Rolling: is the instantaneous axis of rotation tangential?

[tiny-tim]

the instantaneous axis of rotation of a body rolling on a stationary surface must pass through the instantaneous point of contact (C)

it seems obvious that the axis must be tangential at C (to both surfaces), and perpendicular to the instantaneous line of motion of the point of contact

but i can't see how to prove it

any suggestions?

(both surfaces must be continuously differentiable, and they can curve the same way, like an egg rollling inside another egg, or opposite ways, or even have saddle points)

 

[klackity]

I don't think this is true! (Unless you are thinking of a more specific type of case than I am imagining).

Look at this example:

The ideas is to have an elliptical egg (an ellipse rotated about its major axis to form an egg) rolling in a U-shaped channel. The image is of a cross section.

In this case, it seems like the instantaneous axis of rotation is almost orthogonal to the tangent of both surfaces at the point(s) of contact.

It's true that as the surfaces are differentiable, they must be tangent to each-other at the point of contact. But I don't see why the axis of rotation would have to be tangent to both the surfaces.

Here's an example with only one point of contact:

Here's what the picture is trying to represent. A second prior to this drawing, the momentum of the egg was causing it to roll up this wall. It had been rolling "forward" (i.e., into the screen) and upward (along the wall). At this exact moment, however, the rolling upward along the wall has almost stopped (due to gravity), while the rolling forward is continuing. This forward rolling (into the screen) makes the axis of rotation not tangent to the surface.

Or rather, maybe I am using to general a notion of rolling? My notion also allows some spinning (but no sliding). Am I wrong to allow this? Perhaps this makes the "instantaneous axis of rotation" ill-defined? Or perhaps spinning and rolling are linearly independent?

 

[klackity]

I guess in the egg-in-a-channel example, the egg can certainly roll down the channel without sliding. But it cannot possibly roll along an axis tangent to either of the points of contact (because that would imply the other point of contact is sliding).

It seems like this kind of thing could also happen at certain instants as an egg rolls along a wobbly surface.

However, in the egg-in-a-channel example, the line formed by the two points of contact cannot be orthogonal to the surfaces at the point of contacts (or else the egg could not roll without sliding).

That seems to suggest that these notions of rolling and spinning that I'm imaging might actually be able to be separated. In other words, maybe the axis of "spin" is necessarily orthogonal to the surface at the point of contact, and the axis of "roll" is necessarily tangent to the surface at the point of contact, and these two determine an apparent axis. (The "apparent axis" would be the set of points which (if they were part of the egg) would have first derivative 0 at that instant)

I feel like you probably know this stuff better than I do, though...

 

[klackity]

Ah! I think this explains also why a spinning top stays upright!

We imagine the bottom of the spinning top is a differentiable surface. Suppose the spinning top is actually a sphere.

The set of points which will have first derivative zero is determined by two vectors. First, a spin vector, orthogonal to the surface at the point of contact with the ground. Second, a roll vector, tangent to the surface at the point of contact with the ground.

The magnitude and sign of the spin vector determines the angular speed and direction of the spin of the sphere at the point of contact. The magnitude and direction of the roll vector (using some kind of right-hand rule) determines the angular speed of roll at the point of contact, and the axis of roll.

The greater the ratio of the magnitude of the spin vector to the roll vector, the more vertical the apparent axis is.

The roll and spin vectors at the point of contact uniquely determine the first derivative of all the other points in the non-sliding object.

Hmm.. I haven't taken physics since high school, so this stuff is still cool for me!

 

[klackity]

OK, here's what I'm thinking formally.

We define the instantaneous axis of rotation as follows.

First consider two coordinate systems - one which which is rigid with respect to the rolling/spinning object, and one which is rigid with respect to the surface.

Let O be the coordinate system in which the coordinates of a point of the rolling object are always constant. Let S be the coordinate system in which the coordinates of particular point of the surface are always constant.

O and S are two 3 dimensional vector spaces describing the same physical space, but where the origins and basis directions of both are not held fixed relative to one another. (So every point in physical space has coordinates with respect to both O and S at any given time, but the transformation between O and S coordinates is not fixed in time)

Now, Let T_t be the coordinate transformation from O to S. T can be decomposed as the action of a 3x3 orthonormal orientation-preserving matrix M_t followed a translation N_t (which takes the the origin in S coordinates to the origin in O coordinates).

Then the instantaneous axis of rotation at time t is defined to be set of points x where the derivative (with respect to time) of T_t(x) is zero.

As the object and the surface are not allowed to slide, we know there exists at least one point c (the point of contact of the surface and the object) for which d/dt T_t(c) = 0. This means that either d/dt T_t is everywhere zero, or d/dt T_t is a rotation about some line (not necessarily passing through the origin of either coordinate system).

We can see from the rolling egg example that the instantaneous axis of rotation, as I just defined, is not necessarily tangent to the surface at the point of contact. If it is tangent, however, then the object is not spinning. It it is orthogonal to the surface, then the object is not rolling (and only spinning).

