= Solution
Model the cart as a <particle on a fixed plane curve>, remaining constrained to the track. Its vertical <velocity> is $\dot y=y'(x)\dot x$, so its <kinetic energy>, <gravitational potential energy> and <Lagrangian> are
$$
T=\frac m2[1+y'(x)^2]\dot x^2,\qquad
V=mgy(x),\qquad
L=\frac m2[1+y'^2]\dot x^2-mgy.
$$
Consequently the <canonical momentum> conjugate to the horizontal coordinate is
$$
\boxed{p_x=m[1+y'^2]\dot x.}
$$
It is not simply the horizontal component $m\dot x$ of the mechanical <momentum>, because a change in $x$ also moves the cart vertically. Differentiate carefully in the <Euler-Lagrange equation>:
$$
\frac{d p_x}{dt}=m(1+y'^2)\ddot x+2my'y''\dot x^2,
\qquad
\frac{\partial L}{\partial x}=my'y''\dot x^2-mgy'.
$$
Subtracting yields
$$
\boxed{(1+y'^2)\ddot x=-gy'-y'y''\dot x^2.}
$$
The term involving $y''$ accounts for the changing track geometry. An independent check is <conservation of energy>:
$$
H=\frac m2(1+y'^2)\dot x^2+mgy,
\qquad
\frac{dH}{dt}=m\dot x\big[(1+y'^2)\ddot x+y'y''\dot x^2+gy'\big]=0.
$$
Back to article page
