1

A particle P travels in circular path.

It performs circular motion about O with radius r.

Circular motion

s

P

A

If P travels in constant speed, the circular motion is known as uniform.

2

Mathematics – angles

Arc length = circumference x /360o

1 = 2 x / 360o

=180o / ≈ 57o

1 cm

1 cm

1 cm

3

Angle in radian (another unit for angle)

If s = r, then is 1 radian. 1 radian = 180o / 180o = radian 360o = 2 radian

r

r

s

4

Relation between arc length s, angle at centre and radius r is

s

s/r

s = rs = r

5

Describing circular motion Consider a body moving uniformly from A to B in time t

so that it rotates through an angle .

s

B

At

t

sv

time

distance

r

tr

t

rv

Angular displacement (/ rad)* Anticlockwise as positive directionAngular velocity ( / rad s-1) = Angular displacement / time

*Speed of the body

By s = r

6

Speed and angular speed

Consider carts A and B in a mechanical game.

Do they have the same angular speed?

Which of them moves faster?

2 m

1 m

A

B

= 3 rad s-1

7

Describing circular motion

s

B

A

222

r

r

v

rT

Period of the motion

2

T

= distance / speed

Period of the motion

= angular displacement / angular speed

8

Three useful expressions

(1) s = r(2) v = r(3) 180o = radians

9

Example 1Find the angular velocity in radian s-1if a motor makes 3000 revolutions per minute.

Solution:1 revolution = 360o = 2 radians.Angular velocity = 3000 x 2 / 60 = 314 rad s-1

10

Rotational motion and Translational motion

: angular displacement : angular velocity : angular accelerationEquations of motion for uniform

acceleration:

1 = 0 + t

= 0t + ½ t2

12 – 0

2 = 2

v = u + ats = ut + ½ at2

v2 –u2 = 2as

11

Example A turntable is rotating at 1 rev / s initially. A motor is

turned on such that the turntable rotates at an angular acceleration of 0.3 radian s-2. Find the angular displacement covered by the turntable for 5 s just after the motor is turned on.

Solution:0 = 1 rev / s = 2 radian s-1 (∵360o = 2 radians) = 0.3 radian s-2

t = 5 s By = 0t + ½ t2

= (2)(5) + ½ (0.3)(5)2

= 35.2 radians

12

Centripetal acceleration

A body which travels equal distances in equal times along a circular path has constant speed but not constant velocity.

Since the direction of the velocity changes from time to time, the body has acceleration.

13

Centripetal acceleration (Deriving a = v2/r) Consider a body moving with constant speed v in a circle of

radius r. It travels from A to B in a short interval of time t.

vAvB

vvAvB

A

B

r

O

vt

v

t

v

22

raorr

va

Change in velocity = vB – vA = (– vA) + vB = v

Since t is small, will also be small, so v = v .

Acceleration

But v = r, Therefore,

Since is small, v will be perpendicular to vA and points towards the centre.

Hence, the direction of acceleration is towards the centre. i.e. centripetal.

14

Centripetal force Since a body moving in a circle (or a circular arc) is

accelerating, it follows from Newton’s first law of motion that there must be a force acting on it to cause the acceleration.

The direction of this force is also towards the centre, therefore this force is called centripetal force.

By Newton’s second law, the magnitude of centripetal force is

Since the centripetal force displacement, NO work is done by the centripetal force

⇒ K.E. of the body in uniform circular motion remains unchanged.

22

or mrr

vmmaF

15

A stone of mass 2 kg is tied by a string and moving in a horizontal circular path of radius 0.5 m. Find the tension in the string if the speed of the stone is 4 ms-1.

Solution:

Tension

=Centripetal force

= mv2/r

= (2)(4)2/0.5

= 64 N

Smooth table

16

Examples of circular motion

A bob is tied to a string and whirled wound in the horizontal circle.

In vertical direction:

T

mg

)1(cos mgT

)2(sin2

r

vmT

)1(

)2(g

ror

rg

v 22

tantan

In horizontal direction:

17

Rounding a bend

Case 1: Without banking

When a car of mass m moving with speed v goes around a circular bend with radius of curvature r, the centripetal force

f

r

r

vmF

2

The frictional force between the tires and the road provides the centripetal force.

r

vm

2

If limiting friction < , the car slips on the road and it may cause an accident.

18

The coefficients of friction between the tires and the road in rainy days and in sunny days are 0.4 and 0.6 respectively. If a car goes around a circular bend without banking of radius of curvature 50 m, find the maximum speeds of the car without slipping in rainy days and sunny days.

