I A perfectly stiff wheel cannot roll on a stiff floor?

[andrewkirk]

This is again a purely geometrical argument, that has nothing to do with starting the rolling and Newtons Laws. What is missing, is that during rolling the instantaneous rotation center is not a fixed point P.

Because it's a line rather than a point, as the wheel has width? Or because of something else? Can you elaborate?

 

[PeroK]

Rotating a circle $\mathscr C$ that is tangent to the floor (line $l$) at point $P$, around $P$, by angle $\alpha>0$, gives a new circle $\mathscr C_\alpha$ that is tangent, at $P$, to a line $l_\alpha$that passes through $P$ at an angle of $\alpha$ to $l$. Hence $l$ is now a secant that cuts $\mathscr C_\alpha$ at both $P$ and another point $Q$ such that $P$ and $Q$ mark out a sector that subtends angle $2\alpha$ at the centre of $\mathscr C_\alpha$ The chord $\overline{PQ}$ of that sector aligns with the floor $l$, so that the part of the circle outside the chord is below the floor.

Sorry about the lack of a diagram. I am too slow at making them.

It sounds like you are saying that a cycloid is geometrically impossible?!

 

[Nidum]

because of something else? Can you elaborate

I tried in #21 . For a continually rolling wheel the position of the instantaneous centre of revolution is continually moving as well

 

[A.T.]

Because it's a line rather than a point, as the wheel has width? Or because of something else? Can you elaborate?

The circle is not rotating a angle alpha around a fixed point P. The instantaneous center of rotation is moving along the line, and along the circumference of the circle, while the circle rotates.

 

[PeroK]

The circle is not rotating a angle alpha around a fixed point P. The instantaneous center of rotation is moving along the line, and along the circumference of the circle, while the circle rotates.

I think everyone except @andrewkirk gets that!

It's really just old wine in a new bottle. A moving particle is not at any point for a finite length of time. An accelerating particle does not have any specific velocity for a finite length of time. A particle moving in a circle is not moving in any given direction for a finite length of time. And, a rolling wheel is not rotating about any given point for a finite length of time.

And, calculus allows us to have these scenarios as a valid mathematical model.

 

[A.T.]

I think everyone except @andrewkirk gets that!

Sorry, I meant to reply to him.

 

[CWatters]

Point p moves to point q as it rotates.

 

[rcgldr]

A crucial insight from that article is that, when a wheel rolls along a flat surface, its axis of rotation is through the instantaneous point of contact with the ground, not through its axle.

This is a frame of reference issue. If the frame of reference moves at the same speed as the axle (consider a bicyclist observing the bikes wheels rotating), then the wheel rotates about the axle and it's the ground that is moving linearly (the axle has zero velocity relative to this frame).

From the ground frame of reference, the "point of contact" can have the same (instantaneous) velocity as the wheels axle, depending on the definition of "point of contact". For example the similar term "contact patch" of a tire is considered to be moving at the same speed as a vehicle.

 

[andrewkirk]

It sounds like you are saying that a cycloid is geometrically impossible?!

What makes you think that?

 

[Nugatory]

What makes you think that?

Every point on the rim of a translating and rotating ideal rigid wheel describes a cycloid , and there is a straight Iine parallel to the direction of motion such that all the cusps of all the cycloids lie on that line and all other points lie on the same side of the line. That appears sufficient to refute your original argument that some point must lie below the floor - so if we accept that argument we must reject some properties of cycloids, or vice versa.

 

[andrewkirk]

Every point on the rim of a translating and rotating ideal rigid wheel describes a cycloid , and there is a straight Iine parallel to the direction of motion such that all the cusps of all the cycloids lie on that line and all other points lie on the same side of the line. That appears sufficient to refute your original argument that some point must lie below the floor - so if we accept that argument we must reject some properties of cycloids, or vice versa.

I agree with all that as far as the 'refute' sentence. But description of the generation of a cycloid does not require making assumptions that the circle is rotating around the point of contact, whereas the description of commencement of rolling motion that is currently under discussion does.

Indeed, as covered in previous posts, there is no obstacle to the description of constant speed rolling motion, as it is indistinguishable from the motion of a rotating, translating wheel on a frictionless floor, and one can get a cycloid from that motion. It is the commencement of the motion, which involves interactions between forces on the wheel and friction on the ground, for which a model involving perfect circles and floors is currently lacking (in this thread - perhaps not elsewhere).

