What Makes a Sliding Book Stop Both Rotating and Translating Simultaneously?

[sophiecentaur]

"Cycloid" is correct, I think. There is a whole family of cycloids - with and without loops. I looked it up yesterday, aamof.

The above reads quite convincingly and I think the idea of treating all motion as motion makes sense. I've been rather hoping for a null hypothesis argument and your idea could potentially do that. I think a sufficient argument has to clear up the significance of the rigidity of the book.

 

[MikeGomez]

I think a sufficient argument has to clear up the significance of the rigidity of the book.

Agreed. After I wrote that post I was thinking the same thing. I used atoms as an example of the the components which make up the rigid body to point out that they have a single instantanious momentum, but that could still be a little confusing because people tend think of atoms as jiggling around..

I suppose we could use a similar model that breaks down the rigid body into smaller pieces , but not as small as an atom. Maybe clumps of a few hundred atoms or so, but I'm not really sure if that that helps or introduces another confusion factor.

 

[sophiecentaur]

I imagine the way to start would be with two elements.

 

[Dadface]

Following on from post 33 by MikeGomez and post 36 by sophiecentaur imagine the book is a "perfectly" rigid structure. If so if anyone finite part of that structure stopped moving (moving in a straight line, rotating, vibrating or any combination of movements) the rigidity of the structure would ensure that all parts of the book would stop moving in a way that would appear, by observation, to be instantaneous.
The book is not perfectly rigid but is a close approximation to it and ordinary by eye only observations can make it appear that all movements stop at the same time. If there are deviations to this they may be shown up by high speed photography techniques.

 

[D H]

It might help to look at this problem from the perspective of the instantaneous center of rotation. I've just started playing with this mess, so I don't yet know if it's a viable approach; hence the "it might help".

If there exists finite positive bounds d_min and d_max on the distance d between the center of mass and the instantaneous center of rotation is bounded (d_min<d<d_max) while the object is moving, then the object will stop rotating and translating simultaneously. It's a bit trickier if d→0 as v→0 or if d→∞ as ω→0.

 

[bahamagreen]

Following on from post 33 by MikeGomez and post 36 by sophiecentaur imagine the book is a "perfectly" rigid structure. If so if anyone finite part of that structure stopped moving (moving in a straight line, rotating, vibrating or any combination of movements) the rigidity of the structure would ensure that all parts of the book would stop moving in a way that would appear, by observation, to be instantaneous.
The book is not perfectly rigid but is a close approximation to it and ordinary by eye only observations can make it appear that all movements stop at the same time. If there are deviations to this they may be shown up by high speed photography techniques.

Clearly if any single point of the book "...stopped moving (moving in a straight line, rotating, vibrating or any combination of movements)...", then both rotation and translation have stopped, but not necessarily at the same time.

The whole question is whether the book can stop spinning and still translate, or vice versa, before all movement stops.

To look for a reason that spinning and translation stop together leads me to look for something that both motions share in common, and something that is variable wrt both, and something that is differentiable between them to account for the wide range of initial conditions that end in the same result of coincident stopping... the shared friction seems to be that thing.

From a frictional standpoint, greater movement decreases instantaneous friction, right?

For the same translational initial condition, won't an imparted spin increase the distance before stopping? Won't higher initial spin result in longer duration of both trans and spin before stopping?

Likewise, for the same initial spin condition, won't an imparted translation increase the duration of spin? Won't a greater initial trans result in longer duration of both tans and spin?

If the situation was frictionless, both translation and spin would last indefinitely.
If the friction was arbitrarily very high (a magnetic book on a steel plate) both translation and spin duration would approach zero.

The real question here to me is do the translation and spin durations diverge in the middle range of friction?
And if not, what is the relationship between the trans and spin frictions that allows for different initial conditions of combinations of trans and spin magnitudes converge to a common duration for both?

 

[Dadface]

Clearly if any single point of the book "...stopped moving (moving in a straight line, rotating, vibrating or any combination of movements)...", then both rotation and translation have stopped, but not necessarily at the same time.

Possibly so for an imaginary point of infinitesimally small size but I was referring to a point of "finite" size eg a small array of atoms/particles.

