= Cylinder rolling on a freely translating wedge
{title2=$\ddot s=-\frac{2(M_b+M_c)g\sin\alpha}{3(M_b+M_c)-2M_c\cos^2\alpha}$}
For a uniform solid cylinder of mass $M_c$ and a frictionlessly sliding wedge of mass $M_b$, take the horizontal wedge position $X$ and uphill contact coordinate $s$. Contact and <rolling without slipping> leave two <degrees of freedom>. Up to a constant, the <Lagrangian> is $L=\tfrac12(M_b+M_c)\dot X^2+M_c\cos\alpha\dot X\dot s+\tfrac34M_c\dot s^2-M_cgs\sin\alpha$.
The <cyclic coordinate> $X$ gives $(M_b+M_c)\ddot X+M_c\cos\alpha\ddot s=0$. The other <Euler-Lagrange equation> is $\cos\alpha\ddot X+(3/2)\ddot s=-g\sin\alpha$. Elimination proves the displayed acceleration, with $\ddot X=-M_c\cos\alpha\ddot s/(M_b+M_c)$. For $M_b\gg M_c$, the cylinder approaches the fixed-incline acceleration $-(2/3)g\sin\alpha$.
Back to article page
