Circular Motion

What is circular motion?

• Objects that move in a circle experience circular motion.

• I know that’s tough.

• Let’s take a moment and let it sink in…

Now that that is out of the way…

• There are specific features of circular motion that make it different from linear or projectile motion

Constant speed

• An object with a constant speed, which experiences no other forces, will travel in a straight line at that speed infinitely

• Newton’s First Law

Constant speed

Constant speed

• Objects can travel in a circle and maintain a constant speed

Is velocity constant?

• No

• Velocity is speed in a given direction

• Those directions cannot be circular

What is acceleration?

• Acceleration is a change in velocity

• We have so far defined acceleration as a change in speed but it can be a change in direction, also

Circular Motion

• An object traveling in a circle travels at a constant speed but is accelerating

What causes acceleration?

• All changes in velocity are caused by a force

• F = ma

• Newton’s Second Law

What is the force?

• The force keeping the object in its circular path is called a centripetal force

• Centripetal means “center seeking”

Centripetal force

• It is a real force

• It is a contact force

What direction does it point?

• The centripetal force always points towards the center of the circle

What applies the force?

• It depends on the situation

• In general, whatever keeps the item in it’s circular path applies the centripetal force

Example

Example

Example

Are there other forces?

• When you make a turn in your car, what makes you pull to one side?

• When you swing a bucket of water above your head, what keeps the water in the bucket?

What causes that?

• In truth, it is a delicate interplay between the inertia of the item and the acceleration

• It is another force

Centrifugal force

• From the Latin, centrum, “center,” and fugere, “fleeing”

• This is the force that pushes away from the center of the circle

Centrifugal Force

• It is the reaction force that compliments the action of the centripetal force

• Newton’s Third Law

Centrifugal

• The centrifugal force is a fictitious force

• Is it also a contact force

Example

• You have a bucket of water and you are swinging it around above you head. What forces are acting on it and what do they act on?

The two forces

• Remember, we have two forces, the centripetal and the centrifugal

• The centripetal acts on the bucket

• The centrifugal acts on the water

The math

• You knew it was coming

• Math is the language of physics and you need to learn to speak that language

Centripetal acceleration

• There are two equations we can use depending on what we know

The first (and easiest)

• v is the velocity of the object

• r is the radius of the circle

r

vac

2

The second

2

24

T

rac

• T is the time it takes for one full revolution

• r is the radius of the circle

Centrifugal acceleration

• If the centrifugal force arises from Newton’s Third Law and is the equal but opposite reaction to the centripetal force, what is the equation going to be?

Centrifugal acceleration

2

24

T

rac

Or

r

vac

2

Sample problem

• A 1000 kg car enters an 80 meter radius curve at 20 m/s. What centripetal force must be supplied by friction so the car does not skid?

What do we know?

• m = 1000 kg

• r = 80 m

• v = 20 m/s

Find the force

• F = ma = mv2/r

• F = 1000 × (202/80)

• F = 1000 × 400/80

• F = 1000 × 5 = 5000 N

Sample problem

• The centripetal force on a 0.82 kg object on the end of a 2.0 m massless string being swung in a horizontal circle is 4.0 N. What is the tangential velocity of the object?

What do we know?

• m = 0.82 kg

• r = 2.0 m

• Fc = 4.0 N

Find the velocity

• F = ma = mv2/r

• 4.0 = 0.82 × v2/2.0

• 8.0 = 0.82v2

• v2 = 9.76

• v = 3.12 m/s

Sample problem

• A dragonfly is sitting on a merry-go-round 2.8 m from the center. If the centripetal acceleration of the dragonfly is 3.6 m/s2, what is the period of the merry-go-round?

What do we know?

• r = 2.8 m

• a = 3.6 m/s2

Find the period

• ac = (4π2r)/T2

• 3.6 = (4π2 × 2.8)/T2

• 3.6 = 110/T2

• T2 = 31

• T = 5.5 s

Sample problem• A car moving at a 1.08 × 108

m/s (30 km/h) rounds a bend in the road with a radius of 21.2 m. What is the centripetal acceleration on the car and the centrifugal acceleration on the occupants?

What do we know?

• v = 1.08 × 108 m/s

• r = 21.2 m

Centripetal

• a = v2/r

• a = (1.08 × 108)2 / 21.2

• a = 5.50 × 1014 m/s2

Centrifugal

• a = -v2/r

• a = -(1.08 × 108)2 / 21.2

• a = -5.50 × 1014 m/s2