# Minimum Vo for a hoop to get to the top

Tags:
1. Jun 20, 2016

### Cleo James

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

part(b) and (c)
2. Relevant equations
Conservation of momentum
Conservation of angular momentum
Conservation of mechanical energy

3. The attempt at a solution
So first I thought that I could do it by just using the conservation of mechanical energy, but then I realized that since the lower part of the hoop hits the platform, the energy can't be conserved (there's energy lost):

1/2 mv^2 + 1/2 Iw^2 = mgh + Energy lost
mv^2 = mgh + Energy lost

I think that part (b) and (c) is very related, once we know the energy lost, we can determine the v needed, however I don't really understand how to find its value. Any suggestion ??

2. Jun 20, 2016

### haruspex

You are right that KE will not be conserved. What other conservation possibilities are there?
You mentioned two. What is stopping you from trying to use them?

Last edited: Jun 20, 2016
3. Jun 20, 2016

### Cleo James

Well, conservation of momentum:
m1v1 + m2v2 = (m1+m2)v3
But this is if the object experiences inelastic collision, I don't know how to apply it when the other object is just a small bump.
Conservation of angular momentum:
Iw1= Iw2
since the hoop will rotate about the edge of the bump during collision, parallel axis theorem:
mr^2 w1= 2mr^2 w2
w1= 2w2

At this time I can't seem to figure out what to do afterwards.

4. Jun 20, 2016

### haruspex

The problem with those two conservation laws here is that you have an unknown impulse from the step. So you need to find a way of applying one of them in which that impulse makes no contribution.
If you knew the direction of the impulse you could consider momentum orthogonal to it; but you don't.
Angular momentum is always relative to some chosen axis. Can you choose an axis such that the impulse has no moment about it?