A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml23, the initial angular acceleration of the rod will be [figure]
- mgl2
- 32gl
- 3g2l
- 2g3l
Answer
(3) 3g2l
Solution
Weight mg acts at the centre, at a distance l2 from A.
= mg l2
τ = Iα ⇒ mg l2 = ml23
∴ = 3g2l
Answer: (3)