A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))`

Options

A. `(3g)/(2l)`
B. `(2l)/(3g)`
C. `(3g)/(2l^(2))`
D. `mg l/2`

Answer

A

Explanation

Here `tau=Ia
implies(mg)((l)/(2))=((ml^(2))/(3))(alpha)=(3g)/(2l)`