What is the orientation of the vector of friction?

[JrK]

And I measured d1. Again, you forgot something very important: the position of the dot of contact from start to end: from a fixed dot on the circle for example. It is easy to watch the problem: take a circle fixed on the ground. Rotate a wall around the circle without any sliping (there is friction but there is no heating), you can see a distance moved by the distance measured but like the dot of contact moved the same distance: there is no slip, so no heating. Here, I took at start d2 because it is logic for me, but with the construction you will have d1 but you need to take in account the distance moved by the dot of contact (in reference to a fixed dot on a part of the device) and at final I have : d1-(d1-d2).

Note, it doesn't resolve my problem with the needle and the elastic. I don't find the mistake in that different example.

 

[A.T.]

...you forgot something very important: ...

Not important to the work done by friction. All that matters is the relative velocity at contact, which integrated gives you the sum of green and violet lines. Multiplied by force it gives you work:

https://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

 

[JrK]

For me when you asked me to build the case 3/ I rotate the circle of the same angle than the wall rotate around A0 but it is not true ! the circle in the reference of the wall rotates only of a smaller angle. Look at the drawings, the circle rotates only of the angle that will give d2 not d1. So, like it doesn't help me to understand why I see a lack of friction in the movement. And like I drew the example with the needle + elastic to help me. Could you help me to find the mistake inside the example with the needle + elastic please ? Here I don't need any transformation so the mistake could be at the angle.

I rewrite the example:

The circle moves in horizontal translation, the circle doesn't rotate around itself, the red wall rotates around A0, and A0 is fixed to the ground. The circle is always at contact with the red wall. The mass is very low and there is no friction. There is no acceleration/deceleration. The bodies are rigids except the elastic. The force of the elastic is contant : F. I drew the start and the final position, the circle and the red wall are controlled in position with a cylinder and a motor for example, I count the energy I need to give to push the circle and I count the energy stored inside the elastic, I also compute the energy from the force F3 and F4 on the lateral walls. There are 4 objects: the circle and the wall, the elastic and the needle. The needle has a diameter very small: some atoms, to be near the dot of contact, I cannot be at the dot of contact but from start to end with a very small diameter for the needle, the distance from the dot of contact will not change to reach the distance d1 in the calculation of the energy stored inside the elastic. The sum of forces on the needle is 0: the elastic pulls it but the lateral walls prevent it to move closer to the dot B. Please, look the enlargements to watch the needle more farther.

The example:

The example from start to end:

Enlargement of the previous image:

Enlargement of the area where there is the elastic, I drew the start and final position, the elastic is taken by the dots A and B, the dot B is fixed on the red wall, the dot A is fixed on the needle, the needle is near the dot of contact between the circle and the red wall:

To watch the forces of the elastic F1 and F2, I drew the start and final position, at start the elastic has a length at 0, and at final its length is d2:

To watch the needle (the needle is perpendicularly to the screen and its diameter is very small), I drew only the final position, the needle is pulled by the elastic but the lateral walls prevent it to move closer the dot B:

Enlargement, I drew the final position, to watch the forces from the needle to the lateral walls (circle and red wall):

The program :

#include <stdio.h>
#include <math.h>

int main()
{
long double pi=M_PI;

long int N=100000000L;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44.99/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wd=0,wf=0,ix1=0;
long double iy1=0,ix2=0,yi2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,xi1sav=0,yi1sav=0,xsav=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,sumdx=0,dlrc=0,dlpc=0;

long int i;
qy1=D;

for(i=0;i<N;i++)
{
  //a1=(long double)(asup/180.0*pi-ainf/180.0*pi)/N*i;
  qx1=xinf+(xsup-xinf)/N*i;
  a1=atanl(qy1/qx1); if(i==0) a2=a1;
  s=R/tanl(a1/2.);
  px1=qx1-s;
  py1=qy1-R;
  if(i==0 || i==N-1) printf("\npx1=%Lf",px1);
  ix1=px1+R*sinl(a1);
  iy1=py1-R*cosl(a1);
  if(i<1) {xi1sav=ix1;yi1sav=iy1;xsav=px1;asav=a1;}
  //printf("\npx1=%Lf , py1=%Lf , a=%Lf , ix1=%Lf , iy1=%Lf , a1=%Lf",px1,py1,180/pi*atanl((fabsl(iy1-yi2)-fabsl(py1-py2))/(fabsl(ix1-ix2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
  suma+=a1;

