= Eccentric circular-hole disk inertia
{title2=$I_H=\pi\rho(R^4+2R^2r_o^2-r_o^4)/2$}
A uniform <disk> of radius $R$ and area <mass density> $\rho$ has a circular hole of radius $r_o$, whose centre H is displaced by $r_o$ from the original centre C. Set $D=\pi\rho R^2$, $h=\pi\rho r_o^2$ and $W=D-h$. The <parallel axis theorem> gives $I_H=DR^2/2+Dr_o^2-hr_o^2/2$. Removing the material shifts the remaining <centre of mass> G away from H by $CG=hr_o/W$, so $HG=Dr_o/W$ and $I_G=I_H-D^2r_o^2/W$. Thus the rotational component of free rigid-body <kinetic energy> is smaller than fixed-axis <kinetic energy> about H at the same <angular speed>, although total rolling <kinetic energy> also includes translation.
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