EDIT: An example of such a transformation T would be as follows:

We will restrict our case two dimensions. The surface will be the x-axis and S will be the usual coordinates in R² (with origin at (0,0)).

The object will be a sphere of radius 1 resting on top of the x-axis. The O coordinates will take the center of the sphere (which is (0,1) at time t=0) to be the origin. The basis vectors at t=0 will be in the same direction as those of S.

So T₀(x) = M₀(x) + N₀ = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}(x) + \begin{bmatrix}0 \\ 1\end{bmatrix}

We will assume that the circle is rotating clockwise at a rate of 1 radian per second. Thus the matrix M_t = \begin{bmatrix} cost & sint \\ -sint & cost\end{bmatrix}.

For this to happen without the circle sliding, the center of the circle must have moved to the right a distance of 2pi in the time between t=0 and t=2pi seconds.

Thus N_t = \begin{bmatrix}t \\ 1\end{bmatrix}.

If our math is right, then d/dt T_t(x) ought to equal 0 for the point of contact, and for no other points.

We see that d/dt T_t(x) = d/dt M_t(x) + d/dt N_t

= d/dt \begin{bmatrix} cost & sint \\ -sint & cost\end{bmatrix}(x) + d/dt \begin{bmatrix}t \\ 1\end{bmatrix}

= \begin{bmatrix} -sint & cost \\ -cost & -sint\end{bmatrix}(x) + \begin{bmatrix}1 \\ 0\end{bmatrix}.

The point of contact at any time t in O coordinates is the point \begin{bmatrix} cos(t-pi/2) \\ sin(t-pi/2)\end{bmatrix} (the point directly under the center of the circle, which will be moving counter-clockwise in O coordinates as the circle rolls clockwise). Note, we can also write: \begin{bmatrix} cos(t-pi/2) \\ sin(t-pi/2)\end{bmatrix} = \begin{bmatrix} sint \\ -cost\end{bmatrix}.

But then we can verify:

\begin{bmatrix} -sint & cost \\ -cost & -sint\end{bmatrix}\begin{bmatrix}sint \\ -cost\end{bmatrix} + \begin{bmatrix}1 \\ 0\end{bmatrix}

= \begin{bmatrix}-sin(t)sin(t) - cos(t)cos(t) \\ -cos(t)sin(t) + sin(t)cos(t)\end{bmatrix} + \begin{bmatrix}1 \\ 0\end{bmatrix}

= \begin{bmatrix}-1 \\ 0\end{bmatrix} + \begin{bmatrix}1 \\ 0\end{bmatrix}

= \begin{bmatrix}0 \\ 0\end{bmatrix}

Thus the point of contact does in fact have this property, and we see our math was correct.

 

[tiny-tim]

hi klackity! thanks for replying!

I don't think this is true! … I don't see why the axis of rotation would have to be tangent to both the surfaces.
…
… maybe I am using to general a notion of rolling? My notion also allows some spinning (but no sliding). Am I wrong to allow this? Perhaps this makes the "instantaneous axis of rotation" ill-defined? Or perhaps spinning and rolling are linearly independent?

… in the egg-in-a-channel example, the line formed by the two points of contact cannot be orthogonal to the surfaces at the point of contacts (or else the egg could not roll without sliding).

That seems to suggest that these notions of rolling and spinning that I'm imaging might actually be able to be separated. In other words, maybe the axis of "spin" is necessarily orthogonal to the surface at the point of contact, and the axis of "roll" is necessarily tangent to the surface at the point of contact, and these two determine an apparent axis. (The "apparent axis" would be the set of points which (if they were part of the egg) would have first derivative 0 at that instant)

I feel like you probably know this stuff better than I do, though...

no, I've never looked into it before: i don't think very well in 3D, so i needed assistance!

i like your egg-in-a-channel example … it makes it very clear that my original ("obvious") idea was wrong!

also your separation into rolling and spinning

it would be nice to be able to define rolling to exclude spinning, but then your egg-in-a-channel would not be rolling, which i think is contrary to common english usage

so i think the categorisation has to be into rolling-without-spinning and general rolling

rolling-without-spinning would be defined as, in the instantaneous inertial frame of reference in which the point of contact is stationary (necessary to specify if both surfaces are rotating ), there is no relative rotation about the normal direction

in particular, in rolling-without-spinning there is only static friction (and of course rolling resistance from deformation of the surface), but if there is spinning, then there will be some kinetic friction (and loss of energy) …

for example the egg-in-a-channel (unintuitively) must experience some kinetic friction

 

[klackity]

Hmm... maybe we should consider for now just the case of a flat surface.

If the rolling object is a sphere, and there is only rolling-without-spinning, I would initially think that the direction of the axis of rotation is constant (no matter how we tilt or slant the flat surface).

Let's look at an example where the flat surface is tilted. In this case, the edge of the surface furthest away from us has been picked up vertically off the ground:

If we were to roll a ball straight forward, it would eventually lose momentum, and then roll straight back down. The axis of rotation would never change direction (except possibly flipping in the opposite direction, if you want the axis of rotation to be a vector and not just a line).

