Questions dealing with Angular momentum (but maybe kinetic engery)

[Prodan]

I recently had a question on a test that just boggled my mind. I could not find the answer and unfortunatley could not consult a T.A. I have a vague memory of the question. It is a figure skater rotating at angular velocity w = 2 rad/s (not sure if that number is correct). The figure skater is holding a sphere mass 150grams(.15kg) which is located .6 meters from the axis of rotation. The sphere is 1.6m from the ground.
a) What is the net force acting on the small sphere? (I said gravity but was not sure)
b) How long will it take the ball to hit the ground if released?
c) Calculate the Initial Kinetic Energy given the final Kinetic engery is 12000 J. (I think this was the phrasing of last question)

At first I thought of using the
mgh + 1/2I(Inertia)w(Angular vel)^2 = 1/2mv^2(1+ I(Intertia of Centre of Mass) / MR^2)

We are not given intertia though? or did i remember this incorrectly perhaps? Would it make more sense to have intertia?

I am truley stumped, any help would be awesome.

Kevin

 

[blochwave]

a)Well it's rotating. If the only force acting on it was gravity, it wouldn't be rotating. In fact, there's also a normal force on it from her hands since she hasn't dropped it yet, so it's just the centripetal force

b)The rotational motion has no effect on it falling, it's just your typical falling object problem

c)Remember the final kinetic energy is only translational, once she let's go the centripetal force goes away and it's only traveling linearly(EDIT: In the sense that it's not going to be KEtotal=1/2mv^2+1/2Iw^2, just 1/2mv^2, and it'll travel in a parabolic arc like you always see with projectile motion)

Also the moment of inertia of a point mass is just mr^2, where r is its distance from the rotational axis. You should assume it's a point mass, a solid sphere has a different moment of inertia but you're not given a radius of the sphere, so I don't think you assume that