= Solution
\b[Axial moment of inertia.] In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform <mass density> $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi\rho Lr\,dr$. Therefore the <axial moment of inertia of a hollow cylinder> is
$$
I=\int_{R_2}^{R_1}r^2dm
=2\pi\rho L\int_{R_2}^{R_1}r^3\,dr
=\frac{\pi\rho L}{2}(R_1^4-R_2^4).
$$
Since $M=\pi\rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference gives
$$
\boxed{I=\frac12M(R_1^2+R_2^2).}
$$
For a solid cylinder $R_2\to0$, giving \b[$I=MR_1^2/2$]. For a thin cylindrical shell with fixed <mass> and $R_1,R_2\to R$, \b[$I=MR^2$]. The latter limit agrees with every mass element being at radius $R$.
\b[Undamped rolling oscillations.] Let $x$ be axle <displacement> from the <spring> equilibrium. <Rolling without slipping> imposes $|\omega_{\rm roll}|=|\dot x|/R$. The sum of translational and <rotational kinetic energy> is
$$
T=\frac12m\dot x^2+\frac12I\frac{\dot x^2}{R^2}
=\frac12\left(m+\frac I{R^2}\right)\dot x^2.
$$
The <potential energy> is $U=kx^2/2$. Differentiating $T+U$ in the absence of <damping>, or applying the force and torque equations below, gives
$$
\left(m+\frac I{R^2}\right)\ddot x+kx=0.
$$
For the thin shell $I=mR^2$, the effective inertia is $2m$, and the <angular frequency> of the <simple harmonic motion> is
$$
\boxed{\omega_0=\sqrt{\frac{k}{2m}}.}
$$
Static contact <friction> supplies the required <torque> but does no work when the contact point is instantaneously stationary.
\b[Damped force balance.] Take rightward <force> as positive and counterclockwise rotation as positive. Write $f$ for the signed <static friction> at the bottom contact. Then
$$
m\ddot x=-kx-b\dot x+f,\qquad
I\alpha=fR,\qquad\alpha=-\frac{\ddot x}{R}.
$$
Eliminate $f=-I\ddot x/R^2$ and use $I=mR^2$:
$$
2m\ddot x+b\dot x+kx=0,
\qquad
\boxed{\ddot x+B\dot x+Cx=0,\qquad B=\frac b{2m},\quad C=\frac k{2m}.}
$$
Here $b$ has units $\mathrm{kg\,s^{-1}}$, so $B$ has units $\mathrm{s^{-1}}$ and $C$ units $\mathrm{s^{-2}}$. The stated <drag force> is interpreted as acting through the axle, with no additional viscous <torque>. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.
\b[Light damping.] Substitution of $z=Ae^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the <underdamped> regime $B^2<4C$,
$$
p=\frac B2\pm i\sqrt{C-\frac{B^2}{4}}
=\frac b{4m}\pm i\omega_d,
\qquad
\omega_d=\sqrt{\frac{k}{2m}-\frac{b^2}{16m^2}}.
$$
Taking real linear combinations gives the general <damped harmonic oscillator> displacement
$$
\boxed{x(t)=e^{-bt/(4m)}[A_1\cos(\omega_dt)+A_2\sin(\omega_dt)].}
$$
The constants are fixed by the initial <displacement> and <velocity>. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=bx_0/(4m\omega_d)$. Light damping means $b^2\ll8mk$, in which case $\omega_d\simeq\omega_0$ and the envelope decays slowly compared with an oscillation.
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