Minimum Vo for a hoop to get to the top

AI Thread Summary
The discussion focuses on determining the minimum velocity required for a hoop to reach the top of a bump, considering energy loss during the collision with the platform. The conservation of mechanical energy is deemed inadequate due to energy loss, prompting a shift to conservation of momentum and angular momentum. Participants explore the application of these conservation laws, particularly in the context of an inelastic collision and the hoop's rotation about the bump. The challenge lies in accounting for an unknown impulse from the bump, complicating the use of momentum conservation. A suggestion is made to select an axis for angular momentum that negates the impulse's effect, facilitating further calculations.
Cleo James
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Homework Statement


upload_2016-6-20_22-1-39.png
upload_2016-6-20_22-1-50.png

part(b) and (c)

Homework Equations


Conservation of momentum
Conservation of angular momentum
Conservation of mechanical energy

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 ??
 
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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:
haruspex said:
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?
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.
 
Cleo James said:
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.
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?
 

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