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CIRCULAR CIRCULAR MOTIONMOTIONCIRCULAR CIRCULAR MOTIONMOTION

CIRCULAR CIRCULAR MOTIONMOTIONCIRCULAR CIRCULAR MOTIONMOTION

CIRCULAR CIRCULAR MOTIONMOTION

CIRCULAR MOTIONCIRCULAR MOTION

2

r

s

IN RADIANSlength of the arc [ s ] divided by the radius [ r ] subtending the

arc

rs

© 3

22 r

r

IN A FULL CYCLEIN A FULL CYCLE

RADIANSRADIANS

rs

Hence 3600 = 2 Radians

1800 = radians

4

v

v

r

s

A

B

Suppose the distance, s, from A to B takes a time t.r

s rs

tr

t

s

rvv = tangential

velocity

t

= the ANGULAR VELOCITY

[DIVIDING EACH

SIDE BY t]

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v

v

r

rv v = tangential velocity t

= the ANGULAR VELOCITY in

RADIANS PER SECOND

Considering a full circle:

T

2 f 2

T = Periodic time,

f = frequency

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Newton’s 1st Law “I want to go straight on!”

Come back, I am a F

v

Circular motion requires a force, F, towards the centre of the motion

F

v

A CENTRIPETAL FORCE

Something has to provide it, e.g. the tension in a string

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v

v

Angular Velocity

t

Linear Velocityv = r

Centripetal Acceleration

22

rrv

a

Centripetal Force

22

mrrmv

F

SUMMARY

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F

The necessary centripetal force, F is provided by the friction at the tyres

Question If the track has a radius of 50m and the limiting frictional force is 0.5 of the car’s weight, find the car’s maximum speed before it slides off the track

rmv

F2

50

5.02mv

mg

5081.95.0 v v = 15.7 ms-1

MOTION IN A HORIZONTAL CIRCLE

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MOTION IN

A VERTICAL CIRCLE FORCES ACTING: 1. Weight2. Tension in the string

mg

Tv

rmv

F2

rmv

mgT2

Where F is the resultant force

towards the centre

rmv

mgT2

At the bottom the tension has to provide a centripetal force AND support the weight

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MOTION IN

A VERTICAL CIRCLE FORCES ACTING: 1. Weight2. Tension in the string

mg

T

v

rmv

F2

At the top the tension is less than at the bottom, because the weight provides some of the centripetal force

rmv

mgT2

mgr

mvT

2

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A

B10 m

The radius of the track is 10 m

Using g = 10 ms-2, find the g forces at A and B. [Assume no energy is lost]

20m

12

Using EK gain = Ep lost gives velocity at the bottom = 24.5 m s-1

At the bottom

mg

R

v

r

mvmgR

2

This gives R = 70 m

710

70

m

m

mg

RHence

At the top

mg

R r

mvmgR

2

Energy change gives v = 14.1 m-1

R = 10 m and result is 1 g