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

[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.

 

[PeroK]

It occurs to me that if a perfectly rigid wheel cannot roll, then it cannot do anything.

The same practical arguments relating to molecular interactions etc. apply equally to its internal structure and make its existence impossible.

As soon as we say "perfectly rigid wheel" we are in the realm of an idealised mathematical model. Any talk of atoms is absurd in this model, for many reasons.

It's a bit like advancing practical reasons why a point particle, say, cannot move in a perfect circle.

 

[gr71cj5]

Envision: High tinsel steel ball bearing, High tinsel steel plate, no deformation of either will occur, however, apply any force to the bearing either directly (even with a simple magnet to avoid physical contact) or by tilting the metal plate (imposing gravity) and it will roll quite well. No deformation of either surface is needed. In fact, any deformation of either surface simply impedes the motion even if only by absorbing the energy required to deform the surface.

 

[gr71cj5]

Envision: High tinsel steel ball bearing, High tinsel steel plate, no deformation of either will occur, however, apply any force to the bearing either directly (even with a simple magnet to avoid physical contact) or by tilting the metal plate (imposing gravity) and it will roll quite well. No deformation of either surface is needed. In fact, any deformation of either surface simply impedes the motion even if only by absorbing the energy required to deform the surface.

 

[jbriggs444]

Envision: High tinsel steel ball bearing, High tinsel steel plate, no deformation of either will occur, however, apply any force to the bearing either directly (even with a simple magnet to avoid physical contact) or by tilting the metal plate (imposing gravity) and it will roll quite well. No deformation of either surface is needed. In fact, any deformation of either surface simply impedes the motion even if only by absorbing the energy required to deform the surface.

It is not enough to have high tensile strength. If you want a perfectly rigid wheel with an infinitesimal contact patch, you need infinite tensile strength. That is impossible to find in nature.

[Tinsel is the stuff you put on Christmas trees]

 

[rcgldr]

There's no reason that a perfectly rigid wheel on a perfectly rigid surface could not "roll", even with zero friction, if the wheel just happens to be rotating so that it's outer surface matches the speed that it's axis is translating with respect to the surface. Then again, I don't see an issue with the concept of having infinite friction to go along with a perfectly rigid wheel and surface. What if the wheel and the surface were geared (for perfectly smooth motion) to result in the equivalent of infinite friction?

 

[jmolmo]

A little "offtopic", but the question is "Will it roll or not?"

 

[DaveC426913]

Cool.

No, they will not roll.
The thread, the radii and the two touch-points of the wheels together form a trapezoid.
The first thing that rolling would theoretically result in is the lengthening of the longest side (the line between the two touch-points).
That trapezoid cannot be deformed and still maintain symmetry.
Not so sure now.

 

[haruspex]

Cool.

No, they will not roll.
The thread, the radii and the two touch-points of the wheels together form a trapezoid.
The first thing that rolling would theoretically result in is the lengthening of the longest side (the line between the two touch-points).
That trapezoid cannot be deformed and still maintain symmetry.

Well, it does depend on the relationship between the ratio of the radii and the angle of the slope. What if the slope increases to 60 degrees?

 

[kuruman]

Well, it does depend on the relationship between the ratio of the radii and the angle of the slope. What if the slope increases to 60 degrees?

To elaborate, static equilibrium is maintained when the torque due to gravity and the torque due to the tension in the string are in opposite directions. The former is non-zero as long as $\theta < 90^o$. The latter becomes zero when the radius of the inner wheel is large enough so that the position vector from the contact point to the point of application of the tension is horizontal. This happens when $r/R=\cos\theta$. For $r/R>\cos\theta$ the two torques are in the same direction and there can be no equilibrium. Note: Torques are calculated using the point of contact of the wheel with the incline as origin.

 

[Baluncore]

The distance between circle centres is the length of tight string.
It is not the change in height, Sin(30°), but the change in horizontal separation, Cos(30°) that is important.

They will not roll because R/2R = 0.5 string released, is less than Cos(30°) = 0.886 string required between centres.
Roll becomes possible when Cos(slope) = R/2R = 0.5 which will be when the slope becomes > 60°.

 

[A.T.]

Cool.

No, they will not roll.
The thread, the radii and the two touch-points of the wheels together form a trapezoid.
The first thing that rolling would theoretically result in is the lengthening of the longest side (the line between the two touch-points).
That trapezoid cannot be deformed and still maintain symmetry.
Not so sure now.

View attachment 223687

It boils down to whether to string goes above or below the contact points.

 

[A.T.]

Cool.

Can this spool roll down the incline without slipping, as the problem statement suggests?

Found here:
http://www.chegg.com/homework-help/questions-and-answers/spool-move-spool-rests-top-incline-made-two-uniform-disks-radius-r-mass-m-connected-horizo-q24458961

 

[Baluncore]

Can this spool roll down the incline without slipping, as the problem statement suggests?

No. But that is not a genuine solvable problem. It is an advertisement for “advanced physics tutors”. It is designed to disempower and confuse students to the point where they subscribe to a study program. Now being advertised on PF.

 

[jmolmo]

Can this spool roll down the incline without slipping, as the problem statement suggests?

View attachment 223703

Found here:
http://www.chegg.com/homework-help/questions-and-answers/spool-move-spool-rests-top-incline-made-two-uniform-disks-radius-r-mass-m-connected-horizo-q24458961

But, if I pull the string upwards...? Will it roll?

 

[jmolmo]

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.
View attachment 215189 View attachment 215190 View attachment 215191

You're right ... In a rigid body, you can't consider 2 simultaneous "centers of rotation" at the same time (so, a point from which all the other body points motion can be described as circles around that "point")... such body would be no rigid.

The only way you can describe the movement of the points of a wheel if the geometrical lowest point of the wheel is moving along a line, is as the composition of one translation movement that follows that "line", and one rotational movement around the center of the wheel. So, or that "center of rotation" is constantly flipping between the center of the wheel and the geometrical contact point (but never both at the same time), or there's no point in saying that the contact point has been the center of rotation at any time, even if an infinitesimal time is considered.

 

[Baluncore]

But, if I pull the string upwards...? Will it roll?

Define upwards, and what you mean by pull. Consuming string at the fixed anchor point will move the spool up the ramp. Feeding string out from the fixed anchor point will allow the spool to roll. Raising the anchor point to change the angle between the string and the slope will make a big difference at some point.

 

[jmolmo]

Define upwards, and what you mean by pull. Consuming string at the fixed anchor point will move the spool up the ramp. Feeding string out from the fixed anchor point will allow the spool to roll. Raising the anchor point to change the angle between the string and the slope will make a big difference at some point.

If it's an string, I think it's clear what I mean by pull. If you mean how much pull, then image you do very, very little at first (0.01N) ... and then more and more.