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

[.Scott]

There will be only one point at any time that is not moving wrt the table. For a disc the points will lie on a circle where tangential velocity equals minus translation velocity of centre. For a general shape the locus may not be a circle (?) this point is losing no energy from the book.

OK. I misunderstood what you were saying.
Actually, for a disc the point will lie on a spiral that will asymptotically spiral in towards or out towards a circle.

 

[sophiecentaur]

OK. I misunderstood what you were saying.
Actually, for a disc the point will lie on a spiral that will asymptotically spiral in towards or out towards a circle.

Can you be sure of that? Remember, the point is where the tangential equals the translational velocity - whatever the actual value (which will, of course, reduce with time). If the two motions decay together then the locus will be at the same distance from the centre. It would be only if what we are trying to prove were false that the locus curve would spiral inwards / outwards.

 

[sophiecentaur]

@.scott
I'm afraid that I trip out with such long and involved posts. I like to read some Maths, which condenses a model into a recognisable form. Maths was actually invented because it is impossible to hold so many lines of description in your head - if you are not the one who wrote it. Some of the most sophisticated Physics can be boiled down into a very few lines and this problem doesn't approach the complexity of EM theory - which is covered by just four equations.
If your graphics does not 'translate' to equations then it may be that there is something wrong somewhere.

 

[.Scott]

Can you be sure of that? Remember, the point is where the tangential equals the translational velocity - whatever the actual value (which will, of course, reduce with time). If the two motions decay together then the locus will be at the same distance from the centre. It would be only if what we are trying to prove were false that the locus curve would spiral inwards / outwards.

The magic spin to move ratio is about 0.865 - I'm guessing it's really 5/6.
Here's the table. I've set it to show the ratio=0.86 with better resolution. The key is when the "4Sfe/3Tfe" column goes to 1. If you start it at that ratio, it will stay there. Otherwise it will move towards it.

Am I sure? I would say moderately. The only way to be sure about models like this is to run known cases against them. I've done a little of that - but not enough to be more than moderately confident.

