= Solution
For a freely rolling rigid body, decompose the <kinetic energy> into translation of its <centre of mass> G and rotation about G. The <centre of mass after removing material> lies on the side opposite H:
$$
\overrightarrow{CG}=-\frac hW\overrightarrow{CH},\qquad HG=r_o\left(1+\frac hW\right)=\frac{Dr_o}{W}.
$$
Another use of the <parallel axis theorem> gives
$$
I_G=I_H-W(HG)^2=I_H-\frac{D^2r_o^2}{W}.
$$
Thus at the same <angular speed>,
$$
\boxed{T_{\rm rot,G}=\frac12I_G\omega^2<T_H,\qquad T_H-T_{\rm rot,G}=\frac{D^2r_o^2\omega^2}{2W}>0.}
$$
This compares the rotational component, not the total rolling <kinetic energy>. The total also contains $\tfrac12Wv_G^2$ and need not be smaller than $T_H$. Describing all rolling motion as instantaneous rotation about its contact point includes that translational contribution; it is not the rotational component intended by this comparison.
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