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 `ml^(2)//3`, the initial angular acceleration of the rod will be

Options

A. `(mgl)/2`
B. `3/2gl`
C. `(3g)/(2l)`
D. `(2g)/(3l)`

Answer

C

Explanation

Torque about A,
`tau=mgxxl/2=(mgl)/2`
Also `tau=Ialpha`
`therefore` Angular acceleration, `alpha=(tau)/I=(mgl//2)/(ml^(2)//3)=3/2g/l`