| Pivot Pt.| Energy Loss % of Best |           |  d Energy / d Ratio   |
|  (Ts/Ss) | Translate |   Rotate  | 4Sfe/3Tfe | Translate |   Rotate  |
|----------|-----------|-----------|-----------|-----------|-----------|
    0.000,     -0.000,    100.000
    0.100,      9.989,     99.495,     13.281,      0.999,     -0.051
    0.200,     19.902,     97.935,      6.561,      0.991,     -0.156
    0.300,     29.663,     95.235,      4.281,      0.976,     -0.270
    0.400,     39.189,     91.309,      3.107,      0.953,     -0.393
    0.500,     48.391,     86.101,      2.372,      0.920,     -0.521
    0.600,     57.170,     79.612,      1.857,      0.878,     -0.649
    0.700,     65.407,     71.933,      1.466,      0.824,     -0.768
    0.800,     72.952,     63.304,      1.157,      0.754,     -0.863
    0.850,     76.401,     58.769,      1.026,      0.690,     -0.907
    0.851,     76.468,     58.677,      1.023,      0.669,     -0.920
    0.852,     76.534,     58.586,      1.021,      0.660,     -0.909
    0.853,     76.600,     58.496,      1.018,      0.659,     -0.909
    0.854,     76.666,     58.405,      1.016,      0.658,     -0.909
    0.855,     76.732,     58.313,      1.013,      0.665,     -0.921
    0.856,     76.799,     58.220,      1.011,      0.664,     -0.921
    0.857,     76.864,     58.130,      1.008,      0.655,     -0.909
    0.858,     76.930,     58.039,      1.006,      0.654,     -0.909
    0.859,     76.995,     57.948,      1.003,      0.653,     -0.909
    0.860,     77.061,     57.856,      1.001,      0.660,     -0.921
    0.861,     77.127,     57.764,      0.999,      0.659,     -0.921
    0.862,     77.192,     57.673,      0.996,      0.650,     -0.909
    0.863,     77.257,     57.582,      0.994,      0.648,     -0.909
    0.864,     77.321,     57.491,      0.991,      0.647,     -0.909
    0.865,     77.387,     57.399,      0.989,      0.654,     -0.921
    0.866,     77.452,     57.307,      0.987,      0.653,     -0.921
    0.867,     77.516,     57.216,      0.984,      0.644,     -0.909
    0.868,     77.581,     57.125,      0.982,      0.643,     -0.909
    0.869,     77.645,     57.034,      0.979,      0.642,     -0.909
    0.870,     77.710,     56.942,      0.977,      0.649,     -0.921
    0.900,     79.588,     54.210,      0.908,      0.626,     -0.910
    1.000,     84.889,     45.766,      0.719,      0.530,     -0.844
    1.100,     88.136,     40.128,      0.607,      0.325,     -0.564
    1.200,     90.322,     36.021,      0.532,      0.219,     -0.411
    1.300,     91.917,     32.780,      0.475,      0.160,     -0.324
    1.400,     93.132,     30.124,      0.431,      0.121,     -0.266
    1.500,     94.083,     27.895,      0.395,      0.095,     -0.223
    1.600,     94.844,     25.991,      0.365,      0.076,     -0.190
    1.700,     95.465,     24.342,      0.340,      0.062,     -0.165
    1.800,     95.977,     22.897,      0.318,      0.051,     -0.144
    1.900,     96.406,     21.620,      0.299,      0.043,     -0.128
    2.000,     96.769,     20.481,      0.282,      0.036,     -0.114
    2.100,     97.079,     19.460,      0.267,      0.031,     -0.102
    2.200,     97.346,     18.538,      0.254,      0.027,     -0.092
    2.300,     97.578,     17.701,      0.242,      0.023,     -0.084
    2.400,     97.781,     16.938,      0.231,      0.020,     -0.076
    2.500,     97.958,     16.238,      0.221,      0.018,     -0.070
    2.600,     98.115,     15.596,      0.212,      0.016,     -0.064
    2.700,     98.255,     15.002,      0.204,      0.014,     -0.059
    2.800,     98.380,     14.453,      0.196,      0.012,     -0.055
    2.900,     98.491,     13.943,      0.189,      0.011,     -0.051

 

[.Scott]

Here's the code that generated that table:

//
//=====================================================================
// SpinEnegryModel()
//
//  This routine computes the frictional forces acting on a spinning
// and translating disc at one moment in time.  The ratio of the speeds
// of rotation and translation are varied over a wide range.
//  It returns a multiline text report output in the form of a CString.
//
CString SpinEnergyModel()
{
  struct ReportEntry {
    double fRatio;
    double fELossX;
    double fSpinELoss;
  };

  CString s, sRpt;
  ReportEntry reLast, reNew;

  int     nX, nY, nDiv, nYRange, bFirst;
  __int64 nYRange2, nDiv2, nArea;
  double  fR, fX, fY, fX2, fPivot, fPR, fRDiv;
  double  fForceX, fForceY;
  double  fSpeedX, fSpeedY, fSpeed;
  double  fForceXSum, fForceYSum;
  double  fSpinArm, fSpinForce, fSpinELoss;
  double  fSpinELossSum;
  double  fSpinELossMax, fArea;

