(C) The torque $\tau$ about point $A$ is due to the gravitational force acting at the center of mass of the rod,which is at a distance $l/2$ from $A.$

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(C) The torque $\tau$ about point $A$ is due to the gravitational force acting at the center of mass of the rod,which is at a distance $l/2$ from $A.$$\tau = mg \times \frac{l}{2} = \frac{mgl}{2}$Using the relation between torque and angular acceleration,$\tau = I\alpha$,where $I$ is the moment of inertia about $A$ and $\alpha$ is the angular acceleration.Given $I = \frac{ml^2}{3}$,we have:$\alpha = \frac{\tau}{I} = \frac{mgl/2}{ml^2/3} = \frac{mgl}{2} \times \frac{3}{ml^2} = \frac{3g}{2l}$

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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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:

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A
$\frac{mgl}{2}$
B
$\frac{3}{2}gl$
C
$\frac{3g}{2l}$
D
$\frac{2g}{3l}$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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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:

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$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

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