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!


1 Answer 1


enter image description here

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*}

enter image description here


  • $\varphi~$ 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

enter image description here

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

  • $\begingroup$ What is capital theta? Also, I am fairly certain something went wrong with the animation, perhaps a sign error somewhere. With COM movement like that, the outer rim would be turning in the other direction. I've been trying to twist my mind to see if it's a spinning ballerina illusion, but I think there's an actual mistake here. $\endgroup$
    – Erikinho
    Commented Nov 22, 2022 at 15:16
  • $\begingroup$ To clarify, the rotation of the wheel and the orbit around the origin either both have the axis into the plane or out of the plane (though the former of course does not have to be normal to the plane). Perhaps this means that another constraint is needed, that the rim is circular with finite width, but somehow I doubt it. $\endgroup$
    – Erikinho
    Commented Nov 22, 2022 at 15:20
  • $\begingroup$ $~\Theta~$ is the Inertia tensor of the ring $\endgroup$
    – Eli
    Commented Nov 22, 2022 at 15:33

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