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

[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.

 

[.Scott]

The key is that there is a specific spin to speed ratio that is approached asymptotically from either direction. To demonstrate that, all you need to demonstrate is that both the spin and speed friction is a function of the same spin to speed ratio, that each goes from near zero to a much larger value, and that they are both monotonic but one is increasing while the other is decreasing.

Given those constrains, they have to cross at a certain spin to speed ratio - and that that ratio, whatever it is, will be approached asymptotically as the book slows.

When I start that post, I was thinking I would attack the actual integration problem. But I don't have the time. But if you simply draw out the graphics as if you were about to attack that integration, all of the points I listed in the first paragraph become very evidently true.

Anyone who calculates "b" or rather b/r completes the proof.

 

[rcgldr]

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.

I seem to recall that kinetic friction decreases slightly as speed increases at higher speeds, and also the transition from kinetic to static friction at very slow velocity when an object stops sliding is not instantaneous, but a reasonably smooth transition where friction increases during the transition resulting in some jerk (change in rate of acceleration) just before an object stops sliding. I'm wondering how this affects any small portion of the book that has near zero velocity as the book translates and spins. However perhaps the original question is assuming an idealized case where velocity doesn't affect kinetic friction and that the transition from kinetic to static friction when an object stops sliding is instantaneous.

As a real world example, since it's part of the winter olympics, in curling, the stones can be made to curve with a very small amount of induced rotation compared to the induced linear velocity. A clockwise rotation results in a left to right curve.

 

[.Scott]

I did the integrations the dumb way (by having the computer march through an array of 201x201 points superimposed on the disk, so about 40000*pi/4 were evaluated.

The amount of energy in a disk that is spinning with an edge speed "s" has half the energy of a disk that is moving at speed "s", but since they both have the same frictional area they will both have the same total force applied to them.

So I was looking for a situation where the total force retarding the translation was double that retarding spin.

The results were interesting. Drum roll...
When those speeds are equal, the translation friction is at about 84.9% and the spin friction is at about 42.5%.
So that is the condition that the "book" will tend towards. In the case of the circular book, it's when the point on the book that is not actually moving is at the very edge.

I'll post the code in a moment.

 

[.Scott]

  int     nX, nY, nDiv, nYRange;
  __int64 nYRange2, nDiv2, nArea;
  double  fR, fX, fY, fX2, fQ, fB, fBR;
  double  fForceX, fForceY, fForceD, fRDiv;
  double  fForceXSum, fForceYSum, fTurnForce, fTurnForceSum;
  double  fSpinFrictionMax, fArea;

  fR    = 1.0;   // Radius

  for(fQ=0;fQ<=2.5;fQ+=0.05) {
    fB    = fQ;    // Estimated balance point.
    fBR   = fB*fR;
    nDiv  = 100;
    fRDiv = fR/nDiv;
    nDiv2 = (__int64)nDiv*nDiv;

    nArea = 0;
    fSpinFrictionMax = 0.0;

    fForceXSum  = 0.0;
    fForceYSum  = 0.0;
    fTurnForceSum = 0.0;

    for(nX=-nDiv;nX<=nDiv;nX++) {
      //
      // Compute the range of Y values that are within the disk given nX.
      nYRange2 = nDiv2-((__int64)nX*nX);
      nYRange = (int)sqrt((long double)nYRange2);
      fX  = nX*fRDiv;
      fX2 = fX*fX;
      for(nY=-nYRange;nY<=nYRange;nY++) {
        fY = nY*fRDiv;
        nArea++;
        fSpinFrictionMax += sqrt(fX2+fY*fY);
        fForceX = -fY-fBR;
        fForceY = fX;
        fForceD = sqrt(fForceX*fForceX+fForceY*fForceY);
        //
        // Check for zero-friction point.
        if(fForceD<0.001) continue;
        fForceX /= fForceD;
        fForceY /= fForceD;
        fTurnForce = fY*fForceX-fX*fForceY;
        //
        // Accumulate frictional forces.
        fForceXSum += fForceX;
        fForceYSum += fForceY;
        fTurnForceSum += fTurnForce;
      }
    }
    //
    // Compute all the proportional forces.
    fArea    = (double)nArea;
    fForceX  = fForceXSum/fArea;
    fForceY  = fForceYSum/fArea;
    fTurnForce = fTurnForceSum/fSpinFrictionMax;
    //
    // Output fQ, fForceX, and fTurnForce here ...
    ...
  }

 

[MikeGomez]

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.

