Dangling block attached to rotating block

1. Sep 19, 2015

tnedde

I’m an applied math grad student and I wanted to check my conceptual understanding of what I thought was a basic mechanics problem. You have two blocks connected by a weightless, frictionless string passing through a tiny hole in a table. On the table, one block (mass m) rotates without friction about the hole, while the other (mass M) dangles below the hole. The dangling block is at rest.

The original question was: How much work does it take to move the dangling block down a distance d (presumably to a new static equilibrium)? However, if my current understanding is correct, it would be impossible for it to reach a new equilibrium on its own, because the original equilibrium is unstable. As soon as you nudge the dangling block, it should accelerate over time. So the least you could do would be to hold up the block to maintain the equilibrium, but of course this would involve doing negative work to slow it down (the textbook's solution gives a positive answer, which is where the difficulty first started for me).

Does this seem correct? I’m happy to share my reasoning, but I’m curious to have someone else take a crack at it.

2. Sep 19, 2015

Staff: Mentor

If you release the block again it will move to a different position (not the original one and not the one where you released it).
The original equilibrium should be stable.

3. Sep 19, 2015

andrewkirk

According to my calc, because of conservation of angular momentum, the centrifugal force will be proportional to $r^{-3}$. That enables us to do work in the following ways:

1. Pull the dangling block down. Resistance will be encountered and as soon as we stop pulling, it will start moving back up, because the upwards force is more than the block's weight. But we can measure the work done while we were pulling down. It's easiest to imagine this if we pull very slowly, so we can ignore any downwards momentum and associated kinetic energy of the dangling block when we stop pulling.

2. Push the dangling block up. Again we do this very slowly so we can ignore the dangling block's momentum and KE. As soon as we stop pushing, the block will start to move back down, as the centrifugal force will be less than the block's weight. But again we can measure the work done while pushing the block up.

In both cases we will have done positive work.

4. Sep 19, 2015

andrewkirk

Do you think so? I would have though that it would return towards the equilibrium position, overshoot because of its momentum and then enter a (probably non-simple) harmonic motion about that position. But I haven't done any calcs.

5. Sep 19, 2015

Staff: Mentor

Hmm right, "position" is not the right word. It will enter a different orbit.