What is the ratio?

Question

between the rotational kinetic energy about the center of a sphere and the sphere's total kinetic energy in this situation:

The sphere is solid and rolls along a horizontal, smooth surface at a constant linear speed without slipping.

Is it 5/3?

Answer 1

KE(rot) = .5*Iw^2

I = 2/5 mr^2
w = v/r

so, KE(rot) = .5 * 2/5 *mr^2
KE(tot) = KE(rot) + KE(normal)
= .5 * 2/5 *mr^2 + .5 mr^2
= 0.5 * (1+2/5) *mr^2

so the ratio is (1 + 2/5) / (2/5) = 1 + 5/2 = 7/2