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

[tehrv]

Yes, very catchy title. But it also happens to be true! Here's the windup: This question comes from the Halliday and Resnick edition that was used in the 80's. I was a grad student at the time at a very prestigious department and somehow this question came to my attention. I couldn't figure it out. Soon, there was a group of four grad students losing sleep over it. I even took it to a very famous theoretical physicist. He smiled and laughed out loud, because he didn't see an obvious answer. Decades later, I ran into the author of another widely used freshman textbook and he just said well if the famous guy couldn't figure it out, neither could he! So this has bothered me for going on three decades. Hopefully somebody here can put my mind at ease. Remember, this is supposed to be freshman level.

It is a "qualitative" question.

Drum roll...

If you take a flat object (like a book) and slide it along a surface where there is friction, why does it stop translating and rotating at the same time?

That's all there is to it. You can quickly verify that it is true. But why?

I'm sorry if you lose some sleep!

 

[gralla55]

Won't that be because the force of friction works in the opposite direction of the total movement, not seperately for the translational and rotational?

What I mean is, if you take one point on the corner of the spinning book, it has a movement due to the rotation around the center of mass, and one due to the translation of the COM. The total movement for that point is the sum of those, and this is the movement of which friction would be directly opposite.

Say the center of mass of the book is moving with velocity 1 in the x direction, while one corner at that instant is rotating with velocity 4 in the y-direction on top of that. The total velocity vector would be (1,4). Since the negative acceleration due to friction would be the opposite, it would be some vector -a(1,4) / ||(1,4)||, therefore doing proportionaly more work on the element due to rotation (or less if it was lower). Anyway, apply the thought experiment for each point in the book, and each pair of coordinates should reach (0,0) at the same time.

 

[berkeman]

Yes, very catchy title. But it also happens to be true! Here's the windup: This question comes from the Halliday and Resnick edition that was used in the 80's. I was a grad student at the time at a very prestigious department and somehow this question came to my attention. I couldn't figure it out. Soon, there was a group of four grad students losing sleep over it. I even took it to a very famous theoretical physicist. He smiled and laughed out loud, because he didn't see an obvious answer. Decades later, I ran into the author of another widely used freshman textbook and he just said well if the famous guy couldn't figure it out, neither could he! So this has bothered me for going on three decades. Hopefully somebody here can put my mind at ease. Remember, this is supposed to be freshman level.

It is a "qualitative" question.

Drum roll...

If you take a flat object (like a book) and slide it along a surface where there is friction, why does it stop translating and rotating at the same time?

That's all there is to it. You can quickly verify that it is true. But why?

I'm sorry if you lose some sleep!

It's not true for all initial values of spins and linear velocity. As you said, you can quickly verify that it is true.

(Try a slow linear push with a strong spin -- the book spins still after the linear motion stops)

 

[berkeman]

@tehrv -- So a *much* more interesting and challenging question is for you to calculate the initial conditions to have the following things happen:

-1- translation stops before rotation

-2- rotation stops before translation

-3- both stop at the same time

Please show your work...

 

[tehrv]

It is supposed to be a 'qualitative' problem, so I don't think they wanted you to write much. Something like gralla's argument may be on the right track. Berkeman, the high spin case exception you mention is probably because the surfaces are not perfectly flat. Anyway, it is not my question, so it isn't really up to me to tweak it.

We could never figure out what was going on. Really, some brilliant grad students were stumped. I remember one guy was trying to consider a point at infinity. lol.

 

[sophiecentaur]

It's not true for all initial values of spins and linear velocity. As you said, you can quickly verify that it is true.

(Try a slow linear push with a strong spin -- the book spins still after the linear motion stops)

When I first read that I believed you (stands to reason dun nit?). But does it really? Have you (/we) actually measured this or just imagined it? Would we spot a slow forward motion of a fast spinning book? I could believe that we wouldn't. (Are we going to argue with eminent Professors? lol)
I guess it must be to do with the difference between dynamic and static friction and as long as you have motion in one mode or the other, the friction is dynamic in both cases and the retarding force / torque will be tied to the relative speeds between all points on the object and the surface. So static friction kicks in at the same time for both modes. The mode with the higher relative speeds involved will lose energy faster than the mode with the lower relative speeds (Power = force times speed). Arm waving allowed here by the original problem, I'm pleased to say - but the Maths would not be beyond a bright student, I'm sure. It works, I'm sure, if the retardation is proportional to speed and not constant.

