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Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.

For a Lagrangian $\mathcal{L}=T-U$, the Euler-Lagrange equation is $d(\partial \mathcal{L}/\partial \dot{q})/dt=\partial \mathcal{L}/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal{L}=m{\dot{x}}^2/2-k{x}^2/2$ gives $m\ddot{x}+kx=0$ for a spring oscillator.

For a uniform solid sphere, slicing into disks normal to the rotation axis givesUse total dumb-bell mass $M$ and $\ell =L-r$. Since both centres lie on its spin axis, its moment of inertia about that axis is $2M{R}^2/5$. If $\theta$ measures displacement from the bottom of the track, rolling without slipping imposes spin speed $|\dot{\varphi }|=\ell |\dot{\theta }|/r$. The rolling inside a fixed circular cylinder therefore hasThe equation of motion and its small-amplitude limit areWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the given lengths, $\boxed{T=2\pi /\omega \simeq 80.5\mathrm{s}}$. If the centre rises by height $h$, conservation of energy at the bottom givesThe speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.