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

[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.

 

[CWatters]

The assertion that a perfectly stiff wheel cannot roll on a stiff floor is also contrary to experience, in that rigid wheels on rigid surfaces generally have a lower rolling resistance than softer ones.

 

[rcgldr]

Part of the issue here is the point of contact only exists at a specific point for an instant in time. Consider the limit of an n-sided polygon where the center of the polygon moves in the positive x direction at some average velocity v. The point of contact advances as instances in time, but the average rate of advance is the same as the average velocity of the center of the polygon. As the number of sides of the polygon approaches infinity, the average rate of the advance of contact point continues to be the same as the average velocity of the center of the polygon, and the limit of this case is a perfect circle, in which case the point of contact has the same velocity as the center of the circle (both are now moving in a straight parallel lines and at a constant velocity).

My perspective of this is to view the point of contact as a moving point, being the point that the circle is currently in contact with a flat surface, not as a fixed point on the surface of the wheel or as a fixed point on the surface of the flat surface. The point of contact has a velocity, in the same manner that vehicle physics consider a contact patch to be moving at the same speed as the vehicle. Having point of contact with a velocity will not change the perspective of that point for an instant in time.

 

[CWatters]

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. ...snip..
View attachment 215189 View attachment 215190 View attachment 215191

You have shown the point of contact of the polygon on the right hand side. What happens when the polygon changes direction? Presumably you would move it to the left?

Why do you allow the point of contact to move around the perimeter of the polygon but not move around the perimeter of the circle?

 

[Baluncore]

If a polygon can be used as a rolling wheel model, then consider using a square wheel, if a square wheel is unacceptable then all polygons must be unacceptable. Some concepts are useful and some are distractions.

 

[Herman Trivilino]

The assertion that a perfectly stiff wheel cannot roll on a stiff floor is also contrary to experience, in that rigid wheels on rigid surfaces generally have a lower rolling resistance than softer ones.

It's true that the more rigid the materials, the lower the rolling resistance. But it's friction that initiates and maintains the roll, and you cannot have that unless the surfaces deform.

Mathematically, yes, you can have a perfectly rigid circle rotate without slipping along a perfectly straight line, but physically you cannot. I'm not sure which of these points is being debated in this thread.

 

[A.T.]

But it's friction that initiates and maintains the roll

Even on a frictionless surface you could generate rolling by applying the right torque and force combination.

Mathematically, yes, you can have a perfectly rigid circle rotate without slipping along a perfectly straight line, but physically you cannot.

Physically you cannot have perfectly rigid bodies at all. But for many applications it's a acceptable simplification.

 

[A.T.]

if a square wheel is unacceptable then all polygons must be unacceptable

What is acceptable depends on the application.

 

[andrewkirk]

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

Sure, everybody knows that, but I don't see it as having any bearing on the problem. The problem is easily solved by simply interpreting the statement that 'the wheel is rotating around the contact point' to be the statement about relationships of instantaneous linear velocities of different points on the wheel that was made in post 43.

It only remains a problem if we want to interpret the statement as meaning that there is a rotation through a nonzero angle around that point. If we want to make that interpretation, I don't see how replacing the stationary point by the locus of contact points over time helps. I don't even know what it would mean to say that the wheel rotates through a nonzero angle around that locus. Nor can I see any practical benefit to the theoretical work that would need to be done, defining frames of reference etc, to give meaning to that statement.

 

[OCR]

...any bearing...

Lol... nice pun. .

 

[A.T.]

Sure, everybody knows that, but I don't see it as having any bearing on the problem.

The "problem".

 

[russ_watters]

If a polygon can be used as a rolling wheel model, then consider using a square wheel, if a square wheel is unacceptable then all polygons must be unacceptable.

Not that I want to wade into a thread that has probably run its course, but I think it should be obvious that the more "gons," the smoother the roll.

 

[kuruman]

Not that I want to wade into a thread that has probably run its course, but I think it should be obvious that the more "gons," the smoother the roll.

Maybe.
https://www.sciencenews.org/article/riding-square-wheels

 

[Andrew Mason]

For what it is worth: the problem assumes an ideal, non-real, hypothetical situation but relies on an very real, non-ideal and inherently imperfect phenomenon (friction) to still be operating and suggests that this leads to a contradiction. Of course, it does.

