$\operatorname{mr} \omega^{2} \geq \mu m g \text { or } r \geq \frac{\mu g}{\omega^{2}}$

The coin just slips when $\mathrm{r}=\frac{\mu \mathrm{g}}{\omega^{2}}$

For radii $r_{1}$ and $r_{2},$ let angular velocities be $\omega_{1}$ and $\omega_{2}$ respectively.

$\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{\omega_{1}^{2}}{\omega_{2}^{2}}=\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2}$

${\omega_{2}^{2}}=\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2}$

Hence, $r_{1}=4 r,$ when $\omega_{1}=\omega$ and for $r_{2}, \omega_{2}=2 \omega$

$\frac{r_{2}}{4 r}=\left(\frac{\omega}{2 \omega}\right)^{2}=\frac{1}{4}$

$\therefore r_2=r$