= Solution
Take $s$ positive uphill, $X$ positive to the right, and the cylinder's rotation positive anticlockwise. Contact with the fixed-angle surface determines its normal position; no slipping gives $\dot\theta=-\dot s/R$. These two constraints leave $X,s$ as the two independent coordinates of the <cylinder rolling on a freely translating wedge>.
The cylinder centre is at $(X+s\cos\alpha-R\sin\alpha,\ s\sin\alpha+R\cos\alpha)$. Since the angle is constant, its squared <speed> is $\dot X^2+2\dot X\dot s\cos\alpha+\dot s^2$. Including the block's <kinetic energy> and $I_c=M_cR^2/2$, and dropping a constant from the potential, gives
$$
\boxed{L=\frac12(M_b+M_c)\dot X^2+M_c\cos\alpha\,\dot X\dot s+\frac34M_c\dot s^2-M_cgs\sin\alpha.}
$$
The <cyclic coordinate> is $X$. Its conserved <conjugate momentum> is $p_X=(M_b+M_c)\dot X+M_c\cos\alpha\dot s$, the total horizontal <momentum>. The two <Euler-Lagrange equations> are
$$
(M_b+M_c)\ddot X+M_c\cos\alpha\ddot s=0,\qquad
\cos\alpha\ddot X+\frac32\ddot s=-g\sin\alpha.
$$
Eliminating the accelerations, with $D=3(M_b+M_c)-2M_c\cos^2\alpha>0$, gives
$$
\boxed{\ddot s=-\frac{2(M_b+M_c)g\sin\alpha}{D},\qquad
\ddot X=\frac{2M_cg\sin\alpha\cos\alpha}{D}.}
$$
Thus the cylinder rolls downhill and the wedge accelerates rightward. For $M_b\gg M_c$, $\ddot X\to0$ and $\ddot s\to-(2/3)g\sin\alpha$, the ordinary fixed-incline result.
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