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Physics-

General

Easy

Question

A uniform rod $A B$ 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 $fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction$ , the initial angular acceleration of the rod will be

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

The correct answer is: $fraction numerator 3 g over denominator 2 l end fraction$

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
$I equals fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction$
Where $m$ is mass of rod and $l$ is length.
Torque $left parenthesis tau equals I alpha right parenthesis$ acting on centre of gravity of rod is given by
$tau equals m g fraction numerator 1 over denominator 2 end fraction$
or $I alpha equals m g fraction numerator 1 over denominator 2 end fraction$
or $fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction alpha equals m g fraction numerator 1 over denominator 2 end fraction$
or $alpha equals fraction numerator 3 g over denominator 2 l end fraction$

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