a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg.

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a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg.

a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg.

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a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg.

a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg.

2 Answers

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