Rolling coin

It would be quite difficult to explain the answer to this question without images so i would be giving links to images I used for solving the question:

So a simplified figure for starting off with the solution:

Let's first go to the CM frame. As it can be seen in the figure above, 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 $T$ = $2\pi\Omega$ .

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:

$\Omega=2\sqrt{\frac{g}{6 R \tan\theta - 5 r \sin{\theta}}}$

$\Rightarrow T= \pi \sqrt{\frac{6 R \tan\theta - 5 r \sin{\theta}}{g}} \approx 0.446 \text{s}$

Let us consider the triangle relating the base of the quarter, B , center of gravity of the quarter, G, and the center of the circle, C.

Note that angle GBC $= 45^o$ due to the tilt of the quarter, and lengths $BC$ and $BG$ are $R = 4$ cm and $r = 1.2$ cm respectively. Hence, using cosine rule, length $CG = 3.2637 cm$ . Thereafter, using sine rule, angle GCB $= 15.069^o$

Now, we know that the centripetal acceleration required must be along line $CG$ directed at an angle of $15.069^o$ from the horizontal.

Let us now consider the free body diagram of the coin. At the center of gravity, we have the acceleration due to normal contact force , $a_N$ , and the acceleration due to the weight of the coin $g$ . $a_N$ is directed $45^o$ away from the horizontal. Now, we need to resolve the forces to produce one that is along line $CG$

We now set line CG as the new x-axis and its corresponding normal as y-axis, which this new set of orthogonal base tilted $15.069^o$ counter-clockwise. Now, the angle that $g$ makes with the y-axis is $15.069^o$ and the angle that $a_N$ makes with the y-axis is $45^o - 15.069^o = 29.931^o$

With that, resolving forces along y-axis,

$a_N cos(29.931^o) = g cos(15.069^o)$ ---- Equation (1)

Resolving along x-axis,

$a_c = a_N sin(29.931^o) + g sin(15.069^o)$ --- Equation (2)

Hence, we can solve for $a_c$

Using $a_c = r\omega^2$ , and $\omega = \dfrac{2\pi}{T}$ ,

$T = \boxed{0.444s}$

Note : The answer might be slightly off due to round-off errors :X