12/9 circular motion text: chapter 5 circular motion hw 12/9 “rotating drum” hw 12/9...

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

TS,M

WE,M

Stationary Swinging (moving in a circular arc)

What happens to the Tension?

Is there acceleration?

TS,M

WE,M

?If the Tension is greater, then acceleration must point up.

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:

a

What is the acceleration?

v = a if t = 1 second

In one second:What if the speed is doubled?

v

In one second:

a

In one second:

For 2v

For v

4 times the acceleration for twice the velocity, same radius.

In one second:What if the radius is halved?

In one second:What if the radius is halved?

v

In one second:What if the radius is halved?

r/2

ra

2 times the acceleration for half the radius, same velocity.

In one second:

For r/2

For ra

2 times the acceleration for half the radius, same velocity.

4 times the acceleration for twice the velocity, same radius.

For 2v

For va

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 2r, the circumference.The speed, v, is distance ÷ time or 2r/T

Fnet = ma works for centripetal acceleration also and free body diagrams are handled the same way.

The Pendulum

TS,M

WE,M

Stationary Swinging (moving in a circular arc)

TS,M

WE,M

?If the Tension is greater, then acceleration must point up.

a = v2/r and we can get v from energy.

a and Fnet both point up

Example

WE,B

TS,B

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?

y

x

Ty

Tx

Fnet,y = Ty - WE,B = 0 so Ty = WE,B

Fnet,x = Tx = ma = mv2/r

v = 2r/T

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, andb. the bottom of the circle.

Bucket of water problem: at the top

v

mw = 3 kgr = 2 mv = 5 m/s(constant v)

Acceleration points

Draw a FBD of the water.

WE,W

NB,W

Apply Newton’s 2nd law.

Fnet = WE,W + NB,W = ma

Want NB,W, find WE,W and ma

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 = 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

v

mw = 3 kgr = 2 mv = 5 m/s(constant v)

Acceleration points

Draw a FBD of the water.

WE,W

NB,W

Apply Newton’s 2nd law.

Fnet = NB,W - WE,W = ma

Want NB,W, find WE,W and ma

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