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

Let the angular velocity of disc after child jumps off be `omega`
`:.` From conservation of angular mometum
`(I=mR^(2))omega=mvR=Iomega'`
`:. omega'=((I+mR)-omega-mvR)/I`