 

[D H]

Clearly if any single point of the book "...stopped moving (moving in a straight line, rotating, vibrating or any combination of movements)...", then both rotation and translation have stopped, but not necessarily at the same time.

Not true!

Assume the book has non-zero angular velocity. This means the book's instantaneous center of rotation is well-defined. The book has no translational motion if this center of rotation is at the book's center of mass. The book is translating and rotating if the center of rotation is displaced from the center of mass. The book will have a single point that has stopped moving if this displacement leaves the center of rotation still within the bounds of the book.

Think of it along the lines of the concept of an ideal wheel that is rolling without slipping. The velocity of a point on the wheel with respect to the ground is non-zero except one very special set of locations: The set of points on the wheel in contact with the ground.

 

[bahamagreen]

Not clear to me what part "Not true!" is meant to pertain.

A pivot point on the book that is stationary wrt the floor but still the center of rotation is still in rotary motion, right?

Are you saying that friction and/or decelerration is what displaces the COR from the COM? I think I agree.

I'm not sure how "The book will have a single point that has stopped moving if this displacement leaves the center of rotation..." is true if the book is still spinning because that single point may be trans-0 but still rotating.

Likewise with the wheel, the contact point with the ground remains in rotation... If I understand, it sounds like you are considering a non-translating but rotating point to not be moving. Are you defining angular movement as a movement but static point rotation (like a non-translating pivot point) as not a movement?

 

[MikeGomez]

To look for a reason that spinning and translation stop together

No. Try to forget about keeping these separate. Once the book is launched, it simply has motion. Every point that the body is composed of, has one and only one instantanious momenutum vector. That's the trick question part of the trick question.

leads me to look for something that both motions share in common, and something that is variable wrt both, and something that is differentiable between them to account for the wide range of initial conditions that end in the same result of coincident stopping...

What every point shares in common is that they all follow well defined paths. In a pure translating solid body all points have straight line paths. with rotating bodies all points have circular paths. The points of a solid body with a combination of both have well defind paths called cycloids.

Please examine the effect that a single point of solid body which follows a cycloid path, has upon its neighboring points which follow nearly identical (but also well defined) cycloid paths.

 

[MikeGomez]

The velocity of a point on the wheel with respect to the ground is non-zero except one very special set of locations: The set of points on the wheel in contact with the ground.

Perhaps there is an infinitesimally small point on the wheel whose velocity approaches zero, and that is indeed equivalent to saying that there is an infinitesimally small point at the center of a spinning body whose velocity approaches zero, even while the object as a whole remains spinning.

I’m not convinced that a real (physical) component of the rigid body (with dimensions greater than zero) actually stops moving. But even if so, that would be insignificant compared to the random vibration motion of real components such as atoms or electrons etc. At what level do we consider the micro-motions of the components of a rigid body as manifesting into the global motion of the rigid body of which they compose?

My belief is that once we can define the whole rigid body in an adequate way in terms of it’s components, then we find that the motion of a single component can not (by definition) be zero without the motion of the entire body being zero. That’s what it means to be a rigid body.

 

[bahamagreen]

To forget about keeping the spin and translation separate seems to miss the problem's point - to know why both types of motion cease at the same time. The question asks why these motions both continue to have magnitude until the book stops.

I agree that each point has an instantaneous momentum vector wrt the floor, I was the first to mention cycloid paths (I called them epicycles), but that momentum vector may have two components, one from the spin and one from the translation. When the left side of a CCW spinning book's speed of rotation wrt the book's center of rotation is instantaneously equal and opposite the book's center of rotation's translational speed forward. That means a point on that side can "stop" wrt the floor before the whole book stops sliding and spinning, dosn't it?

 

[A.T.]

It is supposed to be a 'qualitative' problem,

My qualitative answer:

For simplicity let's consider a sliding disc and assume sliding friction doesn't depend on speed.