  //
  //  I am presuming a circular book.  "fR" is the radius of that book
  // and I am setting it to 1.  This doesn't really matter because
  // everything will scale to the radius anyway.
  fR = 1.0;
  //
  //  I will be performing discrete integrations across the surface of
  // the book.  "nDiv" is the number of discrete interval I will be
  // using along the radius.  Computation time will be proportional to
  // nDiv*nDiv but precision improves as nDiv is increased.
  nDiv  = 200;
  //
  //  Two values that will be needed in the inside loops (below) are
  // precomputed here.  They are the actual distance of one interval
  // (fRDiv) and the square of nDiv (nDiv2).
  fRDiv = fR/nDiv;
  nDiv2 = (__int64)nDiv*nDiv;
  //
  //  Integrate two initial values, these are not dependent on the
  // spin/translate ratio, so they are computed here, outside the pivot
  // point loop.
  //   nArea:          The area of the circular book surface.
  //   fSpinELossMax:  The force that would have been applied against
  //                friction if there was no translational force.
  nArea = 0;
  fSpinELossMax = 0.0;
  //
  //  Discrete integral along the direction of motion from the
  // trailing edge to the leading edge.
  for(nX=-nDiv;nX<=nDiv;nX++) {
    //
    //  The book is circular and we only want to integrate across
    // points that are on the books surface.  So we compute the range
    // of Y value that will keep us on the book (nYRange).
    //  We are using __int64 foor the intermediate results.  This will
    // allow us to use values of nDiv above 45000 - not that we would
    // want to run the program that long.
    nYRange2 = nDiv2-((__int64)nX*nX);
    nYRange = (int)sqrt((long double)nYRange2);
    //
    //  Compute the X coordinate and its square.
    fX  = nX*fRDiv;
    fX2 = fX*fX;
    //
    //  Compute the book's area by integration so that it will match
    // the other integrals.
    nArea += 1+2*nYRange;
    //
    //  Discrete integral across the direction of motion from the
    // left edge to the right edge.
    for(nY=-nYRange;nY<=nYRange;nY++) {
      //
      // The Y coordinate.
      fY = nY*fRDiv;
      //
      // Integrate the maximum spin energy loss (with no translation).
      // Note:
      //   Both the force and the speed are proportional to the radius,
      //  so the total energy loss will be proportional to the square
      //  the radius.
      fSpinELossMax += fX2+fY*fY;
    }
  }
  fArea = (double)nArea;

  //
  // We skip some steps the first time through the loop.
  bFirst = true;
  //
  // Apply the heading to our report.
  sRpt =
    L"| Pivot Pt.| Energy Loss % of Best |           |  d Energy / d Ratio   |\n"
    L"|  (Ts/Ss) | Translate |   Rotate  | 4Sfe/3Tfe | Translate |   Rotate  |\n"
    L"|----------|-----------|-----------|-----------|-----------|-----------|\n";

  //
  //  The ratio of the rotational speed and the translational speed can be
  // described by specifying the "pivot" as a ratio of the radius, which I
  // calling "pPivot".  When pPivot is zero, there is only spin motion.
  //  The pivot point is the location where there is no relative motion
  // between the book and the surface.  It will always be along a line
  // that crosses through the center of the book and is perpendicular to
  // the translational motion of the book.
  //  When pPivot is 1, the pivot point will be at the circumference of
  // the book.  When pPivot is zero, the pivot point is at the center of
  // the book.
  //  I am varying the pPivot from 0 (book center) to 2.5 (well outside
  // the book so there is much more translation that rotation.
  for(fPivot=0.00; fPivot<=3.00; fPivot+=0.10) {
    //
    //  Look carefully around 0.86.
    if((fPivot>0.81)&&(fPivot<0.995)) {
      fPivot -= 0.099;
      if(fPivot < 0.85) fPivot = 0.85;
      if(fPivot > 0.87) fPivot = 0.90;
    }
    //
    //  Multiplying "fPivot" by the radius gives us the actual distance
    // between the pivot point and the center of the book.
    fPR   = fPivot*fR;
    //
    //  We will be integrating three values:
    //   fForceXSum:  The translational force along the direction of
    //           travel.
    //   fForceYSum:  The translational force perpendicular to the
    //           direction of travel.  This should be zero and is used
    //           as a check.
    //   fSpinELossSum:  The energy loss against the book's spin.
    fForceXSum  = 0.0;
    fForceYSum  = 0.0;
    fSpinELossSum = 0.0;
    //
    //  Discrete integral along the direction of motion from the
    // trailing edge to the leading edge.  These nX/nY loops step
    // through the same values as the nX/nY loops above.
    for(nX=-nDiv;nX<=nDiv;nX++) {
      nYRange2 = nDiv2-((__int64)nX*nX);
      nYRange = (int)sqrt((long double)nYRange2);
      //
      //  Compute the X coordinate and its square.
      fX  = nX*fRDiv;
      fX2 = fX*fX;
      //
      //  Discrete integral across the direction of motion from the
      // left edge to the right edge.
      for(nY=-nYRange;nY<=nYRange;nY++) {
        //
        // The Y coordinate.
        fY = nY*fRDiv;
        //
        //  Compute the direction of the frictional force.  It will be
        // against the direction of motion and can be computed
        // from the relative position of this point to the pivot point.
        fSpeedX = -fY-fPR;
        fSpeedY = fX;
        //
        //  Compute the speed of point (fX,fY) against the surface.
        fSpeed  = sqrt(fSpeedX*fSpeedX+fSpeedY*fSpeedY);
        //
        //  Check for the zero-friction point.  This is important to
        // avoid zero/zero division.
        if(fSpeed<0.001) continue;
        //
        //  I'm using a coefficient of friction of "1", so the absolute
        // value of (fForceX,fForceY) must be 1.
        fForceX = fSpeedX / fSpeed;
        fForceY = fSpeedY / fSpeed;
        //
        //  The turn force is the cross product of the force and the
        // arm to the center of gravity (fX,fY).
        fSpinArm = sqrt(fX2+fY*fY);
        fSpinForce = fY*fForceX-fX*fForceY;
        fSpinELoss = - fSpinArm * fSpinForce;

