Wobbly Wobble

Imgur Choose the CM as the origin. The principal axes are then $\hat{x}$ and $\hat{y}$ , along with $\hat{z}$ pointing into the paper. Assume that the precessional motion is counterclockwise when viewed from above, as indicated in. Consider the setup in the frame rotating with angular velocity $\hat{z}$ . In this frame, the location of the contact point remains fixed, and the coin rotates with frequency ω' around the negative $\hat{y}$ axis. The radius of the circle of contact points on the table is R cos θ. Therefore, the nonslipping condition says that $\omega'R=\Omega Rcos\theta$ , and so $\omega'=\Omega cos\theta$ . Hence, the total angular velocity of the coin with respect to the lab frame is

$\omega=\Omega\hat{z}-\omega'\hat{y}=\Omega(sin\theta\hat{x}+cos\theta\hat{y})-\Omega cos\theta\hat{y}=\Omega sin\theta\hat{x}$

In retrospect, it makes sense that $ω$ must point in the \hat{x} direction. Both the CM and the instantaneous contact point on the coin are at rest, so ω must lie along the line containing these two points, that is, along the $\hat{x}$ axis.

The principal moment around the $\hat{x}$ axis is $I=\frac{mR^{2}}{4}$ . The angular momentum is $L=I{ω}\hat{x}$ , so its horizontal component has length $L_{\perp}=Lcos{θ}=(I{ω})cos{θ}\hat{x}$ . Therefore, $\frac{dL}{dt}$ has magnitude

$\left|\frac{dL}{dt}\right|=\Omega L_{\perp}=\Omega\frac{mR^{2}}{4}(\Omega sin\theta)cos\theta$

and points out of the page.

The torque (relative to the CM) is due to the normal force at the contact point. This normal force is essentially equal to mg, because the CM is assumed to be falling very slowly. Note that there is no sideways friction force at the contact point, because the CM is essentially motionless. The torque therefore has magnitude

$\left|\tau\right|=mgRcos\theta$ and it points out of the page.we finally get, $\omega=2\sqrt{\frac{g}{(Rsin\theta)}}$