Pure rotation:
- Net torque is maximal
- Net force is zero

Pure linear motion:
- Net torque is zero
- Net force is maximal

For the combination:
- Less rotation means less net torque opposing rotation
- Less linear motion means less net force opposing linear motion

 

[MikeGomez]

Just to make the discussion easier, let's say that an idealized rigid body is composed of component parts which always maintain a fixed distance from each other, and that we can refer to these component parts as ‘atoms’ without getting bogged down with details of how “real” atoms actually jiggle about.

I agree that each point has an instantaneous momentum vector wrt the floor, I was the first to mention cycloid paths (I called them epicycles)

Nicely done.

Just for reference here’s a link I found after a quick Google search which shows shapes of the three types of cycloids.

http://www.daviddarling.info/encyclopedia/C/cycloid.html

A point on the outer perimeter of a wheel rolling on the ground traces a path called a cycloid. A point farther from that edge traces out a path called a prolate cycloid, and a point closer to the center of the wheel traces out a path called curtate cycloid. Also, according to the Wikipedia article on cycloids, the term trochoid is used to refer to any of the three types of cycloids, so I suppose it’s best to make that correction and switch to using that terminology from here on.

Every point of a rigid body which is set in motion in a plane, traces out the path of a trochoid. Think of a wheel of an automobile. Every component part (atom) of the wheel is tracing out the path of a trochoid. As per the OP’s question, there is both translation motion (the car is moving forward) and rotation motion (the wheels are spinning). As the car slows down, each atom of the car slows down in velocity along the trochoid path that it traces. Now here is the important part. As the car comes to a complete stop, all atoms following their respective trochoid paths come to a complete stop at the exact same instant in time!

Now don’t misunderstand me. I’m not saying that the physics here is the same as the physics of the atoms of our book which are tracing out trochoid paths (although I’m not saying it isn’t, either). I’m just pointing out that this is a great example of a solid body composed of component parts which trace out trochoid paths and all tend towards zero velocity at the same time.

To forget about keeping the spin and translation separate seems to miss the problem's point - to know why both types of motion cease at the same time. The question asks why these motions both continue to have magnitude until the book stops.

I don't think I missed the problem's point, I think the problem's point makes false assumptions. But anyway, fair enough. We'll just go along with the books assumptions and speak as if rotational motion and translational motion are separate. So back to launching our book again, but this time let's first consider the physics if there were no friction or other outside influences. When we set the book in motion we give it some combination of translational motion and rotational motion, and that (due to its being a rigid body) sets the component atoms along various but well behaved trochoid paths. We could also set the initial motion as rotation only and no translation, or translation only with no rotation, and those two cases would also be valid because their paths (straight line and circular) are just special cases of the general trochoid.

Due to Newton’s first law, all the component atoms will maintain their trochoid paths indefinitely, due to that fact that at this point we still have no friction. So can we change or remove some or all of the translation motion without affecting the rotation motion, or vice versa?

Absolutely. We can apply a force to some chosen location on the book and the book will react according to Newton's 2nd and 3rd laws. The result will be to change its translation motion relative to its rotation motion.

And what is the fate of our ubiquitous trochoid paths that are traced out by the component atoms? They have all changed. Each and every atom now follows a new trochoid path, reflecting the new ratio of translational motion with respect to rotational motion.

Now we add friction, and ask the question: Can friction change or remove some or all of the translation motion without affecting the rotation motion, or vice versa?

The answer is no, and the reason is that friction acts upon each atom in a manner which is proportional to its instantaneous momentum vector. In other words, it slows each atom down in the direction that the atom traces, and in proportion to its velocity along its trochoid path. If there is no change to the shapes of the trochoid paths of the atoms, then there can be no change in the ratio of rotational motion with respect to the amount of translational motion.

 

[A.T.]

The answer is no, and the reason is that friction acts upon each atom in a manner which is proportional to its instantaneous momentum vector. In other words, it slows each atom down in the direction that the atom traces, and in proportion to its velocity along its trochoid path.

Is dynamic friction proportional to velocity?

 

[bahamagreen]

The answer is no, and the reason is that friction acts upon each atom in a manner which is proportional to its instantaneous momentum vector. In other words, it slows each atom down in the direction that the atom traces, and in proportion to its velocity along its trochoid path. If there is no change to the shapes of the trochoid paths of the atoms, then there can be no change in the ratio of rotational motion with respect to the amount of translational motion.

