A uniform rod of mass `2 kg` and length `1 m` lies on a smooth horizontal plane. A particle of mass 1 kg moving at a speed of `2ms^-1` perpendicular to the length of the rod strikes it at a distance `(1)/(4)` m from the centre and stops . Find the angular velocity of the rod about its centre just after the collision (in rad/s)

Options

A. `3 rad s^(-1)`
B. `4 rad s^(-1)`
C. `1 rad s^(-1)`
D. `2 rad s^(-1)`

Answer

A

Explanation

`L_(i)-L_(f)` (about centre of mass of the rod)
`mvr=(ML^(2))/(12).omegarArr omega=(12mvr)/(ML^(2))=((12)(1)(2)(1//4))/((2)(1)^(2))=3 "rads"^(-1)`.