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 m l 2 3 )

Step-by-step solution

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is = m l 2 3 where, m is the mass and l its length. Torque τ = I α acting on centre of gravity of rod is given by τ = m g l 2 also, m l 2 3 · α = m g l 2 α = 3 g 2 l