# Loop-the-loop with potentional and kinetic energy

1. Oct 4, 2014

### Yae Miteo

1. The problem statement, all variables and given/known data

A mass m = 77 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.1 m and finally a flat straight section at the same height as the center of the loop (15.1 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

part 2) What height above the ground must the mass begin to make it around the loop-the-loop?

2. Relevant equations

$$PE=mgh$$
$$KE = \frac {mv^2}{2}$$

and perhaps
$$a_c = \frac {v^2} {r}$$

3. The attempt at a solution

I began by setting KE = PE
$$KE = PE$$
$$mgh = \frac {mv^2}{2}$$

cancel out
$$\frac {v^2}{r} = gh$$

solve for h
$$h = \frac{v^2}{2g}$$

plug in and find h
$$h=7.54999$$

$$22.65$$

However, this is not the correct answer. I feel that I am close, but I am not sure what I am missing.

2. Oct 4, 2014

### Orodruin

Staff Emeritus
Kinetic energy where and potential energy where?

Note that (as stated in the problem) the mass will not make it through the loop if it has zero velocity at the top of the loop. Can you find a condition that must be fulfilled for the mass to make it around?

3. Oct 4, 2014

### Yae Miteo

It needs enough kinetic energy to reach the top, which gets turned into potential energy right when it gets there. That way, it will make it around the loop.

4. Oct 4, 2014

### Orodruin

Staff Emeritus
This is exactly what the problem statement told you will not happen. If it, by some magical means, reached the top with zero velocity, what would happen to it right after it made it to the top?

5. Oct 4, 2014

### Yae Miteo

Hmm... if it had zero velocity at the top, it would stop and fall straight down. I need to develop a different approach.
Perhaps at the top

6. Oct 4, 2014

### Yae Miteo

$$KE \gt PE$$

7. Oct 4, 2014

### Orodruin

Staff Emeritus
Can you perhaps think of a condition on the velocity that must be fulfilled for the mass not to fall down faster than the track is bending?

8. Oct 4, 2014

### Yae Miteo

Centripetal acceleration must be greater than gravitational acceleration.

9. Oct 4, 2014

### Orodruin

Staff Emeritus
Exactly, so this is what gives you the relation between gravitational and centripetal acceleration in the limiting case.

Looking at your OP again, it seems to me that you have a velocity already since you say you plug it in. It also seems part 1 of the problem is missing and that makes me suspect that finding this velocity was part 1. This is important information that should be included. Is this the case? If so, you did have the correct equations, although the argumentation was not clear - which is why I asked for where the potential should be equal to a kinetic energy which was measured where. Total energy is conserved and the kinetic energy at release is zero. If kinetic energy equals potential energy at that point, they will not at a later. The argument you are looking for is that the gain in kinetic energy is equal to the loss of potential energy.