What Causes the Loss of Mechanical Energy in a Frictionless Spinning Cylinder?

[guv]

Homework Statement

Imagine a frictionless horizontal table. On the table a uniform cylinder is spinning with one flat face touching the table at angular velocity $\omega_0$. Through the flat faces of the cylinder, a tiny hole is drilled through. As the cylinder is spinning in place, a pin is inserted through the cylinder very quickly and the cylinder starts to spin around the pin. The horizontal position of the pin does not change (imagine the insertion happens very quickly). Can we apply conservation of energy to find the final angular velocity?

Relevant Equations

N/A

We all know we need to apply conservation of angular momentum here. This necessarily leads to a difference in mechanical energy. Since initial rotational inertial is less than final rotational inertia, there is a loss of mechanical energy. However, I have not been able to convince myself what's doing work to take away the mechanical energy of the disk? Or is it because the problem is set up in a too idealistic way? Thanks,

 

[haruspex]

It doesn't say, but I assume the hole is off-centre.

There is an inelastic impact as the part of the cylinder adjacent to the pin is suddenly brought to a stop.

 

[guv]

What if the disk and the pin are perfectly rigid and the diameter of the pin and the diameter of the hole are identically infinitesimally small?

 

[guv]

I think the problem is the assumption, that the pin can move downward infinitely fast. It cannot, therefore there is a negative work done on the disk as pin slides down the hole while the disk is moving.

 

[jbriggs444]

What if the disk and the pin are perfectly rigid and the diameter of the pin and the diameter of the hole are identically infinitesimally small?

So we would have an infinite pressure and an infinite force acting over an infinitesimal area for an infinitesimal time.

As @haruspex has remarked elsewhere, the correct way to deal with ideal infinities and infinitesimals in physics is by taking the limit of finite quantities as those are increased without bound or decreased toward zero.

If you carry out the limiting process, you will find that every realistic collision with finite holes, finite pins, finite rigidity and finite insertion speeds is either inelastic or leaves some undamped vibrational energy.

 

[haruspex]

I think the problem is the assumption, that the pin can move downward infinitely fast. It cannot, therefore there is a negative work done on the disk as pin slides down the hole while the disk is moving.

Not necessarily. There would, in the real world, be some slack between pin and hole. The insertion could be fast enough to use that. But then you get the collision.

 

[hutchphd]

This is in the same spirit as a number of other electrical (charging a perfect capacitor at fixed V) and mechanical (putting a mass on a spring into a gravity field) examples: the energy goes into (usually macroscopic) oscillations via an elastic interaction which then vibrationally "leaks" the energy into other (usually microscopic) degrees of freedom while we aren't paying attention.