Steady motion with the disk center at rest

The behavior of a spinning disk whose center is at rest can be described as follows.^[6] Let the line from the center of the disk to the point of contact with the plane be called axis ${\displaystyle {\widehat {\mathbf {3} }}}$. Since the center of the disk and the point of contact are instantaneously at rest (assuming there is no slipping) axis ${\displaystyle {\widehat {\mathbf {3} }}}$ is the instantaneous axis of rotation. The angular momentum is ${\displaystyle \mathbf {L} =kMa^{2}\omega {\widehat {\mathbf {3} }}}$ which holds for any thin, circularly symmetric disk with mass ${\displaystyle M}$; ${\displaystyle k=1/2}$ for a disk with mass concentrated at the rim, ${\displaystyle k=1/4}$ for a uniform disk (like Euler disk), ${\displaystyle a}$ is the radius of the disk, and ${\displaystyle \omega }$ is the angular velocity along ${\displaystyle {\widehat {\mathbf {3} }}}$.

The contact force ${\displaystyle \mathbf {F} }$ is ${\displaystyle Mg{\widehat {\mathbf {z} }}}$ where ${\displaystyle g}$ is the gravitational acceleration and ${\displaystyle {\widehat {\mathbf {z} }}}$ is the vertical axis pointing upwards. The torque about the center of mass is ${\displaystyle \mathbf {N} =a{\widehat {\mathbf {3} }}\times Mg{\widehat {\mathbf {z} }}={\frac {d\mathbf {L} }{dt}}}$ which we can rewrite as ${\displaystyle {\frac {d\mathbf {L} }{dt}}={\boldsymbol {\Omega }}\times \mathbf {L} }$ where ${\displaystyle {\boldsymbol {\Omega }}=-{\frac {g}{ak\omega }}{\widehat {\mathbf {z} }}}$. We can conclude that both the angular momentum ${\displaystyle \mathbf {L} }$, and the disk are precessing about the vertical axis ${\displaystyle {\widehat {\mathbf {z} }}}$ at rate

$${\displaystyle \Omega ={\frac {g}{ak\omega }}}$$

(1)

At the same time ${\displaystyle \Omega }$ is the angular velocity of the point of contact with the plane. Let's define axis ${\displaystyle {\widehat {\mathbf {1} }}}$ to lie along the symmetry axis of the disk and pointing downwards. Then it holds that ${\displaystyle {\widehat {\mathbf {z} }}=-\cos \alpha {\widehat {\mathbf {1} }}-\sin \alpha {\widehat {\mathbf {3} }}}$, where ${\displaystyle \alpha }$ is the inclination angle of the disc with respect to the horizontal plane. The angular velocity can be thought of as composed of two parts ${\displaystyle \omega {\widehat {\mathbf {3} }}=\Omega {\widehat {\mathbf {z} }}+\omega _{\text{rel}}{\widehat {\mathbf {1} }}}$, where ${\displaystyle \omega _{\text{rel}}}$ is the angular velocity of the disk along its symmetry axis. From the geometry we easily conclude that:

$${\displaystyle {\begin{aligned}\omega &=-\Omega \sin \alpha ,\\\omega _{\text{rel}}&=\Omega \cos \alpha \\\end{aligned}}}$$

Plugging ${\displaystyle \omega =-\Omega \sin \alpha }$ into equation (1) we finally get

$${\displaystyle \Omega ^{2}={\frac {g}{ak\sin \alpha }}}$$

(2)

As ${\displaystyle \alpha }$ adiabatically approaches zero, the angular velocity of the point of contact ${\displaystyle \Omega }$ becomes very large, and one hears a high-frequency sound associated with the spinning disk. However, the rotation of the figure on the face of the coin, whose angular velocity is ${\displaystyle \Omega -\omega _{\text{rel}}=\Omega (1-\cos \alpha ),}$ approaches zero. The total angular velocity ${\displaystyle \omega =-{\sqrt {\frac {g\sin \alpha }{ak}}}}$ also vanishes as well as the total energy

$${\displaystyle E=Mga\sin \alpha +{\tfrac {1}{2}}kMa^{2}\omega ^{2}=Mga\sin \alpha +{\tfrac {1}{2}}Mka^{2}{\frac {g\sin \alpha }{ak}}={\tfrac {3}{2}}Mga\sin \alpha }$$

as ${\displaystyle \alpha }$ approaches zero. Here we have used the equation (2).

As ${\displaystyle \alpha }$ approaches zero the disk finally loses contact with the table and the disk then quickly settles on to the horizontal surface. One hears sound at a frequency ${\displaystyle {\frac {\Omega }{2\pi }}}$, which becomes dramatically higher, ${\displaystyle {\frac {1}{2\pi }}{\sqrt {\frac {g}{ak}}}{\sqrt {\frac {1}{\sin \alpha }}}}$, as the figure rotation rate slows, ${\displaystyle 2{\sqrt {\frac {g}{ak}}}{\frac {(\sin {\frac {\alpha }{2}})^{2}}{\sqrt {\sin \alpha }}}}$, until the sound abruptly ceases.

Levitation Illusion

As a circularly symmetric disk settles, the separation between a fixed point on the supporting surface and the moving disk above oscillates at increasing frequency, in sync with the rotation axis angle off vertical.