        //
        //  If the speed is less than fR-fPR, we can discard the
        // translation force because it will all cancel out within
        // that circle.
        //  This code was only used for test purposes.  It is now
        // commented out.
        //
        // if(fSpeed<(fR-fPR)) fForceX = 0.0;

        //
        // Accumulate frictional effects.
        fForceXSum += fForceX;
        fForceYSum += fForceY;
        fSpinELossSum += fSpinELoss;
      }
    }
    //
    // Compute all the proportional forces.
    fForceX  = fForceXSum/fArea;
    fForceY  = fForceYSum/fArea;
    reNew.fSpinELoss = fSpinELossSum/fSpinELossMax;
    //
    //  Except for sign, the translational energy loss (fELossX) is
    // proportional to the force.  This is because it is proportional
    // to the overall translational speed (fPR).  So we would
    // multiply by -speed (for energy loss) and then divide by speed
    // (for proportional energy loss).
    reNew.fRatio  = fPivot;
    reNew.fELossX = -fForceX;
    //
    // Output fPivot, fELossX, and fSpinELoss here ...
    s.Format(
      L"%9.3f,%11.3f,%11.3f",
      reNew.fRatio, 100*reNew.fELossX, 100*reNew.fSpinELoss
    );
    sRpt += s;
    //
    if(bFirst) {
      bFirst = false;
      s = L"\n";
    } else {
      double fDeltaR, fDeltaMove, fDeltaSpin, d43RatioSsTs;

      fDeltaR      = reNew.fRatio     - reLast.fRatio;
      fDeltaMove   = reNew.fELossX    - reLast.fELossX;
      fDeltaSpin   = reNew.fSpinELoss - reLast.fSpinELoss;

      s.Format(
        L",%11.3f,%11.3f,%11.3f\n",
        (4.0/3.0) * reNew.fSpinELoss / reNew.fELossX,
        fDeltaMove/fDeltaR, fDeltaSpin/fDeltaR
      );
    }
    sRpt += s;
    //
    reLast = reNew;
  }
  return sRpt;
}

 

[.Scott]

If your graphics does not 'translate' to equations then it may be that there is something wrong somewhere.

I think I translated it successfully. It's just not the most fluent thing I do. Some of those ratios (for example 4/3) are based on simple integrations of circle area - once for spin, once for translation.

