c

frequency)(angular locity angular ve : ω

PHYSICS CHAPTER 6

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ca

ca

ca

caca

ca

Figure 6.5

The centripetal acceleration is defined as the 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.

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 6.5.

PHYSICS CHAPTER 6

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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.

Because of

therefore we can obtain the alternative expression of centripetal

acceleration is

2

2

cT

ra

4

T

rv

2

r

a

2

Tr

c

2

PHYSICS CHAPTER 6

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A motorbike moving at a constant speed 20.0 m s1 in a circular

track of radius 25.0 m. Calculate

a. the centripetal acceleration of the motorbike,

b. the time taken for the motorbike to complete one revolution.

Solution :

a. From the definition of the centripetal acceleration, thus

b. From the alternate formula of the centripetal acceleration, hence

Example 6.1 :

m 25.0 ;s m 20.0 1 rv

r

va

2

c

2

2

cT

ra

4

25.0

20.02

ca

2

25.0416.0

T

2

T

rv

2OR

PHYSICS CHAPTER 6

14

A car initially travelling eastward turns north by travelling in a

circular path at uniform speed as shown in Figure 6.6. The length

of the arc ABC is 235 m and the car completes the turn in 36.0 s.

Determine

a. the acceleration when the car is at B located at an angle of

35.0,

b. the car’s speed,

c. its average acceleration during the 36.0 s interval.

Example 6.2 :

Figure 6.6

PHYSICS CHAPTER 6

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Solution :

a. The period of the car is given by

The radius of the circular path is

Therefore the magnitude of the centripetal acceleration is

s 36.0 m, 235 tsABC

36.044 tT

s 144T

rθsABC

2

π 235 r

2

2

cT

ra

4 2

2

144

1504πca

PHYSICS CHAPTER 6

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Solution :

b. From the definition of the speed, thus

c. 1st method :

By using the triangle method for vector addition, thus the change

in the velocity is given by

s 36.0 m, 235 tsABC

t

s

t

sv ABC

36.0

235v

Cv

Av

AC vvv

2A

2

C vvv

226.536.53 v

θ

θ

PHYSICS CHAPTER 6

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Solution :

Therefore the magnitude of the average acceleration is

and its direction :

s 36.0 m, 235 tsABC

t

vaav

36.0

9.24ava

A

C1

v

vθ tan

6.53

6.53tan 1θ

PHYSICS CHAPTER 6

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Solution :

c. 2nd method :

x-component :

y-component :

s 36.0 m, 235 tsABC

t

vv

t

va AxCxx

xav

36.0

6.530

xava

t

vv

t

va

AyCyy

yav

36.0

06.53

yava

PHYSICS CHAPTER 6

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Solution :

Therefore the magnitude of the average acceleration is

and

s 36.0 m, 235 tsABC

2yav

2

xavav aaa

220.1810.181 ava

xav

yav1

a

aθ tan

0.181

0.181tan 1θ

PHYSICS CHAPTER 6

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A boy whirls a marble in a horizontal circle of radius 2.00 m and at

height 1.65 m above the ground. The string breaks and the marble

flies off horizontally and strikes the ground after traveling a

horizontal distance of 13.0 m. Calculate

a. the speed of the marble in the circular path,

b. the centripetal acceleration of the marble while in the circular

motion.

(Given g = 9.81 m s-2)

Solution :

Example 6.3 :

1.65 m

Before

13.0 m

u

After

u

r =2.00 m

1.65 m

PHYSICS CHAPTER 6

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Solution :

a. From the diagram :

The time taken for the marble to strike the ground is

The initial speed of the marble after the string breaks is equal to

the tangential speed of the marble in the horizontal circle.

Therefore

0 ; yx uuu

29.812

101.65 t

2

2

1gttus yy

m 1.65 ; m 13.0 yx ss

0.58013.0 u

tus xx

PHYSICS CHAPTER 6

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Solution :

b. From the definition of the centripetal acceleration, thus

r

u

r

vac

22

2.00

22.42

ca

PHYSICS CHAPTER 6

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6.3 Centripetal force6.3.1 Equation of centripetal force

From Newton’s second law of motion, a force must be

associated with the centripetal acceleration. This force is

known as the centripetal force and is given by

amFF nett

cc amF

mvmrr

mvF 2

2

c

caa

vrr

va 2

2

c

where cFF

and

and

force lcentripeta :cFwhere

PHYSICS CHAPTER 6

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ca

cF

cF

cF

ca

ca

v

v

v

The centripetal force is defined as a force acting on a body

causing it to move in a circular path of magnitude

and its always directed towards the centre of the circular

path.

Its direction is in the same direction of the centripetal

acceleration as shown in Figure 6.8.

Figure 6.8

r

mvF

2

c

PHYSICS CHAPTER 6

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PHYSICS CHAPTER 6

28

cF

ca v

cF

cF

ca

ca

v

v

v

v

0Fc

0Fc

0ac

0ac

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) speed as show in Figure 6.9.

Note :

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

Figure 6.9

222 mr2

1mv

2

1K

Simulation 6.1

PHYSICS CHAPTER 6

29

As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.

As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion

PHYSICS CHAPTER 6

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Without a centripetal force, an

object in motion continues along a

straight-line path.

With a centripetal force, an object in

motion will be accelerated and change its

direction.

PHYSICS CHAPTER 6

31

Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient.

PHYSICS CHAPTER 6

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PHYSICS CHAPTER 6

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Conical Pendulum

Example 6.4 :

Figure 6.10 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.

