A thin uniform rod of mass $M$ and length $L$ is hinged at one end and is free to rotate in a vertical plane. It is released from a horizontal position. When the rod makes an angle $\theta$ with the vertical, the magnitude of the hinge reaction force is $F$. Find the value of $F$ when $\theta =90^{\circ }$.

Solution

1. At $\theta =90^{\circ }$ (horizontal position), the angular velocity $\omega =0$ since it is released from rest.
2. The torque about the hinge is $$\begin{array}{rl}\tau & =Mg(L/2)\\ & =I\alpha \\ & =(M{L}^2/3)\alpha \end{array}$$, so $\alpha =3g/(2L)$.
3. The acceleration of the center of mass is $$\begin{array}{rl}{a}_{cm}& =\alpha (L/2)\\ & =3g/4\end{array}$$ directed downwards.
4. The vertical equation of motion is $$\begin{array}{rl}Mg-R_y& =M{a}_{cm}\\ & =M(3g/4)\end{array}$$, so $R_y=Mg/4$.
5. Since there is no horizontal acceleration at the start, the horizontal reaction $R_x=0$.
6. Hence the answer is (A).