# Rolling and climbing cylinder

1. May 5, 2015

### mooncrater

1. The problem statement, all variables and given/known data
The question says:
A uniform solid cylinder rolling with angular velocity $\omega$ along a plane surface strikes a vertical rigid wall. With what angular velocity the cylinder begins to roll up the wall because of impulsive blow? It is observed it rolls without sliding after striking the wall.
1. $\omega$/2
2. $\omega$/3
3. $\omega$/5
4. $\omega$/4

2. Relevant equations
$I_1\omega _1=I_2\omega _2$

3. The attempt at a solution
Can we conserve angular momentum along the point at which the cylinder strikes the wall? I don't think so because that point gives an impulse to the cylinder.

2. May 5, 2015

### haruspex

what moment would that impulse have about that point?

3. May 5, 2015

### mooncrater

Since it passes through it, zero?

4. May 5, 2015

Yes.

5. May 5, 2015

### ehild

Friction plays the main role. During the collision, some impulsive force of friction is exerted on the cylinder. That force causes upward acceleration and its torque changes the initial angular velocity. Both the momentum and the angular momentum change.

6. May 5, 2015

### haruspex

Indeed, but that also has no moment about the point of contact, so the OP's proposed method works fine.

7. May 5, 2015

### ehild

The cylinder slides during the contact, it does not roll.

8. May 5, 2015

### haruspex

Not according to the OP.

9. May 6, 2015

### mooncrater

Okay then, I am a little confused about the conditions for applying conservation of angular momentum. Is it that the torque around it should be zero?

10. May 6, 2015

### ehild

When the rolling cylinder hits the wall, it rubs the wall. There is sliding friction. The collision takes some time when the impulsive force and impulsive torque take effect. As a result, the cylinder looses its horizontal velocity and gains an upward velocity, and also its angular velocity changes.

11. May 6, 2015

### ehild

You can apply the the torque equation with respect of a fixed axis or with respect to the CM. Some Δt time is needed to stop the horizontal motion of the cylinder. During that time, the surface of the cylinder slides on the wall. There is no instantaneously fixed axis of rotation.

12. May 6, 2015

### haruspex

I see no reason to assume that. If the normal impulse is J, the frictional impulse due to static friction is up to $\mu_sJ$. If that is enough to provide the vertical velocity consistent with the angular momentum conservation then no slipping.
If we do allow slipping on contact, is there enough information to solve the problem? I doubt it.

13. May 6, 2015

### ehild

Have you solved the problem? Have you got one of the given values?

14. May 6, 2015

### haruspex

Yes.

15. May 6, 2015

### ehild

Well, you are right, angular momentum is conserved with respect to the contact point at the wall. The result is the same obtained with my method.

16. May 6, 2015

### mooncrater

Ok now it's clear to me. Thanks haruspex and ehild. :).