The premise that rolling can occur with the wheel and surface can be in contact over an arbitrarily small distance is not correct. The wheel rolls because of friction and friction requires that the two surfaces overlap somewhat, like two gears meshing together. It is just that this occurs at a microscopic level. If the two surfaces were absolutely smooth down to the molecular level there would be no static friction and, therefore, no rolling at all i.e. no lateral force at the contact point and, therefore, no pivot point.

If one makes the wheel and surface harder and the wheel more perfectly round and the surface more perfectly flat, the contact area is reduced which means that the pressure over the contact area increases. At some point, the pressure exceeds the yield pressure for the material that the wheel and/or surface are made out of so the material breaks down (until the surface area of "contact" increases and the pressure decreases to below the yield pressure).

In summary: no matter what you make the wheel and surface out of, one can never reach an arbitrarily small contact area. But even if you could make them out of some idealised material that does not exist, you would not have any friction and, therefore, no rolling.

AM

 

[Dr.D]

Is this question not very close to the matter of how many angels can dance on the head of a pin?

 

[Herman Trivilino]

@Andrew Mason. You are correct. For the case of what we normally refer to as rolling motion, the surface is supporting the wheel. You can't have that support force without deformation. If you want to look at the limit as the support force approaches zero and just have a rotating wheel next to a flat surface such that the tangential speed of the wheel rim matches the speed of the wheel's center relative to the surface then you can have zero deformation, but that is not what we normally refer to as rolling motion.

 

[PeroK]

The wheel rolls because of friction and friction requires that the two surfaces overlap somewhat, like two gears meshing together. It is just that this occurs at a microscopic level. If the two surfaces were absolutely smooth down to the molecular level there would be no static friction and, therefore, no rolling.

But even if you could make them out of some idealised material that does not exist, you would not have any friction and, therefore, no rolling.

AM

A wheel does not require friction to roll. It will keep rolling through conservation of angular momentum.

The rolling could be initiated by any torque.

 

[Baluncore]

If the two surfaces were absolutely smooth down to the molecular level there would be no static friction and, therefore, no rolling at all i.e. no lateral force at the contact point and, therefore, no pivot point.

In summary: no matter what you make the wheel and surface out of, one can never reach an arbitrarily small contact area. But even if you could make them out of some idealised material that does not exist, you would not have any friction and, therefore, no rolling.

Since when has contact area and not force been important to friction? If a point or line contact forms, the chemical bonds between the two surfaces will stick the two particles or objects together and result in friction, hence torque and rotation.

 

[andrewkirk]

A wheel does not require friction to roll. It will keep rolling through conservation of angular momentum.

The rolling could be initiated by any torque.

The question of how to initiate, for a wheel on a frictionless surface, rolling motion identical to what could occur on a frictionful surface, is interesting. Most pushes on a part of the wheel, or a stiff, weightless handle attached to it, would initiate a translating, rotating motion that did not match any frictionful rolling pattern. It is necessary for the wheel's angular velocity $\omega$ to relate to the linear velocity $v$ of the wheel's centre by the equation $v=\omega R$, where $R$ is the radius of the wheel. For the motion to always match a rolling motion, it is necessary that $\dot v=\dot\omega R$ at all times.

On my calcs, if a force is applied at angle $\alpha$ counter-clockwise of vertical, at polar coordinates $(r,\theta)$ relative to the axle (with $\theta$ being measured as angle to counter-clockwise of the vertical), the following equation must be satisfied
$$I\sin\alpha = rRm\cos(\alpha-\theta)$$
where $I$ and $m$ are the moment of inertia and mass of the wheel.

If we are applying the force to a handle that is at a fixed distance $r$ from the axle, we would need to continuously vary the angle $\alpha$ of our push in order to maintain the motion as rolling-like. This gives $\alpha$ as a function of $\theta$.

Alternatively, if we fix the direction of the applied force as always horizontal, the radius at which it must be applied will vary with $\theta$, being at a minimum when it is applied at a point above the axle ($\theta=0$) and increasing without limit as $\theta\to\pi/2$.

The size of the force makes no difference. It cancels out of all the equations.