I like the approach "...no change in the ratio of rotational motion with respect to the amount of translational motion."

A constant ratio of trans component vector length wrt the floor to rotation component vector length wrt the center of rotation would get the end result of coincident stopping.

From what you also mentioned, it sounds like an observable indication of this would be the preservation of the cyclic aspects... as the spinning and sliding book slows down the cyclic paths are just "drawn" increasingly more slowly, but the actual goemetric form and features of the cycloids' paths ("wavelength") as viewed wrt the floor would be constant as well.

I am beginning to wonder if the role friction plays in this is to actually oppose a curving path... the right and left halves' opposed directions of spin both encounter the same direction of friction, yet there is no curving of the path. But from the cycloid perspective, there may be a balanced compensation between the two sides (cycloid path points' net speed against the floor vs frictional velocity)...

A.T. asked, "Is dynamic friction proportional to velocity?"

I think it is, inversely in this case, and I wonder if that is what answers my curving path question, since it is the part of the cycloid path that has the lessor trans component that appears on the side of the disc that is experiencing the slower relative trans speed wrt the floor... so taking the cycloid path points' speed wrt the floor on each side of the disc, maybe the net friction on each side is balanced as well...?

For CCW; left half is the cycloid's "slow cusp side" with less net motion, with less motion against the friction direction but greater friction (from reduced speed against floor); the right half is the cycloid's "fast non-cusp side" with more net motion, and opposing the friction direction, but with greater speed so less friction.

Left side - slower trans component - motion with friction direction - net friction= F(left)
Right side - higher trans component - motion opposed to friction - net friction= F(right)

It looks like the preservation of the trans/spin ratio would preserve the form of the cycloid wrt the floor, and F(left) = F(right) throughout the duration would prevent the spinning sliding book from taking a curved path...?

 

[MikeGomez]

Is dynamic friction proportional to velocity?

I read in Hyper-physics that "Frictional resistance to the relative motion of two solid objects is usually proportional to the force which presses the surfaces together as well as the roughness of the surfaces." I'll have to edit the last paragraph of post 49 to make that correction.

What do you think about the correctness of the statement that I am making about the trochoid paths of the component parts? I am saying that for the ratio of translational with respect to rotational motion (momentum) to change, the shape of the path traced by the component parts must change.

 

[MikeGomez]

@ bahamagreen I sent post #52 before your post #51 was there.

I think that is correct about what you are saying regarding the balance between the change in curvature of cw vs ccw, but the last sentence confused me when you said "prevent the sliding book from taking a curved path", because we're talking about curved paths. Did you mean to say "prevent the spinning sliding book from changing the curve of its path"?

 

[alva]

Just an idea.
From wiki:

Static friction is friction between two or more solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The coefficient of static friction, typically denoted as μs, is usually higher than the coefficient of kinetic friction.

If you want to take out a nail from a piece of wood you can use brute force and, using a lever, exert a big force on it.

Or you can turn the nail, cw and ccw, using a pliers, and pull out the nail smoothly.

I realize now that the OP question does not talk about how to start a movement but how to stop it. Something is missing in the wiki explanation, something like the minimum velocity to switch from kinetic to static friction.

 

[D H]

The answer is no, and the reason is that friction acts upon each atom in a manner which is proportional to its instantaneous momentum vector.

This is incorrect.

At the freshman physics level, kinetic friction is independent of speed, so long as the speed is not zero. The freshman physics model of friction is that it is a force proportional to weight that is directed against the velocity vector.

And that is the key problem with this problem. Given the velocity $\vec v_p$ at some point p of the book, the velocity at some other point displaced by some quantity $\vec r$ from the point p is $\vec v_p + \vec{\omega} \times \vec r$. Finding the net force due to friction is not a simple task. It is one ugly integral. Finding the net torque is even uglier. Showing that this means angular velocity reaches zero exactly when translational velocity equals zero is hideously ugly.

 

[bahamagreen]

@ bahamagreen ...Did you mean to say "prevent the spinning sliding book from changing the curve of its path"?