 

[willem2]

I guess it must be to do with the difference between dynamic and static friction and as long as you have motion in one mode or the other, the friction is dynamic in both cases and the retarding force / torque will be tied to the relative speeds between all points on the object and the surface. So static friction kicks in at the same time for both modes

I don't think static friction kicks in before the book stops moving completely.
There's no way that a part of the book can stop moving, except for a single point, and the fricton on a single point is 0.
So I don't think static friction is relevant at all.

 

[Brinx]

If the dynamical friction force on a surface element of the book is linearly dependent on the velocity of that surface element, and if we only consider a purely translational motion for simplicity, then the momentum of the book would decay exponentially it seems: the smaller the momentum, the smaller the force that acts to diminish it. This would imply an infinite time interval would pass before the book would come to a complete rest given any nonzero initial momentum.

That would indicate that below some velocity threshold the nature of friction would have to be different, or that we need to consider a different relation between friction and velocity altogether.

 

[sophiecentaur]

I don't think static friction kicks in before the book stops moving completely.
There's no way that a part of the book can stop moving, except for a single point, and the fricton on a single point is 0.
So I don't think static friction is relevant at all.

You are right. Static friction is not relevant - except in as far as it lurks in our subconscious when we think about friction problems.

 

[Office_Shredder]

If the dynamical friction force on a surface element of the book is linearly dependent on the velocity of that surface element,

It's not though, the dynamic friction is constant (just a different constant value from when the book is not moving).


I don't understand the statement of the problem in the context of pushing a book forwards with only linear momentum; certainly that seems possible to do theoretically, or is this example (and where you rotate it with no linear momentum) supposed to be excluded?

 

[tehrv]

It's not though, the dynamic friction is constant (just a different constant value from when the book is not moving).


I don't understand the statement of the problem in the context of pushing a book forwards with only linear momentum; certainly that seems possible to do theoretically, or is this example (and where you rotate it with no linear momentum) supposed to be excluded?

Yes, the two limiting cases are excluded. You just sort of fling the book across the suface so that it both translates and rotates. Maybe somebody has the book (big and green with yellow waves as I recall) and can give us the exact wording.

 

[Brinx]

It's not though, the dynamic friction is constant (just a different constant value from when the book is not moving).

Ah, you're right of course - my apologies. I somehow completely misinterpreted sophiecentaur's post. Dissipation power is proportional to velocity, but dynamic friction force is constant.

 

[jbriggs444]

@tehrv -- So a *much* more interesting and challenging question is for you to calculate the initial conditions to have the following things happen:

-1- translation stops before rotation

-2- rotation stops before translation

-3- both stop at the same time

Please show your work...

The answer is simple and has already been given in this thread. You just have to take the arguments to their logical conclusion.

Answers:

1. The only condition in which translation stops before rotation is when there was no translation to start with.

2. The only condition in which rotation stops before translation is when there was no rotation to start with.

3. In all remaining cases, they both stop at the same time.

Work:

The key observation (from Gralla55) is that for a force that opposes velocity and does not scale with velocity, the component of the force in one direction scales down with the component of velocity in the direction at right angles.

Quantitatively: f_x = f_tot√(1-v_y/v_tot)

Qualitatively, what matters is that when the v_x << v_y, f_x is proportional to v_x. As Brinx has noted, this means that the decay of v_x is approximately exponential.

For sake of clarity, let us assume that this object is sliding from North to South...

Suppose that the rate of translation reaches zero while the rate of rotation is non-zero. It follows that for some time prior to this occurring, almost all of the points on the object would have been rotating east/west much faster than they were translating north/south. [More formally, for any desired value of "almost all" and any desired ratio of "much faster", there is a time after which the rate of translation is small enough that both conditions are upheld]

During this time interval the north/south force is approximately proportional to north/south velocity and rate of decay of translational velocity is exponential. It takes infinitely long for it to reach zero. During this same time interval the opposite condition is not upheld. The resistance to rotation is not reduced significantly by the low translation rate. The rotation rate must be therefore be decreasing approximately linearly and will reach zero in finite time. If the "almost all" and "much faster" conditions continue to be upheld, the rotation rate will reach zero before the translation rate does.

Contradiction. So the rate of translation cannot reach zero first.

The same argument applies if the object is translating much faster than it is rotating. So the rate of rotation cannot reach zero first.

The only consistent resolution is that the two rates must reach zero together (or that one or the other were zero to start). QED.

 

[AlephZero]

The key observation (from Gralla55)...

Do you have a proper reference to this? I can't find anything with Google, except a Gralla who is working in quantum cosmology...