As you can see in the previous post, I have also redone the program to report energy loss instead of force. This hasn't affected the "translate" (or "travel") column because there's a straight speed factor and the energy loss is scaled to speed anyway. But it made a difference with the spin column.

 

[Dadface]

The assumption still seems to be that translational and rotational motion stop at the same time. Where is the evidence to back up this assumption? I suggest that respondents here try out a few preliminary tests for themselves. There are many variables to consider but in the first instance all that is needed is a flat surface and objects that can slide and rotate over it.
I have tried it out with different things and I am not at all convinced that both types of motion stop simultaneously. Try an object or book with a smooth surface and put a lot of spin on it.

 

[voko]

There is quite some literature on the subject. This has even been the subject of at least one doctoral dissertation. The short answer is yes, friction couples rotation and translation and they both stop together. The final ratio between the linear and angular speed depends on the shape and mass distribution in the object in question. Some references:

A. Yu. Ishlinskii, B. N. Sokolov, and F. L. Chernousko, Motion of plane bodies with dry friction, Izv. Akad. Nauk SSSR, Mekh. Tver. Tela, 16 (4) (1981) 17-28

K. Voyenli and E. Eriksen, On the motion of an ice hockey puck, American Journal of Physics 53, 1149 (1985).

S. Goyal, A. Ruina and J. Papadopoulos, Planar sliding with dry friction. Part 1. Limit surface and moment function, Wear 143, 307–330 (1991).

S. Goyal, A. Ruina and J. Papadopoulos, Planar sliding with dry friction. Part 2. Dynamics of motion, Wear 143, 331–352 (1991).

Z. Farkas, G. Bartels, T. Unger and D. E. Wolf, Frictional Coupling between Sliding and Spinning Motion, Phys. Rev. Lett. 90, 248302 (2003). [http://arxiv.org/pdf/physics/0210024.pdf]

Guido Bartels, Mesoscopic Aspects of Solid Friction, PhD thesis, 2006 [http://duepublico.uni-duisburg-essen.de/servlets/DocumentServlet?name=duett-01272006-083621/]

Mark Denny, Comment on “On the motion of an ice hockey puck” by K. Voyenli and E. Eriksen, Am. J. Phys. 74, 554 (2006)

I would recommend reading at least the paper by Farkas et al.

 

[.Scott]

Z. Farkas, G. Bartels, T. Unger and D. E. Wolf, Frictional Coupling between Sliding and Spinning Motion, Phys. Rev. Lett. 90, 248302 (2003). [http://arxiv.org/pdf/physics/0210024.pdf]

I would recommend reading at least the paper by Farkas et al.

Farkas found the asymptotic ratio to be 0.653, a lot less than my 0.865. I believe him.
The key thing is that we both agree that there is a value will always be approached as the disc slows. That, om general, the pivot point draws out a spiral, not a circle, onto the disc.

 

[Dadface]

There is quite some literature on the subject. This has even been the subject of at least one doctoral dissertation. The short answer is yes, friction couples rotation and translation and they both stop together. The final ratio between the linear and angular speed depends on the shape and mass distribution in the object in question. Some references:

A. Yu. Ishlinskii, B. N. Sokolov, and F. L. Chernousko, Motion of plane bodies with dry friction, Izv. Akad. Nauk SSSR, Mekh. Tver. Tela, 16 (4) (1981) 17-28

K. Voyenli and E. Eriksen, On the motion of an ice hockey puck, American Journal of Physics 53, 1149 (1985).

S. Goyal, A. Ruina and J. Papadopoulos, Planar sliding with dry friction. Part 1. Limit surface and moment function, Wear 143, 307–330 (1991).

S. Goyal, A. Ruina and J. Papadopoulos, Planar sliding with dry friction. Part 2. Dynamics of motion, Wear 143, 331–352 (1991).

