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

Answer

D

Explanation

`tau = mg xx (l)/(2)`
`tau= I alpha = (m l^(2))/(3) alpha`
`"mg" (l)/(2) = (ml^(2))/(3) alpha rArr alpha= (3g)/(2l)`