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A uniform rod of mass M and length L is pivoted at one end such that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end of the rod is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle q with the vertical is

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$g \sin θ$

$\frac{g}{L} \sin θ$

$\frac{3 g}{2 L} \sin θ$

$6 gL \sin θ$

answer is C.

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Detailed Solution

The moment of inertia of the uniform rod about an axis through one end and perpendicular to length is

$I = \frac{{ML}^{2}}{3}$

Torque (t = la) acting on the centre of gravity of the rod is given by

$τ = Mg [\frac{L}{2} sinθ]$

or $\frac{{ML}^{2}}{3} α = Mg \frac{L}{2} sinθ$

$α = \frac{3 g}{2 L} sinθ$

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