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A

B

C

Circular Motion

A ball is going around in a circleattached to a string. If the stringbreaks at the instant shown, which

path will the ball follow? 16

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Uniform Circular motion

v

mg

FN

FTr

m

mw

FN = mg

FT = Tension Force

FT = ma

FT = m (v2 / r)

T = mwg as it is at rest.mwg = m v2 / r 

mw= mass

(downward)

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Uniform Circular motion

v

mg

FN

Tr

mw

FT = m (v2 / r)

What is the physical meaning of the formula?(1) What happens if the initial speed v of the bob is forced to move faster?

A greater T is needed, the greater mw, to keep the radius of motion r.

It will slide out to attain a greater r.

(2) What happens if there is energy lost due to friction on the plane?v decreases By T = mv2 / r, r decreases

(3) What happen if an additional mass is added to the weight mw?

M moves down r decreases and T increases v increases

FT

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Uniform Circular motion

Uniform circular motion in a horizontal plane

v

mg

FN 

FTzr

m

M

FT = m (v2 / r)

• Tension suddenly disappears and there is no more centripetal

force

• Circular motion cannot be maintained.

• The bob will move with its velocity tangent to the circle at

the instant the string breaks

v

mg

FN 

FTzr

m

M

v

mg

FN 

FTzr

m

M

v

mg

FN 

FTzr

m

M

Name 3 events as a result of a cablesuddenly breaking:

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Circular Motion Review

When we see an objectcarrying out circularmotion, we know thatthere must be force actingon the object, directedtowards the center of thecircle.

When you look at thecircular motion of a ballattached to a string, theforce is provided by the

tension in the string.

When the force responsiblefor the circular motiondisappears, e.g. by cuttingthe string, the motion will

become linear.

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Circular Motion Review 2

In most cases,the string forcenot only has to

provide the forcerequired forcircular motion,but also the forcerequired tobalance thegravitational

force.

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Circular Motion and its

Connection to FrictionWhen you drive your car arounda corner you carry out circularmotion.

In order to be able to carry outthis type of motion, there must

be a force present that providesthe required accelerationtowards the center of the circle.

This required force is providedby the friction force between the

tires and the road.

But remember ….. The frictionforce has a maximum value, andthere is a maximum speed withwhich you can make the turn.

Required force = Mv2/r.

If v increases, the friction force

must increase and/or the radius

must increase.

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Questions 2 You are driving a car with

constant speed around a

horizontal circular track. On apiece of paper, draw a Free

Body Diagram for the car.

How many forces are acting on

the car? A) 1

B) 2

C) 3

D) 4

E) 5 

25

The net force on the car is 

A. Zero

B. Pointing radially inward

C. Pointing radially outward 

Gravity

Tension

3 forces, Tension(friction), gravity andthe normal force

B: Force of tensionis pointing inward

Normal

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Question 3 Davain sits on the outer rimof a merry-go-round, andDiego sits midway between

the center and the rim. Themerry-go-round makes onecomplete revolution everytwo seconds.– Diego’s velocity is: 

(a)  the same as Davian’s

(b)  Faster than Davian’s 

(c) Slower than Davian’s

Diego Davian

a = V2 /r(C) Diego is slowerthan the velocity of Davian

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Summary

Uniform Circular Motion

–Speed is constant

–Direction is changing

–Acceleration toward center a = v2 / r

–Newton’s Second Law F = ma