I apologise that that was not clear in the opening post. I have made some edits to the opening post (using a different colour to 'track changes') so that other newcomers to the thread are not misled by my faulty original attempt to describe what is puzzling me.

EDIT: It occurs to me that perhaps part of the problem lies in the phrase 'the circle is rotating around the point of contact'. When I examine this phrase closely it seems to lack clarity. I wonder what exactly we mean when we say a shape S is rotating around a point, especially when the point under consideration is different at every time. Can anybody produce a clear definition of what this phrase means?

 

[jbriggs444]

It occurs to me that perhaps part of the problem lies in the phrase 'the circle is rotating around the point of contact'.

Indeed. The wheel never rotates for any finite amount of time about any single fixed point.

 

[andrewkirk]

How about this?

We say that rigid shape* S is 'rotating around point P at time t' if there exists $\omega(t)\in\mathbb R$ such that, for every point particle Q in S, whose position is $(x_Q(t),y_Q(t))$, the velocity is $(-\omega(t) y_Q(t),\omega(t) x_Q(t))$.

The nice thing about this definition is that it doesn't require rotation through any actual nonzero angle to take place.

I now have to do some sums${}^\dagger$ to confirm that the motion of points in a perfectly symmetrical rolling wheel satisfy this definition.

* A rigid shape is a set of point particles (where a point particle is a function from $\mathbb R$ to $\mathbb R^3$, ie from time to space) with the property that the distance between any two of them is time-invariant.

${}^\dagger$ OK that's confirmed. If a wheel of radius R is rolling leftwards with $\omega$ the rate of rotation of the wheel around its axle, the velocity of the axle is $(-\omega R,0)$ and the velocity of a point particle $Q$ with coordinates $(x_Q(t),y_Q(t))$ relative to the point $P(t)$ where the wheel touches the ground, in the inertial frame of the axle, is $(-\omega(y_Q(t)-R),\omega x_Q(t))$, because its displacement vector from the axle is $(x_Q(t),y_Q(t)-R)$. Hence the velocity of that point particle relative to $P(t)$, in the inertial frame of the ground, is the sum of those two velocity vectors, which is $(-\omega\, y_Q(t),\omega\, x_Q(t))$, which satisfies the above criterion

 

[poor mystic]

I've been thinking about rolling motion, helped by @kuruman's excellent Insights article on the topic.
A crucial insight from that article is that, when a wheel rolls along a flat surface, its axis of rotation is through the instantaneous point of contact with the ground, not through its axle.

Thinking about this, I reached a tentative conclusion that a necessary condition for the commencement of rolling to be possible is that either the wheel not be a perfect circle as it sits on the floor OR the ground deforms under the weight of the wheel. Otherwise the wheel could not rotate around the point of contact without part of the wheel going below the floor.

The problem is easily solved by replacing the circular wheel by a regular polygon of n sides. No matter how large n is, there will always be one or two vertices in contact with the floor and, on application of a suitable force to the wheel, it can always pivot around the grounded vertex closest to the forward direction of the applied force. It pivots around that until the next vertex hits the ground, then it starts to rotate around that vertex instead.

A more realistic depiction that makes the commencement of rolling possible is to consider the wheel's shape as a circle with a portion of the bottom part chopped off, along a chord so that the contact with the ground is a line segment (the chord at which the excision takes place) rather than just a point. This allows the wheel to rotate around the foremost part of the chord without 'running into the ground'.

So in practice, the commencement of rolling is possible because the wheel will deform under its own weight (plus that of any load on the axle) to create a flat contact region that allows rolling. With a pneumatic tyre, this is easy to imagine. But even with a metal railway or tram wheel there will be some deformation of the wheel to create a small flat contact patch.

My theory (speculation, rather) is that, without that deformation, the commencement of rolling motion would be impossible.

This would all be much clearer and make much more sense with some diagrams, and I am proposing to make some, and maybe write a short note about this idea if it turns out not to be misconceived. But before I do that, I'd be interested to hear from anybody that has thought deeply about the commencement of rolling motion, if they think the idea is daft because I've missed some important feature, or alternatively if they think it is correct. Perhaps somebody has already written about this. If so it would be good to get a link to that.

