Dynamics and Lagrangian of a ring on a plane So I am exploring my Cyr wheel (not me performing), and I want to try to express its movement in the best possible way. I will not be attempting the impossible, so just the wheel/ring/hoop, and not the wheel+human system.
Assuming no slipping and that the wheel is always in contact with the floor. Radius of ring is much bigger than thickness. (1 m and 1.5 cm)
Defining the $xyz$ coordinate system with the $yz$ plane as the plane of the wheel, and the $x$-axis as the axis of symmetric rotation (rolling). We then also have a spinning motion about the z-axis, and a tilt/fall along the y-axis. Now, I know that this is where my mistake lies, because as the wheel tilts either the coordinate system will not follow the wheel (easiest to visualize) and so x is no longer the symmetric axis, or the coordinate system follows the wheel (maybe easiest to calculate?) and rotation about y loses meaning.
We have
$$
T = \frac{1}{2}mv^2+\frac{1}{2}I_x\omega_x^2+\frac{1}{2}I_y\omega_y^2+\frac{1}{2}I_z\omega_z^2,
$$
$$
V = mgh = mg\sin\phi,
$$
with $\phi$ as the angle between the plane of the wheel and the floor. Since we have no slipping (or rather, no "travelling" slipping, as the wheel can still spin), $v=\omega_xr$.
$I_x=mr^2$, $I_y=I_z=\frac{1}{2}mr^2$, and defining $\omega_x = \dot{\theta}$, $\omega_y = \omega$, $\omega_x = \dot{\phi}$, gives
$$
L = \frac{1}{2}mr^2\left(2\dot{\theta}^2+\frac{1}{2}\omega^2+\frac{1}{2}\dot{\phi}^2\right)-mgr\sin\phi,
$$
yielding
$$
\ddot{\phi} = 2g\cos\phi,\, \ddot{\theta}=0,\, \dot{\omega}=0.
$$
The answers make sense in a naive way, but they provide no information about for example rolling at an angle in a circle or spinning or a combination of the two. Energy lost to friction shouldn't matter in order to produce some illuminating results, right?
I've already seen this question about a coin and this paper about a tilted rolling disk, but I'm afraid it's been too long since uni for me to be able to make the transfer. I'm also guessing there's quite a few similarities to an Euler disc. Any and all help appreciated!
 A: 
this animation is done with those equations of motion
\begin{align*}
&\begin{bmatrix}
   \ddot{\varphi} \\\\
   \ddot{\psi} \\\\
   \ddot{\vartheta} \\\\
   \ddot{x} \\\\
   \ddot{y} \\\\
   \ddot{z} \\\\
 \end{bmatrix}=
 \left[ \begin {array}{c} -{\frac {mg\sin \left( \varphi  \right) \cos
 \left( \vartheta  \right) \rho}{{\rho}^{2}m+{  I_x}}}
\\\\  {\frac {\sin \left( \varphi  \right) mg\sin
 \left( \vartheta  \right) \rho}{\cos \left( \varphi  \right)  \left(
{\rho}^{2}m+{  I_x} \right) }}\\\\  -{\frac {mg\sin
 \left( \vartheta  \right) \rho}{\cos \left( \varphi  \right)  \left(
{\rho}^{2}m+{  I_x} \right) }}\\\\  -{\frac {mg\cos
 \left( \varphi  \right) \sin \left( \vartheta  \right) {\rho}^{2}}{{
\rho}^{2}m+{  I_x}}}\\\\  {\frac {{\rho}^{2}mg\sin
 \left( \varphi  \right) }{{\rho}^{2}m+{  I_x}}}\\\\ 0
\end {array} \right]
\end{align*}

where

*

*$\varphi~$ rotation about the x axis

*$\vartheta~$ rotation about the y axis

*$\psi~$ rotation about the z axis

*$x~,y~,z~$ center of mass position

*$\rho~$ring radius

*$I_x~$ inertia about the x axis

the position of the ring center of mass is a circular motion


with the kinetic energy and potential  energy
$$
 T=\frac{m}{2}\vec{v}^T\,\vec{v}+\frac{1}{2}\vec{\omega}^T\,\Theta\,\vec{\omega}\quad,
 U=-mg\cos \left( \varphi  \right) \cos \left( \vartheta  \right)\\
\vec\omega=   \left[ \begin {array}{c} \cos \left( \vartheta  \right) \dot\varphi -
\cos \left( \varphi  \right) \sin \left( \vartheta  \right) \dot\psi
\\  \sin \left( \varphi  \right) \dot\psi +\dot\vartheta
\\  \sin \left( \vartheta  \right) \dot\varphi +\cos
 \left( \varphi  \right) \cos \left( \vartheta  \right) \dot\psi
\end {array} \right]$$
the non-holonomic rolling conditions
$$e_{Nh}=  \left[ \begin {array}{c} {  \dot{x}}+ \left( \sin \left( \varphi
 \right) \dot\psi +\dot\vartheta  \right) \rho\\  {  \dot{y}}-
 \left( \cos \left( \vartheta  \right) \dot\varphi -\cos \left( \varphi
 \right) \sin \left( \vartheta  \right) \dot\psi  \right) \rho
\end {array} \right]=\vec 0
$$
and additional to keep the z coordinate of the center of mass constant ,the  holonomic constraint equation $~z-\rho=0~$
you can obtain the EOM's
