A child with mass `m` is standing at the edge of a merry go round having moment of inertia `I`, radius `R` and initial angular velocity `omega` as shown in the figure. The child jumps off the edge of the merry go round with tangential velocity `v` with respect to the ground. The new angular velocity of the merry go round is

Options

A. `sqrt((Iomega^(2)-mv^(2))/(I))`
B. `sqrt(((I+mR^(2))omega^(2)-mv^(2))/(I))`
C. `(Iomega-mvR)/(I)`
D. `((I+mR^(2))omega-mvR)/(I)`

Answer

D

Explanation

`L_(i)=L_(f)`
`(I+mR^(2))omega=(mvR)+Iomega'`
`thereforeomega'=((I+mR^(2))omega-mvR)/(I)`