The disk is usually sold as a big and heavy cylindrical steel with chrome on it along with a mirror that has a shallow hole so that the "big coin" stays near the center. You should definitely buy the bestselling $35 Toysmith Euler's disk – which has 251 reviews (it almost looks like the heroes of The Big Bang Theory were using this exact shiny $35 product, or was it this one for $40?) – and also the #1 bestselling fragrange on Amazon.com, the Ivanka Trump spray. The #2 bestselling thing in beauty is the Ivanka Trump Roller Ball, whatever it is. Not bad for a woman who isn't even a real climate skeptic and who teaches her kid Chinese instead of Czech.

The video I embedded at the top explains the key mathematics of Euler's disk. Well, the second part of the video is divided to two half-screens. The intellectually challenged people are more likely to watch the right hand side where the disk is spinning and completes the theater while the smarter viewers tend to look to the left where he describes how it works mathematically.

The spinning takes such a long time that Rajesh almost suffocated because he promised not to breathe while it was spinning, long enough for Penny to lose her interest long before the spinning ends, and long enough for the author of the video at the top to explain how it works. ;-) He drew a vector from the center of the disk to the circumference, a radial vector \(\vec r\). When you observe the spinning of Euler's disk, you see that the direction of \(\vec r(t)\) is changing with time.

It's basically because the actual physical circumference of the disk whose length is \(2\pi R\) is being "drawn" on a shorter circle of circumference \(2\pi r\) which is smaller. You may very accurately assume that the center of mass of the disk sits on a vertical axis at all times – well, it would be true if you managed to fully stabilize the horizontal motion of the disk along the mirror.

Why is the center of mass not oscillating in the horizontal direction? Well, it's because only vertical forces – gravity and the pressure from the mirror – are acting on the disk. There is no horizontal force – which means that the velocity of the center of mass in the horizontal direction remains constant: it is simply not periodically oscillating. This statement isn't quite true because the mirror underneath is a bit curved so the force from the mirror isn't quite vertical and your Euler's disk is slowly moving along the mirror in various directions, too. But we neglect this motion – if you do it well, the motion may be set to zero.

OK, the lower circumference of your Euler's disk is adiabatically moving along a circle of circumference \(2\pi r\) where \(r\) is a bit smaller than \(R\) – even though it approaches \(R\) at the end of the spinning performance. How much smaller \(r\) is relatively to \(R\)? Well, if the angle between the horizontal surface of the mirror and the flat base of the Euler's coin is labeled \(\theta\), you may see a triangle connecting the intersection of the spinning axis with the mirror, the center of mass of Euler's disk, and the only point where the disk touches the mirror.

One of the angles of this triangle is 90 degrees, another one, near the "touching point", is \(\theta\), so you see by the basic trigonometry that \(r=R_+ \cos\theta\). Well, if you were really careful, you must have noticed that I had to use \(R_+\) rather than \(R\) where \(R_+\) is the distance of the

*center of mass*of Euler's disk from the lower circumference – and because the disk is slightly thick, it's a bit longer than the radius \(R\) of the base of the disk (the distance measured along the lower surface of the disk). OK, let's neglect this difference between \(R\) and \(R_+\) for a moment.

We see that one "period" of the spinning tries to wrap the circumference of the disk on the smaller circumference \(2\pi r\) drawn on the mirror but it doesn't quite wrap the whole circumference of the disk. The difference is\[

2\pi R - 2\pi r = 2\pi R (1-\cos\theta) \approx \pi R\theta^2

\] where I Taylor-expanded the cosine, cancelled the leading term, and only kept the subleading term proportional to \(\theta^2\). The deficit we just mention causes the rotation of the radial arrow on the disk, the "wobbling". You may see in what direction it's wobbling. And the ratio of the wobbling and the spin is therefore\[

\frac{2\pi (R-r)}{2\pi r} \approx \frac{1}{\cos\theta} - 1 \approx \frac{\theta^2}{2}

\] where the last approximation is OK for a thin disk and \(\theta\to 0\). In this limit, you see that the wobbling becomes much slower than the spinning because the expression above goes to zero. Now, a funny realization is that the rate of wobbling is constant basically because the angular momentum is conserved (at the end, the whole angular momentum is connected with the rotating direction of the arrow he drew; the vertical component of the angular momentum is conserved because the physical problem is rationally symmetric under rotations around the vertical axis, thanks to Emmy Noether for the idea), and that's why the rate of spinning has to go to infinity for \(\theta\to 0\).

If you return to one of the omissions, \(R\) vs \(R_+\), I think that the effect will be that the ratio above – with \(2\pi R_+\) as the first term in the difference in the numerator – won't go to zero for \(\theta\to 0\). Instead, it will go to a finite positive constant that is an increasing function of the thickness of the disk. Because of the Pythagorean theorem (which actually makes the equation below exact),\[

R_+ = R + \frac{t^2}{4R}

\] where \(t\) is the thickness of the disk. Note that \(R_+/R\) is only greater than one by \((t/2R)^2\) which is very small for a small enough \(t\lt R\). But if you created a thinner Euler's disk, I think that the truly final stage of the spinning would be even faster. Do you agree?

Equivalently, I could have used \(R\) – the radius of the disk's base – everywhere but I would have to use a different angle \(\theta_+\) which is the angle between the horizontal mirror and the line connecting the "touch point" with the

*center of mass*. This angle \(\theta_+\) doesn't quite go to zero at the end of the spinning – and this is why the ratio (and the spinning speed) doesn't quite go to infinity. Again, the corrections go like \(t^2/R^2\).

Do you agree that in the final stages, the motion and sound are almost to the chirping of a black hole merger detected by LIGO? ;-) I hope that the gravitational wave detectors will detect much longer and clearer sounds sometime in the future.

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