 

[PeroK]

The question of how to initiate, for a wheel on a frictionless surface, rolling motion identical to what could occur on a frictionful surface, is interesting. Most pushes on a part of the wheel, or a stiff, weightless handle attached to it, would initiate a translating, rotating motion that did not match any frictionful rolling pattern. It is necessary for the wheel's angular velocity $\omega$ to relate to the linear velocity $v$ of the wheel's centre by the equation $v=\omega R$, where $R$ is the radius of the wheel. For the motion to always match a rolling motion, it is necessary that $\dot v=\dot\omega R$ at all times.

On my calcs, if a force is applied at angle $\alpha$ counter-clockwise of vertical, at polar coordinates $(r,\theta)$ relative to the axle (with $\theta$ being measured as angle to counter-clockwise of the vertical), the following equation must be satisfied
$$I\sin\alpha = rRm\cos(\alpha-\theta)$$
where $I$ and $m$ are the moment of inertia and mass of the wheel.

If we are applying the force to a handle that is at a fixed distance $r$ from the axle, we would need to continuously vary the angle $\alpha$ of our push in order to maintain the motion as rolling-like. This gives $\alpha$ as a function of $\theta$.

Alternatively, if we fix the direction of the applied force as always horizontal, the radius at which it must be applied will vary with $\theta$, being at a minimum when it is applied at a point above the axle ($\theta=0$) and increasing without limit as $\theta\to\pi/2$.

The size of the force makes no difference. It cancels out of all the equations.

a) initiate linear motion by a horizontal force through the centre.

b) initiate rotation by a pair of equal and opposite horizontal forces.

It's clearly and trivially possible to accelerate a wheel to any velocity without rotation and to any angular velocity without linear motion. And, therefore, to have any desired combination of the two.

You need to clear your head!

 

[andrewkirk]

a) initiate linear motion by a horizontal force through the centre.

b) initiate rotation by a pair of equal and opposite horizontal forces.

That is a set of three separate forces, not a single torque, which is what your post above says can initiate rolling motion.

The point is that it has to be a very special torque, not just any torque, to initiate rolling-like motion on a frictionless surface without applying force at multiple points, and I find it interesting to investigate what the nature of that specialness must be.

 

[rcgldr]

If we are applying the force to a handle that is at a fixed distance $r$ from the axle, we would need to continuously vary the angle $\alpha$ of our push in order to maintain the motion as rolling-like. This gives $\alpha$ as a function of $\theta$.

What about applying a horizontal force that always remains horizontal and at a fixed distance above the center of mass? For example, imagine a solid rim or wheel, squeezed between two wheels with vertical axis, and those wheels used to apply a continuous force that remains horizontal and a fixed distance above the center of mass as the target wheel accelerates.

 

[Likith D]

Sure, everybody knows that, but I don't see it as having any bearing on the problem. The problem is easily solved by simply interpreting the statement that 'the wheel is rotating around the contact point' to be the statement about relationships of instantaneous linear velocities of different points on the wheel that was made in post 43.

It only remains a problem if we want to interpret the statement as meaning that there is a rotation through a nonzero angle around that point. If we want to make that interpretation, I don't see how replacing the stationary point by the locus of contact points over time helps. I don't even know what it would mean to say that the wheel rotates through a nonzero angle around that locus. Nor can I see any practical benefit to the theoretical work that would need to be done, defining frames of reference etc, to give meaning to that statement.

I love your interpretation of the rolling motion!
It made me think a lot more about it in many different ways...
I think the most helpful perspective, I found is this;
Think of a point moving in a trochoid...
Now, think of a bunch of points moving in different torchoids (infinitely many such points, actually)... Great, so now we can twiddle with the variety of curves under "trochoid" and place them in a so-and-so position Such that... when we play the entire motion (of the infinite points we have setup) it mimics a rolling circle on a flat plane very perfectly!
So we successfully synced the trochoid movement of all the infinite points to perfectly mimic the rolling circle on a flat plane!
And also from our knowledge about trochoids, none of the points ever make it through the floor they are rolling on...
And lo! we have created a "rolling" motion of a circle without damaging a perfectly smooth (flat) floor!

