Aim: How can we explain circular motion? Do Now: An object travels 5 m/s north and then travels 5 m/s east. Has the object accelerated?

Aim: How can we explain Aim: How can we explain circular motion?circular motion?

Do Now:

An object travels 5 m/s north and then travels 5 m/s east. Has the object accelerated?

Velocity is defined by two things:

•Magnitude (5 m/s)

•Direction (north)

Direction has changed

Therefore velocity has changed

A change in velocity is defined as acceleration

Up to now, we only dealt with linear acceleration (objects traveling in a straight line speeding up or slowing down)

Changing direction – this is centripetal acceleration

In what direction does velocity In what direction does velocity act?act?

Velocity acts tangent to the Velocity acts tangent to the circle made by the object.circle made by the object.

vv

v v

Demo

In what direction does In what direction does acceleration act?acceleration act? Acceleration acts Acceleration acts

inward, towards the inward, towards the center.center.

Is there a force?Is there a force? YesYes Newton’s 2Newton’s 2ndnd law – unbalanced forces law – unbalanced forces

produce accelerationproduce acceleration If there is acceleration, there must be If there is acceleration, there must be

an unbalanced forcean unbalanced force This is centripetal forceThis is centripetal force

In what direction does In what direction does the force act?the force act?

The force acts inward, towards the The force acts inward, towards the center.center.

FcFc

Fc

Fc

Centripetal force and Centripetal force and the centripetal the centripetal acceleration are always acceleration are always pointing in the same pointing in the same directiondirection

Centripetal force is not a separate force

It is whatever force is pointing towards the center of the circle

In this example, the normal force is pointing towards the center of the circle

Therefore the normal force is the centripetal force

What is supplying the centripetal force?

Spinning an object attached to a string in a circle

Tension in the string

Turning a car

Friction between the tires and the road

Ever drive and hit a patch of ice?

No more friction -- no more turn

The car skids in a straight line (tangent to the circle)

Walking in a circle

Friction between your shoes and the floor

Ever try to run in dress shoes and make a sharp turn?

OUCH!

If centripetal force is directed towards the center, why do you feel a “force” pushing you away from the center of the circle when in this motion, like turning in a car?

The object gets pushed away from the center

Remember velocity?

The object wants remain in motion tangent to the circle

The car just gets in the way

This gives the “illusion” of a force that really does not exist!

Water in cup Demo (during lab)

How can we calculate How can we calculate centripetal acceleration?centripetal acceleration?

r

vac

2

Where Where vv is the velocity and is the velocity and rr is the is the

radius of the circle traveled by the radius of the circle traveled by the objectobject

How do we calculate the How do we calculate the centripetal force?centripetal force?

F = ma, so FF = ma, so Fcc = ma = macc

Substitute in aSubstitute in acc =v =v22/r/r

r

mvFc

2

A car whose mass is 500 kg is traveling around a circular track with a radius of 200 m at a constant velocity of 15 m/s.

What is the centripetal acceleration?

Given

m = 500 kg

r = 200 m

v = 15 m/s

ac = ?

r

vac

2

m

smac 200

)/15( 2

2/1.1 smac ac = 1.1 m/s2

What is the centripetal force?

Fc = ?

Fc = mac

Fc = (500 kg)(1.1 m/s2)

Fc = 550 N

What if the velocity isn’t What if the velocity isn’t given?given?

The distance traveled by the The distance traveled by the object is the circumference of object is the circumference of the circle; C = 2the circle; C = 2ππrr

t

r

t

dv

2r

ExampleExampleA 615 kg racing car completes one lap in 14.3 s around a

circular track with a radius of 50.0 m. The car moves at a constant speed.

(a) What is the acceleration of the car?

r

vac

2

Given

m = 615 kg

d = c = 2πr

t = 14.3 s

r = 50 m

ac =?

r

t

d

ac

2

r

t

r

ac

22

m

s

m

ac 50

3.14

)50(22

m

smac 50

)/22( 2

2/7.9 smac

(a)What force must the track exert on the tires to produce this acceleration?

Given

m = 615 kg

d = c = 2πr

t = 14.3 s

r = 50 m

ac = 9.7 m/s2

Fc = ?

Fc = mac

Fc = (615 kg)(9.7 m/s2)

Fc = 5,965.5 N