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.

With this method I won't be concerned with finding net torque. I will simply run the physics for a couple of points (or 3 or 4 points) and check the motion at small increments of time.

Checking translational motion is trivial, because at each slice of time we just compare the current position of the center point with its previous position.

Checking the rotational motion is just as trivial because we just take the arctangent of the angle from the center point to any point, and compare with the previous angle. Technically we don't even need to find the angle, because if there is no rotation there will be no relative change in position between the center point and any of the other points.

 

[campbbri]

Great forum and question! Since this is my first post here, I'd like to confess that I failed the only physics courses I took in college, so feel free to discount my opinions if you'd like.

I don't think the problem is correct. If what the question stipulates is true, then I think it implies the following must be true as well. For these cases I think it's easier to imagine an ice rink than a table, but still assume "perfect" friction surfaces.

CASE A - Take the book and push it forward slightly with no rotation (let's say X velocity) so that it translates forward and stops after one second.

CASE B - Spin the book extremely rapidly (let's say Y RPM) in place so that it takes 100 seconds to stop.

CASE C - Set the book so it goes both forward with X linear velocity and rotates at Y RPM.

The problem as written implies that the book will travel further in CASE C than it would in CASE A. It will necessarily take longer than 1 second to stop translating, if it must stop translating at the same time it stops rotating. It does this even though no additional linear momentum was added.

I don't think that's correct.

 

[.Scott]

Great forum and question! Since this is my first post here, I'd like to confess that I failed the only physics courses I took in college, so feel free to discount my opinions if you'd like.

I don't think the problem is correct. If what the question stipulates is true, then I think it implies the following must be true as well. For these cases I think it's easier to imagine an ice rink than a table, but still assume "perfect" friction surfaces.

CASE A - Take the book and push it forward slightly with no rotation (let's say X velocity) so that it translates forward and stops after one second.

CASE B - Spin the book extremely rapidly (let's say Y RPM) in place so that it takes 100 seconds to stop.

CASE C - Set the book so it goes both forward with X linear velocity and rotates at Y RPM.

The problem as written implies that the book will travel further in CASE C than it would in CASE A. It will necessarily take longer than 1 second to stop translating, if it must stop translating at the same time it stops rotating. It does this even though no additional linear momentum was added.

I don't think that's correct.

It is correct.
To make the RPM and the velocity more comparable, let's say that in case B, points near the outside edges of the book are traveling at velocity E = 100X. In this case, the effective translational friction at the outset of case C will be approximate 1% of what it is at the outset of case A. Even after 25 seconds, it will still be only about 2%. At 50 seconds it will still be spinning, but it will be approaching the V = 2E ratio and will almost ready to stop.

-----------------------

Actually, I can compute the initial conditions based on the information you provided.
The value "E", velocity of points near the edge of the book for case B, must be 50·V, because if they matched (if E==V), then it would only take twice as long for the rotation to stop (ie, 2 seconds).

So for case C:
At the start, E=50V and the translational friction is about 2% of what it was at the start of case A and the rotational friction will be about 100% of what it was.

 

[.Scott]

If anyone finds it useful, this table shows the frictional force at various spin(edge) velocity/book velocity ratios.
The frictional forces are expressed at the percentage of the best possible friction. So for travel friction, it is compared to what it would have been without any spin. And for spin friction, it is compared to what it would have been without any travel.

