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Not sure if this an appropriate venue for this question, please close question as opposed to migrating to Physics SE, because it is not appropriate there, thanks.

If we have a functional given by: $$I=\int_1^2\bigg(L+\lambda f\bigg)\;dt,\tag{1}$$ the Euler-Lagrange equations are given by: $${d\over dt}\bigg({\partial L\over\partial \dot x}\bigg)-{\partial L\over \partial x}=-\lambda{\partial f\over\partial x}.\tag{2}$$ The Lagrangian is $$L={m\over 2}(\dot x^2+\dot y^2)-mgy,\tag{3}$$ $g\gt 0$ and the constraint $$f=y+\cosh x-2=0.\tag{4}$$ We may write the the Lagrangian in terms of the constraint as: $$L={m\over 2}(\dot x^2+\sinh^2 x)-mg(2-\cosh x).\tag{5}$$

This problem concerns finding the point at which a point mass "falls off" a surface defined via the constraint. It is desired to find the value of $x$ such that $\lambda=0$. However my attempt at solving the problem ends at computing the inverse hyperbolic cosine, which is of course undefined. So it seems there is an error? Perhaps someone can show the correct method for finding the solution.

Below is a summary of my work on the problem.

So we have an Euler-Lagrange equation in the variable $x$. Calculating the relevant derivatives gives: $${d\over dt}\bigg({\partial L\over\partial \dot x}\bigg)=m\ddot x;\quad {\partial L\over\partial x}=m\sinh x\cosh x+mg\sinh x.$$ And: $$\lambda{\partial f\over\partial x}=\lambda\sinh x.$$ So the equation of motion is: $$m\ddot x-m\sinh x\cosh x-mg\sinh x=-\lambda\sinh x,$$ from which we can obtain the multiplier as: $$-\lambda={m(\ddot x-\sinh x\cosh x-g\sinh x)\over \sinh x}.$$ From a first integral (conservation of energy) we have: $${m\over 2}\dot x^2-{m\over 2}\sinh^2 x+mg(2-\cosh x)-h=0,$$ where $h$, is a constant. If we differentiate the first integral of the motion we get: $$\ddot x=-\sinh x\cosh x-g\sinh x,$$ which we can substitute into the earlier obtained expression for the Lagrange multiplier to get: $$-\lambda={-2m(\sinh x\cosh x+g\sinh x)\over \sinh x}.$$ Setting $\lambda=0$ we get: $$0=\cosh x+g\implies \cosh x=-g\implies x=\cosh^{-1}(-g).$$ This doesn't seem correct, what to do?

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  • $\begingroup$ One way to troubleshoot this would be to try this method on a problem which you can solve by other means. In particular, the constraint curve $y-\sqrt{1-x^2}=0$ corresponds to the classic problem of a bead sliding down a sphere until it flies off. $\endgroup$ Mar 27 at 17:35
  • $\begingroup$ @Semiclassical Yes, I have tried that, and I do indeed find the correct answer by the method for the case of the sphere. $\endgroup$ Mar 27 at 17:42
  • $\begingroup$ One possible typo: the total Lagrangian was specified as $L+\lambda f$, but the given Euler-Lagrange equations can be rearranged to $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial}{\partial x}(L-\lambda f).$$ That suggests a sign discrepancy. (Whether this actually matters is not evident to me, since either way $\lambda=0$ should not be affected.) $\endgroup$ Mar 27 at 17:50
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    $\begingroup$ Comment to the post (v1): There seems to be a typo in eq. (5). $\endgroup$
    – Qmechanic
    Mar 30 at 9:04

1 Answer 1

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An alternative way. Considering the Lagrangian

$$ L={m\over 2}(\dot x^2+\dot y^2)-mgy+\lambda f(x,y),\ \ \ \ f(x,y) = y+\cosh x-2=0 $$

The movement equations are

$$ \cases{ m x''+\lambda \sinh x = 0\\ m y'' + \lambda + gm = 0 } $$

now deriving $f$ two times regarding $t$ we obtain

$$ \sinh x x''+y''+\cosh x x'^2=0 $$

Solving

$$ \cases{ m x''+\lambda \sinh x = 0\\ m y'' + \lambda + gm = 0\\ \sinh x x'' + y''+ \cosh x x'^2=0 } $$

for $(x'', y'', \lambda)$ we obtain

$$ \left\{ \begin{array}{l} x''=\frac{g \sinh (x)+x'^2 \sinh (x) \cosh (x)}{\sinh ^2(x)+1} \\ y''=-\frac{g \sinh ^2(x)+x'^2 \cosh (x)}{\sinh ^2(x)+1} \\ \lambda =-\frac{g m-m x'^2 \cosh (x)}{\sinh ^2(x)+1} \\ \end{array} \right. $$

Attached a MATHEMATICA script performing a simulation.

parms = {m -> 1, g -> 9.81, y0 -> 2 - Cosh[1/4], x0 -> 1/4};
p = {x[t], y[t]};
L = m/2 D[p, t] . D[p, t] - m g p . {0, 1} - lambda (y[t] + Cosh[x[t]] - 2);
mov = D[Grad[L, D[p, t]], t] - Grad[L, p];
solxy = Solve[Thread[Join[mov, {D[y[t] + Cosh[x[t]] - 2, t, t]}] == 0], Join[D[p, t, t], {lambda}]][[1]];
odes0 = Thread[D[p, t, t] == (D[p, t, t] /. solxy /. parms)];
cinits0 = {x[0] == x0, y[0] == y0, x'[0] == y'[0] == 0} /. parms;
ODE = Join[odes0, cinits0];

tmax = 2;
solode = NDSolve[ODE, {x, y}, {t, 0, tmax}][[1]];
ContourPlot[y + Cosh[x] - 2 == 0, {x, 0, 3}, {y, 3, -10}, AspectRatio -> 4, ContourStyle -> Blue]
Plot[Evaluate[-((g m - m Cosh[x[t]] Derivative[1][x][t]^2)/(1 + Sinh[x[t]]^2)) /. solode /. parms], {t, 0, tmax}, PlotStyle -> Blue]

This script was run for initial conditions $(x_0 = \frac 14, y_0 = 2-\cosh(\frac 14))$. The first plot shows the constraint profile and the second de $\lambda $ evolution along this profile. As can be depicted there is a point for $t \approx 0.906 [s]$ or $x\approx 1.35723$ in which occurs the detachment.

enter image description here

enter image description here

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  • $\begingroup$ Hi, so this implies that the point mass falls off at: $$x=\cosh^{-1}(-g/\dot x^2)?$$ Since $\cosh(x)\gt 0$ for all x, does this mean that the particle does not fall off for any value of $x$? $\endgroup$ Mar 30 at 16:42
  • $\begingroup$ Please. See an edit including a MATHEMATICA script. $\endgroup$
    – Cesareo
    Mar 30 at 18:06

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