Satisfying explanation of Aristotle's Wheel Paradox.

Question

The paradox:

We have a circle and there is another circle with smaller radius. They are co-centeric.

If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.

Unsatisfying Solutions I found:

1. "Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
2. It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)

Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.

Answer 1

The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $\vec V$ of the center $O$ and the velocity $\vec\omega\times\vec{OP}$ of rotation about the center, where $\vec\omega$ is angular velocity (perpendicular to the plane of the wheel).

A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have $$ \vec v_C=\vec V+\vec\omega\times\vec{OC} \quad\text{and}\quad \vec v_{C'}=\vec V+\vec\omega\times\vec{OC'} $$ If $\vec v_C=0$ then $\vec V=-\vec\omega\times\vec{OC}$ and $$ \vec v_{C'}=-\vec\omega\times\vec{OC}+\vec\omega\times\vec{OC'} =\vec\omega\times(\vec{OC'}-\vec{OC})=\vec\omega\times(\vec{CC'}). $$ This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.

Answer 2

Starting with three assumptions

1. Both wheels stay concentric all the way,

2. Both make full circle without sliding,

3. At the end the distance is equal to the perimeter

you get a contradictory result (both wheels have same perimeter). The principle of proof by contradictions tells us that the conjunction of your assumptions is erroneous. Of course, as you pointed out, it makes no sense to reject assumption 3 because one can always go on rolling the wheel until it crosses the right distance.

Therefore, this "paradox" is actually a proof by contradiction that the two wheels cannot simultaneously stay concentric and roll without sliding. If they are to stay concentric all the way, at least one has to slide.

Answer 3

Fix two aligned points (green and red) on the circles. And consider another circle identical to the smaller circle that is attached on top of the bigger circle. Refer to the figure below:

Case 1: If we consider the big circle and top small circle, both rotate with equal velocity and the top small circle will make full round at the red point (at this point the big circle has not made full round yet). Note that for convenience we may consider that the small circle rotates opposite to the big circle.

Case 2: Now if we consider the big circle and the cocentric small circle, the small circle will slide to catch up with the pace of the big circle.

In conclusion, the smaller the smaller circle is, the greater it slides.

Answer 4

Earlier today I was at a construction site where Aristotle’s bulldozer was hard at work. Unlike a normal bulldozer, this odd machine has a second, smaller caterpillar belt on each side, driven by wheels locked to the wheels that drive the main belts. At one point, the undercarriage of Aristotle’s bulldozer came aground on a mound of solid earth, so that even the main belts lost contact with the ground and were spinning freely. A normal bulldozer zipped over to help with the rescue, and, with both the stranded machine and the moving one in my sights, I noticed something I’d never given much thought to before, namely that the bottom of the caterpillar belt of a moving bulldozer is completely stationary with respect to the ground. Bulldozers move by laying down plates of belt in the front and drawing them up in the rear. This is not so different from how a wheel rolls—the point of the wheel that contacts the ground has zero velocity with respect to the ground; the wheel moves, loosely speaking, by laying down a new contact point and drawing up the old one.

Watching Aristotle’s bulldozer as it was being freed, it was clear to me that in the curved parts of the belt the outer and inner belts were moving with the same angular frequency, but in the straight parts, the difference in linear speed was obvious. It reminded me of how a slower runner on an inside track might keep abreast of a faster runner on curves, but lose ground on the straightaway.

Eventually the machine was freed and I got to see it in normal operation, with the outer belt in contact with the ground. As with a normal bulldozer, the belt contacting earth was stationary with respect to the ground. Not so for the smaller belt, however, whose forward motion was evident. (Forward because the backward motion of the belt with respect to the machine was not quite enough to cancel the forward motion of the machine itself, driven by the larger, faster belt.) It was clear that were both belts to contact ground at the same time they would be working at cross purposes. It would not be possible for both belts to be making nice neat tracks in the earth---one or both would be stirring up the ground beneath.