From what you posted, I can't decide between two options:
1. It doesn't make sense
2. I don't understand it
(of course there may be other alternatives, and those two options may not be mutually exclusive )

 

[jbriggs444]

Do you have a proper reference to this?

From what you posted, I can't decide between two options:
1. It doesn't make sense
2. I don't understand it
(of course there may be other alternatives, and those two options may not be mutually exclusive )

The observation from Gralla55 is in post #2 of this thread.

 

[AlephZero]

The observation from Gralla55 is in post #2 of this thread.

OK - I took it the complete argument was from a paper by Gralla.

There is a simpler explanation of why the translational and rotational decelerations will converge to a have consistent relationship between them, independent of the starting conditions and also independent of the details of the friction model.

Suppose the book is translating north, and suppose it has a "low" rotation speed compared with the translation speed. Relative to the table, both the east and west sides of the book are sliding north. So the moments of those frictional forces about the CM of the book will cancel, but the translational friction forces north will add. For fast translation and slow rotation, the friction mostly decelerates the translation, not the rotation.

Now, suppose the rotation speed is "high" relative to the translation. The opposite occurs: one side of the book is sliding north relative to the table, but the other side is sliding south. The translational components of the friction cancel out, but the moments about the CM add. For slow translation and fast rotation, the friction mostly decelerates the rotation, not the translation.

So, for any "reasonable" friction model and any "reasonable shaped" book, the relative translation and rotation speeds will tend to a fixed ratio of each other, and both reach zero at the same time.

(Note to pedants: feel free to argue at length about the definition of "reasonable" in the previous paragraph )

 

[tehrv]

How about this?

The frictional force on each point is opposite to the direction of that point's velocity.

So (by Newton's Third Law) the deceleration on each point is opposite to the direction of that point's velocity.

But that point's velocity can be written as a sum of a translational and a rotational part.

So, we can rewrite the deceleration of each point as a sum of a translational and a rotational part.

Moreover, and this is the key part, we see that the ratio of the magnitudes of the deceleration of the translational part and the deceleration of the rotational part is PROPORTIONAL to the ratio of the magnitudes of their respective velocities. Call this the PROPORTIONALITY CONSTRAINT:

Deceleration of the translational component/Deceleration of the rotational component
=Speed of the translational component/Speed of the rotational component.

For all the points, except a subset with zero total mass, the translational and rotational velocities at the beginning are non-zero.

At each moment, a bit of work gets done on each point. The translational velocity decreases a bit. And so does the rotational velocity. The greater of the two decreases the most. But because of the PROPORTIONALITY CONSTRAINT, the velocities and decelerations all go to zero at the same time.

It's not as though there is some funky relation between the velocities allowing for strange behaviors. Rather there is this simple proportionality constraint.

Needs some polishing, but something like that.

 

[AlephZero]

The frictional force on each point is opposite to the direction of that point's
velocity.

True.

So (by Newton's Third Law) the deceleration on each point is opposite to the direction of that point's velocity.

That is not true for a rigid body. The deceleration depends on the resultant force on "the point", including the internal forces (stresses) in the body. The simplest way to deal with the internal forces correctly, is to find the resultant forces and moments at the center of mass of the body.

(And I think you meant Newton's second law, not the third).

 

[tehrv]

That is not true for a rigid body. The deceleration depends on the resultant force on "the point", including the internal forces (stresses) in the body. The simplest way to deal with the internal forces correctly, is to find the resultant forces and moments at the center of mass of the body.

(And I think you meant Newton's second law, not the third).

Right. I should have known it was too easy. It's been a few decades since I've done any real physics!

And right again. I kept thinking about the force being opposite to the velocity, so I had "opposite" on my tongue.

 

[tehrv]

So can somebody give a complete answer without making reference to other posts in the thread? I'm sorry, but I still don't see a full solution here. :(

 

[bahamagreen]

Think about it backwards... (as a time reversed observation)

1] From rest, the book begins to spin and increase rotation, somewhat after which the COM begins to accelerate
2] From rest, the COM begins to accelerate, somewhat after which the book begins to spin and increase rotation
3] From rest, the book's spin (and increase in rotation) and acceleration of COM begin simultaneously

Just a hunch, but doesn't this suggest that only the third observation (in spite of it being time reversed) doesn't violate some conservation principle(s)...?

 

[Dadface]

Consider the book being made of an infinite number of vanishingly small parts. The rotational and translational motion of each part can be represented by a vector.