Of course, it's based on the C++ code supplied a few posts ago.

travel/     friction
  spin   --(% of best)--
 ratio   travel    spin
-------  ------   ------
  0.00     0.00   100.00
  0.05     5.00    99.81
  0.10     9.99    99.25
  0.15    14.96    98.32
  0.20    19.90    97.02
  0.25    24.80    95.37
  0.30    29.66    93.37
  0.35    34.46    91.03
  0.40    39.18    88.37
  0.45    43.83    85.41
  0.50    48.39    82.16
  0.55    52.84    78.66
  0.60    57.17    74.92
  0.65    61.36    70.99
  0.70    65.40    66.90
  0.75    69.27    62.69
  0.80    72.95    58.42
  0.85    76.40    54.15
  0.90    79.59    49.97
  0.95    82.46    45.99
  1.00    84.88    42.45
  1.05    86.69    39.67
  1.10    88.13    37.36
  1.15    89.31    35.36
  1.20    90.31    33.59
  1.25    91.17    32.02
  1.30    91.91    30.60
  1.35    92.56    29.32
  1.40    93.13    28.15
  1.45    93.63    27.07
  1.50    94.08    26.08
  1.55    94.48    25.16
  1.60    94.84    24.31
  1.65    95.17    23.52
  1.70    95.46    22.77
  1.75    95.73    22.08
  1.80    95.97    21.43
  1.85    96.20    20.82
  1.90    96.40    20.24
  1.95    96.59    19.69
  2.00    96.77    19.18
  2.05    96.93    18.69
  2.10    97.08    18.22
  2.15    97.22    17.78
  2.20    97.35    17.36
  2.25    97.47    16.96
  2.30    97.58    16.58
  2.35    97.68    16.21
  2.40    97.78    15.87
  2.45    97.87    15.53
  2.50    97.96    15.21

 

[A.T.]

CASE A - Take the book and push it forward slightly with no rotation (let's say X velocity) so that it translates forward and stops after one second.

CASE B - Spin the book extremely rapidly (let's say Y RPM) in place so that it takes 100 seconds to stop.

CASE C - Set the book so it goes both forward with X linear velocity and rotates at Y RPM.

The problem as written implies that the book will travel further in CASE C than it would in CASE A. It will necessarily take longer than 1 second to stop translating, if it must stop translating at the same time it stops rotating. It does this even though no additional linear momentum was added.

I don't think that's correct.

The initial amount of linear momentum in CASE A & CASE C is the same, but the rate of it's transfer to the table is different. Let's assume sliding friction doesn't depend on speed, so that only it's direction changes according to the relative velocity vector.

- In CASE A the friction forces of all book parts are opposed to the linear momentum vector, so the net linear force opposing linear motion is maximal.

- In CASE C the friction forces are almost tangential at each point, so the net linear force opposing linear motion is almost zero.

I'm not saying that in reality this works out exactly as claimed even in such extreme cases, but your argument against it is flawed, because it assumes the same net linear forces in CASE A & CASE C.

 

[rcgldr]

... stop translating at the same time it stops rotating ...

I recorded some of the winter olympic curling. In the cases of collision betwen stones, a stone sometimes ends up with a moderate amount of spin after the collision, and the stones forward motion stops before the spin stops. After the forward motion stops but while the stone is still spinning, there's a slight sideways drift, but the forward motion is stopped and if anything, the drift has a slightly curved backwards component.

 

[campbbri]

Thanks Scott and A.T. Those explanations make sense.

 

[A.T.]

I recorded some of the winter olympic curling.

Friction on Ice is quite complex, and cannot be modeled with a constant coefficient. The pressure and friction melt the ice, so it becomes more slippery because of the water film. When something spins almost in place on ice, it might create a small pond where it wants to stay spinning.

I think for the purpose of the original question we should stick to dry sliding friction.

 

[MikeGomez]

I recorded some of the winter olympic curling. In the cases of collision betwen stones, a stone sometimes ends up with a moderate amount of spin after the collision, and the stones forward motion stops before the spin stops. After the forward motion stops but while the stone is still spinning, there's a slight sideways drift, but the forward motion is stopped and if anything, the drift has a slightly curved backwards component.

Changes in spin/translation due to collisions don’t apply here because the external force is acting at a localized location on the stone, not the entire contact surface.

In general I don't think comparison with the dynamics of a curling stone is a good comparison with the book. The ice has an extremely low coefficient of friction, the stone is much heavier, the stone is thrown with much greater forward momentum, the surface contact area is a very thin ring, and the curlers make the friction in the path unbalanced from one side to the other.

