= Solution
Let $\rho=3m/(4\pi R_0^3)$ be the uniform <mass density>. In <spherical coordinates> $(r,\vartheta,\phi)$ 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 therefore
$$
I=\rho\int_0^{R_0}r^4\,dr\int_0^\pi\sin^3\vartheta\,d\vartheta\int_0^{2\pi}d\phi
=\frac{8\pi\rho R_0^5}{15}=\boxed{\frac25mR_0^2}.
$$
This 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> is
$$
\mathbf v_{\rm contact}=a\dot\theta\mathbf e_\theta+\Omega_s\mathbf e_z\times R_0\mathbf e_r
=(a\dot\theta+R_0\Omega_s)\mathbf e_\theta.
$$
The 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 is
$$
\boxed{\dot\psi=\frac{R_1-R_0}{R_0}\dot\theta.}
$$
The 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>:
$$
mg a(\cos\theta-\cos\alpha)=\frac12ma^2\dot\theta^2+\frac12I\left(\frac a{R_0}\dot\theta\right)^2
=\frac7{10}ma^2\dot\theta^2.
$$
On the descent from $0<\alpha<\pi/2$, $\theta$ decreases, so
$$
\dot\theta=-\sqrt{\frac{10g}{7a}(\cos\theta-\cos\alpha)}.
$$
Integrating $dt=-d\theta/|\dot\theta|$ from release to the bottom gives
$$
\boxed{T_R=\sqrt{\frac{7(R_1-R_0)}{10g}}\int_0^\alpha\frac{d\theta}{\sqrt{\cos\theta-\cos\alpha}}.}
$$
The 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>, so
$$
\frac12ma^2\dot\theta^2=mga(\cos\theta-\cos\alpha),\qquad
\boxed{T_S=\sqrt{\frac{R_1-R_0}{2g}}\int_0^\alpha\frac{d\theta}{\sqrt{\cos\theta-\cos\alpha}}=\sqrt{\frac57}\,T_R.}
$$
A 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>. \b[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>.
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