Question

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 ground, is ν. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Answer 1

The linear velocity of girl =
`(mv)/(M)`

Step-by-step explanation:

As no external force is applied , therefore the momentum of the system will remain conserved .

The initial momentum = 0 , because the girl with stone in hand is at rest .

When stone is thrown horizontally , its momentum is = m v

Let the velocity of girl after this throwing of stone is v₀

Thus its momentum is M v₀

Therefore the sum of these two momentum should be zero

Hence m v + M v₀ = 0

or v₀ =
`(mv)/(M)` This will be the linear velocity of the girl

The angular velocity ω =
`(v_0)/(R)` =
`(mv)/(MR)`

here R is the radius of wheel