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# Feynman’s Flying Saucer Explained

To quote Richard Feynman [2]:

“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.1 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.”

Feynman later writes that “The [Feynman] diagrams and the whole business that I got the Nobel Prize for came from that piddling around with the wobbling plate” [2].

Here I offer a quick explanation of Feynman’s observation. Figure 1 gives a summary.

### Watching a Star from the Flying Saucer

Figure 1. To an observer attached to the plate, $$\bf{m}$$ spins around the plate’s $$z$$-axis with angular velocity $$\omega$$. And to the ground observer, this entire picture spins as well — at the rate $$\approx \omega$$ and with angular velocity closely aligned with the $$z$$-axis (assuming small wobble). In short, the combined angular velocity of the $$z$$-axis around the angular momentum direction—i.e., the rate of wobble—is $$\approx 2 \omega$$. This explains Feynman’s observation.
The angular momentum of the plate in flight is fixed because the torque that acts on the plate is zero, neglecting the effect of air. Let $${\bf m}= (m_x, m_y, m_z)$$ be the expression of the angular momentum in the $$xyz$$-frame that is glued to the plate (see Figure 1). As the plate wobbles, $$\bf{m}$$ changes; we will show that $$\bf{m}$$ spins around the $$z$$-axis as in Figure 1, where $$\omega$$ actually turns out to be precisely the angular velocity $$\omega_z$$ of the plate’s spin around the $$z$$-axis. This fact is of independent interest and demonstrates two roles of $$\omega$$: (i) the rate of the plate’s spin (as viewed by someone on the ground), and (ii) the rate of $$\bf{m}$$'s spin around the $$z$$-axis in the eye of the observer stuck to the plate. Postponing the proof that $$\bf{m}$$ moves like in Figure 1, here is an explanation of Feynman’s observation.

### Explaining the 1:2 Spin-to-wobble Ratio

Segment $$AM$$ in Figure 1 rotates at the angular velocity $$\omega \equiv \omega _z$$ relative to the plate, according to the claim in the previous paragraph. And the plate itself rotates with an angular velocity of magnitude $$\approx \omega$$ and closely aligned with the angular momentum if the wobble is small (see Figure 2). The sum of these angular velocities is $$\approx 2 \omega$$; it is the angular velocity of the $$z$$-axis around the fixed direction of the angular momentum in the ground frame. This completes the explanation of the 1:2 ratio; it remains to be proven that $$\bf{m}$$ indeed moves as Figure 1 indicates.

### Explaining the Motion of $$\bf{m}$$ in Figure 1

The plate’s angular velocity $$\boldsymbol{\omega}$$ is related to $$\bf{m}$$ via $$\boldsymbol {\omega} = {\bf I} ^{-1} {\bf m}$$, where $${\bf I} = \hbox{diag} (I_x, I_y, I_z)$$ is the tensor of inertia whose diagonal entries are the moments of inertia with respect to the corresponding axes. Since the moment of inertia around the $$z$$-axis is twice that around the diameter, we have $${\bf I} = I_z\hbox{diag}(\frac{1}{2} ,\frac{1}{2} ,1)$$. Now $$\bf{m}$$ satisfies Euler’s equation

Figure 2. View for the ground observer. The plate’s rotation is added to the rotation of $$AM$$ relative to the plate, yielding the rate of rotation of $$AM$$ in space.
${\bf \dot m} = {\bf m} \times {\bf I} ^{-1} {\bf m}. \tag1$

The derivation of $$(1)$$ is immediate by a creative application of the familiar formula $$\bf {v } = \omega \times {\bf r}$$. Indeed, to the observer on the flying saucer, the surrounding space is rotating with angular velocity $$- \boldsymbol{\omega}$$ and the tip of the angular momentum vector is like a particle affixed to the surrounding space. The apparent velocity of this “particle” is thus given by

$\bf {\dot m} = ( -\boldsymbol{\omega}) \times {\bf m},$

which amounts to Euler’s equations $$(1)$$ upon substitution $$\boldsymbol{\omega} = {\bf I} ^{-1} {\bf m}$$ (a standard derivation of Euler’s equation in [1] or [3], for example, takes a little over half a page).

Finally, Euler’s equations $$(1)$$ become

$\left\{ \begin{array}{l} \dot m_x =- m_y(m_z/I_z) \\ \dot m_y = \ \ m_x(m_z/I_z) \\ \dot m_z=0\end{array} \right. . \tag2$

Now $$m_z/I_z = \omega_z$$ is exactly the angular velocity around the $$z$$-axis, and is constant according to the last equation. Based on the first two equations, the vector $$(m_x,m_y )$$ executes circular motion with angular velocity $$\omega _z$$. This completes the explanation of $$\bf{m}$$’s motion that is sketched in Figure 1.

1 Actually it is the other way around; the wobble is twice as fast as the spin.