Nail-Tinted Glasses

Feynman's Wobble

Snoopy Dance!

I've finally had a breakthrough working on Feynman's wobbling plate problem.

Legend has it that observing a wobbling plate snapped Feynman out of period of burnout - the episode is described in "The Dignified Professor", in Surely You're Joking, Mr. Feynman.

If you throw a plate, the obvious motion will be the spin - the rotation of the plate about its own axis. If you throw the plate slightly imperfectly, you will also observe a wobble - that the plane of the plate rotates around an axis close to, but not exactly equal to, the axis of the spin. It happens that the frequency of the wobble is twice that of the spin.

The puzzle, of course, is to go back to first principles and work out that this is so.

This is a problem with a lot of ellipses. If you look at a dish from above, you have a circle. Tip it toward you slightly, and you see an ellipse - the side to side distance is the same, but front to back is slightly shorter. Similarly, if you look at the plate from the side, you see a line; tipped slightly, and you get a very flat ellipse. Both of these ellipses are spinning (at the same rate), and so have associated angular momenta.

OK, here's the breakthrough. Consider that picture from the top again. You have an ellipse - the major axis is R, the minor axis is Rcosψ. Consider a circle with the same center, and radius R(1+cosψ)/2. You can consider the path of the ellipse to be a simple perturbation about the main path - in other words, this is the wobble. Note that in one period about the circle, you get four crossings - two periods of displacement from the circle to the ellipse.

I use the term displacement here deliberately. How would Feynman have attacked a problem like this? His interest was in the Principle of Least Action. He would have wanted to derive a solution from a Lagrangian description of the mechanics, perhaps by applying a perturbation to the radius:
R(t) = R(0) + δ(t)

If I'm dealing with squares of R(t), I can count on a 2δ falling out somewhere. I don't quite have the right Lagrangian to follow through with this (I said I had a breakthrough, not a solution).

But being able to see that it will come out is a great comfort.

September 21, 2003 2:48 PM | TrackBack