A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m andm are attached to it at distance b and 3b respectively from the pivot. The rod is held horizontal and then released. The angular acceleration of the rod at the instant it is released is :

Options

A. bg
B. `(b)/(g)`
C. `gb+5`
D. `(5g)/(11b)`

Answer

D

Explanation

Here torque due to two masses 2m and m is
`tau=2mgxxb+mgxx3b`
`=5mgb`
M.I. of the two masses
`I=2mb^(2)+9mb^(2)`
`=11mb^(2)`
Now `alpha=(tau)/(I)=(5mgb)/(11mb^(2))`
`=(5g)/(11b)`