 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 $I_c=M_c{R}^2/2$, and dropping a constant from the potential, givesThe 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 areEliminating the accelerations, with $D=3(M_b+M_c)-2M_c{\cos }^2\alpha >0$, givesThus the cylinder rolls downhill and the wedge accelerates rightward. For $M_b\gg M_c$, $\ddot{X}\to 0$ and $\ddot{s}\to -(2/3)g\sin \alpha$, the ordinary fixed-incline result.