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A uniform rod AB of length $l$ and mass $m$ is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is $\dfrac{ml^2}{3}$, the initial angular acceleration of the rod will be
A
$\dfrac{mgl}{2}$
B
$\dfrac{3}{2}gl$
C
$\dfrac{3g}{2l}$
D
$\dfrac{2g}{3l}$
Detailed Solution
The weight $mg$ of the rod acts at its centre of mass, at a distance $\dfrac{l}{2}$ from A.
With the rod horizontal, the weight will produce a torque about A: $\tau = mg \times \dfrac{l}{2}$
Also, $\tau = I\alpha$, where $I = \dfrac{ml^2}{3}$ is the moment of inertia about A and $\alpha$ is the angular acceleration.
$\therefore \dfrac{ml^2}{3}\,\alpha = mg \times \dfrac{l}{2}$
$\alpha = \dfrac{mgl}{2} \times \dfrac{3}{ml^2}$
$\Rightarrow \alpha = \dfrac{3g}{2l}$