A thin uniform rod of mass $M$ and length $L$ is hinged at one end and released from a horizontal position. When the rod makes an angle $\theta$ with the vertical, the magnitude of the hinge reaction force is:

Solution

1. By conservation of energy, the angular velocity $\omega$ at angle $\theta$ is given by $\frac{1}{2}I{\omega }^2=Mg(\frac{L}{2}\cos \theta )$, where $I=\frac{M{L}^2}{3}$, yielding ${\omega }^2=\frac{3g\cos \theta }{L}$.
2. Using torque $\tau =I\alpha$, we find the angular acceleration $$\begin{array}{rl}\alpha & =\frac{Mg(L/2)\sin \theta }{M{L}^2/3}\\ & =\frac{3g\sin \theta }{2L}\end{array}$$.
3. The hinge reaction components are $$\begin{array}{rl}{F}_x& =M{\omega }^2(L/2)\\ & =\frac{3}{2}Mg\cos \theta \end{array}$$ and $$\begin{array}{rl}{F}_y& =Mg-M\alpha (L/2)\\ & =Mg-\frac{3}{4}Mg\sin \theta \end{array}$$.
4. The magnitude of the reaction force is $$\begin{array}{rl}F& =\sqrt{F_x^2+F_y^2}\\ & =\sqrt{(\frac{3}{2}Mg\cos \theta {)}^2+(Mg-\frac{3}{4}Mg\sin \theta {)}^2}\end{array}$$, which simplifies to $\frac{Mg}{4}\sqrt{81{\cos }^2\theta +{\sin }^2\theta }$.
5. Hence the answer is (D).