# Calculus and perpetual motion machines

I’ve always been fascinated with perpetual motion machines, since I was a small child. I even built one when I was 6 or 7– the idea was that you’d have a magnet on the end of a rotating arm like the minute hand of a clock and you’d put a bunch of other magnets arranged in a circle around the perimeter, say at every number on the clock. My theory was that if you started the minute hand with a push, it would be attracted to the nearest number’s magnet (true) and then momentum would carry it right past that magnet (also true) until it came close enough to the next magnet to be pulled in by that magnet, and so forth… clearly, the arm would spin forever.

If only I’d understood symmetry, I would have realized before I actually built it that you wouldn’t get any net energy from passing a magnet on the clock face. The arm just oscillates back and forth until friction slows it to hover over the magnet on the face. This was painfully obvious once you tried it.

My next idea was to replace each magnet with an electromagnet and turn the electromagnets off once the hand passed. This would break the symmetry and allow you to keep the arm spinning (as long you added power to the electromagnets). I was briefly convinced that this was a genius invention until my father the electrical engineer gently broke the news to me that this was basically the operating principle for electric motors.

I got thinking about all this today when I came upon this great webpage about overbalanced wheels, written by Donald Simanek at Lock Haven University. He poses the following interesting calculus question:

Suppose you have a circular wheel with three chambers and 3-fold symmetry. Suppose that when the chambers are empty, the center of mass of the wheel is at the center of the circle. Now suppose you add the same volume of liquid to each chamber.

Is the wheel balanced, or will it move?

Simanek links to a nice argument by Alex Pannis, which says

That’s an easy one… The center of gravity will always be lower than the center of the wheel. So, it would require an external force to move it [continually] in either direction.

So far, so good. If we fill each chamber completely, the center of mass is still at the origin (after all, it was at the origin when the chambers had constant density 0). And then if we take some liquid out of each chamber we’ve removed mass from the top (but not the bottom), so the center of mass has to move down.

Since the center of gravity will always be lower [than the rotation axis], independent of the position of the wheel (and the shape of the volume occupied by the liquid), the wheel will stay in equilibrium, no matter what its initial position.

This part seems less clear to me. After all, the center of mass could have moved “down and right” or “down and left” when we took liquid out of the top of each chamber– maybe as we rotate the wheel, the center of mass describes some loop in the plane (always staying below the height of the center of the circle, of course). So the wheel doesn’t have to be in equilibrium just because the center of mass is below the origin– you’d have to convince me that the center of mass is always in the same place.

So how about it? Any clever answers? You should, of course, be able to generalize to n-fold symmetry, though n-even is probably too easy. 🙂