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))
Answer & explanation
Correct answer: option 1
\(\tau = I \alpha\)
mg x \(\frac{l}{2}\) = \(\frac{ml^2}{3}\)(α)
α = \(\frac{3g}{2l}\)