1. A rotating vector which increases in length from the centre of rotation outwards.
2. A vector which always points in the same direction

Consider one of the small parts for example at one of the edges of the book. Because of frictional forces the lengths of the vectors will reduce with time and these reductions will be synchronous. As an example consider any instant when both vectors are parallel and in the direction of translational motion. At this instant the lengths of both vectors must be the same or we would have the impossible situation where the small part is moving at two different speeds at the same time.
Using the same reasoning for all small parts of the book, it can be deduced that the book stops spinning when the rotating vectors reduce to zero size and at that instant the vectors representing translational motion also reduce to zero size. So the book will stop moving altogether.

NOTE After writing the above and going for a drive I realized there is a (basic) error in my reasoning. I'm not changing it but letting it stand as an example of the right for everyone to be dopey at times. I'm going to think about it again.

 

[sophiecentaur]

So can somebody give a complete answer without making reference to other posts in the thread? I'm sorry, but I still don't see a full solution here. :(

I would say that, as OP, it is up to you to act as 'Chairperson of the meeting' and do the summing up yourself. The thread is full of ideas so you have plenty to go on. What have you gained from it all so far?
We are very mean, on PF, when it comes to spoon feeding people. I'm sure there's enough here for you to be getting some useful thoughts.

 

[bahamagreen]

Dadface,

I thought you had it there for a second, but I'm doubting it now...

It looks to me if your rotation vector points from the book's COR toward the point in question, that direction is not the direction the point is moving (not the rotation direction movement)... that vector direction is 90 degrees behind the direction the point is going with respect to the COR, so when the translation and rotation vectors are parallel in the same direction, at that moment the point does have two speeds, but 90 degrees apart, so they are components of another vector that would be the net motion of the point with respect to the surface on which the book is sliding and spinning...

If you advance the rotation vector's direction definition forward 90 degrees so that its direction does indicate the point's direction of movement with respect to the COR, then when the rotation and translation vectors are parallel in the same direction the point still has two speeds... but one is the translation with respect to the surface, and the other is the point's speed with respect to the the COR. So these are just components that add.

If you add them wrt the surface, the "parallel in the same direction" case is the point acellerating about the side edge in the direction of motion wrt the surface (for example, if the rotation is CCW and the translation is in the 12 o'clock direction, the advanced 90 degree rotation vector and the translation vector will both be pointing to 12, and the point at that moment will be at 3 o'clock experiencing max acelleration and speed wrt to the surface the book is sliding and spinning across.

What I'm getting at is that the two vectors are independent because each is defined wrt a different frame, one the COR and the other the floor or surface.

 

[Dadface]

Thanks Bahamagreen. Please see the note I added to my post. I was expecting more comments similar to yours. I'm going to think about it some more. My approach considering vectors might have something to it that can be developed correctly or it might lead down a blind alley.

 

[bahamagreen]

I think my time reversed observation with respect to conservation approach was the blind alley. :)

I've spent a few minutes sliding a spinning book across the floor and it looks like the spin and slide stop at the same time... but I noticed something else. Spinning CCW, I expected the slide to curve to the right... like a baseball or a huricane, but all my trials look like the book always slides straight and never curves.

Maybe your vector approach can reveal a relationship between friction and speed conspiring to balance on both sides of the COR...?

... why does the spinning book path not curve?

 

[sophiecentaur]

I think my time reversed observation with respect to conservation approach was the blind alley. :)

I've spent a few minutes sliding a spinning book across the floor and it looks like the spin and slide stop at the same time... but I noticed something else. Spinning CCW, I expected the slide to curve to the right... like a baseball or a huricane, but all my trials look like the book always slides straight and never curves.

Maybe your vector approach can reveal a relationship between friction and speed conspiring to balance on both sides of the COR...?

... why does the spinning book path not curve?

Why should it? For every bit of the book moving to the left, there is another bit moving to the right. Even if the mass distribution is not uniform (or, even the 'roughness') there will always be equal forces in each lateral direction.

Earlier in the thread, the notion of exponential decay came up but it seems to have been ignored.
1. If you assume that the book is rigid then, if any part is moving then all parts must be moving.
2. The speeds of any of the parts will be decaying exponentially. This means that, until stiction occurs, no part will reach zero speed. So I think it is sufficient to use that argument to show it cannot stop spinning at a different time to stopping linear motion.

 

[bahamagreen]

Why should it? For every bit of the book moving to the left, there is another bit moving to the right. Even if the mass distribution is not uniform (or, even the 'roughness') there will always be equal forces in each lateral direction.

