Eccentric circular-hole disk inertia ($I_H=\pi \rho (R^4+2R^2{r}_o^2-r_o^4)/2$)

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 $I_H=D{R}^2/2+D{r}_o^2-h{r}_o^2/2$. Removing the material shifts the remaining centre of mass G away from H by $CG=h{r}_o/W$, so $HG=D{r}_o/W$ and $I_G=I_H-D^2{r}_o^2/W$. 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.