If the stone had a flat contact surface and the ice was perfectly flat and smooth with nobody messing with it, then that might be an interesting experiment, but still overkill. Sorry to be grumpy, but to me it’s almost like having a discussion about the velocity of a bullet and going off on a tangent about the relativistic effects or something else that is not pertinent.

You toss the book on the table and the rotation and translation stop at the same time.
If there is any minute difference it is well below our ability to perceive it as humans.

Try this. Toss the book so that it translates for one second. Then toss it so that is spins for 5 seconds. Now do both at the same time: toss it so it will stop translating in one second, while at the same time giving it enough spin so that it will stop spinning in 5 seconds. You can’t do it, and the question is why.

 

[uumlau]

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.

Yes. There has to be an acceptable qualitative explanation that, even if you cannot write out all the math, proves the behavior from first principles.

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.

This is close to the correct approach. It can be refined by noting the following two principles:

• for the frictional force to STOP rotation, it must be 100% symmetrical around the center of mass
• for the frictional force to STOP translation, it must be 100% against the direction of motion of the center of mass (the net force of friction must "point at" the center of mass)

I shouldn't have to prove the above points, but I can if someone would care to question them.

The proof of why both translation and rotation have to stop at the same time is that neither of these conditions can be met while the book is both rotating and translating. If the book is translating, any torque will be around a node that is some distance from the center of mass (the distance as described by D H, above). Similarly, if the book is rotating, the net force on the book due to friction cannot be pointed straight at the center of mass.

So here are the three cases:

1. If rotating only, the node sits at the center of mass and stays there.
2. If translating only, there is no node, because there is no rotation by which to determine where the node is.
3. If both rotating and translating, the node gradually moves from some point away from the center of mass (even beyond the extent of the book, if it's a fast translation/slow rotation), inexorably toward the center of mass. The book MUST stop when the node reaches the center of mass, due to both rotation and translation needing to approach that point continuously.

The most difficult part of my argument to justify for others is the requirement that both rotation and translation must stop at the same time, since simply asserting it merely begs the question.

The reason is the two principles I started with: the conditions for stopping rotation and translation are mutually exclusive. They are only met at that infinitesimal point of time when both rotation and translation are stopped, or in the case that there is only rotation or the case that there is only translation.

 

[sophiecentaur]

I still think the most relevant factor in all of this is that only one point on the book is ever stationary at anyone time. This refers to the spinning and translating condition. There is exponential (instantaneous) rate of decay (with distance) so no single point (except the Zero velocity point - which is constantly shifting about, of course) will lose its energy before any other. So you cannot choose more than the 'zero velocity' point that will also have zero velocity.

I don't think this argument is circular (not a pun) but I don't see how more than one point could actually be stationary. Considering the trajectories of all the points on the surface, they will take some of them to the zero velocity condition (only the ones for which ωr = |v|). None of the other points can be stationary.

Does anyone see what I am getting at?

 

[Delta Kilo]

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.

Well, I think I actually did prove it, at least for a uniform ring.

If we introduce $p = \log(\frac{v}{\omega})$, $q= v \omega$, then it can be shown that
$p' = -\sqrt{\frac{2}{q}} \sinh p \int\limits_0^{2\pi} \frac{\sin \alpha}{\sqrt{\cosh p - \sin \alpha}} d\alpha$
It is easy to show that the integral is always > 0, and so the sign of $p'$ is always the opposite of $p$. This means, starting from a finite value of p, $|p|$ always decreases with time. Which in turn means, when $q → 0$ (that is when either $v$ or $\omega$ or both → 0), $p$ and therefore the ratio $\frac{v}{\omega}$ remains finite, so both stop at the same time.

 

[.Scott]

How about this?
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.

This is not true.
In the simple case that I examined, where the book is in the shape of a circular disk, the translational and rotational parts did not remain constant.

Instead they approached a specific ratio. In terms of energy, the translate to rotate ratio was 2 to 1. In terms of speed, where the rotational speed is defined as the relative speed of a point on the circumference to the point in the middle, that ratio was 1 to 1.

If the ratio starts higher than this, the translational friction is more than twice the rotational friction, so the translational energy is reduced disproportionately faster.

