Initial conditions have been set up so that a coin of radius r rolls around in a circle, as shown in fig. The contact point on the ground traces out a circle, of radius R, and the coin makes a constant angle $\theta$ with the horizontal. The coin rolls without slipping (assume that the friction with the ground is as large as needed). What is the angular–velocity $\Omega ,$ of the circular motion of the contact point on the ground? Where (R = 6m, r = 2m, $\theta ={53}^0)$

$\left(g = 10 m/s^2\right)$

 

$\omega =\Omega \widehat{z}-{\omega }^{/}{\widehat{x}}_3=\Omega \widehat{z}-\left(\frac{R}{r}\right)\Omega {\widehat{x}}_3$

But $\widehat{z}=\sin \theta {\widehat{x}}_2+\cos \theta {\widehat{x}}_3,$ so we can write   in terms of the principal axes as $\omega =\Omega \sin \theta {\widehat{x}}_2-\Omega (\frac{R}{r}-\cos \theta ){\widehat{x}}_3$

The principal moments are $I_3=\left(\frac{1}{2}\right)m{r}^2,$ and $I_2=\left(\frac{1}{4}\right)m{r}^2$

The angular momentum is $L=I_2{\omega }_2{\widehat{x}}_2+I_3{\omega }_3{\widehat{x}}_3,$ so its horizontal component has length $L_{\perp }=I_2{\omega }_2\cos \theta -I_3{\omega }_3\sin \theta ,$ with leftward taken to be positive. Therefore, the magnitude of $\frac{dL}{dt}$ is 

$\left|\frac{dL}{dt}\right|=\Omega L_{\perp }=\Omega (I_2{\omega }_2\cos \theta -I_3{\omega }_3\sin \theta )$

$=\Omega [\left(\frac{1}{r}m{r}^2\right)\left(\Omega \sin \theta \right)\cos \theta -\left(\frac{1}{2}m{r}^2\right)(-\Omega (\frac{R}{r}-\cos \theta ))\sin \theta ]=\frac{1}{4}mr{\Omega }^2\sin \theta (2R-r\cos \theta )$

With a positive quantity corresponding to $\frac{dL}{dt}$ pointing out of the page (at the instant shown). The torque (relative to the CM) comes from the force at the contact point. There are two components to this force. The vertical component is mg, and the horizontal component is $m(R-r\cos \theta ){\Omega }^2$ leftward, because the CM moves in a circle of radius $(R-r\cos \theta ).$ The torque therefore has magnitude. $\left|\tau \right|=mg\left(rco\theta \right)-m(R-r\cos \theta ){\Omega }^2\left(r\sin \theta \right)$

Without of the page taken to be positive. Equating this $\left|\tau \right|$ with the $\left|\frac{dL}{dt}\right|$ in eq. gives ${\Omega }^2=\frac{g}{\frac{3}{2}R\tan \theta -\frac{5}{4}r\sin \theta }$