Answer
$\omega = \sqrt{\frac{4g}{3R}}$
Work Step by Step
We can find the moment of inertia of the system.
$I = mR^2+ \frac{1}{2}mR^2 = \frac{3}{2}mR^2$
We can use conservation of energy to find the angular speed.
$KE = PE$
$\frac{1}{2}I\omega^2 = mgR$
$\frac{1}{2}(\frac{3}{2}mR^2)\omega^2 = mgR$
$\frac{3}{4}R\omega^2 = g$
$\omega = \sqrt{\frac{4g}{3R}}$