  if(i>0)
  {
   sumdx+=(fabsl(px1-px2)-fabsl(ix1-ix2))*cosl(a1);

   l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+yi2*yi2))/2.;
   xf=ix2+l*sinl(a1)*fabsl(a2-a1);
   yf=yi2-l*cosl(a1)*fabsl(a2-a1);
   af=fabsl(atanl((iy1-yf)/(ix1-xf)));
   wf+=fabsl(px2-px1)*cosl(af);
   wd+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cosl(a1-af);
   //printf("\nl=%Lf , xf=%Lf , yf=%Lf , af=%Lf , diffa=%Lf",l,xf,yf,180/pi*af,180/pi*(a1-af));
   wp+=fabsl(sinl(a1-af))*l*fabsl(a2-a1);
   dlpc+=(px2-px1)*sinl(a1);
   dlrc+=l*fabsl(a2-a1);

  }
  px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  ix2=ix1;  yi2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\ndlp=%Lf , dli=%Lf, dlpc=%.18Lf , dlrc=%.18Lf, diff=%.18Lf",px1-xsav,ix1-xi1sav,dlpc,dlrc,(dlpc-dlrc));
printf("\nsuma=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",(asav-a1)*180/pi,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}

The results:

The energy to move the circle: lg*cos(44.5°)*F = d1*F (in the computation I calculate the integral)
The difference of energy stored inside the elastic: d2*F
The energy needed from F3 is equal to the energy recovered from F4

d1 is different of d2

 

[A.T.]

For me when you asked me to build the case 3/ I rotate the circle of the same angle than the wall rotate around A0 but it is not true.

That's just a coordinate transformation to the rest frame of the wall. In such a coordinate transformation you have to rotate everything in the same manner.

 

[JrK]

That's just a coordinate transformation to the rest frame of the wall. In such a coordinate transformation you have to rotate everything in the same manner.

I drew, with your method, the rotation of a wall around a fixed circle without slipping but with friction. The energy from the friction is 0. The energy to rotate the wall around the circle is 0 (I supposed no acceleration, no decceleration and no mass). It is logic, no energy from friction, no energy to rotate the wall. But with your method I have the energy from friction at L*F with L the length of the friction your method sees. Because your method forget to take in account the movement of dot of contact relatively to a fixed dot on the circle. Something is wrong with your method, for me it is the dot of contact that it didn't take in account. The rotation of the circle I drew in the case 3/ relatively to the wall exists if the dot of contact is as far as the full rotation, the dot of contact is the dot where there is the force of friction. Anyway, if your method works, it must work for a basic rotation of a wall around a fixed circle, it is not the case.

 

[A.T.]

I drew, with your method, the rotation of a wall around a fixed circle without slipping but with friction. The energy from the friction is 0. The energy to rotate the wall around the circle is 0 (I supposed no acceleration, no decceleration and no mass). It is logic, no energy from friction, no energy to rotate the wall. But with your method I have the energy from friction at L*F with L the length of the friction your method sees.

No, my method gives 0 slip distance for pure rolling, if you apply it correctly. Please read the instructions I gave for the two elastics (which are just proxies for the green and violet contact areas you drew in post #179):

You use two elastics, one for each body, with the ends attached to that body at the initial and final contact points. The slipping distance depends on how the elastics meet at the final contact point:

- If they meet at 180° (straight angle) you add their lengths (special case where you can use a single elastic).
- If they meet at 0° (acute angle) you subtract the shorter from the longer length.

Note that even the more general case works only for monotonic motion, not for back and forth sliding which you would have to do piece-wise. The elastics have to stay in contact with the bodies and cannot take short cuts through air. So for concave surfaces you have put them on the inside of the body, and for changing curvature sign you have to do it piece-wise.

 

[JrK]

I used the method of the dot of contact fixed on the ground. I tested that method with another movements and each time I found the energy from the friction equal to the energy needed to move the object(s). I measured the distance moved by the objects compared to the last position. For my example with the friction, I have:

The distance moved by the red wall relatively to the start position of the red wall is d2. The distance moved by the circle in translation is d1-d2. But the dot of contact is fixed, in position and in rotation, so I need to rotate the device like that:

And the distance moved by the circle is well 0. So the distance of friction is d2.

