e06 circular motion

8/7/2019 E06 Circular motion

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Circular motion Mechanics p1/10

Circular motion

6.1 Kinematics

When a body is traveling in along the circumference of a circle, it has

= Angle moved = Angular displacement ( ) in radians

s = Length of arc subtending angle

r= Radius of the circular path

t= Time taken to moves and

They are related by

rs=

Linear speed/velocity

dt

dsv =

Angular speed/velocity

dt

d=

Angular acceleration

2

2

dt

d

dt

d ==

If it is a uniform circular motion i.e. constant speed,

t

sv =

t

=

rv=

f= Frequency = Number of revolutions per unit time

T= Period = Time for one complete revolution

Tf

1=

fT

22

==

r

s


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EXAMPLE 61An object is fixed at a point in HK. Is the Earth center

coincides with the center of rotation of the object? Find the

linear speed of the object.

Given the radius of the Earth is 6380 km and the latitude of

HK is 22.5o N.

SOLUTION

No, the center of rotation o of the object is drawn below.

v = r = (REcos22.5o) = 6380 1000(cos22.5o)

= 2 /(24 3600)

v = 428 m s1

6.2 Centripetal acceleration

Uniform circular motion = moving with constant speed along the

circumference of a circle constant speed in a straight line.

Mathematical derivation

Let v be the uniform speed of the point mass moving around the

circle. is the angle of the mass turned from point A to point B

within a small time t. Then arc lengths = r so v = r.

Vectors vA and vB are the velocity of mass at point A and B

respectively.

The change in velocity v = vBvA. in vector form.

Graphically, since the angle between OA and OB is . The angle

between vA and vB is also .

As t0, 0, vvA. and vvB.

The direction of v is towards centre, i.e. centripetal.

22.5o

o

22.5o

A

Br

v

A

vB

O

vA

vB

v


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The magnitude of v, written as v, will be

As t0, 0,

v = v

t

v

t

va

=

=

As t0,

r

vrva

2

2===

EXAMPLE 62(a) Is the centripetal acceleration constant for a point mass

performing a uniform circular motion ?

(b) Is the centripetal acceleration directly or inversely

proportional to the radius of the circle ? Illustrate your

answer by an example.

SOLUTION(a) No, although its magnitude is constant, its direction is

always changing with time. Indeed, displacement, velocity

and acceleration are all rotating with constant magnitude.

(b) If is constant, a will be directly proportional to r.

For instance, objects at different latitudes of the earth.

However, ifv is constant, a will be inversely proportional

to r. For instance, a car turns around at a corner.

6.3 Centripetal force

To maintain the circular motion, a resultant force is required to

accelerate the object i.e. to change the direction of the velocity of the

object. Therefore F 0 (not in equilibrium).

Also, the net force must be always perpendicular to the velocity of the

object; otherwise, the speed of the object will be increased.

Hence, this force must be radial out or in.

However, as the momentum of the object is changed its direction

inwards continually so the force (acceleration) is centripetal.

The centripetal force required to produce a is

2

2

mrr

mvF ==

The centripetal force is the net force Facting on an object of mass m,

which performs uniform circular motion with speed v and radius of

curvature r.

Experiment to verifyF= mr2

v v

v


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Set up the experiment as shown.

Measure the mass m of the cork by electronic weighing machine.

Measure the length of the stringL from the top of the glass tube to the

centre of mass of cork with the help of the paper marker.

Swing the cork to move in a horizontal circle until the marker is close

to the bottom of the glass tube.

When the angular speed of the cork is steady, measure the time P

for 50 revolutions of the cork. Then find the angular speed by =

100 /P.

The tension Tof the string is given by the known weight hanged.

Repeat the experiment by using changing the known weight.

Sets of values ofTand mL2 are compared.