As I watched for a while longer I got to see how the machine operated when driven by the smaller belt. The larger belts fell into deep ruts briefly and lost traction, but the smaller belts, which protrude laterally from the machine, managed to contact earth. This slowed the bulldozer’s progress since the inner belts don’t move as fast. Getting up close I could see the reverse motion of the larger belts (the faster reverse motion of the belts more than cancelling the forward motion of the bulldozer).

Watching all this made me curious what Aristotle (or whichever of his disciples wrote Mechanica) might say about his namesake machine.

Answer 5

Its really quite simple and doesnt even need math to show..

Take any given point on the outside larger circle as it moves along..an do the same for the smaller one...neither one travels in a straight line from point A to point B....and they dont travel the same distance. Its seems like they do but they dont..Thats they key..

. For proof..stick a pencil in the outermost part of the larger circle and one in the smaller one..rotate the circle and notice the paths traced out..

They will not be straight lines but curved lines..and the curved line not the same length for both..Bigger the circle, the longer the path traced out from A to B.

Answer 6

In my view, the crux of the paradox is the concept of slipping (or sliding), which seems straightforward at first glance, but is not.

First let us imagine a wheel on a surface that is spinning on its axle but is otherwise motionless. In this scenario, it is obvious that the wheel is slipping on the surface, and we are not tempted to think that, just because the single point of contact of the wheel with the surface has zero length, the circumference of the wheel is zero. But what is tempting, at least for me, is to define "slipping" as the lack of a one-to-one correspondence between the points on the perimeter of the wheel and the points on the surface that the wheel is in contact with. That is, the wheel is slipping because many points on the perimeter of the wheel correspond to a single point on the surface.

But now imagine that we are spinning the wheel very fast (with uniform rotational speed) while simultaneously moving the wheel across the surface at some very slow uniform speed. Intuitively, the wheel is still slipping on the surface. But if we idealize the situation mathematically by modeling the wheel as a circle in $\mathbb{R}^2$ and the surface as a line segment in $\mathbb{R}^2$, and we limit the wheel to a single turn on its axle, then there is a one-to-one correspondence between the points on the circumference of the wheel and the points on the surface. So is the wheel really slipping?

Galileo seems to have been the first to emphasize the usefulness of considering discrete approximations to the wheel paradox. If we replace the fast-spinning wheel with a fast-spinning $n$-gon for large $n$, then we see that there is no way to set up a one-to-one correspondence. At one moment in time, one of the sides is in full contact with the surface; then the center of the $n$-gon lifts slightly and a vertex drags (slips) on the surface for a while before the next side lands on the surface, overlapping with the part of the surface that was previously in contact with the previous side. Importantly, if we keep the rotational and (horizontal) translational speeds fixed, the total amount of overlap does not go to zero as $n$ goes to infinity. This indicates that using one-to-one correspondence to define slipping is not a good idea; it does not behave well under taking limits of discrete approximations.

As mentioned by others, the "right" way to define slipping is in terms of the velocity of the point of the wheel that is in contact with the surface. We define "no slipping" to mean that this velocity is zero. This definition does behave well when we take limits of discrete approximations. With this definition of slipping, the resolution of the paradox is easy: we are no longer tempted to identify the distance traveled by the center of the wheel with the circumference of the circle just because we can form a one-to-one correspondence between the points on the circumference and the points on the surface; we know that we also have to perform a velocity calculation. We find that the lengths match up if and only if the velocity is zero, and the paradox is resolved.

One final remark: the explanation above applies most directly to the version of the wheel paradox in which the inner wheel is rolling without slipping and the outer wheel is slipping. If the outer wheel is rolling without slipping and we are puzzled about the inner wheel, then the technique of considering a discrete approximation shows us that the inner $n$-gon is "skipping" points on the surface, and again, the total length of the skipped points does not go to zero as $n$ goes to infinity. There is a beautiful Up and Atom video that demonstrates this point clearly.