 

[A.T.]

I used the method of the dot of contact fixed on the ground. I tested that method with another movements

No idea what method you mean. If you mean the method I proposed, then please provide the same type of image as in post #179, with the same frame, markings, colors etc.

 

[JrK]

The method:

1/ I take the dot of contact fixed
2/ I measure the distance from the dot of contact from start to end of the first object, it is the path of the dot of contact on the first object
3/ I measure the distance from the dot of contact from start to end of the second object, it is the path of the dot of contact on the second object
4/ The algebraic sum is the distance of friction

The distance 'da' is the distance of the path of the dot of contact, it is a lack of distance of friction because the direction is the direction of the wall but in the same time the wall rotates CW (because the dot rotates CCW but the dot is fixed) and that distance is 'db' the direction is an augmentation of the distance of friction, so I need to add 'db'. With the absolute value of 'da' is equal to the absolute value of 'db', so da+db=0. The distance I have at final is the distance from the dot of contact with the wall and it is d2.

 

[A.T.]

This is not the method I explained to you (see post #179):
- your grey wall is intersecting the circle, but it should always be tangential to it
- the path of the contact on the wall should be on the wall, not on the circle
- the paths of the contact on both objects should meet at the final contact location

 

[JrK]

The method you explained works only for some examples not all. I would like a general method. I used the method of the dot of contact fixed on the ground but like I need to transform the geometry I would like to have a direct method. I find this one, easier, and it is possible to think directly with the drawing.

I fixed a dot C on the end of the red arm, I measure the distance from the dot of contact i1 to C and I reported that distance from the dot of contact i2 (red dotted line), I can measure the distance the red arm slipped WITH the rolling: d3. And d3=9.3, I need to add the distance 'db' because the circle rotates CCW relatively to the red arm, the sum is 9.3+1.05=10.35 but I need to subtract the distance the of rolling 1.05 the sum is 9.3 it is d2 (d1 corrected with the integral of the cosine angle is 10.28).

Why I subtracted the rolling: think with a circle fixed on the ground, rotate the wall around the circle of one turn without any slipping: the distance d3 is equal to the perimeter of the circle: there is no distance of friction. Here, without any slipping of friction, d3 would be at 1.05 (the angle is 15°, the radius at 4). If I measured d3=1.05, there is no energy from friction. If I measure 'x' I need to subtract the 1.05 to 'x'.

 

[A.T.]

The method you explained works only for some examples not all.

It works fine for your original scenario (see post #179). Which other scenario do you want to analyze where it fails? Show me how you applied it.

 

[JrK]

Here is one way to determine the slipping distance relevant for energy dissipated by friction, using an elastic, that works for your specific case (but not in general!):

Attach one end to the circle at the initial contact point, and the other end to the wall at the initial contact point. The initial length is zero, the final length is the slipping distance.

It is what you said. Could you think with my last method please ? it is easy and we don't need to transform the drawing.

 

[A.T.]

It is what you said.

So which other scenario do you want to analyze where my method fails?

Could you think with my last method please ? it is easy and we don't need to transform the drawing.

You don't need to transform for my method either. It just makes it easier to avoid the errors you keep making.

Friction depends on the relative motion of the objects, which is the motion of one object in the rest frame of the other object. We have done the rest frame of the wall in post #179. But you keep bringing up the wall rolling around the circle, so why don't you apply my method in the rest frame of the circle now. Then compare the result to the rest frame of the wall from #179.

 

[JrK]

So for which other scenario do you want to analyze where my method fails?

This one for example:

When I drew your method, I take the wall and the circle at the final position and I rotate all the group to compare with the start position, but the circle doesn't rotate in reality so I need to replace it like I drew for your method. If I rotate the circle, all is fine, but in that case I need to do the same job in my drawing because here like in my example the circle doesn't rotate :

And I measure d2. I'm agree, friction depends of the relative motion but I think you forgot the rotation of the wall around the circle. My method I drew is the method I used in #133.

In the example I drew here, you could apply too the method with the dot fixed on one end of the red wall: the dot C. It is easier to think with the drawing fixed.

 

[A.T.]

