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. `(3g)/(2l)`
B. `(2g)/(3l)`
C. `mg""(l)/(2)`
D. `(3)/(2)gl`

Answer

A

Explanation

`:' tau =I alpha :. (mg)""(l)/(2)=((ml^(2))/(3)) alpha implies alpha=(3g)/(2l)`