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?

Respuesta :

The linear velocity of girl = [tex]\frac{mv}{M}[/tex]

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 = [tex]\frac{mv}{M}[/tex]   This will be the linear velocity of the girl

The angular velocity ω = [tex]\frac{v_0}{R}[/tex] = [tex]\frac{mv}{MR}[/tex]

here R is the radius of wheel