A Rolling Quarter A U.S quarter is rolling on the floor without slipping in such a way that it describes a circular path of radius $R=4 \text{cm}$. The plane of the coin is tilted at an angle of $\theta=45^{∘}$ with respect to the horizontal plane. Find the coin's period $T$ in seconds, that is, the time it takes for the coin to go around the circle of radius $R$. The radius of a U.S quarter is $r=1.2 \text{cm}$.
 A: This is a well known problem. So, I will try to solve it in the general case. For a somewhat more detailed answer, please look at Chapter 9 of David Morin's  Introduction to Classical Mechanics book. Also, since this looks like a homework, whenever you think you can do the rest of the problem yourself, stop reading and actually do it yourself!

Let's first go to the CM frame. As it can be seen in the figure below, I will illustrate the principal axis by $x_1$(into the page), $x_2$ and $x_3$.

Let $\Omega$ be the angular velocity of the attaching point of the coin with the floor around the center. This means $T=\frac{2\pi}{\Omega}$.
Using the non-slipping condition, one will find out that the total angular momentum is:
$$\vec{\omega}=\Omega \hat{z}- \frac{R}{r}\Omega \hat{x}_3$$
Now writing $\hat{z}$ in the principal coordinates as $\sin{\theta}\hat{x}_2+\cos{\theta}\hat{x}_3$ we will have:
$$\vec{\omega} = \Omega \sin{\theta} \hat{x}_2 - \Omega \left(\frac{R}{r}-\cos{\theta} \right)\hat{x}_3$$
The principal moments are:
$$I_1=I_2=\frac{mr^2}{4} \ \text{   and   } \; I_3=\frac{mr^2}{2}$$
so the angular momentum will be:
$$\vec{L}= \frac{mr^2}{2}\left(\frac12\Omega \sin{\theta} \hat{x}_2 - \Omega \left(\frac{R}{r}-\cos{\theta} \right)\hat{x}_3\right)$$
Only the horizontal component of $\vec{L}$ is changing:
$$|\frac{d \vec{L}}{dt}|=\Omega L_{\perp}=\frac{1}{4}mr\Omega^2\sin\theta\left( 2R-r\cos\theta \right).$$
Now we have to calculate the torque relative to CM, as well. The torque comes from the forces at the contact point, the horizontal force is $m(R-r\cos\theta)\Omega^2$ and the vertical force will be simply $mg$.
$$|\vec \tau|= mg(r \cos θ) − m(R − r \cos θ)Ω^2 (r \sin θ)$$
Using $|\vec\tau|=|\frac{d\vec L}{dt}|$, we can find:
$$\Omega=2\sqrt{\frac{g}{6 R \tan\theta - 5 r \sin{\theta}}}$$
$$ \Rightarrow T= \pi \sqrt{\frac{6 R \tan\theta - 5 r \sin{\theta}}{g}} \approx 0.446 \text{s}$$