I think that humans have a better (and a more agreeable) intuition of understanding resultant of infinitely many things happening (like integrating the motion of infinite number of particles) rather than conceive the happenings at an infinitesimal time gap regarding geometry at infinitesimally small space scale (infinitely zoomed up vision of geometry of stuff/space).
I can say this from my attempts to REALLY believe that perpendicular force does not change the speed of a particle, as a interpretation of it got me believing otherwise... even though the cold hard math for uniform circular motion by differentiating position vectors clearly said otherwise (it was only later that I found a way to drill the geometrical intuition for that result into me - which I won't talk about here)
I digress
Hope that help! Please do tell me your opinion!

 

[PeterO]

If you are thinking about rolling motion, and buoyed by someones article on the topic and reach the conclusion that something cannot start rolling, but we know a wheel CAN start rolling, then either you are either thinking incorrectly or have mis-interpreted the article.
Even if you can come up with a mathematical analysis that shows the wheel cannot start rolling (but we know it can) - something is wrong with your maths.
Peter

 

[Clausen]

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

NO! This is not nonsense. It is a perfectly valid question and you don't close a thread just because you don't agree with the question posed.

 

[Clausen]

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.

That is exactly the point! A rolling wheel will not have a contact point that is stationary wrt the ground. The OP is asking about a perfectly rigid wheel on a perfectly rigid surface. In other words, there is no deformation of the wheel or the surface, and the wheel is only making contact with the surface at a single point.

If the wheel rotates, the adjacent point of the wheel that is turning through the rotation can only set down in the exact same spot on the surface as the point that is lifting off of the surface and this follows for all points on the wheels circumference.

IIn such a case, the wheel spins in place, but does not, and cannot roll, unless there is deformation.

This should be common sense!

Of course, there is no such thing as a perfectly rigid wheel on a perfectly rigid surface, but this still serves as an interesting thought experiment about what is involved in a wheel rolling on a surface; there must be some deformation, however small, for the wheel to roll, otherwise it just spins in place.

 

[jbriggs444]

Even if you can come up with a mathematical analysis that shows the wheel cannot start rolling (but we know it can) - something is wrong with your maths.

Rather than saying that something is wrong with the maths, there might be something wrong with the translation between math and the real world. The model of a perfectly circular and perfectly rigid "wheel" interacting with a perfectly flat and perfectly rigid "road" fails to accurately reflect the behavior of a real world wheel on a real world road. Real world wheels are neither rigid nor circular. Real world roads are neither rigid nor flat.

An argument that a perfectly rigid and perfectly circular wheel on a perfectly rigid and perfectly flat road would experience no friction and therefore never start rolling is plausible and is not falsifiable by physical experiment -- we have no way to perform a real world test. However, that is not the argument made in post #1.

An argument that the motion of a perfectly rigid and perfectly circular wheel about an instantaneous axis of rotation at a perfectly rigid and perfectly flat road surface must involve interpenetration of the wheel with the road can be made. That argument is also not falsifiable by experiment. It is falsifiable by careful examination of the mathematics.

 

[jbriggs444]

If the wheel rotates, the adjacent point of the wheel

There is no such thing as an "adjacent point" on a wheel. [By "adjacent point", I expect that you refer to two points next to each other on the wheel's surface].

 

[Clausen]

Yes, of course that is exactly what I meant.

 

[jbriggs444]

Yes, of course that is exactly what I meant.

It is a provable property of the real numbers (and, accordingly, of points on the circumference of an ideal wheel) that between any two distinct points there is a point between them. It follows that there is no such thing as a pair of "adjacent" points.

 

[Eric Bretschneider]

I find this rather lengthy thread curious. If your frame of reference is the flat surface then the wheel rotates about the contact point. Thus intuition says that the wheel can not be perfectly rigid or it would deform the flat surface and not roll.

If you choose your frame of reference as the center of the wheel then you have an entirely different situation. The flat surface moves and the wheel can be perfectly rigid. The wheel only rotates about its center.

I think its great to look at problems from different viewpoints, but sometimes a little flexibility in thinking makes solving the problem much simpler.

