A girl of mass `M` stands on the rim of a frictionless merry-go-round of radius `R` and rotational inertia `I` that is not moving. She throws a rock of mass `m` horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the rock. relative to the ground, is `v`. After, the linear speed of the girl is.
.

Options

A. `(mvR^2)/(I + MR^2)`
B. `((m + M)v R^2)/(I + MR^2)`
C. `(mvR^2)/(I+ (M + m)R^2)`
D. `(mvR^2)/(I + (M - m)R^2)`

Answer

A

Explanation

(a) The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merry-go round is `(I + MR^2) omega`, which we will take to be positive. The final angular momentum we associate with the thrown rock is negative : `- mRv` where `v` is the speed (positive by definition) of the rock relative to the ground.
Angular momentum conservation leads to
`0 = (I + MR^2) omega = mRv rArr omega = (mRv)/(I + MR^2)`
The girl's linear speed is `R omega = (mv R^2)/(I + MR^2)`.