A thin circular ring M and radius r is rotating about its axis with a constant angular velocity `omega`. Four objects each of mass m are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be...........

Options

A. `((M-4m)omega)/(M+4m)`
B. `(Momega)/(4m)`
C. `(Momega)/(M+4m)`
D. `((M+4m)omega)/(M)`

Answer

C

Explanation

`I_(1)` = moment of inertia of ring + moment of inertia of arrangement of four object each of mass m
`I_(1)=Mr^(2)+(4m)r^(2)`
According to the conservation of angular momentum
`=[("Initial angular"),("momentum"),("of ring")]=[("Total angular"),("momentum of"),("system after"),("arrangment of object")]`
`Iomega=I_(1)omega_(1)`
`therefore Mr^(2)omega=(Mr^(2)+4mr^(2))omega_(1)`
`therefore Momega=(M+4m)omega_(1)`
`therefore` New angular velocity `omega_(1)=(Momega)/(M+4m)`