= Eccentric wheel with missing-mass kinetic energy neglected
{title2=$\ddot s+\frac{2hgr_o}{3DR}\sin(s/R)-\frac{2(D-h)g}{3D}\sin\alpha=0$}
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 $T\simeq3D\dot s^2/4$. On a slope of <angle> $\alpha$, its <potential energy> is $U=-(D-h)gs\sin\alpha-hgr_o\cos(s/R)$ 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.
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