Particle moving around the inside of a semicone - how to model its position up incline? An inverted hollow cone (cut in half) is set up with its vertex at the origin $O$ and an angle $\alpha$ between the horizontal ($x$ and $y$-axes) and the cone (so $\alpha$ close to $0$ would be a flat circle while close to $\pi/2$ would be a cylinder).
A particle is then launched and moves around the inside of the semicone (slipping), we could assume friction is negligible. Of course, there is a specific initial speed $v$, angle $\alpha$, and starting location of particle up along the incline of the cone $s$ which would produce circular motion if specific conditions are met. However, at differing initial conditions the motion will oscillate around some equilibrium (never reaching such). I have looked into the Lagrangian equations one could form for the particles based on a cylindrical coordinate system, and while this is all good I still cannot understand how to clearly model some aspects of the particle's motion
I'd like to be able to quantitatively model the displacement of the particle up the incline $s$ as a function of its angular displacement around the semicone $\phi$ ($0\leq\phi\leq\pi$), so that I could say, for example, at $\phi=\pi/4$, the ball will be $s$ cm up the incline (depending on $\alpha, v,$ and $s_{\textrm{initial}}$).
I know that this would have to do with formulating the Lagrangian of the particle, but then I have gotten stuck on having a bunch of differential equations with derivatives whose values are unknown to me.
How can this be represented in a calculable way, preferably not involving simulations? (I'm very new to DEs and Lagrangians, sorry...)
 A: In polar coordinates, the constraint fixing the particle on the cylinder reads :
$$z = r\tan(\alpha)$$
Therefore, the kynetic energy is (setting $m = 1$ for simplicity) :
$$T = \frac{1}{2}(\dot r^2 + \dot z^2+r^2\dot \theta^2)=\frac{1}{2}((1+\tan(\alpha)^2)\dot r^2 + r^2\dot\theta^2)$$
The potential energy is :
$$V = gz = gr\tan(\alpha)$$
Therefore, the Lagrangian is :
$$L = \frac{1}{2}((1+\tan(\alpha)^2)\dot r^2 + r^2\dot\theta^2)-mgr\tan(\alpha)$$
Using this, we can derive the Euler-Lagrange equations :
\begin{align}
(1+\tan^2(\alpha))\ddot r &= g\tan(\alpha)-r\dot \theta^2 \\
\frac{d}{dt}(r^2\dot\theta)&=0
\end{align}
The second equation is the conservation of angular momentum. It implies that there is a cosntant $C$ such that $r^2 \dot \theta  = C$. Then, plugging this into the other equation, we get a differential equation for $r$ :
$$(1+\tan^2(\alpha))\ddot r -g\tan(\alpha) + \frac{C}{r^3} = 0 $$
This is equivalent to a particle with mass $1$ and moving in an effective potential :
$$V_{\text{eff}}(r) = \frac{1}{1+\tan^2(\alpha)}\left(g\tan(\alpha)r + \frac{C}{2r^2}\right)$$
I don't think that there is an analytical solution to this equation. However, we can derive some qualitative properties of this potential. For example, for a given value of $C$, the radial coordinate will oscillate around the equilibrium $r_* = \sqrt[R]{\frac{C}{g\tan(\alpha)}}$, which is the radius of the corresponding circular orbit.
