Physics - Rotation Question with Solution | TestHub

When the string is cut, the rod rotates about the hinge at P. The torque is provided by the weight of the rod acting at its center of mass.

The distance of the center of mass from the hinge P is $\frac{L}{2}$.

Torque about P is given by:
$\tau =Mg\times \frac{L}{2}$

The moment of inertia of the rod about the hinge P is:
$I=\frac{M{L}^2}{3}$

Using the relation $\tau =I\alpha$, where $\alpha$ is the angular acceleration:
$Mg\frac{L}{2}=\frac{M{L}^2}{3}\alpha$

Solving for $\alpha$, we get:
$\alpha =\frac{3g}{2L}$

Answer: $\frac{3g}{2L}$