Question

A coin spinning about its axis of symmetry with angular frequency $\omega$ is set down on a horizontal surface. After it stops slipping, it rolls away with velocity $v=-\frac{\omega R}{n}$, find n.

Answer 1

Take the coordinates as shown. Before the coin stops slipping, the frictional force is f $=\mu$mg, where $\mu$ is coefficient of sliding friction.
Let x${}_c$ be the x-cordinate of the centre of mass of the coin. The equation of motion of the coin before it stops slipping are
$m{\ddot{x}}_c=-\mu$mg
I$\ddot{\theta }=-\mu mgR$
where m and R are respectively the mass and radius of the coin and, $I=\frac{1}{2}m{R}^2$
Integrating and using initial conditions
${\dot{x}}_c=$0 $\dot{\theta }=\omega$ at $t=0$
$\Rightarrow {\dot{x}}_c=-\mu gt$
$\dot{\theta }=\omega -\frac{2\mu gt}{R}$
when the coin rolls without slipping, we have ${\dot{x}}_c=-\dot{\theta }R$
Let this happen at time t, the the above gives
$-\mu gt=-\omega R+2\mu gt$
$\Rightarrow t=\frac{\omega R}{3\mu g}$
At this time , the velocity of the centre of mass of the coin is
${\dot{x}}_c=-\mu gt=-\frac{1}{3}\omega R$
$\Rightarrow$
$n=3$