The levitation illusion results when the disk edge reflects light when tilted slightly up above the supporting surface, and in shadow when tilted slightly down in contact. The shadow is not perceived, and the rapidly flashing reflections from the edge above supporting surface are perceived as steady elevation. See persistence of vision.

The levitation illusion can be enhanced by optimizing the curve of the lower edge so the shadow line remains high as the disk settles. A mirror can further enhance the effect by hiding the support surface and showing separation between moving disk surface and mirror image.

Disk imperfections, seen in shadow, that could hamper the illusion, can be hidden in a skin pattern that blurs under motion.

Moffatt

In the early 2000s, research was sparked by an article in the April 20, 2000 edition of Nature,^[7] where Keith Moffatt showed that viscous dissipation in the thin layer of air between the disk and the table would be sufficient to account for the observed abruptness of the settling process. He also showed that the motion concluded in a finite-time singularity. His first theoretical hypothesis was contradicted by subsequent research, which showed that rolling friction is actually the dominant factor.

Moffatt showed that, as time ${\displaystyle t}$ approaches a particular time ${\displaystyle t_{0}}$ (which is mathematically a constant of integration), the viscous dissipation approaches infinity. The singularity that this implies is not realized in practice, because the magnitude of the vertical acceleration cannot exceed the acceleration due to gravity (the disk loses contact with its support surface). Moffatt goes on to show that the theory breaks down at a time ${\displaystyle \tau }$ before the final settling time ${\displaystyle t_{0}}$, given by:

${\displaystyle \tau \simeq \left[\left({\frac {2a}{9g}}\right)^{3}{\frac {2\pi \mu a}{M}}\right]^{1/5}}$

where ${\displaystyle a}$ is the radius of the disk, ${\displaystyle g}$ is the acceleration due to Earth's gravity, ${\displaystyle \mu }$ the dynamic viscosity of air, and ${\displaystyle M}$ the mass of the disk. For the commercially available Euler's Disk toy (see link in "External links" below), ${\displaystyle \tau }$ is about ${\displaystyle 10^{-2}}$ seconds, at which time the angle between the coin and the surface, ${\displaystyle \alpha }$, is approximately 0.005 radians and the rolling angular velocity, ${\displaystyle \Omega }$, is about 500 Hz.

Using the above notation, the total spinning/rolling time is:

${\displaystyle t_{0}={\frac {\alpha _{0}^{3}M}{2\pi \mu a}}}$

where ${\displaystyle \alpha _{0}}$ is the initial inclination of the disk, measured in radians. Moffatt also showed that, if ${\displaystyle t_{0}-t>\tau }$, the finite-time singularity in ${\displaystyle \Omega }$ is given by

${\displaystyle \Omega \sim (t_{0}-t)^{-1/6}}$

Experimental results

Moffatt's theoretical work inspired several other workers to experimentally investigate the dissipative mechanism of a spinning/rolling disk, with results that partially contradicted his explanation. These experiments used spinning objects and surfaces of various geometries (disks and rings), with varying coefficients of friction, both in air and in a vacuum, and used instrumentation such as high speed photography to quantify the phenomenon.

In the 30 November 2000 issue of Nature, physicists Van den Engh, Nelson and Roach discuss experiments in which disks were spun in a vacuum.^[8] Van den Engh used a rijksdaalder, a Dutch coin, whose magnetic properties allowed it to be spun at a precisely determined rate. They found that slippage between the disk and the surface could account for observations, and the presence or absence of air only slightly affected the disk's behavior. They pointed out that Moffatt's theoretical analysis would predict a very long spin time for a disk in a vacuum, which was not observed.

Moffatt responded with a generalized theory that should allow experimental determination of which dissipation mechanism is dominant, and pointed out that the dominant dissipation mechanism would always be viscous dissipation in the limit of small ${\displaystyle \alpha }$ (i.e., just before the disk settles).^[9]

Later work at the University of Guelph by Petrie, Hunt and Gray^[10] showed that carrying out the experiments in a vacuum (pressure 0.1 pascal) did not significantly affect the energy dissipation rate. Petrie et al. also showed that the rates were largely unaffected by replacing the disk with a ring shape, and that the no-slip condition was satisfied for angles greater than 10°. Another work by Caps, Dorbolo, Ponte, Croisier, and Vandewalle^[11] has concluded that the air is a minor source of energy dissipation. The major energy dissipation process is the rolling and slipping of the disk on the supporting surface. It was experimentally shown that the inclination angle, the precession rate, and the angular velocity follow the power law behavior.

On several occasions during the 2007–2008 Writers Guild of America strike, talk show host Conan O'Brien would spin his wedding ring on his desk, trying to spin the ring for as long as possible. The quest to achieve longer and longer spin times led him to invite MIT professor Peter Fisher onto the show to experiment with the problem. Spinning the ring in a vacuum had no identifiable effect, while a Teflon spinning support surface gave a record time of 51 seconds, corroborating the claim that rolling friction is the primary mechanism for kinetic energy dissipation. Various kinds of rolling friction as primary mechanism for energy dissipation have been studied by Leine^[12] who confirmed experimentally that the frictional resistance of the movement of the contact point over the rim of the disk is most likely the primary dissipation mechanism on a time-scale of seconds.