If we roll it up at an angle, though, it seems like it would follow a parabolic path (I'm not sure if it will be exactly parabolic, though). The red arrow is the velocity of the point of contact. We expect the axis of rotation to be perpendicular to the velocity vector (or at least roughly so).

My hunch is that for the ball to follow this parabolic path, there is some spinning component induced in the ball (similar to the spinning component induced in the egg as is rolls down the channel). So the question is: is the ball also picking up some spin along the way, as the axis of rotation changes direction?

If the answer is yes, then it may not be natural to consider spin-less rolling. If the answer is no, then that suggests that spin-less rolling is a natural thing to consider (and then it makes sense to ask whether it occurs for non-flat surfaces too).

I just did a little experiment with the materials I had at my disposal, and it seems like the answer is no: it appears as though the ball doesn't ever spin relative to the surface.

[Actually, it "appears" as though the ball is spinning clockwise (opposite of what I would expect) towards the top of the parabola. This is just an optical illusion, though. As the ball rolls, the markings on it blur into circles, and these apparent circles appear to abruptly change direction near the top of the parabola when the ball stops and starts again. But the ball itself is not spinning.]

I messed around with it a bit too, tilting and jerking around this bit of cardboard and trying to induce the ball to spin. I couldn't do it. So I think you must be right that rolling-without-spinning is a very natural case to consider (and not just a special circumstance).

Even, for example, if the surface is a funnel, I am convinced now that although a ball rolling in it will be driven into tighter and tighter circles, it is only ever rolling-without-spinning.

_____________________________________________

Ah! I think I have another example. Have you ever spun a quarter? Eventually it falls over, and starts to wobble around on its edge. At anyone time, only one point is touching the table. And the point of contact is moving around the edge of the quarter. If there is no spinning involved at this time, then I think we can say that each point on the perimeter of the quarter is just moving up and down over the same spot on the table (there is no spinning or sliding involved).

I think that this fits the definition of rolling-without-spinning.

The velocity vector of the point of contact always lies tangent to the table (as the path of the point of contact will always lie on the table surface).

Now here's an interesting fact. The center point of the quarter (George Washington's nose, maybe) will remain perfectly fixed as this quarter rolls along its edge. Thus the first derivative of its position is always 0.

We also know that the first derivative of the position of the point of contact is always zero (for any situation in which there isn't sliding). Thus the points of the quarter which have first derivative zero are those which lie on the line formed by the center of the quarter and the point of contact.

But the center of the quarter doesn't lie on the table, so the axis of rotation can't lie on the table surface.

EDIT: Here's is my hypothesis. Let c(t) be the position of the point of contact at time t (i.e., the path the object would make on the surface if it were covered in wet paint). Let a(t) be the axis of rotation (in vector form) at time t. (Note: a(t) is attached to the point c(t)). Your prediction is that a(t) will always lie tangent to the surface, if there is no spin. I predict, instead, that a(t) will always be orthogonal to c'(t) (= dc/dt) where both are considered vectors in 3-space attached to the point of contact c(t).

I think we both can agree that c'(t) will lie tangent to the surface. But I do not think that a(t) need be tangent to the surface (even if there is no spin). But actually, I think that both our hypotheses are wrong. Here's what I think: Even if there is no spin, none of the following always hold:
(i) a(t) will always be orthogonal to the c'(t)
(ii) a(t) will always be tangent to the surface.

Counter-example to (i). Imagine we are rolling a giant screw-threaded log. If the spiral screw threading wraps around the log 10 times, then there will be 10 points touching the flat plane. Pick one, and let's follow it while we roll the log. As we roll the log straight forward, the point we chose will go diagonally (forward and to the left, maybe). Thus c(t) is a diagonal line, and so c'(t) is a diagonal vector. But clearly the axis of rotation a(t) is perfectly horizontal - left to right, along the length of the log. By changing the slant of the threading of the log, we can change c'(t) without affecting a(t) at all, even though they both lie tangent to the surface.

And the quarter example seems to be a counter-example to (ii).

So really, I'm not sure what to believe...

 

[klackity]

Here is the coin example:

The red side is the underside of the coin. The blue side is on top. The orange dot is the center of the coin, which happens to remain fixed.

The point of contact is moving counterclockwise along the surface.

However, the coin itself is not spinning. The points of the coin are really just moving up and down as the coin rolls around its edge.

As the axis of rotation contains the orange dot, and the point of contact, it cannot lie on the table.

Here is the threaded log example:

The log is rolling straight forward. However, the chosen point of contact (marked as an orange dot) traces out a diagonal line.

But the axis of rotation must contain all the points of contact, and thus must be parallel to the log (and must lie tangent to the surface).

This implies that c'(t) is not orthogonal to the axis of rotation a(t).

The question is, are these examples subtly wrong? I would have thought that either (i) or (ii) above would have been true, but if these examples are right, then neither is...