= Particle on a fixed plane curve
{title2=$(1+y'^2)\ddot x=-gy'-y'y''\dot x^2$}
A <point mass> constrained to the fixed graph $y=y(x)$ has <velocity> $(\dot x,y'\dot x)$. Under uniform downward <gravity>, its <Lagrangian> is $L=m(1+y'^2)\dot x^2/2-mgy$, so its <canonical momentum> is $p_x=m(1+y'^2)\dot x$, not the horizontal mechanical <momentum>. The <Euler-Lagrange equation> gives the equation in the title, and the conserved <energy> is $m(1+y'^2)\dot x^2/2+mgy$. A straight segment with constant slope $s$ has constant horizontal <acceleration> $-gs/(1+s^2)$; zero curvature does not imply zero <acceleration>. A cart held to a track satisfies this <mechanical constraint> only while the assumed contact or retaining mechanism remains effective.
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