Rotational Kinetic Energy of a sphere

[kelslee28]

Homework Statement

A small sphere, with radius 1.6 cm and mass 5.2 kg, rolls without slipping on the inside of a large fixed hemispherical bowl with radius 0.82 m and a vertical axis of symmetry. It starts at the top from rest. What is the kinetic energy of the sphere at the bottom?
What fraction of its kinetic energy at the bottom (in percent) is associated with rotation about an axis through its center of mass?

Homework Equations

KE = 1/2 mv²
KE_rot = 1/2Iw²
V_f² = V_i² + 2a(y_f-y_i)

The Attempt at a Solution

I found the answer to the first question. My final equation was KE = mg(r₂ - 0.5r₁). r₁ being the ball and r₂ is the ramp.
My attempt at the second question was to find the KE and then subtract that from the total KE. When I came up with an equation for KE it was the same exact thing as the equation for my total KE, so I couldn't really subtract them to find the KE_rot.
I knew that KE = 1/2 mv² so I used V_f² = V_i² + 2a(yf-yi) to find V² and got that V² = 2g(r1-0.5r2). Plugging this into the KE equation gives me the same equation [KE= mg(r[SUB]2[/SUB] - 0.5r₁)]

PLEASE HELP! Thanks!

 

[srmeier]

you found the total kinetic energy at the bottom. Because the sphere is rolling what is the total kinetic energy equal to?

also note:

\omega=\frac{v}{r}

 

[kelslee28]

I can't use the constant acceleration equation because it's not in freefall. What I needed to do, which now I have figured out, is solve for v using the fact that mg(R-r) = 1/2mv^2 + 1/2Iw^2 where w = v/r

Then use that V to solve for the rotational kinetic energy. Divide this by the total kinetic energy and multiply by 100 to get percent.