Take for example Zeno's paradox. If I want to cross a 10ft room, I first take a (big) step that covers half of the distance to the other side, then another step that covers half the remaining distance and then another step that covers half the remaining distance and so on. It will take an infinite number of steps to get to the other side of the room. There is the mathematical argument that you can sum an infinite series and get a finite number.

There is also the more mundane argument that if my intent was to go twice the distance then my first step would take me to the other side of the room. Zeno allows you to travel half the distance regardless of what the distance is. Change your frame of reference and Zeno's paradox disappears - actually the paradox is that it is not self consistent.

 

[Clausen]

It is a provable property of the real numbers (and, accordingly, of points on the circumference of an ideal wheel) that between any two distinct points there is a point between them. It follows that there is no such thing as a pair of "adjacent" points.

That’s great! So, we have a perfectly rigid wheel on a perfectly rigid surface such that there is only point contact being made between the wheel and the surface. Now, the wheel spins very slightly* CW, such that the point on the wheel that was touching the surface moves CW and upwards so that it no longer touches the surface. There WILL be another point on the wheel that moves into the exact same position as the original point, touching the exact same spot on the surface.

What do you call that second point on the wheel? I would call it the adjacent point to the first point. Obviously, you disagree, so what do you call it?

Note: the wheel did NOT roll, it can’t without deforming and it has not deformed, it just spun in place.

Thanks!

· Will you next post ask me to define “very slightly”?

 

[Clausen]

I find this rather lengthy thread curious. If your frame of reference is the flat surface then the wheel rotates about the contact point. Thus intuition says that the wheel can not be perfectly rigid or it would deform the flat surface and not roll.

If you choose your frame of reference as the center of the wheel then you have an entirely different situation. The flat surface moves and the wheel can be perfectly rigid. The wheel only rotates about its center.

I think its great to look at problems from different viewpoints, but sometimes a little flexibility in thinking makes solving the problem much simpler.

Take for example Zeno's paradox. If I want to cross a 10ft room, I first take a (big) step that covers half of the distance to the other side, then another step that covers half the remaining distance and then another step that covers half the remaining distance and so on. It will take an infinite number of steps to get to the other side of the room. There is the mathematical argument that you can sum an infinite series and get a finite number.

There is also the more mundane argument that if my intent was to go twice the distance then my first step would take me to the other side of the room. Zeno allows you to travel half the distance regardless of what the distance is. Change your frame of reference and Zeno's paradox disappears - actually the paradox is that it is not self consistent.

These sort of problems are very interesting, even though there is no such thing as a perfectly rigid wheel or a perfectly rigid surface, what you can glean from thinking about this is the fact that any real wheel must deform in order to roll. You would be surprised how many people do not understand that!
If you drive a wheel with a belt, it is the belt that deforms, not the wheel, and the wheel can only be spun by the belt, it cannot roll! If the wheel translates on the belt, it is due to sliding while spinning, not rolling.

 

[jbriggs444]

There WILL be another point on the wheel that moves into the exact same position as the original point, touching the exact same spot on the surface.

You seem to be assuming that the wheel is spinning with an instantaneous center of rotation at its center.

A wheel rolling without slipping on a stationary surface does not rotate about its center. Instead, the instantaneous center of rotation is at the instantaneous contact point with the floor.

Edit: As far back as response #17, @A.T. noted that the instantaneous center of rotation will depend on a choice of reference frame. Here I have chosen a reference frame by referring to the surface as "stationary".

What do you call that second point on the wheel? I would call it the adjacent point to the first point.
Obviously, you disagree, so what do you call it?

A different point on the wheel. It is clearly not adjacent if there is a non-zero distance between the two points.

Note: the wheel did NOT roll, it can’t without deforming and it has not deformed, it just spun in place.

Sorry, you are assuming the conclusion there. In fact, a wheel can roll without deforming.

 

[jbriggs444]

If one wishes to address this question with proper mathematical rigour, one should probably start by defining what is meant by a ideal rigid wheel rolling on an ideal flat surface without slipping. Dotting all the i's and crossing all the t's make this a painful process.

Let us work in two dimensions -- on an x-y coordinate grid. The grid counts as "space". The x-axis running left to right down the center counts as the "floor". This x-y space is the set of locations where various points on the wheel may momentarily exist.