My original impression of the problem was that there were two problems to answer - one, why do the slide and spin of the book stop at the same time; and two, why doesn't the path of the center of mass of the book curve (curve to the left for CW and right for CCW)? I was thinking that one side half of the book would be spinning "against" the surface friction "harder" than the other side, and this would form a couple that would displace the center of rotation away from the center of mass... and cause the book's translational path to curve off to the side.

I still think this imbalance of friction direction between the left and right side is still happening, but the imbalance is the speed of the surface between the book and the floor, but the way friction varies inversely with surface speed in this case (I think?), the two principles seem to be conspiring to net out to even... so the ratio of spin to trans is staying undisturbed and constant... so the cycloid forms are "drawn" on the floor as identical repeating patterns of the same size and shape... just slowing down, everything in proportion, until the book stops.

Now I'm thinking of how to take a disc and mount some felt tip markers through it and see what kind of patterns it would draw on a surface if I gave it a sliding spin... or better, a time lapse photo of LEDS mounted on top, or something... there must be an elegant way to verify if the cycloids maintain there self same repeating pattern, size, shape, and "wavelength" etc. as the thing slows down.

 

[MikeGomez]

This is incorrect.

At the freshman physics level, kinetic friction is independent of speed, so long as the speed is not zero. The freshman physics model of friction is that it is a force proportional to weight that is directed against the velocity vector.

Thank you. I went to edit and correct that last paragraph of post #49, and the edit button seems to have disappeared. Weird.

And that is the key problem with this problem. Given the velocity $\vec v_p$ at some point p of the book, the velocity at some other point displaced by some quantity $\vec r$ from the point p is $\vec v_p + \vec{\omega} \times \vec r$. Finding the net force due to friction is not a simple task. It is one ugly integral. Finding the net torque is even uglier. Showing that this means angular velocity reaches zero exactly when translational velocity equals zero is hideously ugly.

That’s too bad. It would have been nice to mathematically either verify or falsify the premise that translational and angular motion stop at the same time. My feeling is still that they do, for the exact reason that fundamentally they are the same thing, and that it is total momentum that is being affected by friction, regardless of what combination of linear momentum or angular momentum.

My current thinking is that concentrating on the time factor will help a lot. How long does it take to slow down to zero velocity given x total initial linear momentum, and how long to slow down to zero velocity given x amount of total angular momentum. If the calculated times are the same, that would be very promising, and the next step would be to show that a combination of both will take the same amount of time to slow down to zero velocity.

 

[MikeGomez]

I still think this imbalance of friction direction between the left and right side is still happening, but the imbalance is the speed of the surface between the book and the floor, but the way friction varies inversely with surface speed in this case (I think?)

I’m afraid there is little hope for this. A.T. and DH (not to mention dozens of online sites) point out that friction does not depend on velocity. Friction is complex thing, and technically velocity can be a factor, but I think you will find that the magnitude of this effect is sure to be insignificant.

Now I'm thinking of how to take a disc and mount some felt tip markers through it and see what kind of patterns it would draw on a surface if I gave it a sliding spin... or better, a time lapse photo of LEDS mounted on top, or something... there must be an elegant way to verify if the cycloids maintain there self same repeating pattern, size, shape, and "wavelength" etc. as the thing slows down.

If you are seriously thinking about doing that, that would be fantastic!

 

[anorlunda]

Im afraid that we have been talking past each other because of sloppy language, regarding two meanings of the word stop. Let me coin some definitions to distinguish the two.

Full Stop: Consider a single point. If the coordinate X is fixed and all orders time derivatives of X are zero (velocity=0, acceleration=0, ...), than that point is in Full Stop. If the book is rigid, then it is easy to see that if any point is in Full Stop, then all points must be in Full Stop.

Momentary Stop: A pendulum stops twice in each cycle, although swinging continues. A Momentary Stop at a point happens when the first time derivative of the total velocity is zero, although acceleration and higher derivatives are non-zero. Momentary Stops of some points on the book must necessarily occur during the time evolution before the moment of Full Stop. Specifically, there should be a cycloid locus of points in Momentary Stop when the translational velocity is sufficiently slow.

