Kinetic Energy in Pure Rolling

Since pure rolling is a combination of translational and rotational motion happening simultaneously, its total kinetic energy is simply the sum of a translational part and a rotational part:

If the rolling object has mass $M$, center-of-mass velocity $v_{CM}$, moment of inertia about the center of mass $I_{CM}$, and angular velocity $\omega$, then

With center of mass as the reference. Writing $I_{CM}=MK^2$ ($K$ = radius of gyration) and using the pure-rolling condition $v_{CM}=R\omega$ (so $\omega=v_{CM}/R$):

With the point of contact as the reference. Exactly the same total kinetic energy can equivalently be obtained by treating the rolling object's motion as a momentary pure rotation about the instantaneous point of contact, $O$: $KE=\frac12I_O\omega^2$, where, by the parallel axis theorem, $I_O=I_{CM}+MR^2=MK^2+MR^2$. Substituting $\omega=v_{CM}/R$ once again:

the identical result — confirming that both descriptions of pure rolling (§5.6.1) genuinely give the same physics.

Splitting the total kinetic energy. From $KE=\frac12Mv_{CM}^2\left(1+\dfrac{K^2}{R^2}\right)$ against $KE_{TRANS}=\frac12Mv_{CM}^2$ and $KE_{ROT}=\frac12Mv_{CM}^2\left(\dfrac{K^2}{R^2}\right)$, the three quantities are always in the fixed ratio …