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

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(B) When the string is cut,the only torque acting on the rod about the hinge $P$ is due to the gravitational force acting at the center of mass of the

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

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(B) When the string is cut,the only torque acting on the rod about the hinge $P$ is due to the gravitational force acting at the center of mass of the rod.The gravitational force $Mg$ acts at a distance of $L/2$ from the hinge $P$.The torque $	au$ about the hinge $P$ is given by:$	au = Mg 	imes \frac{L}{2}$The moment of inertia $I$ of the rod about an axis passing through its end $P$ is:$I = \frac{ML^2}{3}$Using the rotational analog of Newton's second law,$	au = I\alpha$,where $\alpha$ is the angular acceleration:$Mg 	imes \frac{L}{2} = \left(\frac{ML^2}{3}ight) \alpha$Solving for $\alpha$:$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$

$A$ 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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