Eccentric wheel with missing-mass kinetic energy neglected ($\ddot{s}+\frac{2hg{r}_o}{3DR}\sin (s/R)-\frac{2(D-h)g}{3D}\sin \alpha =0$)

Model an eccentric-hole wheel as a complete uniform disk for kinetic energy only, subtracting the absent disc's potential energy. If $s=R\theta$ and $\theta$ is its hole-centre direction measured clockwise from upward vertical, rolling without slipping gives $T\simeq 3D{\dot{s}}^2/4$. On a slope of angle $\alpha$, its potential energy is $U=-(D-h)gs\sin \alpha -hg{r}_o\cos (s/R)$ up to a constant. Substitution into the Euler-Lagrange equation proves the displayed equation. Treating the hole's kinetic energy exactly introduces angle-dependent inertia, so this simplified equation must retain its explicit modeling assumption.