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We are going downhill on a path $$y=a(1-x^2),$$ and I need to calculate the constraint force-position function. What I've done is this:

The lagrangian of the system is $$L=\frac{1}{2}(\dot{x}^2+\dot{y}^2)-mgy+\lambda(t)(y-a(1-x^2)),$$ so the 2 equation are $$m\ddot{x}=2a\lambda x$$ and $$m\ddot{y}=-mg+\lambda.$$ Based on the equations, the constraint force is $F_y=\lambda$ and $F_x=2a\lambda x$, but I can't calculate the value of $\lambda.$ How should I do it?

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closed as off-topic by sammy gerbil, Jon Custer, John Rennie, Kyle Kanos, Aaron Stevens Nov 16 '18 at 13:35

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  • $\begingroup$ Why not just find the normal force? $\endgroup$ – PiKindOfGuy Nov 7 '18 at 21:00
  • $\begingroup$ Isn't the constraint force just the normal force? $\endgroup$ – PiKindOfGuy Nov 7 '18 at 21:02
  • $\begingroup$ @PiKindOfGuy I think it is. But how would you calculate it? $\endgroup$ – J. Doe Nov 7 '18 at 21:03
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    $\begingroup$ You have 2 equations for 3 unknowns. If you take the second time derivative of the constraint equation , you get one more equation to solve your problem $\endgroup$ – Eli Nov 7 '18 at 23:26
  • $\begingroup$ Hints: 1. First solve for $x$: Energy conservation yields $E/m = (1+ (2ax)^2)\dot{x}^2/2+ga(1-x^2).$ What are the initial conditions? 2. Next solve for $y$ and $\lambda$. $\endgroup$ – Qmechanic Nov 8 '18 at 8:44