A uniform rod AB of length l and mass m is free to rotate about point A. rod is released from r

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6.System of Particles and Rotational Motion

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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 $ml^2/3$, the initial angular acceleration of the rod will be

A

$\frac{{mgl}}{2}$

B

$\frac{3}{2}gl$

C

$\;\frac{{3g}}{{2l}}$

D

$\;\frac{{2g}}{{3l}}$

(AIPMT-2006) (AIPMT-2007)

Solution

Torque about $A,$

$tau = mg \times \frac{1}{2} = \frac{{mgl}}{2}$
Also$\,\tau = I\alpha $
$alpha = \frac{\tau }{I} = \frac{{mgl/2}}{{m{l^2}/3}} = \frac{{3g}}{{2\,l}}$

Standard 11

Physics

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