What are the curves of contact on a convex body $B$ rolling down an inclined plane? Assume $B$ is smooth, and there is sufficient friction to prevent slippage.

Certainly, one can develop a geodesic to a straight line on a plane by rolling $B$ so that the geodesic is the point of contact, but it doesn't appear that this in general would be the point of contact in the physical situation of free rolling under gravity. It is for a sphere, which will roll along great circles. And an ellipsoid should roll along its three simple closed geodesics (although there are significant instabilities with all but the shortest closed geodesic—I'd prefer to ignore stability issues). But for other shapes, I imagine that an off-center center of gravity (and perhaps rotational momentum?) will cause the rolling to deviate from a geodesic. (But I am uncertain of this. Please correct me if I'm wrong!)

Assuming this is correct (that rolling curves are not always geodesics), what are the conditions that determine if a curve $\gamma$ is a rolling curve: $\gamma$ is the trace of the point of contact between $B$ and an inclined plane as it rolls, from some initial position? Perhaps: If $p \in \gamma$ is a point of contact, then (a) the normal vector $N$ of $\gamma$ at $p$ must be perpendicular to the plane, and (b) the center of gravity of $B$ must lie in the osculating plane of $\gamma$ at $p$ (the plane containing $N$ and the tangent $T$ vector at $p$)?

Have what I christened rolling curves been studied in the literature? If so, under what name? My searches have been unsuccessful. Can you think of shapes outside of {sphere, ellipsoid, cylinder} where the rolling curves can be determined? Thanks for any thoughts or pointers!

  • $\begingroup$ springerlink.com/content/w2273555w336u976 $\endgroup$ Jul 16, 2010 at 12:58
  • $\begingroup$ @Steve: Thanks for reference to "On conjugation of solutions to two integrable problems: Rolling of pointed body in the plane." Definitely relevant, although he specifically attends to "the horizontal plane." Haven't penetrated it entirely, and it gives a trail of related references in Russian journals. Thanks! $\endgroup$ Jul 16, 2010 at 13:20
  • $\begingroup$ Just for the record, a new paper appeared on this topic: "Rolling Manifolds: Intrinsic Formulation and Controllability," by Yacine Chitour, Petri Kokkonen, arxiv.org/abs/1011.2925 . $\endgroup$ Nov 15, 2010 at 14:52

1 Answer 1


The rolling motion of a convex symmetric body on a horizontal plane is a classical problem. In the symmetric case, Chaplygin was the first who showed that the full equations of motion can be reduced to a linear integrable system of two ODEs. A modern exposition of Chaplygin's results can be found in the very recent book by Cushman, Śniatycki and Duistermaat.

The problem of rolling motion on an inclined plane is, in general, nonintegrable (this problem was studied, in particular, by V.V. Kozlov in 1990s).

As for the tracing trajectories of the point of contact, you might be interested in this article (also available on arXiv) and the references therein. The authors discuss the case of a disc (i.e. a convex body of revolution) rolling on a horizontal plane. From the Introduction:

It appears that the point of contact performs the composite bounded motion: it periodically traces some closed curve which rotates as a rigid body with some constant angular velocity about the fixed point...

  • $\begingroup$ @Andrey: Great! This is just what I need. I like this quote from the "Dynamics of Rolling Disk" paper: "We also present various types of trajectories which are traced by the point of contact in the body-fixed and relative frames of references since they have curious forms which are difficult to predict"! Some of their "curious forms" are illustrated in Fig.8--neat! Thanks! $\endgroup$ Jul 16, 2010 at 14:14
  • $\begingroup$ Thanks for adding the Kozlov pointer on inclined planes. I see that Kozlov and Federov have a new book, "A Memoir on Integrable Systems," springer.com/mathematics/analysis/book/978-3-540-59000-2 . $\endgroup$ Jul 16, 2010 at 15:45
  • $\begingroup$ Yeah, but it looks like it hasn't been published yet. $\endgroup$ Jul 16, 2010 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.