For the pivoted slender rod of length l as shown in figure, the angular velocity as the bar reaches the vertical position after being released in the horizontal position is

$$\begin{gathered} when the systm is released loss in P.E for\frac{3 l}{4}part is \frac{3mg}{4}\frac{3l}{8} and gain in P.E. for l/4 part is \frac{mg}{4}\frac{l}{8} ; \\ net loss in P.E is \frac{9mgl}{32}-\frac{mgl}{32}=\frac{mgl}{4}= increase in roattional K.E ; \\ \frac{1}{2}I{\omega }^2=\frac{mgl}{4} here I about an axis \frac{l}{4} from the center is \frac{m{l}^2}{12}+\frac{m{l}^2}{16}=\frac{7m{l}^2}{48} ; \\ soving for \omega we get \omega =\sqrt{\frac{24g}{7l}} \end{gathered}$$