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 Incoming links: Particle on a fixed plane curve

Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is

\overset{y}{˙} = y^{'} (x) \overset{x}{˙}

T = \frac{m}{2} [1 + y^{'} (x)^{2}] \overset{x}{˙}^{2}, V = m g y (x), L = \frac{m}{2} [1 + y^{'2}] \overset{x}{˙}^{2} - m g y .

Consequently the canonical momentum conjugate to the horizontal coordinate is

p_{x} = m [1 + y^{'2}] \overset{x}{˙} .

It is not simply the horizontal component

m \overset{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}}{d t} = m (1 + y^{'2}) \overset{x}{¨} + 2 m y^{'} y^{''} \overset{x}{˙}^{2}, \frac{\partial L}{\partial x} = m y^{'} y^{''} \overset{x}{˙}^{2} - m g y^{'} .

Subtracting yields

(1 + y^{'2}) \overset{x}{¨} = - g y^{'} - y^{'} y^{''} \overset{x}{˙}^{2} .

The term involving

y^{''}

accounts for the changing track geometry. An independent check is conservation of energy:

H = \frac{m}{2} (1 + y^{'2}) \overset{x}{˙}^{2} + m g y, \frac{d H}{d t} = m \overset{x}{˙} [(1 + y^{'2}) \overset{x}{¨} + y^{'} y^{''} \overset{x}{˙}^{2} + g y^{'}] = 0.