= Solution
Take $q=\ddot x$ positive when the hanging <mass> descends and the other <mass> moves uphill. For the sliding block, <Newton's second law> gives $mg-T=mq$ and $T-Mg\sin\theta=Mq$. Therefore
$$
\boxed{q=\frac{g(m-M\sin\theta)}{m+M},\qquad
T=\frac{mMg(1+\sin\theta)}{m+M}.}
$$
For the <pulley coupled to a rolling disc>, tension and weight act through its centre; the contact normal is perpendicular to the slope. <Static friction> provides the <torque> required for <rolling without slipping>. Define signed $F$ down the slope, so $F>0$ for uphill <acceleration>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-ia2014-disc-forces.png]
{title=Original force diagram for an uphill-accelerating disc: centre-applied tension and weight, contact normal and downslope static friction}
The rotational equation and rolling constraint give $Fa=I q/a$, hence $F=Mq/2$. The translational equations become $T-Mg\sin\theta-F=Mq$ and $mg-T=mq$, giving
$$
\boxed{q=\frac{2g(m-M\sin\theta)}{2m+3M},\qquad
T=\frac{mMg(3+2\sin\theta)}{2m+3M},\qquad
F=\frac{Mg(m-M\sin\theta)}{2m+3M}.}
$$
Thus \b[the threshold remains $m>M\sin\theta$]: the <moment of inertia> increases the effective <mass> but does not change the sign of the driving <force>. For downhill <acceleration> $F$ reverses. The <normal force> is $Mg\cos\theta$; <static friction> does no work at the stationary contact point and sufficient roughness is assumed.
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