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I recently wondered whether there is a simple proof that circular wheels maximise mechanical efficiency. By this I mean:

Show that for a wheel with a given width and cross-sectional area, the circular wheel reaches the bottom of an incline faster than a wheel with any other shape.

This sounds like a rather elementary problem in classical mechanics but I haven't found a proof of this fact in any of the classical mechanics texts that I have read.

Note 1: I assume that the distribution of mass is uniform across the wheel so comparable wheels will have equal mass.

Note 2: Dry friction is assumed to be present.

  • $\begingroup$ What are your requirements for "wheel"? A spherical wheel would reach the bottom faster, but you cannot specify the width and cross section independently. $\endgroup$ – alex_d Oct 12 '16 at 7:45
  • $\begingroup$ @alex_d I'm assuming that the wheel has uniform cross-sectional area and pre-defined width. The cross-sectional geometry need not be constant. $\endgroup$ – Aidan Rocke Oct 12 '16 at 8:08
  • $\begingroup$ Aidan : You are a mathematics undergraduate. Surely you have tried to solve this problem. (Some of your earlier questions were very mathematical.) What attempt have you made on this one? $\endgroup$ – sammy gerbil Oct 12 '16 at 22:16
  • $\begingroup$ @sammygerbil I have some thoughts on the matter, but from a mathematical perspective this question is not trivial. I'll need to take time to think about it this weekend. However, if I do find a mathematical answer, can I post it on this forum? I haven't found any mathematically rigorous answers on this forum but for this question, I think that anything short of a mathematically rigorous answer would be insufficient. $\endgroup$ – Aidan Rocke Oct 13 '16 at 20:33
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    $\begingroup$ Yes, you can post an answer to your own question. The more rigorous the better. I agree, mathematically it is not trivial. $\endgroup$ – sammy gerbil Oct 13 '16 at 20:38

A complication arises as the shape gets further from circular symmetry. When it gains sufficient angular velocity, any non-circular object can 'jump' off the incline. See A jumping cylinder on an inclined plane. On leaving the plane the object becomes a projectile. Whether this speeds the descent or not might be difficult to determine.

Another complication is that the circle is the only shape which has neutral stability. It will roll continuously on any incline, no matter how small the angle. For all other shapes, the stability is positive. On a horizontal plane or small incline it will rock about an equilibrium position. It will only start rolling continuously if the angle is large enough.


The width is probably meant into the third dimension, like thickness? This means that we can just look at a 2D problem with a surface mass density $\rho$.

The question does not state whether there is friction involved. If there was no friction, every body would slide down exactly the same. This way the question is not interesting in any way.

Since all the “wheels” are supposed to have the surface area and width, they all have the same mass. So one can reword the question perhaps like this: If we redistribute the mass from a circular shape to something else, what changes? The inertia tensor will change. We are only interested in the one moment of inertia which is along the rotation axis. This is given by the following: $$ I = \iint \mathrm d^2 x \, \rho (\vec x) \, |\vec x^2| = \int \mathrm d x_1 \int \mathrm dx_2 \,\rho (\vec x) \, (x_1^2 + x_2^2) =\int \mathrm d r \, r\int \mathrm d \phi \,\rho (r, \phi) \, r^2 \,.$$ We assume $\rho$ to be constant within the surface and zero outside of it. And there you can see why the circle is the best option: If you have a compact circle, the factor $r^3$ will only become as large as the radius of the circle. If you distort the same surface area to an ellipse, for some angles $\phi$ you will not have that large of $r^3$, that “saves” you a bit of inertia. However, for other angles you need to integrate over $r$ further, and that $r^3$ bit becomes “expensive”.

So no matter how you deviate from the circle with the constraint that the total surface area is the same, you will have make $I$ larger than before. Therefore the circle has the smallest moment of inertia and can rotate the fastest given the same driving moment.

An ellipse with the same surface area has a larger circumference. Therefore it does not need to rotate as fast as the circle to cover the same distance. However, I think that the effect of $I$ becoming larger will outweigh the reduction in revolutions needed to reach the bottom of the slope.

  • $\begingroup$ Awesome! No calculus of variations in sight, only elegant, simple observations and deductions. $\endgroup$ – WetSavannaAnimal Oct 12 '16 at 11:01
  • $\begingroup$ Yes, but an incomplete proof. Your last sentence is an assumption, not a deduction. $\endgroup$ – sammy gerbil Oct 12 '16 at 16:25
  • $\begingroup$ Indeed, it is only a heuristic argument to why it would make sense for the circular shape to be the most efficient. A full derivation would probably use variational calculus, though, and I do not have much experience with it. $\endgroup$ – Martin Ueding Oct 12 '16 at 16:34
  • $\begingroup$ @Martin Ueding Dear Martin, I'd appreciate if you take a look at this (and if possible answer it): physics.stackexchange.com/questions/287730/… $\endgroup$ – user.3710634 Oct 21 '16 at 2:42

The statement is false as stated. The following shape will be faster:

enter image description here

I don't prove this fact, but it's hopefully clear that by choosing the distribution of holes I can make the average density as a function of $r$ the same as that of a sphere, which you know is faster.