Earlier in the thread, the notion of exponential decay came up but it seems to have been ignored.
1. If you assume that the book is rigid then, if any part is moving then all parts must be moving.
2. The speeds of any of the parts will be decaying exponentially. This means that, until stiction occurs, no part will reach zero speed. So I think it is sufficient to use that argument to show it cannot stop spinning at a different time to stopping linear motion.

I'm struggling, but I'll try...

Last part first;
Point 1. Yes, parts are rotating wrt the COR, and parts are moving wrt the floor (which means wrt the floor some parts may be accelerating in epicyclic loops or waves...?)

Point 2. No, or rather yes all parts wrt the COR, and yes the COR wrt to the floor, but not necessarily all parts exponentially wrt to the floor. With respect to the floor, there will be parts making epicyclic loops/waves that wrt the floor will be oscillating accelerations...?

First part;
Assume for simplicity the book is a disc. If you just spin the disc, the center of rotation and the center of mass coincide. The friction magnitude is symmetric around both the COR and COM.

If you just slide the disc, there is no rotation because the degree to which the friction acts on the right and left sides is balanced... the friction's center of effort coincides with the COM, so there is no couple.

But, when the disc is spinning and sliding, the friction under one side is greater than the other (right side for CCW, left side for CW).

In the case of CCW, I was thinking this shifting of the friction's COE to the right of the COM of the disc makes a couple between that COE and the COM. The result is a torsion that attempts to push the COR to the right of the COM. The push of the COR shifting to the right of the COM in a rigid disc will be expressed as the path of the COM turning to the right...?

 

[tehrv]

I would say that, as OP, it is up to you to act as 'Chairperson of the meeting' and do the summing up yourself. The thread is full of ideas so you have plenty to go on. What have you gained from it all so far?
We are very mean, on PF, when it comes to spoon feeding people. I'm sure there's enough here for you to be getting some useful thoughts.

While I completely respect the prevailing attitude on PF, I consider this thread a special case because:

1) I have done very little real physics in decades unless you include arguing with the wife about the efficacy of cheap diswashing detergent vs. expensive dishwashing detergent.

2) Along with some other very sharp folks, I spent a fair bit of time on this problem a long time ago to no avail.

3) I have thought about this problem on and off for decades.

4) I haven't seen anything yet in this thread that is particularly promising. For example, invoking anything "exponential" seems incorrect as this comes from an early chapter in classical mechanics text where almost no exponential equations/arguments appear. However, if the "exponential" reasoning were sufficiently qualitative, then perhaps it might be OK.

5) My real hope was that this would sufficiently bother and annoy much more qualified and practiced physicists as much as it has bothered me and somebody would produce a concise clear explanation around which a consensus of approval would form.

6) No consensus has yet formed!

:D

 

[Dadface]

It's an interesting question but I would like to see better experimental evidence that the spin and rotational motion do actually stop at the same time. I have just tried it with a pencil and in some cases the pencil continues moving after the spin seems to have stopped.However it mainly seems to be a rolling rather than a sliding movement.

 

[sophiecentaur]

It's an interesting question but I would like to see better experimental evidence that the spin and rotational motion do actually stop at the same time. I have just tried it with a pencil and in some cases the pencil continues moving after the spin seems to have stopped.However it mainly seems to be a rolling rather than a sliding movement.

That extra degree of freedom probably means that the test is not 'fair'. I think the condition would need to be very well controlled to produce a conclusive experimental proof.

 

[anorlunda]

I can think of possible explanation, but not one at can be analyzed by the laws of motion. That is because friction is not a fundamental force, but rather an average inter-molecular behavior for countless molecules and lattices of unknown type.

As the translational and angular velocities approach zero, there must be a point at every given time where the translational and angular velocities on one side of the book cancel. On the opposite side of the book those velocities would add. At the point where net velocity is zero, static friction takes over. The location of that point moves from instant to instant. I visulize a zone of static-dominated friction, with the area of the static zone expanding from an infitessimal point to a zone as large as the whole book within a short elapsed time. If static friction is much stronger than dynamic friction, the book should "stall" abruptly in a nonlinear avalanche of static zones.

Of course the whole premise depends on the ratio of dynamic/static friction as the velocity approaches zero. I would expect some continuous function, but I don't know of any laws of physics that govern it.

In the case of pure translation, there is no point where velocities cancel. In the case of pure rotation, the velocity in the center is zero, but extra static friction at the center point has zero moment arm for torque. Only in the case of translation plus rotation can there be an off-center point with zero net velocity.