Z. Farkas, G. Bartels, T. Unger and D. E. Wolf, Frictional Coupling between Sliding and Spinning Motion, Phys. Rev. Lett. 90, 248302 (2003). [http://arxiv.org/pdf/physics/0210024.pdf]

Guido Bartels, Mesoscopic Aspects of Solid Friction, PhD thesis, 2006 [http://duepublico.uni-duisburg-essen.de/servlets/DocumentServlet?name=duett-01272006-083621/]

Mark Denny, Comment on “On the motion of an ice hockey puck” by K. Voyenli and E. Eriksen, Am. J. Phys. 74, 554 (2006)

I would recommend reading at least the paper by Farkas et al.

I can't take the work of Farkas et al and others seriously yet because because their findings do not conform to simple observations I have made. Where is the experimental evidence that both types of motion stop simultaneously?
I have done simple tests on my desk top and in many cases with different objects the motions do seem to stop at the same time. But not in all cases. I am able to put a lot of spin on a light weight, smooth covered book at the edge of my desk whilst sliding it towards the middle of my desk. Each time it seems to stop sliding long before it stops spinning. try it for youself.

 

[voko]

The Farkas paper deals with a uniform disk specifically. Bartel's thesis is more generic, and he shows that for some mass distributions the terminal motion may be pure rotation. See Fig. 3.10.

 

[Dadface]

Just tried it several times with a disc shaped lid from a storage jar and in each case both motions did seem to stop at the same time. Will try different variations later.

 

[MikeGomez]

Just tried it several times with a disc shaped lid from a storage jar and in each case both motions did seem to stop at the same time. Will try different variations later.

That's because your desk surface and/or the book (or disk) is not perfectly flat.

It just takes a very slight imbalance in the smoothness of the contact surface to throw the whole experiment off. Unless you can verify the smoothness of both the book and desk with high precision, the heavier book experiments are more valid because they are not quite as sensitive to small imperfections in smoothness.

This "catching" effect is even more pronounced if you are using a lid or any kind of ring shape instead of book or flat disk shape. Friction pressure gets localized instead of being spread out over the entire object.

 

[.Scott]

I can't take the work of Farkas et al and others seriously yet because because their findings do not conform to simple observations I have made. Where is the experimental evidence that both types of motion stop simultaneously?
I have done simple tests on my desk top and in many cases with different objects the motions do seem to stop at the same time. But not in all cases. I am able to put a lot of spin on a light weight, smooth covered book at the edge of my desk whilst sliding it towards the middle of my desk. Each time it seems to stop sliding long before it stops spinning. try it for youself.

The Farkas paper does include experiments. They used CD discs, data side down.

 

[Dadface]

The Farkas paper does include experiments. They used CD discs, data side down.

Yes I know. I scanned the paper. The initial observations I referred to were made were made with a book whose surface was smooth enough for me to set it spinning fairly fast whilst getting it moving with translational motion. I have since tried other things, including CD discs, but was unable to get these spinning as fast as I was able to get the book spinning.
Quite often the spinning /moving discs seemed to stop and then move momentarily in a different direction. There were also times when air pressure seemed to affect and slow down the movement. I assume this was due to the smoothness and close contact of the surfaces
There are so many variables to consider and pressure now seems to be another one.

 

[.Scott]

There are so many variables to consider and pressure now seems to be another one.

The high spin situation is especially sensitive to anything that varies from the ideal. For translational motion, you're creating a near frictionless surface, so any local valleys in the surface will affect a slow moving object. Any dust or smudge on the table or the disc could cause a snag. The spin will cause more heat along the edge of the disc than in the middle - giving the disc a slight saddle shape. Also, if the spin is not strictly horizontal, there may be a bit of precession.

 

[A.T.]

Friction on Ice is quite complex, and cannot be modeled with a constant coefficient.

Here is more on the physics of curling:

https://www.youtube.com/watch?v=7CUojMQgDpM

 

[sbstratos79]

hehe, I skipped pages 3-7 of this thread. But didn't jbriggs444 nail it? :/ His/Her explanation seemed pretty convincing(to me atleast)