The wheel has its own coordinate system. To avoid confusion, let us use i and j for these coordinates. We can lay this out with (0,0) in the center. If the wheel has radius r then its circumference will be the set of points (i,j) such that $\sqrt{i^2+j^2} = r^2$ When we refer to points "on the wheel", we are referring to points either on the circumference or in its interior.

We want the system to evolve over time. So we introduce a real-valued parameter t for time.

The momentary "position of the wheel" is a function that maps a point on the wheel at a particular time to a position in space at that time: $\vec{P(t,i,j)} => \vec{(x,y)}$. We may refer to this function component-wise as P_x(t,i,j) and P_y(t,i,j)

If we ask whether the wheel can roll without slipping, that means that we ask whether there is a function P such that:

1. P is continuous and differentiable. That term is rigorously definable. Roughly speaking, it means that for nearby tuples (t₁, i₁, j₁) and (t₂, i₂, j₂) will map to nearby tuples (x₁, y₁) and (x₂, y₂). Further, roughly speaking, this means that every point on the wheel moves smoothly. It does not jump around. Every point on the wheel has a well defined velocity at all times.

2. For every t there is exactly one point (i,j) on the wheel such that P_y(t,i,j) = 0

Roughly speaking, this means that the wheel rests on the floor.

2a. For every t and every point (i,j) on the wheel, P_y(t,i,j) >= 0

Roughly speaking, this means that no point on the wheel extends below the floor.

3. The point (i,j) at time t on the wheel whose existence is guaranteed by (2) has $\frac{d\vec{P}(t,i,j)}{dt} = \vec{0}$

This means that the point on the wheel that touches the floor is momentarily at rest: The wheel is rolling without slipping.

4. For every t and every pair of points (i₁,j₁) and (i₂,j₂) it is always the case that $\sqrt{(i_1-i_2)^2+(j_1-j_2)^2)} = \sqrt{(P_x(t,i_1,j_1)-P_x(t,i_2,j_2))^2+(P_y(t,i_1,j_1)-P_y(t,i_1,j_1))^2}$

Roughly speaking, this means that the wheel rotates rigidly.

5. There is a time t and a point on the wheel (i,j) such that $\frac{d\vec{P}(t,i,j)}{dt}$ is non-zero.

This means that the wheel can't just sit there motionless forever.

If I am not clumsily mistaken, it should be immediately clear that P(t,i,j) = (rt+icos(t)+jsin(t),r+jsin(t)-icos(t)) is such a function.

Anyone who has taken a course in linear algebra and/or topology could probably pose this a heck of a lot more compactly and elegantly. But I've never taken formal courses in either.

 

[Baluncore]

Friction is a function of FORCE, not a function of contact AREA or pressure.
That is one fundamental misunderstanding of friction being displayed in this thread.

Friction is fundamentally a surface chemistry bonding effect between dynamically sliding surfaces. It is the chemistry, not the shape of the surfaces that is critical. Friction exists when any two atoms pass close enough for there to be a temporary chemical bond. There is no requirement that there be dynamic deformation for there to be friction.

“Static friction” or the term “stiction”, describes the state of topological lock and chemical bond between static surfaces. There is no relative sliding movement, so there is no force acting over any distance of movement, so there is no frictional energy dissipation. Adhesion between contact surfaces is broken when the static friction threshold is overcome, when the system becomes dynamic.

“Sliding friction” takes over once relative movement begins, the real rules of friction only come into play when energy is dissipated at a sliding contact. If there is no relative movement there is no energy dissipative sliding friction.

Squaring the circle comes to mind, either this is mathematics or engineering physics, is it a circle or a wheel? The fields of study can never meet since the gaps between the atoms forming weak bonds between materials prevent the mathematical continuous movement of a point.

It seems this thread is confusing the adhesion of stiction with the concept of friction between sliding surfaces.
Statements of the form "Either there is friction or the wheel will spin" demonstrates the absolute misuse of terms.

 

[andrewkirk]

@jbriggs444 That's the sort of thing I had in mind as a way of formalising a description of the rotational motion of a rigid body in terms of all the points - effectively regarding it as an uncountable collection of point particles, all maintaining fixed distances from one another.

