# dynamics of a rotating coin

Sometimes when I’m bored in a waiting area, I would take a coin out of my pocket and spin it on a table. I never really tried to figure out what was going on. But, recently I wondered about a few things:

1. Can I determine the critical rotational speed that determines the instant when vertical displacement is involved?
2. Assuming that the initial rotational impulse is known, can I predict the exact location where the coin lands?
3. Can two coins with the same dimensions but different mass stop spinning at the same instant assuming that they both receive the same rotational impulse?

More precisely :

We assume that a coin of dimensions $R$ and height $h$ where $R >> h$ is initially standing on a flat surface prior to receiving a rotational impluse in a plane parallel to the surface. Further, we assume that the coin has mass $M$, with uniform mass distribution. Empirically, I observed that there are two regimes:

1. Sliding contact with the flat surface, while the rotational speed $||\dot{\theta}||\geq c$ where $c$ is a constant which can be determined.
2. Rolling contact without slipping, when $||\dot{\theta}|| < c$

Here’s my reasoning so far:

1. Assuming small rotational speeds, which is reasonable, drag is proportional to the first power of rotational speed:

\begin{aligned} F_D = C_d \dot{\theta} \frac{4 \pi}{3 \pi} \end{aligned}

where $C_d$ is the drag coefficient and $\frac{4 \pi}{3 \pi}$ is the distance of the centroid of the semicircular half of the coin from the coin’s center of gravity. Now the work done by the $F_D$ is proportional to the distance travelled by the centroid on both halves of the coin, so the total energy dissipated at time $t$ is given by:

\begin{aligned} \Delta E(\theta, \dot{\theta},t) = 2 \int_{0}^{t} F_D \theta \frac{4 \pi}{3 \pi}= 2 C_d \big(\frac{4 \pi}{3 \pi}\big)^2 \int_{0}^{t} \dot{\theta} \theta dt= C_d \big(\frac{4 \pi}{3 \pi}\big)^2 {\theta (t)}^2 \end{aligned}

So the energy dissipated is just an explicit function of the angle $\theta$ and the Hamiltonian is given by:

\begin{aligned} H(\theta, \dot{\theta}) = \frac{1}{2} I \dot{\theta}^2+mg\frac{h}{2}-\Delta E(\theta) \end{aligned}

The equations of motion are then given by:

\begin{aligned} \dot{Q}=\frac{\partial H}{\partial \dot{\theta}} = I \dot{\theta} \end{aligned}

\begin{aligned} \dot{P}=-\frac{\partial H}{\partial \theta} = 2 C_d(\frac{4R}{3 \pi})^2 \theta(t) \end{aligned}

2. It’s not clear to me how I should interpret the equations of motion but my hunch is that the first phase has ended when the kinetic energy vanishes. This happens when the dissipated energy equals the kinetic energy:

\begin{aligned} \frac{1}{2} I \dot{\theta}^2=C_d \big(\frac{4 \pi}{3 \pi}\big)^2 {\theta (t)}^2 \end{aligned}

If we let $C_1 = \frac{I}{2}$ and $C_2 = C_d \big(\frac{4 \pi}{3 \pi}\big)^2$, the solutions I found are of the form:

\begin{aligned} \frac{\dot{\theta}}{\theta} = \sqrt{\frac{C_2}{C_1}} \end{aligned}

But, this is problematic as the solution is meant to be a unique $\theta$. However, I assume that this difficulty can be resolved and I think it might be due to a problem that occurred earlier.

Now, it remains to explain why the coin enters a second phase and begins to roll without slipping instead of simply stopping completely. One reason has to do with experimental conditions. In practice the surface on which the coin spins is never completely flat, and the initial rotational impulse is never completely planar. A second reason is the minimum total potential energy principle which states that a system tries to minimise its total potential energy in order to maximise stability.

3. I could try to explain what happens in the second phase where the coin is rolling without slipping but there are already signs that my mathematical arguments for the first phase might have an error.

However, assuming that there are no dissipative forces the total energy is given by:

\begin{aligned} E = MgRsin(\alpha) + \frac{1}{2}I \Omega^2 sin^2(\alpha) \end{aligned}

where $\alpha$ is the angle of inclination with respect to the vertical and $\Omega$ is the precession rate. From this point I can go a bit further but I think this is a good place to stop for now.

Remark : I have also asked the first two questions on the physics stackexchange and I look forward to having a precise answer to all three questions by the end of this week.

References :