Q. 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 ml²/3, the initial angular acceleration of the rod will be
A) mgl/2
B) (3/2)gl
C) 3g/(2l)
D) 2g/(3l)
Explanation
[Solution from source] Torque of gravity about A is mg(l/2) = Iα = (ml²/3)α, so α = 3g/(2l). mgl/2 is the torque itself, not the angular acceleration. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.