Answer
$\omega = 6.3~rad/s$
Work Step by Step
We can find the rotational inertia:
$I = (0.20~kg)(0.50~m)^2+(0.20~kg)(0.50~m)^2+(0.20~kg)[(\sqrt{2})(0.50~m)]^2$
$I = 0.20~kg~m^2$
We can find the magnitude of the change in gravitational potential energy:
$\Delta U = (0.20~kg)(9.8~m/s^2)(0.50~m)+ (0.20~kg)(9.8~m/s^2)(0.50~m)+ (0.20~kg)(9.8~m/s^2)(1.0~m)$
$\Delta U = 3.92~J$
We can use conservation of energy to find the angular speed:
$\Delta U = \frac{1}{2}~I~\omega^2$
$\omega^2 = \frac{2~\Delta U}{I}$
$\omega = \sqrt{\frac{2~\Delta U}{I}}$
$\omega = \sqrt{\frac{(2)(3.92~J)}{0.20~kg~m^2}}$
$\omega = 6.3~rad/s$
