Answer
$\omega = 11.4~rad/s$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia:
$I = \frac{1}{2}MR^2+Md^2$
$I = \frac{1}{2}(20~kg)(0.10~m)^2+(20~kg)(0.050~m)^2$
$I = 0.15~kg~m^2$
From the initial position to the lowest position, the center of mass falls a distance of $5.0~cm$
We can use conservation of energy to find the angular speed at the lowest position:
$\frac{1}{2}I~\omega^2 = mgh$
$\omega^2 = \frac{2~mgh}{I}$
$\omega = \sqrt{\frac{2~mgh}{I}}$
$\omega = \sqrt{\frac{(2)~(20~kg)(9.8~m/s^2)(0.050~m)}{0.15~kg~m^2}}$
$\omega = 11.4~rad/s$
