# Coin rolling problem

1. Aug 26, 2012

### VortexLattice

1. The problem statement, all variables and given/known data

At what speed should a coin of radius a and mass m roll to stay upright? Assume the coin is uniform, a thin disk, almost upright, and rolling on a perfectly rough horizontal surface, on the surface of the earth.

3. The attempt at a solution

I've found a solution using the Virial Theorem, but I really don't buy it. It just uses the potential energy of the coin (mga, because it's almost upright) and the kinetic energy from the rolling (both translational and rotational) and relates them, allowing you to find v. That doesn't seem quite right to me for some reason, though:

$\omega = v/a$
$I = ma^2/2$
$T = mv^2/2 + I\omega^2/2 = mv^2/2 + mv^2/4 = 3mv^2/4$
$V = mga$ (since θ is essentially 0)
$V \propto r^1$
→$2<T> = 1<V> \rightarrow v^2 = (2g/3a) \rightarrow v = \sqrt{2g/3a}$

Seems sketchy to me, this just doesn't seem like a legitimate use for the virial theorem.

Anyway, I know there must be a good solution for this using actual mechanics. Here's a free body diagram with some other details:

Here's what I've got so far, but there are a lot of as yet unjustified assumptions. I'm pretty sure it has to be rolling in the arc of a circular path. If it wasn't, it seems like there would be absolutely nothing just stopping it from falling down from gravity as it rolls.

So, I realized that since the coin is spinning as it's going around, if we look in the frame moving with the coin, there's a coriolis force:

$\vec F_c = -2\vec \Omega \times \vec v$

Where Ω is the rotational velocity of the frame and v is the velocity of a point on the disk. Because the ground is "perfectly rough", we can treat the point where the coin's touching the ground as not slipping with respect to the axis (that the arc is around) it's rolling around. So, that point is an axis the coin can "rotate around". Due to the coriolis force, there should be a force on the top and bottom sides of the coin (because their velocities are in opposite directions), of the same magnitude. However, because the bottom of the coin has a shorter lever arm, its torque is less.

Thus, the equilibrium condition I get is

$\tau_t = \tau_b + \tau_g$

The problem is, $\tau_g = mga sin(\theta)$ where θ is the very small angle between the coin and the vertical. Also, there's still the radius of the coin's path from its axis in this equation. The way I got around this was by saying that a perpendicular line drawn from the center of the coin to the ground hits the spot it's rotating around, giving me the relation $sin(\theta) = a/r$, which cancels both r and θ. However, I don't see any reason why this should necessarily be the case...it seems like the coin could theoretically be almost flat to the ground but going in a path with a fairly large radius.

Does my solution sound right? Can anyone enlighten me?

Thanks!

Last edited: Aug 27, 2012