If the ratio starts lower than this, the rotational friction is more than half the translational friction, so the rotational energy is reduced disproportionately faster.

---- edit to add ---
I just realized the post I am responding to is not recent.
Still, it shows that the first attempt by the OP was almost valid. The only problem being that the situation is better than he supposed.

 

[AlephZero]

This is close to the correct approach. It can be refined by noting the following two principles:

• for the frictional force to STOP rotation, it must be 100% symmetrical around the center of mass
• for the frictional force to STOP translation, it must be 100% against the direction of motion of the center of mass (the net force of friction must "point at" the center of mass)

I shouldn't have to prove the above points, but I can if someone would care to question them.

It would be interesting to see the proofs. Both the assertions see to be wrong. The first one seems obviously wrong, if you spin a wheel and stop it with a friction force applied at one point on the rim.

 

[uumlau]

It would be interesting to see the proofs. Both the assertions see to be wrong. The first one seems obviously wrong, if you spin a wheel and stop it with a friction force applied at one point on the rim.

A clarification of my initial points:

• for the frictional force to STOP rotation without also stopping translation, it must be 100% symmetrical around the center of mass
• for the frictional force to STOP translation without also stopping rotation, it must be 100% against the direction of motion of the center of mass (the net force of friction must "point at" the center of mass)

My apologies for not being more clear in my initial post. I'm not trying to "always be right" by modifying my explanation, but communicating qualitative ideas is an imprecise art. I appreciate your feedback in that it helps clarify things all around. If it were an easy problem, this thread would be very short.

In your counter-example, there is also a fixed axle exerting a force that exactly counters any translational force on the wheel. I'm regarding the book as uncoupled to anything other than the frictional force on the surface. If the book were nailed down with some sort of axle on the surface, the OP problem could not be posed. In the problem as posed, to stop rotation without stopping translation requires a rotationally symmetric application of forces.

Given a rotating/translating book, the rotation is not around the center of mass (it is, in the book's frame of motion, but not ours), but around a moving point some distance from the path followed by the center of mass, direction and distance depending on the the ratio between the angular spin and the translational velocity.

If there were no translation, that center of rotation would be at the same point as the center of mass. If there were no rotation, that point would be infinitely far from the center of mass.

It is the fact that the point is a finite distance from the center of mass that forces both rotation and translation to stop simultaneously. The simplest intuitive case to consider is when that point is underneath the book, but not coincidental with the center of mass. In this case, there is an asymmetric set of forces on the book, exerting a torque around that point of rotation (not the center of mass), and also decelerating the translation (because the forces aren't symmetrical around the center of mass).

Only that single point of rotation, underneath the book, feels no net force. In order for the book to stop rotating, but keep translating, that center of rotation must instantaneously move from underneath the book to infinitely far from the book. In order for the book to stop translating, but keep rotating, the net force must instantaneously move from where it is (exerting a force offset from the center of mass, slowing both rotation and translation) to the center of mass.

If (contrary to my intuitive interpretation in my earlier post) the point of rotation never reaches the center of mass, then both rotation and translation must cease simultaneously. If it continuously approaches the center of mass (or a point infinitely far), then the condition that rotation and translation must stop simultaneously applies for the duration until reaching either limit, which, by arguments of continuity, implies that rotation and translation must stop simultaneously AT either limit. I'm beginning to suspect that the center of rotation doesn't move unless the frictional force from the surface is uneven (due to normal irregularities in the surface). I'd have to sit down and do the math to figure out if it does, but I'm refraining from doing so because I prefer to hone the qualitative explanation, for now. (And because I'm lazy. )

tl; dr version:

If the book is both rotating and translating, there is no way for it to stop just one or the other without some kind of discontinuity in the motion, as indicated by the location of the center of rotation.

 

[Stochastic13]

Isn't is simply due to loss of kinetic energy i.e. the book is moving (rotating and translating) due to kinetic energy that friction opposes. Since friction is not a conservative force, the kinetic energy is lost and since the book is on the flat table there is not potential energy so when the kinetic energy ends then the movement (rotation and translation) stops at the same time.