How Much Initial Kinetic Energy is Rotational in a Rolling Sphere?

[G-reg]

Homework Statement

A hollow sphere of radius 0.45 m, with rotational inertia I = 0.026 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 11° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 80 J.
(a) How much of this initial kinetic energy is rotational?

Homework Equations

KEi = KEf
KE = (1/2)mv^2 + (1/2)Iw^2
w = v/r

The Attempt at a Solution

I believe that with what I'm given I can solve for the mass of the object with rotational inertia and use that to find v? I also know that I'll have set one side of the equation equal to 80 and then solve for (1/2)Iw^2. I'm not 100% sure if finding v, then finding w and then finding (1/2)Iw^2 is the correct thought process..

I'd really appreciate any help at all for this one

 

[Troels]

Hello there :)

First of all, the proper symbol for Kinetic Energy is T - not KE :)

The first thing you should do is to find the angular velocity. Do this by substituting v for \omega r in the equation for the total kinetic energy. This will give you the angular velocity as a function of r, I, m and T.

Next substitute this expression into the equation for rotational kinetic energy; that is

T_\mathrm{rot}=\frac{1}{2}I\omega ^2

- that will give you the rotational kinetic energy as a fraction of the total kinetic energy. And yes, you will need to find the mass as well. You can find it from the Moment of inertia, like you had reasoned

[Answer: T_\mathrm{rot}=\frac{I}{I+mr^2}T]