= Solution
For the complete uniform disc, the surface density is $\sigma=M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi\sigma r\,dr$, so its contribution to the axial <moment of inertia> is $r^2dm$. Hence
$$
\boxed{I_{\rm disc}=2\pi\sigma\int_0^ar^3\,dr=\frac12Ma^2.}
$$
This derives the <moment of inertia of a uniform disc> rather than assuming it.
Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\tfrac12(M/4)(a/2)^2=Ma^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ and
$$
\boxed{I_A=\frac12Ma^2-\frac1{32}Ma^2=\frac{15}{32}Ma^2=\frac58m_Aa^2.}
$$
The <moment of inertia of a uniform annulus> gives the same value directly.
Let $A_c$ be the downhill centre-of-mass <acceleration> and let $F$ denote the magnitude of the uphill <static friction>. Translation, rotation about the centre, and <rolling without slipping> give
$$
m_AA_c=m_Ag\sin\alpha-F,\qquad I_A\dot\omega=Fa,\qquad A_c=a\dot\omega.
$$
Thus $F=I_AA_c/a^2=(5/8)m_AA_c$. Eliminating $F$ gives the <rolling acceleration with rotational inertia>
$$
\boxed{A_c=\frac{g\sin\alpha}{1+5/8}=\frac8{13}g\sin\alpha.}
$$
Substitution gives the contact forces
$$
\boxed{F=\frac{15}{52}Mg\sin\alpha\quad\text{up the plane},\qquad N=\frac34Mg\cos\alpha.}
$$
The <normal force> balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the <torque> that increases its rolling angular speed. For a plane with $0\leq\alpha<\pi/2$, the assumed no-slip motion requires a coefficient of <static friction> satisfying $\mu_s\geq F/N=(5/13)\tan\alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational <kinetic energy>.
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