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

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Solution

## Step 1: Given dataThe length of the rod is $L$The mass of the rod is $M$DiagramStep 2: Formula UsedTorque is the product of force and distance. Torque is defined as, $\stackrel{\to }{\tau }=\stackrel{\to }{S}×\stackrel{\to }{F}$, where, F is the force and S is the distance.For a body rotating with acceleration $\alpha$, then the torque exerted on the body is $\tau =I\alpha$, where, I is the moment of inertia of the body. The moment of inertia of a rod about an axis passing through the one end is $\frac{M{L}^{2}}{12}$ where M= mass of rod, L= length of rod Step 3: CalculationsFinding the angular accelerationWhen the string is cut, due to gravity a torque will be acted on the rod. This torque is on the rod is $\stackrel{\to }{\tau }=\stackrel{\to }{L}×\stackrel{\to }{F}$,We know due to rotation torque is $\tau =I\alpha$. considering the balancing torque at point Q due to angular acceleration.So, $FL=I\alpha \phantom{\rule{0ex}{0ex}}or\alpha =\frac{FL}{I}...............\left(1\right)$We know, that the moment of inertia of a rod about an axis passing through the one end is $\frac{M{L}^{2}}{3}$.From equation (1)$\alpha =\frac{FL}{I}=\frac{Mg\frac{L}{2}}{\frac{M{L}^{2}}{12}}=\frac{3g}{2L}\phantom{\rule{0ex}{0ex}}or\alpha =\frac{3g}{2L}.$Therefore, the initial angular acceleration of the rod is $\frac{3g}{2L}$.

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