A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is : (Moment of inertia of rod about A is ml²/3)

Question

Dr. Abubacker · Accepted Answer

[Solution from source] The weight acts at the centre, l/2 from A, giving torque τ = mgl/2. With I = ml²/3, α = τ/I = (mgl/2)/(ml²/3) = 3g/(2l). The expression mg(l/2) is the torque itself, not the angular acceleration. Answer (1).

Q. A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is : (Moment of inertia of rod about A is ml²/3)

Explanation

Reviewed by Dr. Abubacker

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