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

