Problem solving Euler-Lagrange equations of a particle constrained to a spherical spiral

Question

Problem

I want to calculate the time it takes for a particle living in a spherical spiral to fall under de force of gravity down to the bottom. So far I've sketched the procedure but when I tried to solve the equations, they've seemed too complicated to solve analytically so I'm stucked. Let me introduce my attempt:

First I've calculated the expression of the curve in parametric form $\alpha(u)=(x(u),y(u),z(u))$: $$x(u)=\frac{R \cos(u)}{\sqrt{1+\kappa^2u^2}}\\ y(u)=\frac{R \sin(u)}{\sqrt{1+\kappa^2u^2}}\\ z(u)=\frac{-R\kappa u }{\sqrt{1+\kappa^2u^2}}$$ where $R$ is the radius of the sphere and $\kappa$ is some constant which encodes how much the spiral curls. I've used $u$ as the parameter of the curve. Now I want to solve the Euler-Lagrange equations, and I wanted to re parameterize the curve with respect to the length parameter($s(u)=\int_{u_0}^u |\alpha'(t)|dt$) to make use of the fact that our problem reduces to 1-D. The Lagrangian would take the form $\mathcal{L}=\frac{m}{2}\dot{s}^2-mgz(u)=\frac{m}{2}\dot{s}^2-mgz(s)$, where in the last step we simply reverse $s(u)\rightarrow u(s)$. Therefore the Euler-Lagrange equations would be: $$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{s}}-\frac{\partial \mathcal{L}}{\partial s}=0=\ddot{s}-g\frac{\partial z(s)}{\partial s}\rightarrow s(t)$$ And finally substituting $s(t)$ in $\alpha(s)$, I would have the equation of motion.

Where I'm stucked

Since the equations are so complicated I tried using python to solve them, but I can't even solve for $s(u)$, and even if I could, I don't think I could find the inverse $u(s)$.

What I'm asking

It´s been a long time since I used Python or Euler-Lagrange equations, so maybe I did something wrong in the procedure or even in the coding.

Is my attempt correct?

If it is, is there a way I can solve the problem numerically?

Should I tackle the problem in any other way?

Thanks

Answer 1

Should I tackle the problem in any other way?

yes, take u the new generalized coordinate, thus the kinetic energy is

$$T=\frac{m}{2}\left(x'(u)^2+y'(u)^2+z'(u)^2\right)\,\dot{u}^2$$

where $'=\frac{\partial}{\partial u}$

$$T=\frac 1 2 \,{\frac {{R}^{2}{{\dot{u}}}^{2}m \left( {{\it \kappa}}^{2}{u}^{2}+1+{{ \it \kappa}}^{2} \right) }{ \left( 1+{{\it \kappa}}^{2}{u}^{2} \right) ^{2}} } $$

and the potential energy

$$U=m\,g\,z(u)$$

with EL you get this equation of motion:

$$\ddot{u}-{\frac {{{\dot{u}}}^{2}{\kappa}^{2}u \left( 1+{\kappa}^{2}{u}^{2}+2\,{ \kappa}^{2} \right) }{ \left( 1+{\kappa}^{2}{u}^{2} \right) \left( { \kappa}^{2}{u}^{2}+1+{\kappa}^{2} \right) }}+{\frac {\sqrt {1+{\kappa} ^{2}{u}^{2}}\kappa\,g}{R \left( {\kappa}^{2}{u}^{2}+1+{\kappa}^{2} \right) }}=0 \tag 1$$

for numerical simulation you have to transfer equation (1) to first order differential equation:

with: $y_1=\dot{u}\quad $ and $y_2=u$ you obtain:

$$\left[ \begin {array}{c} \dot{y}_1\\\dot{y}_2 \end {array} \right] =A $$

where :

$$A= \left[ \begin {array}{c} -{\frac {\kappa\, \left( \sqrt {1+{\kappa}^{ 2}{y_{{2}}}^{2}}+\sqrt {1+{\kappa}^{2}{y_{{2}}}^{2}}{\kappa}^{2}{y_{{2 }}}^{2} \right) g}{ \left( 1+{\kappa}^{2}{y_{{2}}}^{2} \right) \left( {\kappa}^{2}{y_{{2}}}^{2}+1+{\kappa}^{2} \right) R}}-{\frac { \kappa\, \left( -{y_{{1}}}^{2}{\kappa}^{3}{y_{{2}}}^{3}-{y_{{1}}}^{2} \kappa\,y_{{2}}-2\,{y_{{1}}}^{2}{\kappa}^{3}y_{{2}} \right) }{ \left( 1+{\kappa}^{2}{y_{{2}}}^{2} \right) \left( {\kappa}^{2}{y_{{2}}}^{2}+ 1+{\kappa}^{2} \right) }}\\ y_{{1}}\end {array} \right] $$

edit

simulation result

$\kappa=0.4\,,R=1\,,u(0)=20,D(u)(0)=0$

I stop the simulation if $u(t)=0\quad (z(u)=0)\quad $ and get $t=9.9$[s]