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 ml^⅔

A uniform rod of length l and mass m is free to rotate in a vertical plane about A.

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Answer:

Here given the torque about A is τ=mg(l/2)

We know that 

τ=lα=mg(l/2)

⇒α=3g/2l

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