# Rotating a Perfectly Smooth Cylinder

1. Nov 27, 2006

### Bmmarsh

Consider the following model of a perfectly smooth cylinder. it it a ring of equally spaced, identical particles, with mass M/N, so that the mass of the ring is M and its moment of inertia MR², with R the radius of the ring. Calculate the possible values of the angular momentum. Calculate the energy eigenvalues. What is the energy difference between the ground state of zero angular momentum, and the first rotational state? Show that this approaches infinity as N-->oo. Constrast this with the comparable energy for a thick "nicked" cylinder, which lacks the symetry under the rotation through 2pi/N radians. This exemple implies that it is impossible to set a perfectly smooth cylinder in rotation, which is consistent with the fact that for a perfectly smooth cyinder such a rotation would be unobservable.

I've seen this question asked before, but no one offered a solution. And now that I have had this problem assigned to me, I figured I'de check to see if anyone has come up with a solution yet.

Conceptually, I understand that a perfectly smooth cylinder cannot rotate because it would be unobservable, but how would one go about showing this mathematically? Perhaps showing that it would take an infinite amount of energy to set such a cylinder in motion would be easier--however, I still wouldn't know how to mathematically derive this result.

Thanks for at least reading this =)
Any help would be appreciated.

Last edited: Nov 27, 2006
2. Nov 27, 2006

### Epicurus

I don't really understand. Is this a classical or quantum cylinder?

3. Nov 27, 2006

### Bmmarsh

I'm sorry--this would be a quantum cylinder.

4. Nov 27, 2006

### Epicurus

And i suppose that thick necked cylinder has some mass density which is uniform?