It is found that T mL2 with errors

Theory

Free body diagram of the cork

Resolve components or Vectors diagram

y-component:

In equilibrium, F= 0

Tcos = mg

x-component:

Uniform circular motion, F= ma

Tsin = mr2

Also, (a lot of students miss this step)

r=LsinThen

T= mL2

Small

cork of

mass m

Thin

glass

tube

marker

Weight W

Light string

L

T

mg

Tmg

F= ma =

mr2


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EXAMPLE 62What are the major sources of error of the above experiment

?

SOLUTION

The friction between the glass tube and the string.

It is difficult to keep the cork performing a horizontal

uniform circular motion.

EXAMPLE 63Compare the velocity of a body having uniform accelerated

linear motion and a body having uniform circular motion.

SOLUTION

The body moves in a straight line. The magnitude of

velocity is always changing but the direction of velocity is

either unchanged or changed once only.

The body moves in a circle. The magnitude of velocity is

constant but its direction is always changing.

There is no centrifugal force acting on the body having circular

motion. It is a fictitious force felt by body in accelerated ref. frame.

6.4 Uniform circular motion

Satellites or moon moving around the earth are examples of uniform

circular motion (studied in Gravitation). Also, object performing

horizontal circular motion can be a uniform circular motion.

Consider an object performing horizontal uniform circular motion, the

net force acting on it should be pointed towards the centre of the

circle horizontally. Hence, its force diagram must be of the form

As seen from the diagram, the point mass moving into paper, at least

one more force Q must be acted on the object in order to balance the

weight vertically and provide centripetal force horizontally.

Vertically, F= 0

Q cos = mgHorizontally, F= ma

Q sin = mv2/r= mr 2

mg

F

Q

r

centre of

the circle

mg

F

Q


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EXAMPLE 64Car turning on a level road A car of mass 1600 kg

travelling at a constant speed 20 m s1 around a flat, circular

track of radius 190 m. Name the force which provides the

necessary centripetal force for turning and find it value.The diagram shows the car running out of the paper and

turning right.

SOLUTION

Force diagram of the car (as a point mass)

The centripetal force is provided by the friction between

the car and the ground.

Vertically, F= 0

N= W

Horizontally, F= ma

f= mv2/r= 1600 202/190 = 3368 N

EXAMPLE 65This question is the extension of the example 64.

What is the work done by the friction which provides the

necessary centripetal force for turning ? Explain your

answer.

SOLUTION

The work done by the friction is zero because there is no

sliding when the car turns around so the friction for turning

is a static friction.

EXAMPLE 66Car turning on a banked road It is not safe to turn using

friction because the friction may be lowered for different

road conditions. A banked road can be ideal for a car to

turn without using friction as shown.

Find the value for the ideal angle if the speed of the car

of mass m is v and the radius of the bend is r.

SOLUTION

Force diagram of the car (as a point mass)

The centripetal force is provided by the horizontal

component of the normal force.

Vertically, F= 0

Ncos = W= mg

Horizontally, F= ma

Nsin = mv2/r

Hence, tan = v

2

/gr.However, one ideal angle is for one speed only.

weight

normal force

friction

weight

normal force


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EXAMPLE 67Turning of aeroplane An aircraft in straight, level flight

experiences a lifting force Uat right angles to its wings, and

the force balances its weight Was shown. This lifting force

increases with increasing the speed of the aircraft.How can the aircraft turn ? What is the relationship

between the speed v of the aircraft and the radius rof the

circle ?

The aircraft is moving out of the paper.

SOLUTION

The aircraft must tilt to an angle so the horizontal

component of the lifting force provides the centripetal

force. The weight is now supported by the vertical

component of the lifting force, which must be increased by

increasing the speed of the aircraft.

The aircraft turns right as shown.

Vertically, F= 0

U2cos = W= mg

Horizontally, F= ma

U2sin = mv2/r

Hence, tan = v2/gr.

(FPI 94) Examples 10, 11, 12

If the force acting towards the center of circle is not equal to the

required centripetal force (mr2), the body will fail to perform

circular motion.

EXAMPLE 68The figure shows the path of a car turns a corner. Draw on

the same diagram the expected paths of the car if the

resultant force towards the center is now

(a) greater than the centripetal force required, and

(b) less than the centripetal force required.

