A coin resting on a rotating turntable just begins to slip when...

The condition for the coin to just slip on the rotating turntable is that the required centripetal force is provided by the maximum static friction.

$m r \omega^2 = \mu m g$

$r = \dfrac{\mu g}{\omega^2}$

This implies that $r \propto \dfrac{1}{\omega^2}$.

Let the initial angular velocity be $\omega_1 = \omega$ and the initial distance be $r_1 = 4 \text{ cm}$.

When the angular velocity is doubled, $\omega_2 = 2\omega$.

Using the proportionality $r_1 \omega_1^2 = r_2 \omega_2^2$, we get:

$4 \times \omega^2 = r_2 \times (2\omega)^2$

$4 \omega^2 = r_2 \times 4\omega^2$

$r_2 = 1 \text{ cm}$