# Work, Power, and Energy

1. Nov 7, 2005

### Erwin Schrodinger

Hmm I thought I was pretty good at this stuff but I guess not. I'd really appreciate it if someone could show me how to solve this problem. Does it involve anything more than work, power, and energy concepts?

A small mass m slides without friction along the looped apparatus shown. If the object is to remain on the track, even at the top of the circle (whose radius is r), what minimum height h must it be released?

2. Nov 7, 2005

3. Nov 7, 2005

### Erwin Schrodinger

What do you mean just energy? Sorry I don't understand your response. All I could come up with was mgh = (something), but I don't know what to equate it to.

4. Nov 7, 2005

### whozum

Just energy methods are needed. The answer I'm getting is given right in the picture.. to "loop the loop" you need enough kinetic energy when youre at the bottom of the circle to get up to the top, so mg(2r) needs to be smaller than the KE of the object at the bottom. So the object has to get atleast that KE from falling from a height h. What is h?

5. Nov 7, 2005

### Erwin Schrodinger

Hmm okay I think I understand.

For the sliding down the slope part:

mgh = 1/2mv^2

Then for the loop:

1/2mv^2 > mg(2r) so the ball continues to move when its at the top of the loop

Then:
mgh > mg(2r)
h > 2r

Is that correct?

6. Nov 7, 2005

### whozum

Yes sir. It asked for a minimum height so you don't need a >

7. Nov 7, 2005

### Erwin Schrodinger

So I should just write h = 2r then? Great! Thanks for the help.

8. Nov 8, 2005

### Galileo

Sorry to interrupt, but that is not correct. If h=2r, then the ball will reach the top of the loop with zero velocity and fall down. It will in reality fall down before it reaches the top, since it has motion in the horizontal direction as well.

To complete the loop, the ball must have enough kinetic energy at the top so it makes a circular trajectory and not fall down in a parabolic path, which it does when it looses contact with the loop.
To maintain circular motion, you need centripetal force, which is provided in this case by the normal force of the loop and by gravity. Can you figure out what velocity the ball should have at the top so that gravity will provide the necessary centripetal acceleration?

9. Nov 8, 2005

### kp

you have both kinetic energy (1/2 mv^2) as well as potential energy (mg2r) at the top of the loop.

10. Nov 8, 2005

### whozum

I can't believe I ****ed that up. Is 5/2 the correct coefficient?

11. Nov 8, 2005

### Galileo

Yup. blahdiblah