In post #32 in this thread I argued for the significance of Momentary Stop. I assume that static friction is significantly different than dynamic friction. Static friction applies anyplace on the cycloid locus of Momentary Stop. Therefore, we can not dismiss static friction.

Let us also say that there is a small velocity ε below which friction transitions from dynamic to static. For any non-zero ε, the cycloid locus of Momentary Stop, become a region of finite area. In the final moment before Full Stop, that area must necessarily increase to 100% of the book's area. This I feel, is a spoiler to the problem. The final moment before Full Stop, is dominated by nonlinearities related to the dynamic-static friction transition. That spoils any linear analysis of moments prior to the final moment.

 

[MikeGomez]

Execellent. Just to give a further visual picture, if we imagine that example about the point on that wheel in contact with the ground having momentary zero velocity, that would be the momentary stop at apex of the cycloid, As we watch a racing car whizzing by us at 200 mph, we see that the point tracing its cycloid path isn't on the verge of coming to a full stop.

Let us also say that there is a small velocity ε below which friction transitions from dynamic to static. For any non-zero ε, the cycloid locus of Momentary Stop, become a region of finite area. In the final moment before Full Stop, that area must necessarily increase to 100% of the book's area. This I feel, is a spoiler to the problem. The final moment before Full Stop, is dominated by nonlinearities related to the dynamic-static friction transition. That spoils any linear analysis of moments prior to the final moment.

It’s true about the effects of static friction vs dynamic friction directly before the book stops, but I don’t think that is a spoiler for the problem. That time period ε that you mention directly before the object stops is extremely small and can be neglected for the most part. When we toss the book on the table it slides for a couple of seconds and then when it stops, it stops abruptly with a jerk. That is only a fraction of a second.

 

[anorlunda]

It’s true about the effects of static friction vs dynamic friction directly before the book stops, but I don’t think that is a spoiler for the problem. That time period ε that you mention directly before the object stops is extremely small and can be neglected for the most part.

The time is indeed very short, but the velocities are correspondingly small at the beginning of this period. Say that we were on track for translational and rotational velocities to reach full stop simultaneously, and also that that the final moment is only one millisecond long. It would take only a tiny perturbation to spoil the simultaneous arrival of both velocities at zero. My argument is that the nature of the nonlinearities, not the laws of motion will break the tie in the last moment. Which way will it break? It depends on the nonlinearities.

I feel bad raising a spoiler because I think that this was an exceptionally fun physics problem. Thanks to the OP.

 

[AlephZero]

Finding the net force due to friction is not a simple task. It is one ugly integral. Finding the net torque is even uglier. Showing that this means angular velocity reaches zero exactly when translational velocity equals zero is hideously ugly.

If somebody wants to try this, I suggest you first try it for a thin circular ring, and then for a circular disk. That will make the integrals easier.

If the assertion about simultaneous stopping isn't true in those cases, it probably isn't true for an book whose shape is an arbitrary rectangle either.

 

[tehrv]

This is incorrect.

At the freshman physics level, kinetic friction is independent of speed, so long as the speed is not zero. The freshman physics model of friction is that it is a force proportional to weight that is directed against the velocity vector.

And that is the key problem with this problem. Given the velocity $\vec v_p$ at some point p of the book, the velocity at some other point displaced by some quantity $\vec r$ from the point p is $\vec v_p + \vec{\omega} \times \vec r$. Finding the net force due to friction is not a simple task. It is one ugly integral. Finding the net torque is even uglier. Showing that this means angular velocity reaches zero exactly when translational velocity equals zero is hideously ugly.

How about looking at upper and lower bounds of some kind?

If you draw pictures of the velocity vectors, it is pretty easy to see that if it is spinning quickly most frictional work goes into slowing the spin, while if it is translating quickly most frictional work goes into slowing the translation.

 

[Delta Kilo]

If somebody wants to try this, I suggest you first try it for a thin circular ring... That will make the integrals easier.

