Question

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))`

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

Answer 1

Correct Answer - B
Torque on rod
= moment of weight of the rod about `A`
`tau = mg xx (l)/(2)`
As `tau = I alpha = mg (l)/(2) :. (ml^(2))/(3) alpha = mg (l)/(2)`
`alpha = (3g)/(2l)`