Minimum Vo for a hoop to get to the top

In summary: 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.
  • #1
Cleo James
2
0

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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  • #2
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:
  • #3
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.
 
  • #4
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?
 

Related to Minimum Vo for a hoop to get to the top

What is the concept of "Minimum Vo for a hoop to get to the top"?

The concept of "Minimum Vo for a hoop to get to the top" refers to the minimum initial velocity that is required for a hoop to successfully reach the top of a loop-the-loop track without falling off.

How is the minimum Vo calculated for a hoop to get to the top?

The minimum Vo is calculated using the conservation of energy principle, where the kinetic energy of the hoop at the top of the loop is equal to its potential energy at the bottom of the loop. This equation can be rearranged to solve for the minimum initial velocity.

What factors affect the minimum Vo for a hoop to get to the top?

The factors that affect the minimum Vo include the radius of the loop, the mass of the hoop, and the force of gravity. A smaller radius, lighter hoop, and weaker gravity will result in a lower minimum Vo.

Why is it important to understand the minimum Vo for a hoop to get to the top?

Understanding the minimum Vo for a hoop to get to the top is important in designing and building safe and successful loop-the-loop tracks. It ensures that the hoop has enough initial velocity to complete the loop without falling off.

Can the minimum Vo be exceeded for a hoop to get to the top?

Yes, the minimum Vo is the minimum required velocity for the hoop to successfully complete the loop. If the hoop has a higher initial velocity, it will still be able to complete the loop but may experience more force and stress on the track and the hoop itself.

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