A uniform rod AB of length l and mass m is fr | vedclass

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(A) The torque $	au$ acting on the rod about point $A$ is due to the weight of the rod acting at its center of mass,which is at a distance of $\frac{

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

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(A) The torque $	au$ acting on the rod about point $A$ is due to the weight of the rod acting at its center of mass,which is at a distance of $\frac{l}{2}$ from $A$.$	au = mg \left(\frac{l}{2}ight)$Using the rotational form of Newton's second law,$	au = 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:$mg \left(\frac{l}{2}ight) = \left(\frac{ml^2}{3}ight) \alpha$Solving for $\alpha$:$\alpha = \frac{mg(l/2)}{ml^2/3} = \frac{mg l}{2} \cdot \frac{3}{ml^2} = \frac{3g}{2l}$

$A$ 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 $\frac{ml^2}{3}$,the initial angular acceleration of the rod will be:

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