If I am not clumsily mistaken, it should be immediately clear that P(t,i,j) = (rt+icos(t)+jsin(t),r+jsin(t)-icos(t)) is such a function.

I don't think you're mistaken, but it's not immediately clear to me. However it looks promising, so I'll try to perform the service of working through the calcs to show this is the case.
But, being inherently lazy, I'm going to change to local polar coordinates because I think that will be easier. Let the time-$t$ location of a point on the wheel have polar coordinates $r(t),\theta(t)$ in a coordinate system whose origin is the axle and whose x-axis is parallel to the floor. The frame for those polar coordinates moves as the axle does.

We also use a global, stationary frame of reference, with Cartesian coordinates, to describe the absolute location of a point on the wheel. Let the origin be the initial axle position and the x-axis be parallel to the floor.

Each point particle $P$ in the wheel has a locus that is a function $f_P$ from $\mathbb R$ to $\mathbb R^2$ giving the points global coordinates at each time. Let's define functions $r_P,\theta_P,x_p,y_P$ that give, respectively, the local polar radial coordinate, local polar angular coordinate, global Cartesian $x$ coordinate and global Cartesian $y$ coordinate of the particle in terms of time.

To keep things simple, consider a wheel that is translating horizontally at a constant speed $v$, and at the same time rotating around its axle with angular velocity $\omega$. It's straightforward to extend this to non-constant motion, but we'd need to introduce integrals, which would be a bit messy.

Applying the definitions of rotation and translation, we see that the locus functions are:

\begin{align*}
r_P(t)&=r_P(0)\\
\theta_P(t)&=\theta_P(0) + \omega t\\
x_P(t) &= vt + r_P(t) \cos \theta_P(t)= vt + r_P(0) \cos (\theta_P(0)+\omega t)\\
y_P(t) &= r_P(t) \sin \theta_P(t)= r_P(0) \sin (\theta_P(0)+\omega t)
\end{align*}

Then the time-$t$ distance between two particles $P$ and $Q$ will be
$$\sqrt{(x_P(t)-x_Q(t))^2 + (y_P(t)-y_Q(t))^2}$$

This will remain constant iff its square remains constant, which means its square has a zero time derivative. The time derivative of the square is:

\begin{align*}
\frac d{dt} [ (x_P(t)&-x_Q(t))^2 + (y_P(t)-y_Q(t))^2 ]
=
\frac d{dt} [ (vt + r_P(0) \cos (\theta_P(0)+\omega t)) - (vt + r_Q(0) \cos (\theta_Q(0)+\omega t)))^2 \\
&\quad\quad\quad\quad+ ((r_P(0) \sin (\theta_P(0)+\omega t)) - (r_Q(0) \sin (\theta_Q(0)+\omega t)))^2 ]\\
&=
\frac d{dt} [ (r_P(0) \cos (\theta_P(0)+\omega t) - r_Q(0) \cos (\theta_Q(0)+\omega t)))^2 \\
&\quad\quad\quad\quad+ ((r_P(0) \sin (\theta_P(0)+\omega t)) - (r_Q(0) \sin (\theta_Q(0)+\omega t)))^2 ] \\
&=
\frac d{dt} [ (r_P(0)^2 +r_Q(0)^2
-2 r_P(0) r_Q(0) \cos (\theta_P(0)+\omega t) \cos (\theta_Q(0)+\omega t)\\
&\quad\quad\quad\quad-2 r_P(0) r_Q(0) \sin (\theta_P(0)+\omega t) \sin (\theta_Q(0)+\omega t)\\
&=
-2 r_P(0) r_Q(0)
\frac d{dt} (
(\cos \theta_P(0) \cos \omega t - \sin \theta_P(0) \sin \omega t)
(\cos \theta_Q(0) \cos \omega t - \sin \theta_Q(0) \sin \omega t)\\
&\quad\quad\quad\quad+
(\sin \theta_P(0) \cos \omega t - \cos \theta_P(0) \sin \omega t)
(\sin \theta_Q(0) \cos \omega t - \cos \theta_Q(0) \sin \omega t)
)\\
&=
-2 r_P(0) r_Q(0)
\frac d{dt} (
\sin \theta_P(0) \sin \theta_Q(0) + \cos \theta_P(0) \cos \theta_Q(0))\\
&=0
\end{align*}

So the shape does indeed remain rigid.

