12/9 Circular Motion Text: Chapter 5 Circular Motion HW 12/9 “Rotating Drum” HW 12/9 “Rotating Disk” These two are for practice (will not be collected) and they also review friction forces. Course Evaluation today

The Pendulum Stationary Swinging (moving in a circular arc) TS,M WE,M ? If the Tension is greater, then acceleration must point up. What happens to the Tension? Is there acceleration?

Find the acceleration a = v/t as usual Draw vi, vf, and v as usual (Place v’s tail to tail and draw tip to tip.

In one second: v

In one second: v

In one second: v What is the acceleration?

In one second: v = a if t = 1 second a What is the acceleration?

In one second: What if the speed is doubled? v

In one second: a

In one second: For v For 2v 4 times the acceleration for twice the velocity, same radius. For v For 2v

What if the radius is halved? In one second:

What if the radius is halved? In one second: v

What if the radius is halved? In one second: 2 times the acceleration for half the radius, same velocity. r a r/2

a a In one second: For r For v For r/2 For 2v v2 So: ac = r 4 times the acceleration for twice the velocity, same radius. 2 times the acceleration for half the radius, same velocity. For r For v a a For r/2 For 2v So: ac = r v2

Centripetal Acceleration perpendicular to the velocity, points towards the center of the circle ac = v2/r v is the instantaneous velocity tangent to the path, r is the radius.

Some Equations and Definitions The period, T, is the time for one revolution. The distance for one revolution is 2r, the circumference. The speed, v, is distance ÷ time or 2r/T Fnet = ma works for centripetal acceleration also and free body diagrams are handled the same way.

The Pendulum Stationary Swinging (moving in a circular arc) TS,M WE,M ? If the Tension is greater, then acceleration must point up. a and Fnet both point up a = v2/r and we can get v from energy.

Example A 1kg ball is swung in a horizontal circle at constant speed at the end of a string. Draw a FBD. Which way does a point? Fnet,y = Ty - WE,B = 0 so Ty = WE,B Fnet,x = Tx = ma = mv2/r v = 2r/T y x TS,B Ty Tx WE,B

What about the centrifugal force? There is no such force, regardless of what Mr. Wizard says.

Bucket of water problem You swing a bucket of water (m=3kg) in a vertical circle at constant speed of 5 m/s. The radius of the circle is 2 m. What is the normal force by the bottom of the bucket on the water at: a. the top of the circle, and b. the bottom of the circle.

Bucket of water problem: at the top v Acceleration points Fnet = WE,W + NB,W = ma Draw a FBD of the water. mw = 3 kg r = 2 m v = 5 m/s (constant v) Want NB,W, find WE,W and ma Apply Newton’s 2nd law. WE,W = mg = 3(9.8) = 29.4 N ma = mv2/r = 3(52)/2 = 37.5 N Fnet = 37.5 = 29.4 + NB,W NB,W WE,W NB,W = 8.1 N If you swing slow enough the water will come out. How slow do you have to swing? Slow enough so that NB,W becomes zero.

Bucket of water problem: at the bottom Acceleration points Fnet = NB,W - WE,W = ma Draw a FBD of the water. mw = 3 kg r = 2 m v = 5 m/s (constant v) Want NB,W, find WE,W and ma NB,W Apply Newton’s 2nd law. WE,W = mg = 3(9.8) = 29.4 N ma = mv2/r = 3(52)/2 = 37.5 N Fnet = NB,W - 29.4 = 37.5 NB,W = 66.9 N WE,W v