Solution:

Limiting friction is attained to achieve maximum speed.

rgvr

vmmg 2

2r

vmF

2

By

1.1410504.0

3.1710506.0

In rainy days,Maximum speed without slipping =

In sunny days,Maximum speed without slipping =

ms-1 (50.9 km h-1)

ms-1 (62.4 km h-1)

Maximum speed without slipping = rg

19

Rounding a bend Case 2 with banking

mg

r

R

In order to travel round a bend with a higher safety speed, the road is designed banked.

The centripetal force does not rely on friction, but provided by the horizontal component of the normal reaction.

)1(cos mgR

)2(sin2

r

vmR

gr

v 2

tan:)1(

)2(

tangr

In vertical direction:

In horizontal direction:

The safety speed of rounding the bank is

which can be increased by increasing the degree of banking of the road.

20

The figure above shows a car moving round a corner with a radius of 8 m on a banked road of inclination 20o. At what speed would there be no friction acting on the car along OA? Solution:

)1(20cos mgR

)2(8

20sin2

v

mR

In vertical direction:

In horizontal direction:

4.58

20tan:)1(

)2( 2

vg

vms-1

Note: If the car travels at a higher speed say 6 ms-1 round the above corner, some frictional force would act on the car to prevent the car from slipping upward along the road provided that the limiting friction is not exceeded i.e. Rf

21

Aircraft turning in flight

mgmg

LL L L coscos

L L sinsin

In straight level flight, the wings provide a lifting force L that balances its weight mg. i.e. L = mg.

mgmg

LL

To make a turn at v with radius r, the flight banks and the horizontal component of the lifting force (L sin ) provides the centripetal force for circular motion.

straight level flight turning in flight

22

Aircraft turning in flight

mgmg

LL L L coscos

L L sinsin

To make a turn at v with radius r, the flight banks and the horizontal component of the lifting force (L sin ) provides the centripetal force for circular motion.

turning in flight )1(sin2

r

vmL

)2(cos mgL

The weight of the aircraft is supported by the vertical component of L. i.e.

23

Find the angle of inclination of the wings of an aircraft which is traveling in a circular path of radius 2000 m at a speed of 360 km h-1.

(1) / (2):tan = v2 / rgtan = 1002 / (2000 x 10) = 26.6o

mg

L Lcos

Lsin

Solution:Speed of the aircraft = 360 /3.6 ms-1 = 100 ms-1

Resolve horizontally,L sin = mv2 / r --- (1)

Resolve vertically, L cos = mg --- (2)

24

Cyclist rounding a corner The frictional force

provides the centripetal force.

By the law of friction, the maximum speed of the cyclist without slipping is given by

r

vmmgf

2

grv

C.G.

mg

Rf

r

However, f also has a moment about C.G which tends to turn the rider outwards.

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The rider must lean inwards so that the moment of f is counterbalanced by the moment of R about C.G..

For no overturning of the cyclist, C.G.

Take moment about C.G..:

C.G.

mgR

f

r

h

a In vertical direction:

)1(Rafh

)2(mgR In horizontal direction:

)3(2

r

vmf

Sub (2) and (3) into (1):

gr

v

h

a

mgahr

vm

2

2

gr

v 2

tan ∴

For given v and r, the cyclist must bend at the correct angle )(tan2

1

gr

v

in order not to overturn.

26

The rotor

mg

f

R

r

vmR

2

mgf

A mechanical game in amusement parks.

It consists of an upright drum, inside which passengers stand with their backs against the wall.

The drum spins at increasing speed about its central vertical axis. Hence, the centripetal force required for circular motion also increases.

The centripetal force is provided by the normal reaction from the wall. i.e.

Therefore, the normal reaction increases as the spinning speed increases.

As normal reaction increases, the weight of the passenger can be balanced by the frictional force. i.e.

When the floor is pulled downwards, the passenger will not fall but remains stuck against the wall of the rotor.

27

It is given that the radius of the drum is 2 m and the coefficient of static friction between clothing and the wall is 0.4 m. Find the minimum speed v of the passenger before the floor is pulled downwards?

Solution:

mg

f

R

)1(2

r

vmR

)2(mgf

)3(Rf

r

vmmg

2

4.0

210v

In horizontal direction:

In vertical direction:

By the law of friction:

Sub. (1) and (2) into (3):

The minimum speed required is

7.07 ms-1

gr

v 2

28

Looping the loop At C, the centripetal force is provided by the normal reaction and

the weight of the object.

A

ru

v

R mg

C

BD

If the object does not leave the track, R 0.

