1. 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

2. 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?

3. 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.

4. In one second: v

5. In one second: v

6. In one second: v
What is the acceleration?

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

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

9. In one second: a

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

11. What if the radius is halved?
In one second:

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

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

14. 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

15. 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.

16. 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.

17. 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.

18. 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

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

20. 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.

21. Bucket of water problem: at the topv
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 = 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.

22. 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 = 37.5
NB,W = 66.9 N
WE,W v