What is the Speed of a Falling Object Attached to a Rotating Spherical Shell?

[CoreanJesus]

Homework Statement

A uniform spherical shell rotates about a vertical axis on frictionless bearings. A light cord passes around the equator of the shell, over a light, frictionless pulley, and is attached to a small object that is otherwise free to fall under the influence of gravity. Calculate the speed of the object after is has fallen a distance h from rest. Use the work-energy theorem.

Homework Equations

Well the Work-energy Theorem :P and the Rotational Inertia for a Sphere and disk

The Attempt at a Solution

So assuming that the rotational inertia is given, I used the right equations but one step confuses me. I set the Gravitational potential energy of block m equal to the kinetic energy of the block m, rotational kinetic energy of M and I.
To take into account the kinetic energy of the disk, I tried to do the same thing I did for the sphere but it didn't work. Am I missing something?
The answer is

 

[ehild]

The images do not show.
Are the masses of the sphere, pulley, and the falling object given?

 

[CoreanJesus]

No all variables
and the link should work... (hopefully)

 

[ehild]

This is the figure (just copy and paste)

Determine the speed of the falling object in terms of the mass and radius of the sphere, the moment of inertia and radius of the pulley and the mass of the falling object. You have to know how the linear speed of the falling object is related to the angular speeds of the sphere and pulley.

 

[CoreanJesus]

mgh=1/2 I(s)*w^2 + 1/2 I(p)*w^2 + 1/2 mv^2 right?

 

[ehild]

mgh=1/2 I(s)*w^2 + 1/2 I(p)*w^2 + 1/2 mv^2 right?

Not quite. The string moves with certain linear velocity, and it does not slip neither on the pulley nor on the sphere. So the linear speed of the perimeters is the same as the speed of the falling object, but the angular speeds are different. How is the angular speed and linear speed related?

 

[CoreanJesus]

V=rW?

 

[ehild]

V=rW?

Yes, V=radius times ω. The radius of the sphere is R, that of the pulley is r.

 

[CoreanJesus]

But the answer that I get using this method is wrong...
I don't get 3I as the denominator

 

[ehild]

But the answer that I get using this method is wrong...
I don't get 3I as the denominator

The given answer can be wrong. What did you get?

 

[haruspex]

But the answer that I get using this method is wrong...
I don't get 3I as the denominator

I get the given answer. Please post your working.

 

[ehild]

But the answer that I get using this method is wrong...
I don't get 3I as the denominator

3I is not the denominator. What did you use as the moment of inertia for the spherical shell?

 

[haruspex]

3I is not the denominator. What did you use as the moment of inertia for the spherical shell?

I assumed CoreanJ meant 3I in the denominator.

 

[CoreanJesus]

So I set mgh=1/2 I(s)*w^2 + 1/2 I(p)*w^2 + 1/2 mv^2
So mgh=(1/2)(2/3)mR^2(V/R)^2 + (1/2)(1/2)(IR^2)(V/R)^2 + (1/2)mV^2
Simplifying 6mgh=2mV^2 + (3/2)IV^2 + 3mV^2 And I know that this can't simplify into the proper answer...
Why does r remain for the kinetic energy of the the disk as I'm assuming that must be the case for the answer to be what it is... And what is wrong with my equation for the inertia for a pulley? is it not a solid disk?

 

[haruspex]

(2/3)mR^2

M, not m.

(1/2)(IR^2)

I is the moment of inertia of the pulley, not the mass.
Also, you have the wrong angular speed for the pulley. The radius is r, not R.

 

[CoreanJesus]

Wait WHAT? so the I that's given by the pulley is the inertia given to you?

 

[haruspex]

Wait WHAT? so the I that's given by the pulley is the inertia given to you?

Yes. You do not know its mass, or whether it is a uniform disk.
Also note the edit I just made at the end of my previous post.

 

[CoreanJesus]

Well... That makes this problem easier and solvable :P Thank you for the clarification!

 

[haruspex]

Well... That makes this problem easier and solvable :P Thank you for the clarification!

Ok.
The question statement should have made it clear, but the use of the symbol I was a clue, and dimensional analysis of the supposed answer makes it certain.