A thin circular ring of mass M and radius r is rotating about its ais with an angular speed `omega`. Two particles having mas m each are now attached at diametrically opposite points. The angular speed of the ring will become

Options

A. `omega(M)/(M+2m)`
B. `omega(M+2m)/(M)`
C. `omega(M-2m)/(M+2m)`
D. `omega(M)/(M+m)`

Answer

A

Explanation

As there is no external torque
So, angular momentum remian conserved
`I_(1)omega_(1)=I_(2)omega_(2)`
`(MR^(2))omega=(MR^(2)+2mR^(2))omega`
`omega=(M omega)/(M+2m)`