Thanks to the OP for the thought provoking question.

 

[MikeGomez]

You might consider that linear momentum and angular momentum are two sides to the same coin. Some people think of linear momentum as a special case of angular momentum. I prefer to think of angular momentum as a special case of linear momentum. Of course angular momentum can be a mathematical convenience in many cases, but in other cases like this puzzle of yours, perhaps it is best that it be shunned, despised, and cast out as unworthy of being considered a phenomenon on its own right.

Imagine a rigid spinning body that is composed of many massive components which are held together by spokes of negligible mass. What happens if we release the bonds (cut all the spokes)? All the massive pieces immediately fly off in a straight line directly opposed from the center of mass.

All the separate pieces have linear momentum, but what happened to the so called angular momentum that the original object possessed before we broke the bonds? Did it disappear? Transform? Or perhaps it never really existed. Hmm.

The reason I bring up this subject is that the puzzle as presented is a kind of trick question where there are assumptions in the asking of the question, and making the mistake of accepting the assumptions of the question as posed, is what confounds us from finding a solution. These assumptions are 1: linear momentum and angular momentum are separate and independent physical phenomenon, and 2: linear and angular momentum are expected to behave completely independently from each other in a problem such as this.

For the purpose of this discussion the book is considered to be a solid object. It doesn’t stretch or bend or compress much. It is composed of many micro-objects (let’s say atoms) and their behavior combines to form the overall behavior of the book. That is to say that all the atoms interact with each other according to Newton’s laws, and together they manifest as the entity (and associated motion) that is the book.

It should be clear that at every instant in time each individual atom has one and only one instantaneous momentum vector. It has a magnitude and direction, and we might think of the total momentum vector as a sum of separate linear and angular momentum vectors, but my whole point here is, well, don’t do that!

Let’s toss the book across the table.

Linear translation only:
-The path taken by individual components is described as a straight line.
-Each component atom of the book has the same instantaneous linear momentum.
-It seems intuitive to us that since the book as a whole acts as a single solid object, no individual component which makes up the book may have zero velocity unless the entire book has zero velocity.

Rotation only:
-The path taken by individual components is described by circles of various radii.
-Each component atom at a given distance from the center of book has the same magnitude of instantaneous linear momentum, with differing direction. Components at differing distances from the center have differing magnitudes.
-Various components have various momentum magnitudes. The atoms closer to the center of the book move slower than the atoms farther from the center of the book, yet somehow it still seems very intuitive to us that since the book as a whole acts as a single solid object, no component of the book can reach zero velocity without the entire book reaching zero velocity.

So far a pattern has emerged in each case that the individual components each follow a well defined path, and even though the momentum of the components may differ, it seems intuitive to us that since the object is solid as a whole, no individual component may reach zero velocity without the entire object reaching zero velocity.

Given this pattern that we see, and keeping in mind the idea of only linear instantaneous momentum of individual components, it would be reasonable to suspect that the case for rotation combined with translation would be the same, and that no individual component may reach zero velocity without the entire object reaching zero velocity.

I’m going to call the path taken a "cycloid". I’m not sure if that’s one hundred percent accurate, but that doesn’t matter, since it is only a name given to the path. The path taken by individual components is described by cycloids of various shapes. The cycloids of components at a given distance from the center of the book are exactly like each other, with the exception that they have various phase angles. The cycloids of components of differing distances are slightly different in shape (but nearly identical to neighbors).

Since the book is solid, no individual component may reach zero velocity without the entire book reaching zero velocity.

This holds true for any solid object. It also holds true for more degrees of freedom. For example we would see the same behavior in 3D if we could apply friction uniformly throughout.

Tehrv, you say you don’t want references to previous posts, but now I invite you to revisit post #2, which is the very first reply. It is accurate.

 

[tehrv]

Imagine a rigid spinning body that is composed of many massive components which are held together by spokes of negligible mass. What happens if we release the bonds (cut all the spokes)? All the massive pieces immediately fly off in a straight line directly opposed from the center of mass.

I don't have time right now for a full response, but this is incorrect. The pieces fly off in straight lines, but those straight lines don't intersect the center of mass.

 

[tehrv]

I did make a bit of progress today. If you consider a translating+spinning object and another identical object that is only translating with the same velocity but not spinning, the two objects start with the same translational energy and the frictional forces doing the work of stopping the translations are the same. In both cases, this force should be: mu*m*g So those two objects should stop translating at the same time! (Hope that is right! :) )