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Integrating coaxial cylindrical shells givesRolling without slipping has spin speed $u/a$, so $K=m{u}^2/2+I{u}^2/(2a^2)=3m{u}^2/4$.

For the rolling-cylinder loop contact threshold, set $\ell =R-a$, the radius of the centre's circular path, and let $\theta$ run from the loop bottom. The centre starts $h$ above its lowest centre position. Its height gain from the bottom is $\ell (1-\cos \theta )$. Energy conservation givesAt the top, contact requires $N=m(u_t^2/\ell -g)\ge 0$, hence $u_t^2\ge g\ell$. Inserting the energy relation proves the limiting conditionThe same condition keeps $N\ge 0$ elsewhere on the circle, sinceis smallest at the top.

For the centre's acceleration, the inward normal component is $u^2/\ell$ and the tangential component is $-(2g/3)\sin \theta$. If $A=h/\ell -1\ge 7/4$ and $x=\cos \theta$,This convex quadratic has its largest endpoint value at $x=1$. The maximum on the circular loop is at the bottom, directed vertically upward. For the limiting release height,The approach track's curvature is unspecified, so a global bound for an arbitrary approach shape cannot be proved from the sketch alone; the result above concerns the specified circular part and the usual smooth approach interpretation.

The finite-friction qualification of a marginal rolling loop also matters: the contact threshold presumes the stated no-slip idealization. At equality, near the top, required tangential friction vanishes linearly with angular distance from the top but $N$ vanishes quadratically. Thus any finite coefficient of static friction requires a release height above this limiting bound to maintain rolling everywhere; no coefficient is provided. This does not change the requested ideal contact criterion.