You might have more luck if you restrict the shape to be convex. But to make a well-formed problem you still have to specify what you mean by "reaches the bottom of the incline faster". For example, if you mean the time taken for the contact point to move from the top to the bottom of the slope then a long stick with the same length of the slope which has nearly fallen over will clearly be the fastest. (This might sound like a joke, but it shows that if you want a mathematically rigorous solution you have to make a mathematically precise problem!)

There are other issues too... if there are no dissipative forces and the slope is long enough then any non-circular object will eventually start bouncing down the slope rather than rolling. How do you want to handle these cases?

  • $\begingroup$ Your proposed shape contravenes Note 1 : uniform distribution of mass. But I agree about the bouncing problem. Likewise, non-circular shapes will not start to roll until the slope exceeds a certain value. $\endgroup$ – sammy gerbil Oct 17 '16 at 19:58
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    $\begingroup$ @sammy gerbil It isn't clear to me that the shape contravenes note 1. I read that as requiring that the density is uniform, rather than a requirement that the global mass distribution is 'uniform'. If the latter was the case then a circle would be the only valid shape to test. $\endgroup$ – Penguino Oct 17 '16 at 20:39
  • $\begingroup$ @sammy gerbil I, like Penguino interpreted note one to mean a constant cross-sectional mass density, in which case my shape is fine. In fact, I can't see an interpretation of that note which rules out my shape and still allows other non-circular shapes like ellipses. Can you explain your interpretation to me? $\endgroup$ – Mark A Oct 18 '16 at 1:03
  • $\begingroup$ My interpretation is that the space inside the rim which touches the incline should be filled with material of constant density - no gaps. However, this is not clearly stated in the question - another issue for the OP to sort out. In any case, a circle with the same filled area as yours but without holes will have a smaller diameter and therefore a smaller MI. $\endgroup$ – sammy gerbil Oct 18 '16 at 2:07
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    $\begingroup$ Yes, I apologise, you are correct. A solid disc of the same mass would have a slightly smaller MI about the centre, and significantly (but not proportionately) smaller MI about the rim (parallel axis theorem), but a proportionately smaller torque due to smaller radius - which outweighs the smaller MI about the rim. $\endgroup$ – sammy gerbil Oct 20 '16 at 19:12


Imagine that the slope is arbitrarily small. A circular wheel takes an arbitrarily long time to reach the bottom.

Now consider an highly-elliptical wheel that reaches the bottom in 1/2 rotation (so the orientation is the same at top and bottom and you aren't cheating). Start with its long axis almost vertical and leaning a bit downhill. Quite rapidly it topples downhill, rolls past its short axis, and continues rotating until it is standing upright at the bottom of the hill.

How much tilt it can have at the beginning and still rotate past upright at the bottom (and keep on rolling) depends on how arbitrarily small the slope is and the aspect ratio of the ellipse, but it is easy to intuit that the tumbling ellipsoid will be faster than the circle under reasonable circumstances.


When it is coming down the motion of the wheel of any shape is one dimensional. Degree of freedom of two dimensional object is one when no torque is acting . In this case of if we consider free fall then only acceleration is gravitational acceleration which will not apply torque here. By using the two dimensional object we can utilize maximum energy on bringing down wheel .

If we consider a wheel of n polygonal shape and distance d and mean velocity v . Then we have time t=s/v or, t=na/v where a is the length of arm of the polygon. In our case a and v are constant so if n tends to zero then t tends to zero or (n-1) vertices of the polygon tends to zero. The only geometric shape which is close to that requirements is a circle. Since there is no existence of 2d object so circle with thickness able to withstand different environmental resistance and reach bottom steadily is used.

I hope this at least takes us closer to the solution


Any shape other than a circle will have its centre of gravity moving up and down as it rolls down the slope, thus it will convert some of the attained kinetic energy back into potential energy. The circle doesn't lose any kinetic energy like that so it clearly is faster.

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    $\begingroup$ If the centre of gravity moves up and down, some energy will be converted back and forth between potential and kinetic. No energy is lost. When the CG has fallen through the same height, the shape will have the same KE as the circle. The average speed might be lower, but it is not obvious. Can you prove it? $\endgroup$ – sammy gerbil Oct 12 '16 at 14:40
  • $\begingroup$ @sammygerbil - yes, it is obvious. $\endgroup$ – Suzu Hirose Oct 12 '16 at 20:00

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