Announcements!

• Extra credit posted this coming Sunday--get your team together!• Mon/Tue-Circular applications of Newton’s Laws•Good examples in the book!• Fuddrucker’s tomorrow (Tuesday) night•Test this Thursday/Friday--practice test posted now!

An object moving in a circle of radius r w/ constant speed v has an acceleration directed toward the middle of the circle, with a magnitude

∆l θr v1v2r

€

ac =v2

r

Centripetal Acceleration

“Kinematics” = how things move

“Dynamics” = why things move

Objects moving in circles are accelerating (centripetally), so the next question is: “What causes centripetal acceleration?

The answer: centripetal force.

Centripetal Force!

Fcentripetal

Centripetal Force = the “center-seeking” force necessary to keep an object moving in a circle.

It’s important to understand that centripetal force is a general name used to describe any force(s) that keep an object moving in a circle, just like we use Fnet generally to describe the forces that accelerate an object linearly. The actual forces that make an object move centripetally are due to things that we’ve already discussed: FTension for ropes, cords, strings, etc., Fgravity for graviational force, Ffriction for... friction, and (in some odd cases) Fnormal, etc.

Centripetal Force

Centripetal Force = the “center-seeking” force necessary to keep an object moving in a circle.

Centripetal Force

€

F = ma∑

ac =v 2

r

Fc = mac =∑ mv 2

r

ExampleAt the beginning of a hammer throw, a 5 kg mass is swung in a horizontal circle of 2.0 m radius, at 1.5 revolutions per second. What is the centripetal force required to keep this object moving in a horizontal circle? (Ignore the vertical effect of gravity.)

Example: A 1000 kg car on a flat road is traveling at 14 m/s.

a. On a curve of radius 50m, what centripetal force will be necessary to keep the car on the road?

b. If the µ static for this road is 0.60, will the car make the turn?

c. If the µ static for the road is 0.20, will the car make the turn?

d. What is the maximum speed the car can have and still make the turn?

Cool problem

Example: A 1000 kg car on a flat road is traveling at 14 m/s.

d. How high should we bank the turn if we want the car to be able to stay on with no friction?

Advanced Cool Problem

Example: A small body of mass m is suspended from a string of length L which makes an angle θ with the vertical. The body revolves in a horizontal circle. Find the speed of the body and the period (time) of one revolution in terms of L, θ, and fundamental constants.

How the heck...? Problem

Vertical Circles

Vertical CircleObjects that are traveling in vertical circles are treated exactly the same as objects traveling in horizontal circles: the sum of the centripetal forces adds up to allow the object to accelerate centripetally, and thus, travel in a circle. (∑Fc=mac)

Vertical Circles

Vertical CircleOne obvious challenge is that the Tension in the rope (in this example) is no longer the only force that is a factor in the ball’s circular motion--the force of gravity now has to be taken into account.

Vertical Circles

Vertical CircleExample: A 1.8-kg ball is being

swung in a vertical circle, at the end of a 1.2-m long rope.

a. What is the minimum velocity the ball can have at the top of its circular path?

b. What is the Tension in the rope as the ball swings past the bottom of the path at 5.0 m/s?

Vertical Circles

Vertical CircleFg FTension a. What is the minimum velocity

the ball can have at the top of its circular path?

Vertical Circles

Vertical CircleFg FTension a. What is the minimum velocity

the ball can have at the top of its circular path?

€

Fc = mac∑

Fg + FTension = m(v2

r)

mg + 0 = m(v2

r)

v = rg = (1.2m)(9.8m /s2)

v = 3.43m /s

Vertical Circles

Vertical CircleFg FTension b. What is the Tension in the

rope as the ball swings past the bottom of the path at 5.0 m/s?

Vertical Circles

Vertical CircleFg FTension

€

Fc = mac∑

−Fg + FTension = m(v2

r)

−mg + FTension = m(v2

r)

FTension = m(v2

r) + mg

FTension = (1.8kg)((5m /s)2

1.2m) + (1.8kg)(9.8m /s2) = 55.1N

b. What is the Tension in the rope as the ball swings past the bottom of the path at 5.0 m/s?

So... what is centrifugal force?

Centrifugal force is an “apparent” force--a fake force!--that we mistakenly think pulls an object away from the center of the circle. There is no force pulling the ball outward!!!

If you’re holding on to the string attached to the ball while it goes in a circle, it’s true that your hand feels an outward pull: this is due to Newton’s 3rd Law (your hand pulls on the ball to keep it moving in a circle, the ball pulls back on your hand).

Centrifugal Force

It may help to think about what happens when we let go of the string: does the ball go flying away from the center of the circle, due to the mysterious, fake, centrifugal force?

No! It continues to travel in a straight line from the point where it was let go.

The fake centrifugal force is due to ball’s inertia.

Centrifugal Force

Earlier, we said that an object moving in a circle can have radial and tangential accelerations.

These obviously result from radial and tangential forces.

Calculate the radial and tangential forces acting on the billiard ball, in terms of m, v, r, and Ø (angle from vertical).

Non-Uniform Circular Motion

€

ra =

r a t +

r a r

€

rF =

r F t +

r F r