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 two things:

  1. Can I determine the critical rotational speed that determines the instant when vertical displacement is involved? (Per Floris' clarification, 'vertical displacement' could be understood as 'tilting over'.)
  2. Assuming that the initial rotational impulse is known, can I predict the exact location where the coin lands?

More precisely :

We assume that a coin of dimensions $R=\text{radius and } h=\text{height}$ 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 and that air resistance is much more important than kinetic friction due to the surface on which the coin spins.

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 what I've attempted so far:

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

\begin{equation} F_D = C_d \dot{\theta} \frac{4 \pi}{3 \pi} \tag{1} \end{equation}

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{equation} \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\tag{2} \end{equation}

So the energy dissipated is just an explicit function of the angle $\theta$:

\begin{equation} \Delta E(\theta) = C_d \big(\frac{4 \pi}{3 \pi}\big)^2 {\theta (t)}^2\tag{3} \end{equation}

So the Hamiltonian is given by:

\begin{equation} H(\theta, \dot{\theta}) = \frac{1}{2} I \dot{\theta}^2+mg\frac{h}{2}-\Delta E(\theta) \tag{4} \end{equation}

The equations of motion are then given by:

\begin{equation} \begin{cases} \dot{Q} = \frac{\partial H}{\partial \dot{\theta}} = I \dot{\theta} \\ \dot{P}=-\frac{\partial H}{\partial \theta} = 2 C_d(\frac{4R}{3 \pi})^2 \theta(t) \end{cases} \tag{5} \end{equation}

  1. 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{equation} \frac{1}{2} I \dot{\theta}^2=C_d \big(\frac{4 \pi}{3 \pi}\big)^2 {\theta (t)}^2 \tag{6} \end{equation}

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{equation} \frac{\dot{\theta}}{\theta} = \sqrt{\frac{C_2}{C_1}} \tag{7} \end{equation}

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. My argument is that in practice the surface on which the coin spins is never completely flat, and the initial rotational impulse is never completely planar.

  1. In the second phase, assuming that there are no dissipative forces the total energy is given by:

\begin{equation} E = MgR\sin(\alpha) + \frac{1}{2}I \Omega^2 \sin^2(\alpha) \tag{8} \end{equation}

where $\alpha$ is the angle of inclination with respect to the vertical and $\Omega$ is the precession rate. I can go further but the analysis becomes a lot more complicated due to the role of air resistance.

This leads me to two questions:

  1. $\theta$ is monotonically increasing as a function of time so shouldn't we find a unique $\theta$ that determines the instant when vertical displacement is involved? (Note: the method I use is to equate kinetic energy and dissipated energy)
  2. Beyond imperfect experimental conditions, can the second phase be explained by the minimum potential energy principle? (i.e. there's a theoretical reason why the coin doesn't simply stand still)

    References :

  3. http://puhep1.princeton.edu/~kirkmcd/examples/rollingdisk.pdf

  4. http://www.eulersdisk.com/PRE56610.pdf
  • 1
    $\begingroup$ I see that you do not include the drag offered by air. You might find this interesting: spinning ring versus spinning disc .<br> I haven't followed through to the links mentioned there but most probably you will find what you were looking for. Either way it will offer a fresh perspective. Cheers! $\endgroup$ Jan 8, 2017 at 17:29
  • $\begingroup$ There's a discrepancy in your assumptions: friction can't be so small as to be negligible compared to air drag, yet be large enough to bring the coin into a no slip condition. I believe it's actually the friction that will disrupt a perfectly vertical axis of rotation of a perfectly vertical coin, so neglecting it will prevent you from being able to solve when/how the coin starts tipping. $\endgroup$
    – Rick
    Oct 11, 2017 at 11:43
  • 2
    $\begingroup$ I am trying to understand your question - is it "when and why does the coin tip over"? You say "vertical displacement is involved" which I think is just a complicated way of saying "tips over". Right? $\endgroup$
    – Floris
    Dec 22, 2017 at 14:35
  • $\begingroup$ @Floris I didn't want to be misunderstood but I'd say your interpretation is correct. $\endgroup$
    – user29305
    Dec 23, 2017 at 4:15

1 Answer 1


Both of your questions are answered as part of the thorough analysis of reference 3, but here is a summarized answer:

  1. No, there is not a unique and precise "instant" when the coin goes from vertical to tipped. If by vertical you mean oscillating about a stable vertical equilibrium, however, then there is a threshold angular velocity (see equation 65 in reference 3 for the frictionless case) above which a small oscillation is stable and below which it is not. In practice the coin will visibly change behavior near this rotation rate, but the precise instant is determined by the minutia of the tiny oscillations about a perfectly vertical spin.

  2. In a sense the second phase is due to imperfect conditions, as can be illustrated by using a foriegn coin with a thick square edge and ending with a stationary coin on edge, exactly where it was spinning. However there is a great deal of principle and order describing the process by which a tiny deviation from the vertical grows into the spiraling roll which you see as phase two. The equations outlining this are set up in reference 3, but they are fundamentally similar to those describing a precessing top -- an easier starting place for understanding inertia tensors than the spinning coin which is also already conveniently explained step by step in hundreds of places on the internet.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.