Answer
$$\omega=\sqrt{7.5 \frac{g}{L}}
$$
Work Step by Step
$$
\begin{aligned}
& T_1+V_1=T_2+V_2 \\
& 0+m g L=\frac{1}{2} m v_G^2+0 \\
& v_G=\sqrt{2 g L} \\
& H_1=H_2 \\
& m \sqrt{2 g L}\left(\frac{L}{2}\right)=\frac{1}{3} m L^2\left(\omega_2\right) \\
& \omega_2=\frac{3}{2} \frac{\sqrt{2 g L}}{L} \\
& T_2+V_2=T_3+V_3 \\
& \frac{1}{2}\left(\frac{1}{3} m L^2\right) \frac{9(2 g L)}{4 L^2}+0=\frac{1}{2}\left(\frac{1}{3} m L^2\right) \omega^2-m g\left(\frac{L}{2}\right) \\
& \frac{3}{4} g L=\frac{1}{6} L^2 \omega^2-g\left(\frac{L}{2}\right) \\
& \omega=\sqrt{7.5 \frac{g}{L}}
\end{aligned}
$$
