A circular ring of mass `m` and radius `R` rests flat on a horizontal smooth surface as shown in figure. A particle of mass `m`, and moving with a velocity `v`. Collides inelastically `(e=0)` with the ring the angular velocity with which the system rotates after the particle strikes the ring is

Options

A. `(v)/(2R)`
B. `(v)/(3R)`
C. `(2v)/(3R)`
D. `(3v)/(4R)`

Explanation

`C_(1)C_(2)=(R)/(2)`
Where `C_(2)=` combined centre of mass of ring and particle.
From conservation of angular momentum about `C_(2)`
`L_(i)=L_(f)`
`therefore(mv.R)/(2)=Iomega`
`=[mR^(2)+(mR^(2))/(4)+(mR^(2))/(4)]omega`
solving this equation we get,
`omega=(v)/(3R)`.