Yeah , right :)

Assuming dry friction, the force at a point is $d\vec{F} = -\rho \vec{D}P dS$, where $\vec{D} = \frac{\vec{V}}{\left|V\right|}$ is the instantaneous direction of motion at a point.
Linear and angular accelerations are:
$M \dot{v} = \int d\vec{F} = \int \vec{D} \rho P dS$,
$I \dot{\omega}= \int (\vec{r} \times d\vec{F}) = \int (\vec{r} \times \vec{D}) \rho P dS$

The instantaneous velocity field is
$\vec{V}(r) = \vec{v} + (\vec{\omega} \times \vec{r})$ where $v, \omega$ are translation and rotation velocities.
In coordinates, assuming $\vec{v} = \begin{bmatrix}v \\ 0\end{bmatrix}$ and $\vec{r} = \begin{bmatrix}r \cos \alpha \\ r \sin \alpha\end{bmatrix}$:
$\vec{D} = \frac{1}{\sqrt{v^2 + (\omega r)^2 - 2 v \omega r \sin \alpha}}\begin{bmatrix} v - \omega r \sin \alpha \\ \omega r \cos \alpha \end{bmatrix} = \frac{1}{\sqrt{k^2 + r^2 - 2 k r \sin \alpha}}\begin{bmatrix} k - r \sin \alpha \\ r \cos \alpha \end{bmatrix}$
where $k = \frac{v}{\omega}$

We are interested in the behaviour of $k$ when $v, \omega→0$. Let's look at the sign of $\dot{k}$.
$\dot{k} = \left(\frac{v}{\omega}\right)'_t = (\frac{\dot{v}}{\dot{\omega}} - k) \frac{\dot{\omega}}{\omega}$

For a ring of radius $r=1$:
$\frac{\dot{v}}{\dot{\omega}} = f(k) = \frac{\int\limits_0^{2 \pi}{\frac{ k - \sin \alpha} { \sqrt {k^2 + 1 - 2 k \sin \alpha}}}d\alpha }{\int\limits_0^{2 \pi}{\frac{ 1 - k \sin \alpha} { \sqrt {k^2 + 1 - 2 k \sin \alpha}}}d\alpha}$

The plot of $f(k)-k$ looks like an S-curve with roots at $k= 0$, $k=1$ and $k=∞$, with $f(k)>k$ when $k<1$ and $f(k)<k$ when $K>1$. Which would indicate $k=1$ is a point of stable equilibrium. This corresponds to a case where the ring 'rolls' with the point (0,r) being the instantaneous center of rotation. Which makes sense as even a small change of v/w ratio reverses the direction of friction force at that point.

For non-rotationally symmetric shapes things quickly become ugly. For a start, the whole thing is not even guaranteed to go in a straight line. And the forces depend on orientation so there is no steady state to speak of.

 

[D H]

I've run into the same wall of ugliness you have run into, Delta Kilo, with pretty much the same ugly elliptical integrals.

One difference between your analysis and mine: I looked at your $\dot k$ from the perspective of $\dot k = k\left( \frac {\dot v} v - \frac {\dot \omega}\omega \right)$.

 

[MikeGomez]

I've run into the same wall of ugliness you have run into, Delta Kilo, with pretty much the same ugly elliptical integrals.

One difference between your analysis and mine: I looked at your $\dot k$ from the perspective of $\dot k = k\left( \frac {\dot v} v - \frac {\dot \omega}\omega \right)$.

I think the idea of starting with a ring was pretty good, but what about making it even simpler like just two points on opposite sides of the ring?

Can anyone give the physics of two point masses separated by massless rod?

I've looked around a bit and haven't found the right thing, just blocks on inclines, pulley problems, gravitation problems, billiard balls colliding, etc. It seems like it would be a trivial physics problem to able to set a system of two identical point masses (separated by a massless rod) in motion, and then apply the external force of friction in opposition to their direction of motion.

 

[A.T.]

I think the idea of starting with a ring was pretty good, but what about making it even simpler like just two points on opposite sides of the ring?

I'm not sure if this idea helps. The net force/torque will depend on the orientation of the two points w.r.t the linear velocity. It seems more complicated than the ring where force/torque depend only on v & omega.

 

[AlephZero]

With only two point masses, you lose the rotational symmetry of the system, so the behaviour would be different.

