Use Conservation of Energy:

1. Initial Position (Horizontal): $K_i=0$. $U_i=Mg(L/2)$ (taking the pivot as vertical reference, center of mass is $L/2$ above ground).
2. Final Position (Vertical): $U_f=0$. $K_f=\frac{1}{2}I{\omega }^2$.

• Moment of inertia for a rod pivoted at one end: $I=\frac{1}{3}M{L}^2$. Equation: $Mg\frac{L}{2}=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\omega }^2$ $Mg\frac{L}{2}=\frac{1}{6}M{L}^2{\omega }^2$ ${\omega }^2=\frac{6MgL}{2M{L}^2}=\frac{3g}{L}$ $\omega =\sqrt{\frac{3g}{L}}$.