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The rim-mass wheel has moment of inertia $I=M{R}^2$. A uniform solid cylinder has axial moment $\frac{1}{2}m{S}^2$, so the roller of mass $M/2$ has $J=M{S}^2/4$. No slip gives equal contact speeds, $R\omega =S\sigma$, when $\omega ,\sigma$ denote angular-speed magnitudes. ThereforeWith parallel signed axes the two angular velocities have opposite signs. Their kinetic energies add regardless of that convention: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,Separating this linear equation givesFor 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 $\Gamma =\gamma (R/S{)}^3$,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. HenceFor a nonzero fan drag this is less than $\Omega$, and as $\gamma \to 0$ 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 ${\sigma }^2$.