That is clear if you consider the friction forces when the rod is perpendicular to the direction of motion: if $\omega r /2 < v$, there is no torque component of the friction because both masses are moving in the same direction as the $v$ vector.

On the other hand, n > 2 point masses should be a rotationally symmetrical system. This is the same as evaluating the integrals using the trapezium rule with n integration points.

But I don't expect it would give any more insight than Delta Kilo's result, which seems to confirm my hand-waving conjecture in an earlier post.

You might like to try it for n = 4, which is probably easier math than n = 3.

 

[.Scott]

Okay. Let me try this one out. If you don't mind, I will deal with this situation:
* The book I will use will be in the shape of a circular disk of radius "r" meters .
* It's translational speed will be "t" meters per second.
* It's rotational speed will be "s" cycles per second in the clockwise (CW) direction. CCW would be specified as s<0.
* The actual value for the frictional force is not important - except that it must be a positive value. We will call this "f" and we don't need to further define it.
* I'm going to set up a Cartesian coordinate system centered on the center of the book. Positive x will be in the direction of the the translational motion and positive y will be at a right angle to the left of that motion.

If the translational speed (t) is ignored, the speed along the perimeter of the book would be 2πrs.

But with both translational and rotational speeds, there will be a point (perhaps within the boundary of the book and perhaps not) where the t and s balance out to zero.
This "quiet" point will lie at a distance "q" from the center of the book such that 2πqs=t, thus q=t/(2πs).
This q-point will be directly to the receding side of the center of the book. For example, if the book is traveling from North to South and turning clockwise, then the q-point will be just to the West of the book's center. In our coordinate system, the q-point will be at (0,-q).

One way of describing the instantaneous motion of the book relative to the table top is that it is not translating at all, simply spinning at rate "s" around point "q".

I described my x,y axis above. Except at the q-point, the magnitude of the frictional force at any point on the book will be "f". Although the coordinates relative to the center of the book are (x,y), the coordinate relative to the q-point would be (x,y+q). So the direction of the force at any point (x,y) will be counter to the spin and therefor (-y-q,x). If someone wants an exact formula for the frictional force at any point on the book, it is f·(-y-q,x)/√(x²+y²+q²-2qy). But that not required to demonstrate the point.

The key is the relationship between the absolute value of "q" and "r". In fact, the ratio |q|/r is all that is needed to determine what portion of the total friction is being used to retard the translational speed and what portion is being used to retard the rotational speed.

If |q|>r, then the quiet point falls outside the boundary of the book and every point on the book is contributing to retarding the translation speed and the affect on the rotational speed will be slight. As this quiet point falls within the boundary of the book from |q|=r to q=0, there are more and more points that are actually attempting to accelerate "t" while at the same time there becomes a disk or radius r-|q| that applies a full retarding force to the spin. Once q=0, there is no longer any affect on the translational velocity and a maximum effect of the rotational speed.

So there are two specific functions: one for the translational retarding force Ft(q/r) with an absolute value that monotonically increases from 0 to 1 as q/r increases from 0 to infinity; and one for the rotational retarding force St(q/r) with an absolute value that monotonically decreases from 1 to 0 as q/r increases from 0 to infinity.

So there must be a point "b" where these functions cross. If the q-point falls outside the b-point, the translational velocity will be reduced in greater proportion that the rotational - moving the q point towards that b-point. If the q-point falls inside the b-point, the rotational velocity will be reduced in greater proportion that the translational - again moving the q point towards that b-point. No matter what the starting conditions are, as long as there is some spin and some translation, the ratio of spin to translation will continuously move closer to some certain value until that ratio becomes 0/0 and both stop.

 

[AlephZero]

Okay. Let me try this one out...

I think we already got to that point, about 50 posts ago.

Arm-waving math + Arm-waving physics still equals Arm-waving solution, IMO.

The thing you haven't proved (and to be fair, neither did I, nor Delta Kilo) is

No matter what the starting conditions are, as long as there is some spin and some translation, the ratio of spin to translation will continuously move closer to some certain value until that ratio becomes 0/0 and both stop.