AIPMT PRELIMS 2006 Physics Angular Acceleration MCQ Question
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))
3g/2l
2l/3g
3g/2l²
mg(l/2)
Correct Answer
Detailed Explanation
The initial angular acceleration can be found using the formula α = τ/I, where τ is the torque due to gravity and I is the moment of inertia. The torque is mg(l/2) and the moment of inertia is (ml²/3), leading to α = 3g/2l.
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