# Mechanics - Uniform Hoop and Angular Velocity

1. Nov 26, 2006

### Asrai

A heavy uniform hoop of mass M and radius R rolls in a straight line on level ground without slipping. It encounters a step of height h, where h<R perpendicular to its path. On striking the corner of the step, the hoop does not rebound and has sufficient energy to just surmount it. If the angular velocity of the hoop about the corner just after impact is omega, show that:

omega = ((g*h)^0.5)/R

I've solved this problem using conservation of energy, saying that the sum of the linear and angular kinetic energy before the impact is equal to the potential energy the hoop gains by surmounting the corner. The thing is, while the maths works out fine, I'm wondering while the hoop is still moving at all? If it's just got enough energy to surmount the corner, surely it would be stationary and not have any angular velocity?

Additionally, I'm wondering whether the omega that I've calculated is just the omega before the impact; the angular velocity that's necessary to surmount the corner.

2. Nov 26, 2006

### OlderDan

There seems to be a contradiction in your statements. If all the kinetic energy were converted to potential energy, the hoop would no longer be moving, as you have observed. So you cannot possibly have gotten a non-zero value for ω from your calculation.

The no-rebound condition is quite artificial, but if you take that as a condition and assume the hoop stays in contact with the step there will be a torque acting on it. I doubt energy conservation is going to work for you here because the hoop would have to bounce if energy were conserved.

3. Nov 26, 2006

### Asrai

Oh dear.

I thought this problem sounded dodgy... But how do you then calculate omega, by approaching the problem using torque? What I don't understand it that it hast just the right energy to surmount the counter but still has angular velocity after arriving at the top- how can that possibly fit together?

4. Nov 26, 2006

### OlderDan

I need to look at this again. I think I misinterpreted it. You are probably correct that what you calculated was the ω it had before it left the ground and climbed the step. Once the impact has ocurred, energy can be conserved as long as there is no slipping. I'll assume there is a clamp that grabs the wheel when it hits but is free to rotate.

OK. That's what it is. ω is the angular velocity it has before it leaves the ground and it somehow stays attached to the corner of the step and rotates to the top where it comes to rest. Could never happen in the real world, but that's OK.

Last edited: Nov 26, 2006
5. Nov 26, 2006

### Asrai

I don't know about the clamp- that is the literal transcription of the problem as it is in my booklet (except that it then goes on about some other calculations). Nevertheless, if my calculations concering omega were correct (the omega before it left the ground), how can the omega after the impact that's given in the problem possibly be the same? That would mean that all the linear kinetic energy is transformed to potential energy but that angular kinetic energy is conserved- is that even physically possible?

6. Nov 26, 2006

### OlderDan

I think you misinterpreted the problem as well, which is what got me thinking it was something other than it is. When it gets to the top, the hoop is not moving. The ω calculation is the ω it has before it leaves the ground, not when it gets to the top. At the top it is at rest.

7. Nov 26, 2006

### Asrai

But wouldn't the angular velocity change? In this problem, I though h to be constant, so the angular velocity would remain constant throughout the surmounting? Also, can I use conservation of energy or do I have to use torques to get to the answer?

8. Nov 26, 2006

### OlderDan

The angular velocity does change. It goes from the value you calculated to zero as the hoop climbs the step. You can use conservation of energy for this process. The pivot point is the corner of the step. You have to assume the hoop does not slip and that it stays in contact with the step the whole time.

9. Nov 27, 2006

### Asrai

Ok, so I've got that bit of the problem. It then goes on:

Consider motion before and after impact and use conservation of angular momentum to show that the hoop rolls with a velocity V before impact as follows: V = beta*R*omega where beta = (2R)/(2R-h)

Now, I don't even know where to start! V is obviously bigger before the impact than after it; and V after the impact is just (g*h)^0.5. If I do all the calculations for the angular momentum, I just get V = (g*h)^0.5 as well, so I'm thinking I'm missing a detail. Is there really no external torque acting on the system?

10. Nov 27, 2006

### OlderDan

Before impact, the hoop has an angular velocity related to its linear velocity (no slipping). After impact, it has a new angular velocity that was calculated in the first part of the problem. The force of impact from the corner of the step cannot change the angular momentum about that contact point. (There is no torque about that point until the hoop leaves the ground and gravity is no longer balanced by the normal force from the ground.)

Write an expression for the angular momentum relative to the contact point before impact using the initial angular and linear velocities. Write another expression for the angular momentum just after impact using the previously calculated angular velocity right after impact. Equate these expressions and use the linear-angular velocity relationship to solve for the angular and linear velocities just before impact.

It works. I did it.

11. Nov 27, 2006

### Asrai

Maybe I'm just too thick to really get that problem, but how do you do that? I've got L = M*R*(gh)^0.5 for the angular velocity right after impact; but how do I get to the expression for angular velocity right before the impact? I know that omega' = V/R, but that's about it; if I then use L = I*omega... well, it doesn't work.

12. Nov 27, 2006

### OlderDan

You need the moment of inertia about the contact point to calculate the L after the collision. You are off by a factor of 2. Before the collision, the angular momentum about the contact point can be calculated as the angular momentum of the CM plus the angular momentum about the CM. For that you will need the moment of inertia about the CM and L written in terms of the linear and angular velocities converted to an expression in terms of one of those velocities.

13. Nov 27, 2006

### Asrai

Thank you so much! I'm still not quite sure whether everything I've done is quite right, but everything makes a lot more sense now!