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A uniform rod of length $l$ and mass $m$ is free to rotate in a vertical plane about A. The rod, initially in horizontal position, is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is $\dfrac{ml^2}{3}$) :
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A
$\dfrac{3g}{2l}$
B
$\dfrac{2l}{3g}$
C
$\dfrac{3g}{2l^2}$
D
$mg\dfrac{l}{2}$
Detailed Solution
The weight $mg$ of the rod acts at its centre, at a distance $\dfrac{l}{2}$ from A.
Torque of the weight about A when the rod is horizontal: $\tau = (mg)\left(\dfrac{l}{2}\right)$
Here $\tau = I\alpha$, with $I = \dfrac{ml^2}{3}$
$\Rightarrow (mg)\left(\dfrac{l}{2}\right) = \left(\dfrac{ml^2}{3}\right)(\alpha)$
$\alpha = \dfrac{mgl}{2} \times \dfrac{3}{ml^2}$
$\Rightarrow \alpha = \dfrac{3g}{2l}$
Torque of the weight about A when the rod is horizontal: $\tau = (mg)\left(\dfrac{l}{2}\right)$
Here $\tau = I\alpha$, with $I = \dfrac{ml^2}{3}$
$\Rightarrow (mg)\left(\dfrac{l}{2}\right) = \left(\dfrac{ml^2}{3}\right)(\alpha)$
$\alpha = \dfrac{mgl}{2} \times \dfrac{3}{ml^2}$
$\Rightarrow \alpha = \dfrac{3g}{2l}$