Coin on rotating table

The coin slips on a rotating table when the frictional force is equal to the centripetal force $F$ , which is given by $F=mR\omega^2$ , where $m$ is the mass of the coin, $R$ the distance from the center of the rotating table and $\omega$ is the angular velocity of the rotating table.

Therefore, for the two cases,

$F=mR_1 \omega_1^2 = m R_2 \omega_2^2$

$\Rightarrow R_1 \omega_1^2 = R_2 \omega_2^2$

Since $R_1=4r$ and $\omega_2=2\omega_1$ , we have:

$(4r) \omega_1^2 = R_2 (4\omega_1^2)$

$\Rightarrow R_2 = \boxed {r}$