A small bead of mass `m` moving with velocity `v` gets threaded on a stationary semicircular ring of mass `m` and radius `R` kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

Options

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

Answer

C

Explanation

Using conservation of angular momentum about `O` we get
`mvR=(mR^(2)+mR^(2))omega, omega=v/(2R)`