KE of sphere???? help:)

A spherical object (with non-uniform density) of mass 47 kg and radius 0.29 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is 0.34M R2.
The object’s rotational kinetic energy about its own center is what fraction of the object’s total kinetic energy? I know the object has translational and rotational KE and I need to divide it by total KE, but I don't know how!:(

Translational KE = ½ * mass * linear velocity^2
Rotational KE = ½ * I * ω^2
I = 0.34 * m * r^2
Rotational KE = ½ * (0.34 * m * r^2) * ω^2
Rotational KE = 0.17 * m * r^2 * ω^2



We need to convert linear velocity to angular velocity, so we can add the translational KE and rotational KE together and determine the total kinetic energy
Linear velocity = angular velocity * radius = ω * r
Translational KE = ½ * m * (ω * r)^2
Translational KE = ½ * m * ω^2 * r^2

Total KE = (½ * m * ω^2 * r^2) + (0.17 * m * r^2 * ω^2)
Total KE = m * r^2 * ω^2 * (½ + 0.17) = m * r^2 * ω^2 * 0.67

Fraction = Rotational KE / Total KE
Fraction = (0.17 * m * r^2 * ω^2) ÷ (m * r^2 * ω^2 * 0.67)
m * r^2 * ω^2 cancels!

Fraction = 0.17 / 0.67

I am sorry that I made a mistake on my previous attempt at this problem!

Let me show you the normal way to show exponents here.
0.34M R^2 where the ^ is shift+6

The ball travels a distance 2*pi*R in time t, so the velocity v = 2*pi*R/t.
KEt = (1/2)*M*v^2 = 2*M*(pi*R/t)^2
The ball rotates through 2*pi radians in time t so the angular velocity w = 2*pi/t
KEr = (1/2)*I*w^2 = 0.17*M*R^2*(2*pi/t)^2 = 0.68*M*R^2*(pi/t)^2

rotational kinetic energy / total kinetic energy = r/tot
r/tot = (0.68*M*R^2*(pi/t)^2) / (0.68*M*R^2*(pi/t)^2 + 2*M*(pi*R/t)^2)
r/tot = 0.68 / 2.68 I think, check my algebra.