Thank you

EDIT 6 Nov 2017: I realized after some of the discussion below that the real difficulty was not in explaining rolling motion, but in explaining the commencement of rolling motion by application of a force to the wheel. That is corrected in later posts below. But in order to avoid wasting the time of newcomers to the thread, who don't have time to read the whole thing, and who otherwise might spend the time to help by writing explanations of the constant-speed motion of a translating, rotating circle - which is already clearly understood, I have added words in this green font above to make clear that it is only the commencement of motion that is under investigation.

Hi :-)
This question reminds me of the problem that ancient mathematicians had with Achilles and the Tortoise, because their problem required the expansion of mathematical understanding to encompass the calculus - in other words the paradigm within which they attempted to get a grip on the problem was unsuitable for such work.
Nowadays, a philosopher could point out that the rate of progress is the pertinent fact in the discussion, and easily demonstrate that Achilles must win the race on that basis. I think that the correct approach here is to admit that the centre of rotation is at a height r with position also mapped at the tangent point on the ground.

 

[andrewkirk]

Hi :-)
This question reminds me of the problem that ancient mathematicians had with Achilles and the Tortoise, because their problem required the expansion of mathematical understanding to encompass the calculus - in other words the paradigm within which they attempted to get a grip on the problem was unsuitable for such work.
Nowadays, a philosopher could point out that the rate of progress is the pertinent fact in the discussion, and easily demonstrate that Achilles must win the race on that basis. I think that the correct approach here is to admit that the centre of rotation is at a height r with position also mapped at the tangent point on the ground.

It is natural to see parallels with that problem, which is from Zeno of Elea (not to be confused with Zeno of Citium, who was the founder of Stoicism). Zeno wrote many paradoxes, most of which appeared to suggest that motion was impossible. Calculus - which was not invented until about more than two millenia after Zeno - can resolve all of those paradoxes.

But I don't think that calculus is the answer to this problem of rotation. I am now fairly sure that the difficulty in this case (I wouldn't call it a paradox) is that saying things like 'the wheel is rotating around the point of contact with the ground' lead one (or at least it leads me) to think that rotation by some nonzero angle around that point occurs. But in fact there is no rotation through any nonzero angle about that point. Rather, the statement that it is 'rotating about that point' is simply saying something about the relationships between the instantaneous velocities, relative to that point, of all the points in the wheel at an instant in time.

I still haven't yet worked out how to connect the commencement of rotation to Newton's Laws. I suspect that going via d'Alembert's principle may be a fruitful path, because that is designed to deal with systems that have external constraints, as this does. But I am now fairly confident that, equipped with the insight of the previous paragraph (details in post 43) I will get there.

 

[rcgldr]

The wheel never rotates for any finite amount of time about any single fixed point.

A frame of reference issue. Consider the case where the wheel's axis is the frame of reference, for example a wheel rolling on a treadmill that is moving at constant speed, using the ground as a frame of reference, and with the wheel's axis having zero velocity relative to the ground. The wheel constantly rotates about it's fixed (wrt ground) axis.

 

[PeroK]

But I don't think that calculus is the answer to this problem of rotation. I am now fairly sure that the difficulty in this case (I wouldn't call it a paradox) is that saying things like 'the wheel is rotating around the point of contact with the ground' lead one (or at least it leads me) to think that rotation by some nonzero angle around that point occurs. But in fact there is no rotation through any nonzero angle about that point. Rather, the statement that it is 'rotating about that point' is simply saying something about the relationships between the instantaneous velocities, relative to that point, of all the points in the wheel at an instant in time.

That's exactly what it is means. The instantaneous velocity of each point of the wheel represents rotation about the point touching the surface (which is instantaneously at rest).

There is no reason to assume that the point of rotation is the same for a finite time interval. Each point on the rim is only instantaneously the centre of rotation.

 

[Nidum]

Can someone explain to me where in all of this discussion there is any problem that needs solving ?

 

[Herman Trivilino]

Consider the case where the wheel's axis is the frame of reference, for example a wheel rolling on a treadmill that is moving at constant speed, using the ground as a frame of reference, and with the wheel's axis having zero velocity relative to the ground. The wheel constantly rotates about it's fixed (wrt ground) axis.

But that point is the center of the wheel, not the point where the rim contacts the treadmill's surface. The rotation you describe occurs for finite nonzero amounts of time, whereas the other doesn't.

 

[rcgldr]

The wheel never rotates for any finite amount of time about any single fixed point.

A frame of reference issue. Consider the case where the wheel's axis is the frame of reference, for example a wheel rolling on a treadmill that is moving at constant speed, using the ground as a frame of reference, and with the wheel's axis having zero velocity relative to the ground. The wheel constantly rotates about it's fixed (wrt ground) axis.