The minimum speed at C to just complete the loop is given by

r

vmmgR

2

r

vmmg

2

)1(grv

By conservation of energy, the minimum speed at A for the object to complete the loop is given by

mghmvmu 22

2

1

2

1)2(422 grvu

Sub. (1) into (2): grugrgrgru 5542

29

gr5If the ball passes point A with speed , the ball will fall from the track.

ru

v

R mg

C

BD

gr5 , the ball will complete the loopIf the ball passes point A with speed

and the remaining centripetal force is provided by the normal reaction.

Notice that:

30

Weightlessness Weightlessness means the weight of an object is equal

to zero. This happens at a place where there is no gravitational field (g = 0).

We are aware of our weight because the ground exerts an upward push (normal reaction) on us.

If our feet are completely unsupported, for example, in a free falling lift, we experience the sensation of ‘weightlessness’.

An astronaut orbiting the earth in a space vehicle with its rocket motors off also experiences the sensation of ‘weightlessness’.

31

An astronaut orbiting the earth An astronaut orbiting the

earth in a space vehicle with its rocket motors off also experiences the sensation of ‘weightlessness’. Why?Why?

The weight of the astronaut provides the centripetal force and the walls of the vehicle exert no forceno force on him.

mg

R =R = 0 0

v

r

i.e. r

vg

r

vmmg

22

Note: Note: gg < 10 ms < 10 ms-2-2 because the astronaut is now far away because the astronaut is now far away from the surface of the earth.from the surface of the earth.

32

Summary

Weightlessness Experience the sensation of ‘weightlessness’

W = 0 W > 0 but R = 0

At a place where there is no gravitational field

(g = 0).

In a free falling lift. In a space vehicle with its

rockets off orbiting the earth. Normal reaction R = 0

33

Centrifuge (離心機 )

Centrifuges separate solids suspended in liquids, or liquids of different densities.

The mixture is in a tube, and when it is rotated at high speed in a horizontal circle, the less dense matter moves towards the centre of rotation.

On stopping the rotation, the tube returns to the vertical position with the less dense matter at the top. Cream is separated from milk in this way.

34

Working principle of a centrifuge

Consider the part of liquid between A and B inside the tube.

The pressure at B is greater than that at A. This provides the necessary centripetal force acting inwards.

For this part of liquid, the force due to pressure difference supplies exactly the centripetal force required.

If this part of liquid is replaced by matter of small density or mass, the centripetal force is too large and the matter is pushed inwards.

On stopping the rotation, the tube returns to the vertical position with the less dense matter at the top.

pA pB

BA

pB > pA

35

H.W. Chapter 2 (1, 2, 3, 5, 6(a)(b)(c) ) Due date: 27/11 Day 1 1. A particle moves in a semicircular path AB of

radius 5.0 m with constant speed 11 ms-1. Calculate

(a) the time taken to travel from A to B ( = 22/7)

(b) the average velocity,

(c) average acceleration

A 5 m 5 m B

11 ms-1

36

2 A turntable of a record player makes 33 revolutions per minutes. Calculate

(a) its angular velocity in rad s-1,

(b) the linear velocity of a point 0.12 m from the centre.

33 rev / min

v

180o = radians

v= r

37

3 What is meant by a centripetal force? Why does such a force do no work in a circular orbit?

(a) An object of mass 0.5 kg on the end of a string is whirled round a horizontal circle of radius 2.0 m

with a constant speed of 10 ms-1. Find it angular velocity and the tension in the string.(b) If the same object is now whirled in a vertical circle of the same radius with the same speed, what are

the maximum and minimum tensions in the string? A

BT = mv2/r

T + mg = mv2/r

T – mg = mv2/r

38

Referring to the above diagram, find the normal reaction acting on the ball at A, B and C if the speed of the ball at A is

When the ball is at A,

gr6

ru

vc

R mg

C

BD RRBB

mgmg

Centripetal force:RA - mg = mu2/rRA = mg + m(6gr)/r = 7mg

When the ball is at B, By conservation of energy,

½ mu2 = ½ mvB2 + mgr

vB2 = u2 – 2gr = 6gr – 2gr = 4gr

Centripetal force = RB = mvB2/r

RB = m(4gr)/r = 4mg

A

RRAA

mgmg

When the ball is at C, By conservation of energy,

½ mu2 = ½ mvc2 + mg(2r)

vc2 = u2 – 4gr = 6gr – 4gr = 2gr

Centripetal force: RC + mg = mvB

2/rRC = m(2gr)/r – mg = mg

vB