= Solution
The rim-mass wheel has <moment of inertia> $I=MR^2$. A uniform solid cylinder has axial moment $\tfrac12mS^2$, so the roller of mass $M/2$ has $J=MS^2/4$. No slip gives equal contact speeds, $R\omega=S\sigma$, when $\omega,\sigma$ denote angular-speed magnitudes. Therefore
$$
\boxed{I=MR^2,\quad J=\frac{MS^2}{4},\quad \frac{\sigma}{\omega}=\frac RS.}
$$
With parallel signed axes the two <angular velocities> have opposite signs. Their <kinetic energies> add regardless of that convention:
$$
\frac12I\omega^2+\frac12J\sigma^2
=\frac12\left[I+J(R/S)^2\right]\omega^2
=\frac12K\omega^2,\qquad \boxed{K=\frac54MR^2.}
$$
The <reflected inertia of a no-slip wheel and roller> must be used because accelerating the wheel also accelerates the roller. The transmitted contact forces redistribute energy without dissipating it. For an applied wheel <torque> $T$, the input power is $T\omega$, so the <work-energy theorem> gives $K\omega\dot\omega=T\omega$, or $K\dot\omega=T$; this also holds at rest by continuity or the constrained equation of motion.
Assume the driving constants $Q,\Omega$ are positive. With the speed-dependent <torque>,
$$
K\dot\omega=Q(1-\omega/\Omega),\qquad\omega(0)=0.
$$
Separating this linear equation gives
$$
\boxed{\omega(t)=\Omega\left(1-e^{-Qt/(K\Omega)}\right).}
$$
For the fan, its resisting roller <torque> dissipates power $\gamma\sigma^3$. Expressing this in the wheel coordinate gives a resisting wheel torque $\gamma(R/S)^3\omega^2$. Thus, with $\Gamma=\gamma(R/S)^3$,
$$
K\dot\omega=Q(1-\omega/\Omega)-\Gamma\omega^2.
$$
The right-hand side is positive at zero, strictly decreasing for nonnegative $\omega$, and has a unique positive zero; the wheel starting at rest approaches this terminal angular speed. Hence
$$
\boxed{\omega_{{\max}}=\frac{2Q}{Q/\Omega+\sqrt{(Q/\Omega)^2+4\gamma Q(R/S)^3}},\qquad
v_{{\rm rim,max}}=R\omega_{{\max}}.}
$$
For a nonzero fan drag this is less than $\Omega$, and as $\gamma\to0$ it tends to $\Omega$. The cubed radius ratio is essential: one factor arises from converting roller power to wheel torque, in addition to the two factors in $\sigma^2$.
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