SOLUTION

6.5 Non-uniform circular motion

For non-uniform circular motion, the formula for centripetal force can

still be applied to the centripetal component of the net force.

However, there is a tangential component of the net force to change

the speed of the object.

Object performing vertical circular motion is a non-uniform circular

motion except at its highest and lowest positions.Only the highest and lowest positions are considered in AL course.

W

U

F

W

U2

W

U2

centre

of circlepath

corner


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Consider an object performing vertical uniform circular motion, the

net force acting on it should be pointed towards the centre of the

circle vertically when the object are at A or B.

r

A

B

Positive direction should be chosen as the direction towards centre.

At A, as the centre of circular motion is above the object, there must

exist an extra upward forcePsuch that

vr

mg

centre of the motion

A

P

since F= ma

r

mvmgP

2

=

The upward forcePmust be greater than mg, so the net force provides

the necessary centripetal force. The body can perform uniform

circular motion momentarily at the lowest position.

For instance, the tension of a swinging pendulum is the greatest when

it reaches its lowest position.

At B, the weight can help providing centripetal force.

centre of the motion

r mg

Bv

QR

Downward force Q or Upward force R may or may not exist in the

force diagram.


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There are three cases for the circular motion:

1 Weight equals the centripetal force required.

2 Weight is less than the centripetal force required. Q exists.

3 Weight is greater than the centripetal force required.R exists.

For case one, F= ma

r

mvmg

2

=

It happens only when the speed of the object is v = (gr) at B.

The object is just able to move through the top of the vertical circle

using its own weight without the aid of other forces.

For case two, F= ma

rmvmgQ

2

=+

It happens when the speed of the object is greater than v = (gr) at B.

For instance, an up-side-down cart moves through the top of a loop,

for which the normal force of the track acts on the cart in downwards

direction.

If there is no Q, such as a very fast moving car on the convex part of a

road, the car will jump off the road!

For case three, F= ma

r

mvRmg

2

=

It happens when the speed of the object is slower than v = (gr) at B.

For instance, on the convex part of a road, a car moves through it

safely. However, as the normal force acting on the car is less than

mg, the passengers inside the car feels themselves lighter.

If there were no R, such as swinging a bucket of water slowly, the

water inside the bucket falls down. The necessary centripetal force is

small and equals to the weight of part of the water so the other part of

the water falls down.

EXAMPLE 69Looping the loop The passengers on a cart of a roller

coaster can loop the loop without falling downwards at the

top of the loop. If the radius of the loop is 10 m, find the

minimum speed of the cart when it is at the top of the loop.

SOLUTION

Force diagram of the cart.

Minimum speed happens whenN= 0

mg= mv2/r

v = (gr) = 10 m s1.

mg

N


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6.6 Centrifuges

A machine uses for separating solids suspended in liquids or liquids

of different densities. It is also applied in the operation of the spin-

dry portion of many washers.

The mixture is in tubes and when it is rotated in high speed in a

horizontal circle the less dense matter moves towards the centre of

rotation. On stopping the rotation, the tubes return to the vertical

positions with the less dense matter at the top.

Liquid pressure gradient

Consider a liquid inside a test tube, a pressure gradient is set up from

the surface of the liquid to the bottom of the test tube. The force due

to excess pressure P = P2 P1 on any part of the liquid m

balances its weight as shown.

Theory of centrifuges (qualitatively only)

As the tube rotates, a pressure gradient is set up increasing from A to

B. It is because the closed end B provides the necessary centripetal

force on the liquid for rotation, while liquid surfaceA does not.

Therefore, the force due to excess pressure P on any part of the

liquid m supplies exactly the centripetal force required.

However, if this part of the liquid is replaced by a matter of smaller

density ( hence of smaller mass ), the force due to excess pressure willbe greater than the centripetal force required for the smaller mass..

Then less dense matter will be pushed inwards in a spiral path.

P1

P2

m

A

P1

P2

mcenter of

rotation B


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e06 circular motion

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