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

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(C) The weight of the rod acts at its center of mass,which is at a distance of $l/2$ from the pivot point $A$. This weight creates a torque about poin

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

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(C) The weight of the rod acts at its center of mass,which is at a distance of $l/2$ from the pivot point $A$. This weight creates a torque about point $A$.The torque $	au$ is given by:$	au = mg 	imes \frac{l}{2}$According to Newton's second law for rotation:$	au = I\alpha$Substituting the given moment of inertia $I = \frac{ml^2}{3}$ and the torque:$mg 	imes \frac{l}{2} = \left(\frac{ml^2}{3}ight) \alpha$Solving for the angular acceleration $\alpha$:$\alpha = \frac{mg 	imes l/2}{ml^2/3} = \frac{mg 	imes l}{2} 	imes \frac{3}{ml^2} = \frac{3g}{2l}$

$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

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