A uniform rod of length `l` and mass `m` is free to rotate in a vertical plane about `A` as shown in Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is

Options

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

Answer

D

Explanation

(d) The moment of inertia of the uniform rod about and axis through one end and perpendicular to its length is `I = (ml^2)/(3)` where `m` is mass of rod and `l` is legth. Torque `(tau = I prop)` acting on cetre of gravity of rod is given by
`tau = mg (l)/(2)`
or `I prop = mg (l)/(2)`
or `(ml^2)/(3) prop = mg (l)/(2)`
or `prop = (3g)/(2 l)`.
.