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. `(a)sqrt((Iomega^(2)-mv^(2))/(I))`
B. `(b)sqrt(((I+mR^(2))omega^(2)-mv^(2))/(I))`
C. `(c)(Iomega-mvR)/(I)`
D. `(d)((I+mR^(2))omega-mvR)/(I)`

Explanation

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