A rod PQ of mass M and length L is hinged at | vedclass

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(D) When the string is cut,the rod will rotate about the hinge point $P$ due to the torque produced by its weight.Let $\alpha$ be the initial angular

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

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(D) When the string is cut,the rod will rotate about the hinge point $P$ due to the torque produced by its weight.Let $\alpha$ be the initial angular acceleration of the rod.The torque $	au$ about point $P$ is given by $	au = I\alpha$,where $I$ is the moment of inertia of the rod about one end.$I = \frac{ML^2}{3} \quad ...(i)$Also,the torque due to gravity acting at the center of mass (at distance $L/2$ from $P$) is:$	au = Mg 	imes \frac{L}{2} \quad ...(ii)$Equating $(i)$ and $(ii)$:$\frac{ML^2}{3} \alpha = Mg \frac{L}{2}$$\alpha = \frac{MgL}{2} 	imes \frac{3}{ML^2}$$\alpha = \frac{3g}{2L}$

$A$ 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 the figure. When the string is cut,the initial angular acceleration of the rod is

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