I am referring to this:

Within a week I was in the cafeteria and some guy, fooling around, throws a plate in the air. As the plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on the plate going around. It was pretty obvious to me that the medallion went around faster than the wobbling.

I had nothing to do, so I start to figure out the motion of the rotating plate. I discover that when the angle is very slight, the medallion rotates twice as fast as the wobble rate - two to one. It came out of a complicated equation! Then I thought, ``Is there some way I can see in a more fundamental way, by looking at the forces or the dynamics, why it's two to one?''

I don't remember how I did it, but I ultimately worked out what the motion of the mass particles is, and how all the accelerations balance to make it come out two to one.

Anyone knows how to derive the two-to-one relationship?

  • $\begingroup$ Take a look at this article. The authors present an elementary explanation for the two-to-one ratio of wobble to spin frequencies. UPDATE: In case you don't have easy access to the AJP, here it is: article. $\endgroup$ Sep 26 '11 at 17:45
  • $\begingroup$ Also, an older version of the same idea here: ruina.tam.cornell.edu/research/topics/miscellaneous/… (Shameless self promotion.) Note, we now know that this was NOT Feynman's approach. He had a simple diagram with angular momentum vectors and such. Not as beautiful, IMHO. $\endgroup$
    – user75231
    Mar 12 '15 at 15:54
  • $\begingroup$ @Physicsworks, can you post your comment as an answer, so that I can mark it as an accepted answer? $\endgroup$
    – Graviton
    Mar 13 '15 at 3:26
  • $\begingroup$ @Physicsworks and Graviton: actually that comment was originally posted as an answer and was changed to be a comment because it does not constitute an answer. If the now-deleted answer was edited to include the relevant information from the article (so that one doesn't have to click on any links to understand it), then it can be undeleted and you can accept it then. $\endgroup$
    – David Z
    Mar 13 '15 at 5:24

While I cannot possibly claim to know what Feynman did, here is a fairly-simple way of deriving the 2:1 ratio (which incidentally Feynman remembered backwards in that quote; the wobble is faster than the spin).

The viewpoint I'll take is that it's a given that the spin rate and wobble rate are constant, then ask why they are related. We'll work with a thin plate and a small wobble angle, $\theta$.

If the wobble is around a vertical axis, the spin axis (symmetry axis) of the plate points mostly vertically, but off to the side by an angle $\theta$. If the spin is much faster than the wobble, the direction of the angular momentum is almost exactly aligned with the spin axis. As time goes on, the wobble carries the spin axis around and so the angular momentum precesses. This is impossible because angular momentum must be conserved.

Evidently, the wobble must be just fast enough that the angular momentum of the wobbling adds to the spin angular momentum so the total angular momentum of the entire disk is vertical; this way we get no precession as the disk wobbles.

By symmetry, the plate has a principal axis along its spin axis (moment of inertia $I_s$), and the other two principal axes are degenerate, in the plane of the disk (moment of inertia $I_p$). Let's write the angular velocity of the plate as $\vec{\omega} = \vec{\omega}_s + \vec{\omega}_p$, denoting angular velocity components along the spin axis and in the plate. The wobble angular velocity will be denoted $\omega_w$.

We would like to ensure that the horizontal components of total angular momentum are zero. The spin part of the angular velocity gives horizontal angular momentum $L_{s,h} = \omega_s I_s \sin\theta \approx \omega_s I_s \theta$

There is also a horizontal component from the angular momentum in the plane of the disk, which is $L_{p,h} = \omega_p I_p \cos\theta \approx \omega_p I_p$

These must have equal magnitude to sum to zero, so $\frac{\omega_s \theta}{\omega_p} = \frac{I_p}{I_s}$

Finally, we need to relate the angular velocity component in the plane to the angular velocity of the wobble. One way to do this is to note that for a radius r, the left hand edge of the disk (ignoring the spinning) wobbles up and down a distance $r \theta$ over the course of a wobble, and so has a maximum speed of $r\theta \omega_w$. The maximum speed induced by the angular velocity in the plane is $r \omega_p$, so $\omega_p = \theta \omega_w$.

We find

$\frac{\omega_s}{\omega_w} = \frac{I_p}{I_s}$

A quick integral gives that this ratio is the factor of two that Feynman found.

As this is a famous anecdote, there are many solutions based on finding the equations of motion or otherwise analyzing the dynamics already online. I thought it would be nice to add a simple view of the factor of two based solely on a conservation law.

I originally wrote this answer on Quora.


Here's another method using acceleration (which only uses intro level physics):

Let $\omega$ be the spin angular velocity, $\Omega$ the wobble angular velocity, $y$ the height of a particle on the plate as it goes up and down with the wobbling, $r$ the distance between the particle and the plate's center, and $\theta$ the angle of wobble at the location of the particle. The particles feels a centripetal force, and a component of that force acts vertically, accelerating it up and down:

$$m\omega^2 r \sin(\theta)=-ma_y$$ $$a_y=-\omega^2y$$

As the plate spins, the particle oscillates between the high and low point of the plate (and the motion is simple harmonic because $y$ and $a_y$ are proportional). But as the plate wobbles, those high and low points rotate, so the motion of the particle is based on the difference between the angular velocities: $$y=A\sin((\Omega-\omega)t)$$ $$a_y=-A(\Omega-\omega)^2\sin((\Omega-\omega)t)$$ $$a_y=-(\Omega-\omega)^2y$$

Combining the two equations gives:


This is satisfied if $\Omega$ is 0 (representing no wobbling, or the plate just spinning), and if $\Omega=2\omega$.


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