Rotating a Perfectly Smooth Cylinder

• Bmmarsh
In summary, the model of a perfectly smooth cylinder consists of a ring of equally spaced, identical particles with a mass of M/N and a moment of inertia of MR². The possible values of angular momentum and energy eigenvalues can be calculated for this model. The energy difference between the ground state and the first rotational state approaches infinity as N-->oo, indicating that it is impossible to set a perfectly smooth cylinder in rotation. This is contrasted with a thick "nicked" cylinder, which lacks symmetry under rotation through 2pi/N radians. The example suggests that a perfectly smooth cylinder cannot be set in rotation due to the fact that it would be unobservable. The mathematical derivation of this result is still unclear, but it is possible that
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:
I don't really understand. Is this a classical or quantum cylinder?

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

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

1. What is a perfectly smooth cylinder?

A perfectly smooth cylinder is a three-dimensional geometric shape that has a circular base and straight, parallel sides. It is defined as smooth because it has a uniform surface without any bumps, ridges, or imperfections.

2. How do you rotate a perfectly smooth cylinder?

To rotate a perfectly smooth cylinder, you would need to apply a force or torque to it in order to make it spin around its central axis. This can be done manually by pushing or pulling on the cylinder, or it can be achieved using a motor or other mechanical device.

3. What are the properties of a perfectly smooth cylinder?

A perfectly smooth cylinder has several properties, including a constant radius, a fixed height, and a circular cross-section. It also has a constant angular velocity when rotating, and its surface is frictionless.

4. How is the motion of a perfectly smooth cylinder affected by external forces?

Since a perfectly smooth cylinder has no friction, it will continue to rotate with a constant angular velocity unless acted upon by an external force. If a force is applied to the cylinder, it will accelerate or decelerate its rotation depending on the direction of the force.

5. What are some real-world applications of rotating a perfectly smooth cylinder?

Rotating perfectly smooth cylinders can be found in many real-world applications, such as engines, turbines, and motors. They are also used in various manufacturing processes, such as metal rolling and polishing. Additionally, perfectly smooth cylinders are used in scientific experiments to study rotational motion and its effects.

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