= Solution
{c}
Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta=0$ initially. For <rolling without slipping>, the contact point has zero <velocity>, so $\dot x-b\dot\theta=0$. Integrating the <rolling constraint> gives
$$
\boxed{x=b\theta,\qquad \ddot x=b\ddot\theta.}
$$
Let $F$ be the uphill <static friction>. Along the inclined <plane>, <Newton's second law> for the <centre of mass> gives
$$
M\ddot x=Mg\sin\varphi-F.
$$
The weight passes through the <centre of mass>, and the <normal reaction> has its line of action through it. Thus only <static friction> supplies <torque> about $G$. The <angular velocity> is along the cylinder's symmetry axis, so the <inertia tensor> gives <angular momentum> $C\dot\theta$ in that direction. The <torque> equation is therefore
$$
C\ddot\theta=bF,\qquad F=\frac{C}{b^2}\ddot x.
$$
Eliminating <static friction> and inserting $C$ gives
$$
\left(M+\frac{C}{b^2}\right)\ddot x=Mg\sin\varphi,
\qquad
\boxed{\left(1+\frac{a^2+b^2}{2b^2}\right)\ddot x=g\sin\varphi.}
$$
The extra term represents the rotational <inertia> that must be accelerated along with the translational <mass>.
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