A slender rod of mass m and length L is pivoted about a horizontal axis through one end and released from rest at an angle of 30^o above the horizontal. The force exerted by the pivot on the rod at the instant when the rod passes through a horizontal position is

The angular velocity of the rod about the pivot when it passes through the horizontal position is given by

$\begin{array}{l}\mathrm{mg}\times \frac{\mathrm{L}}{2}\sin {30}^{\circ }=\frac{{\mathrm{mL}}^2}{3}\times \frac{{\mathrm{\omega }}^2}{2}\\ \mathrm{\omega }=\sqrt{\frac{3\mathrm{g}}{2\mathrm{L}}}\end{array}$

Radial acceleration of the centre of mass (as centre of mass is moving in a circle of radius L/2) is given by 

           ${\mathrm{a}}_{\mathrm{r}}={\mathrm{\omega }}^2\frac{\mathrm{L}}{2}=\frac{3\mathrm{g}}{4}$

Torque about pivot, in the horizontal position, is $\mathrm{\tau }=\mathrm{mg}\frac{\mathrm{L}}{2}=\mathrm{I\alpha }$

          $\mathrm{\alpha }=\frac{\mathrm{mgL}/2}{{\mathrm{mL}}^2/3}=\frac{3\mathrm{g}}{2\mathrm{L}}$

Tangential acceleration of the centre of mass, ${\mathrm{a}}_{\mathrm{t}}=\frac{\mathrm{L}}{2}\mathrm{\alpha }=\frac{3\mathrm{g}}{4}$

Draw the FBD of the rod at an instant when it passes through the horizontal position. Use Newton's second law of equation.

               $\begin{array}{l}{\mathrm{R}}_1={\mathrm{ma}}_{\mathrm{r}}=\frac{3\mathrm{mg}}{4}\\ \mathrm{mg}-{\mathrm{R}}_2=\mathrm{m}\times {\mathrm{a}}_{\mathrm{t}}=\frac{3\mathrm{mg}}{4}\\ {\mathrm{R}}_2=\frac{\mathrm{mg}}{4}\end{array}$

So, reaction force by the pivot on the rod, $\mathrm{R}={\mathrm{R}}_1^2+{\mathrm{R}}_2^2=\sqrt{10}\mathrm{mg}/4$ at an angle of ${\tan }^{-1} \left({\mathrm{R}}_2/{\mathrm{R}}_1\right)\left[={\tan }^{-1} (1/3)\right]$ with the horizontal.