Say I have a detached bicycle wheel and I roll it down an infinite, flat, asphalt inclined plane, in air (ie there will be a terminal velocity). When, if ever, will the wheel fall over?

I've been wanting to try this and see, but I don't know of a suitable hill near my house. My second question is if someone can recommend such a hill in the Trenton, NJ, USA area.

  • 2
    $\begingroup$ Make sure the wheel is really heavy and there is something fragile strategically placed on the bottom :-) $\endgroup$ Sep 22, 2013 at 2:13

2 Answers 2


Cute question! This is not really an answer, but it's too long for a comment.

I have a metal ring that I keep my keys on, and I did some experiments rolling it down a book at various angles. I also tried a quarter.

Empirically, the result seems to depend on the angle $\theta$ at which the book is inclined, and there seem to be three regimes. For $\theta<\theta_{c1}$, the wheel falls over. For $\theta_{c1}<\theta<\theta_{c2}$, the wheel rolls without slipping. For $\theta>\theta_{c2}$, it seems to slip immediately and fall over, although it's a little hard to tell because the motion is so fast.

I would conjecture that the critical angles $\theta_{c1}$ and $\theta_{c2}$ are determined by expressions something like $\tan^{-1}(k\mu_s)$, where $k$ is a unitless constant that is different for $\theta_{c1}$ and $\theta_{c2}$ and that also depends on whether the wheel is a ring, solid disk, or whatever.

Looking forward to seeing a more definitive answer.

  • $\begingroup$ Using a smaller wheel is a really good idea. It reduces the need for a long hill. I'm particularly interested in the middle regime, whether it is actually stable. I have a 2.1 m table that I inclined at 3°; with a little starting push the quarter will reach the end without falling. Unfortunately I don't think it reaches terminal velocity and it is going too fast for me to tell if it is "going to fall" eventually. $\endgroup$
    – Owen
    Sep 22, 2013 at 2:11
  • $\begingroup$ @Owen: I think the general condition for stability is something of the form $v>v_s=\sqrt{cgr}$, where, e.g., $c=1/4$ for a ring or hoop (see physicsforums.com/showpost.php?p=2303868&postcount=6 ). So once the wheel reaches this speed, it should be stable forever. I would think that there's some characteristic time it takes for a stationary wheel to tip over (which would be something like $\sqrt{r/g}$), so the question would be whether it reaches $v_s$ in less than this time. $\endgroup$
    – user4552
    Sep 22, 2013 at 3:52
  • $\begingroup$ I'm wondering if the explanation on that page is the full story. When I roll the quarter down the table, it often starts veering off to either the left or right, and doesn't show signs of getting back on course -- if it veers off too far it might "wipe out" and fall down. My table isn't long enough to tell. $\endgroup$
    – Owen
    Sep 22, 2013 at 23:26

First off, great question.

Ok I have crunched the numbers. I made a 5 degree of freedom model of the rolling disk/wheel which includes, 2 variables $x$, $y$ for the location on the incline plane (with positive $x$ downwards and $y$ across the plane), $\psi$ for the orientation of the wheel ($\psi=0$ when wheel is facing downwards), $\phi$ for the tilt of the wheel from the incline, and finally $\theta$ the spin of the wheel (with $\dot{\theta}=\omega$).

Included is the incline plane angle $\gamma$ and gravity $g$, and the rolling radius of the wheel $r$. The no-slip condition is described as $\omega r = \dot{x} \cos \psi + \dot{y} \sin \psi $ (the velocity along the $\psi$ direction).

The equations simplify when I use $\dot{x} = \omega r \cos\psi + v_{slip} \sin \psi$ and $\dot{y} = \omega r \sin \psi - v_{slip} \cos \psi $, to get the motion of the orientation and tilt as

$$ \begin{aligned} \ddot{\psi} & = \frac{2 \dot{\phi} ( \dot{\psi} \sin \phi - \omega)}{\cos \phi} + \frac{\sin\phi}{\cos^2 \phi} \left( \frac{2 \dot{\psi} v_{slip}}{r} - \frac{2 g \cos \psi \sin \gamma}{r} \right) \\ \ddot{\phi} & = \frac{ \cos\phi \left( \left( \frac{1}{2} I_{disk} \dot{\psi}^2 - m r^2 \dot{\phi}^2 \right) \sin\phi + I_{disk} \omega \dot{\phi} \right)}{\frac{1}{2} I_{disk} + m r^2 \sin^2 \phi} + \frac{g\, m\, r \sin\phi \cos \gamma}{\frac{1}{2} I_{disk} + m r^2 \sin^2 \phi} \end{aligned} $$

where $I_{disk}$ is the mass moment of inertia about the spin axis, and $\frac{1}{2} I_{disk}$ the mass moment of inertia about any other axis of the wheel.

So is the above stable? There are two scenarios that I can come up with.

  1. Stable when $\dot{\psi}=0$ and $\dot{\phi}=0$. This can only be true if $\psi=0$ or the orientation is always "downhill", but with a small disturbance it will throw it off. So not stable on this condition.
  2. Stable when $\dot{\psi}$ and $\dot{\phi}$ oscillate about zero with appropriate corrective twists $\ddot{\psi}$ and $\ddot{\phi}$ developing. I can only examine this case when $\phi \approx 0$ and $\phi \approx 0$. This has the solution of $$ \dot{\psi} =B \sin(2 \omega t) -A \cos(2 \omega t) \\ \dot{\phi} = B \cos(2 \omega t)+A \sin(2 \omega t) $$

Any coefficients of $A$, $B$ work when $\phi = 0$ and $\phi = 0$, but beyond that no constants would work. Nevertheless it shows that under some conditions the orientation and tilt couple in an stable oscillation, also known as a wobble to anyone who was rolled a coin on a table.

I think this is a far as anyone can take it without employing numerical methods, and I think this would make for a great thesis for someone.

PS. I used a MATLAB script with the symbolic toolbox to find this solution, with some custom code I have developed for 3d kinematics and the equations of motion for rigid bodies.


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