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?

The linear velocity of girl =

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 =   This will be the linear velocity of the girl

The angular velocity ω = =

here R is the radius of wheel

No it doesnt because it can be any number.

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.