To forbid slipping, we set $v=\omega R$ where $R$ is the wheel radius. That makes the instantaneous speed of the lowest point of the wheel in the global Cartesian frame always zero.

On reflection, what this proves is not something about physics, but rather the mathematical fact that rotation and translation are isometries - ie transformations that, when applied to a set of points, preserve the shape and size of that set.

To incorporate physics into the analysis, we need to introduce one or more forces and - using the rigidity as a set of constraint forces - apply d'Alembert's principle (I think). That's beyond the scope of this post, which is already too long.

 

[A.T.]

Friction is a function of FORCE, not a function of contact AREA or pressure.

But a non zero friction force acting over zero area implies infinite shear stress and pressure, which is unphysical.

Friction exists when any two atoms pass close enough for there to be a temporary chemical bond.

When you model it as a bunch of atoms interacting over small distances via position dependent forces, then it's not a prefect rigid circle anymore.

There is no requirement that there be dynamic deformation for there to be friction.

If a road atom exerts a force on the outermost tire atom, then that tire atom will change it's position relative to the other tire atoms. That is deformation (strongly simplifed).

 

[Baluncore]

But a non zero friction force acting over zero area implies infinite shear stress and pressure, which is unphysical.

There cannot be a non-zero friction force if there is zero contact. You are neglecting the fact that the frictional force acts parallel with, not perpendicular to the surfaces in contact.

When you model it as a bunch of atoms interacting over small distances via position dependent forces, then it's not a prefect rigid circle anymore.

I agree, it is not a mathematical circle, it is then a real wheel. Never the twain shall meet.

If a road atom exerts a force on the outermost tire atom, then that tire atom will change it's position relative to the other tire atoms. That is deformation (strongly simplifed).

Again your thinking is perpendicular to the real surface of contact. The frictional force will act parallel with the surface, it will pull atoms sideways in the plane surface which will not significantly change the flatness of the road.

The title of this thread is “A perfectly stiff wheel cannot roll on a stiff floor?”. If it is perfectly stiff then deflection is impossible and you are discussing the mathematics of a circle rolling on a line, or a cylinder rolling on a plane. The title precludes reality from discussion by invoking an impossible contradiction. If the title was “A perfect circle cannot roll on a line?” then the question would be mathematical and tractable.

The engineering physics of a wheel rolling on a surface acknowledges that rolling resistance will consume energy. Rolling resistance is not friction. The weak chemical bonds that form when the wheel contacts the road must be later broken for the wheel to roll forwards. That is an inefficient process and requires energy. During the period in contact the surfaces adhere, they are not sliding, friction is not involved.

Why do you need to confuse adhesion with friction?
Why must the irrelevant perpendicular deflection of the surfaces be considered?

 

[A.T.]

...you are discussing the mathematics of a circle rolling on a line...

This is what the thread starter is discussing.

 

[Robert Morphis]

The engineering physics of a wheel rolling on a surface acknowledges that rolling resistance will consume energy. Rolling resistance is not friction. The weak chemical bonds that form when the wheel contacts the road must be later broken for the wheel to roll forwards. That is an inefficient process and requires energy. During the period in contact the surfaces adhere, they are not sliding, friction is not involved.

Rolling resistance is mostly the deformation of the wheel / surface, not to mention the friction in the axle.

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.

That is one valid method of analyzing the situation. Another is to consider it as a combination of translational and rotational motion.

In the referenced article he starts with an arbitrary definition about rotation and the application of force, then he "proves" his point by applying force to the center of the spool while pretending to apply it elsewhere by using the pvc pipe.

My theory (speculation, rather) is that, without that deformation, the commencement of rolling motion would be impossible.
[...]
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.
EDIT 2: 18 Nov 2017: There are now diagrams of what this is talking about, in this post.

Even for a "perfect" wheel and surface while the physical contact is a line of length equal to the width of the wheel the chemical bonding Baluncore refers to above does not take place just at that line, but also at some (very small) distance from that line.

Friction will occur, the wheel will rotate, and, even if it is pushed hard enough so that it starts both sliding and rotating, eventually the wheel will roll.