Question

A rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in figure. When string is cut, the initial angular acceleration of the rod is:

Answer 1

Torque on the rod = moment of weight of the rod about P

$T=Mg\frac{L}{2}......\left(i\right)$

Moment of inertial of rod about P

$I=\frac{M{L}^2}{3}......\left(ii\right)$

AS $T=I\alpha$

From equation (i) and (ii), we get

$Mg\frac{L}{2}=\frac{M{L}^2}{3}\alpha$

$\therefore \alpha =\frac{3g}{2L}$