(Given g =9.81 m s2)

6.3.2 Examples of uniform circular motion

Figure 6.10

PHYSICS CHAPTER 6

34

Solution :

a. The free body diagram of the bob :

b. i. From the diagram,

5.00 ;m 10.0 ;kg 80.0 θlm

gm

θT

θT cos

θT sin

0 yF

mgθT cos

ca

PHYSICS CHAPTER 6

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The centripetal force is contributed

by the horizontal component of the

tension.

Solution :

b. ii.

5.00 ;m 10.0 ;kg 80.0 θlm

cx FF

r

mvθT

2

sin

θl

mvθT

2

sinsin

r

ll

rθ sin

θlr sin

m

θTlv

2sin

80.0

5.00sin10.07882

v

PHYSICS CHAPTER 6

36

Solution :

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

iii. From the definition of the radial acceleration, hence

5.00 ;m 10.0 ;kg 80.0 θlm

T

rv

2

T

θlv

sin2

T

5.00sin10.020.865

θl

va

2

rsin

5.00sin10.0

0.8652

ra

r

va

2

r

PHYSICS CHAPTER 6

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Centre of

circle

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

motorcycle, bicycle, etc…)

Example 6.5 :

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.

(Given g = 9.81 m s-2)

Solution :

a. The free body diagram of the car :

gm

N

f

0.70 ;m 20 ;kg 2000 μrm

ca

Picture 6.1

PHYSICS CHAPTER 6

38

Solution :

b. From the diagram in (a),

y-component :

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

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

Therefore

0.70 ;m 20 ;kg 2000 μrm

0yF mgN

r

mvF

2

x

r

mvf

2

μrgv r

mvμmg

2

PHYSICS CHAPTER 6

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T

gm

r

ca

Motion in a horizontal circle

Example 6.6 :

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.

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 :

a. The free body diagram for the ball :

Hz 2.00 ;m 1.10 ;kg 0.150 frlm

PHYSICS CHAPTER 6

40

Solution :

b. i. The linear speed of the ball is given by

Therefore the centripetal acceleration is

ii. From the diagram in (a), the centripetal force enables the ball

to move in a circle is provided by the tension in the string.

Hence

Hz 2.00 ;m 1.10 ;kg 0.150 frlm

rfT

rv

2

2

2.001.102v

r

va

2

c 1.10

13.82

ca

ccx maFF cmaT

PHYSICS CHAPTER 6

41

Motion in a vertical circle

Example 6.7 :

A small remote control car with mass 1.20 kg moves at a constant

speed of v = 15.0 m s1 in a vertical circle track of radius 3.00 m as

shown in Figure 6.12. Determine the magnitude of the reaction

force exerted on the car by the track at

a. point A,

b. point B.

(Given g = 9.81 m s2)

m 3.00

v

v

A

B

Figure 6.12

PHYSICS CHAPTER 6

42

Solution :

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

1s m 15.0 ;m 3.00 ;kg 1.20 vrm

gm

AN

ca

r

mvF

2

r

mvmgN

2

A

3.00

15.01.209.811.20

2

AN

PHYSICS CHAPTER 6

43

Solution :

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

1s m 15.0 ;m 3.00 ;kg 1.20 vrm

BN

ca

r

mvF

2

r

mvmgN

2

B

3.00

15.01.209.811.20

2

BN

gm

PHYSICS CHAPTER 6

44

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)

Example 6.8 :

v

v

Figure 6.13

PHYSICS CHAPTER 6

45

Solution :

a. The constant speed of the rider is

The free body diagram of the rider at the top of the circle :

s 10 ;m 8 ;kg 60 Trm

T

rv

2 10

82πv

ca

gm

tN

r

mvF

2

r

mvNmg

2

t

8

5.03609.8160

2

tN

PHYSICS CHAPTER 6

46

Solution :

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

s 10 ;m 8 ;kg 60 Trm

ca

gm

r

mvF

2

r

mvmgN

2

b

8

5.03609.8160

2

bNbN

PHYSICS CHAPTER 6

47

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 s1 as

shown in Figure 6.14. By the aid of the free body diagram,

determine the tension in the string at points A, D and E.

(Given g = 9.81 m s-2)

Example 6.9 :

Figure 6.14

A

D

E 3.0 m s1

3.0 m s1

3.0 m s1

PHYSICS CHAPTER 6

48

Solution :

The free body diagram of the sphere at :

Point A,

Point D,

1s m 0.3 ;m 55.0 ;kg 0.5 vrm

ca

r

mvF

2

r

mvmgTA

2

0.55

3.05.09.815.0

2

AT

A

gmAT

ca

D

gm

DT

r

mvTD

2

0.55

3.05.02

DT

PHYSICS CHAPTER 6

49

Solution :

The free body diagram of the sphere at :

Point E,

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.

1s m 0.3 ;m 55.0 ;kg 0.5 vrm

ca r

mvmgTE

2

0.55

3.05.09.815.0

2

ET

E

gm

ET

PHYSICS CHAPTER 6

50

Exercise 6.2 :

Use gravitational acceleration, g = 9.81 m s2

1. A cyclist goes around a curve of 50 m radius at a speed of

15 m s1. 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 s2; 1.02 kN; 24.6

PHYSICS CHAPTER 6

51

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 s1; 0.416 s

Figure 6.14

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

of a sphere as shown in Figure 6.14.

If the coefficient of static friction is s

= 0.60, calculate the angle would

the mass start sliding.

ANS. : 31

m

θ

O

PHYSICS CHAPTER 6

52

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 s1

Figure 6.15

PHYSICS CHAPTER 6

53

THE END…

Next Chapter…CHAPTER 7 :

Gravitation