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A $\operatorname{rod} P Q$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kepts 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
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The correct answer is:
$\frac{3 g}{2 L}$
Torque on the rad = Mament of weight of the rod about $\rho$
$\tau=m g \frac{1}{2}$
$\because$ Moment of inertia of rod about
$P=\frac{M L^2}{3}$
As $\tau=1 \alpha$
From Eqs. (i) and (ii), we get
$\begin{aligned}
& M g \frac{L}{2}=\frac{M L^2}{3} \alpha \\
\therefore \alpha & =\frac{3 g}{2 L}
\end{aligned}$
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