When the end M is released and rod makes angle $$\alpha$$ with horizontal, the displacement of centre of mass is $${L \over 2}\sin \alpha $$.

Now, we know $$mg{L \over 2}\sin \alpha = {1 \over 2}I{\omega ^2}$$

Here, $$I = {{m{L^2}} \over 3}$$; therefore,

$$mg{L \over 2}\sin \alpha = {1 \over 2}{{m{L^2}} \over 3}{\omega ^2}$$

$$ \Rightarrow {\omega ^2} = {{3g\sin \alpha } \over L} \Rightarrow \omega = \sqrt {{{3g\sin \alpha } \over L}} \Rightarrow \omega \propto \sqrt {\sin \alpha } $$