But that point is the center of the wheel, not the point where the rim contacts the treadmill's surface. The rotation you describe occurs for finite nonzero amounts of time, whereas the other doesn't.

I only mentioned that as an example where a perfectly circular wheel does rotate about a fixed point for a finite amount of time, if the wheels axis is chosen as the basis for a frame of reference.

Getting back on the original topic, there is still the issue of considering a wheel as the limit of a n sided polygon as n approaches infinity. For n less than infinity, then the n sided polygon will rotate about the contact point for some finite period of time, until it transitions to the next contact point. The average velocity of the rate of advance from contact point to contact point will be the same as the average velocity of the wheel's axis with respect to the ground, regardless of n.

 

[OCR]

How about this?

This ??

 

[Laurie K]

If you are interested in a SR based approach to relativistically rolling rings see. https://www.physicsforums.com/threads/rolling-rings-in-sr.920107/

 

[andrewkirk]

I promised some pictures earlier, and I've finally made them. Here are three pictures, showing a wheel rotating around a point of contact with the ground. The first is a perfectly circular wheel, the second is a wheel with the bottom flattened (eg a tyre compressing under the weight of the load) and the third is a polygon.

We see that, if the wheel rotates around the foremost point of contact in the direction of travel (marked) through a non-zero angle, it all works OK for the polygon and the flattened wheel, but not for the perfect circle, which has to go below the floor.

The answer arrived at in the thread above is that there is no rotation through any angle about that point. Rather, the statement that the wheel is rotating about that point is just a description of the instantaneous relative velocities of points on the wheel.

 

[CWatters]

If the polygon is allowed to lift itself up why not the round wheel?

 

[CWatters]

This all seems nonsense to me? Time to retire this thread?

 

[Baluncore]

"A perfectly stiff wheel cannot roll on a stiff floor?" Mathematically it can roll perfectly well.
Has anyone here mentioned the fixed point theorem?

In reality, neither the wheel nor the floor will be perfectly stiff. There will be elastic deflection of each. If there was no elastic deflection the contact area would be zero, the stress infinite and the materials would be momentarily crushed to carry the load. Being perfectly stiff does not prevent a material being infinitely brittle.

Heinrich Hertz was a realist, (applied mathematician), who studied the stress and deflection of bearing surfaces in contact with rolling balls and cylinders. https://en.wikipedia.org/wiki/Contact_mechanics

 

[NFuller]

We see that, if the wheel rotates around the foremost point of contact in the direction of travel (marked) through a non-zero angle, it all works OK for the polygon and the flattened wheel, but not for the perfect circle, which has to go below the floor.

No, because the contact point is also moving along the circumference of the wheel. Look at your other drawings, you allowed the contact point to move in order for the wheel to stay above the floor in those cases, so why not for the circular wheel as well?

 

[A.T.]

Here are three pictures, showing a wheel rotating around a point of contact with the ground. The first is a perfectly circular wheel,

This is rotation around a point stationary w.r.t. ground. The contact point of a rolling wheel is not stationary w.r.t. ground.

 

[256bits]

but not for the perfect circle, which has to go below the floor.

It's that part I do not get, and I have the following comment on the argument.
Considering that a circle is an n-polygon ( where n approaches infinity in the mathematical sense, or a very large number in the real sense ), there is always a picture that can be drawn that resembles the 2nd or third. The 'next' point of contact should always be the endpoint of a chord.

Perhaps the problem is not rolling per se, but points of contact between surfaces of varying curvature.
Here I have flipped the one picture upside down. Here we see a circle of very large radius ( so that the surface approaches a flat plane ) rolling on a another surface of negative curvature ( a circle ). I would tend to think that the "plane" does not penetrate the circle as it moves around. Other scenarios of interacting surfaces of positive and negative curvature could be investigated, as either 'chord length model' or 'continuum model'.

At least that is how I see the disclosed problem. Not sure if that is what you had in mind.

 

[A.T.]

We see that, if the wheel rotates around the foremost point of contact in the direction of travel (marked) through a non-zero angle, it all works OK for the polygon and the flattened wheel, but not for the perfect circle, which has to go below the floor.

Note that your polygon will go below the floor too, when you fail to shift the rotation center forward (rotate around the previous corner). This is what you do with the circle here.