This one for example:

- There are no markings on the wall, to see how it moves. Is there slippage or pure rolling?
- The contact areas are missing (the green and violet lines in post #179)

 

[JrK]

- There are no markings on the wall, to see how it moves. Is there slippage or pure rolling?
- The contact areas are missing (the green and violet lines in post #179)

It is a pure rolling.

I added the information:

 

[A.T.]

It is a pure rolling.
View attachment 262418

Then top left is not "my method", but a completely different scenario (pure sliding). Bottom left looks OK, so let's apply my method:

- If they meet at 180° (straight angle) you add their lengths (special case where you can use a single elastic).
- If they meet at 0° (acute angle) you subtract the shorter from the longer length.

They meet at 0° (acute angle), and their lengths are equal, so the difference is zero. Which is the correct slip distance for pure rolling.

Note that in your original scenario (see post #179) they meet at 180° (straight angle), so you have to add their lengths, to get the slip distance.

 

[JrK]

Then top left is not "my method", but a completely different scenario (pure sliding).

It is the method you gave to find the length of sliding in my example at start, look at the drawings, I grouped the circle and the wall, I rotated the group and I never rotate the circle alone. Yes, here I used a pure rolling to show the method doesn't work for the rolling. You can use a part of rolling, your method will not count the rolling. Yes, bottom left is ok, and if I apply the method, it gives that for my example:

And the green measured d2.

 

[A.T.]

Then top left is not "my method", but a completely different scenario (pure sliding).

It is the method you gave to find the length of sliding in my example at start, look at the drawings, I grouped the circle and the wall, I rotated the group and I never rotate the circle alone.

No, you did not. In the top left image the grey circle has the same orientation as in the middle right (original frame). So you have obviously not rotated the grey circle with the black wall. You have only rotated the black wall. You did not transform correctly.

Yes, bottom left is ok,

Yes, here you transformed correctly and it gives the correct result of zero slip distance, for this pure rolling scenario.

and if I apply the method, it gives that for my example:

What is this? What is "my example"? In your original scenario the circle was moving horizontally. If you are introducing a new scenario, then define it properly.

 

[JrK]

What is this? What is "my example"? In your original scenario the circle was moving horizontally. If you are introducing a new scenario, then define it properly.

No, it is the same example, the circle move in horizontal translation and the wall rotates around A0. It is when I rotated the group around A0. Remember the image #147, you replied at #156 and I work on the image:

https://www.physicsforums.com/attachments/cas2-png.261958/

For the pure rolling, I forgot the fixed element, in my example it was A0, so we take A0 fixed in reference, here it is the circle fixed on the ground, so I have:

And if I don't rotate the circle alone after rotate all the group, I can measure a length of friction (I drew a small green but it is at 0) with the magenta curved part.

 

[A.T.]

No, it is the same example, the circle move in horizontal translation and the wall rotates around A0. It is when I rotated the group around A0. Remember the image #147, you replied at #156 and I work on the image:

https://www.physicsforums.com/attachments/cas2-png.261958/

Yes, that old one is correct for your original scenario:

What you posted now in #199 is wrong:

Look at the grey circle. Is it really so hard for you to see the difference?

For the pure rolling...

View attachment 262433

No, this is nonsense again. The correct image for pure rolling is bottom left in the old version:

 

[JrK]

Yeah ! you're right ! Thanks a lot :) I understood my mistake I saw a rolling where there was not. And your method works with a rolling and a slipping, I tested. Just for fun, could you help me in the same example with the needle I resumed at the message #183 ? it is the same but I replace the friction by an elastic attached between the needle and the red wall. It is more a problem of calculation here because the needle is at the position of the dot of contact. Even the lateral forces are high (with a needle of 1 atom, the lateral forces F3 and F4 are around 70000 higher than the force of the elastic).

 

[A.T.]

I saw a rolling where there was not.

What you have in your original scenario is the opposite of rolling: Relative to the wall the rotation of the circle is in opposite direction to roll rotation. And that increases the slip distance.

You can capture all cases with this formula (rest frame of the wall, wall is horizontal, with the circle above it):

slip_distance = | right_circle_translation - CW_circle_rotation * circle_radius |

In your scenario CW_circle_rotation is negative (because the rotation is CCW), so that subtraction makes the slip_distance greater.

could you help me in the same example with the needle I resumed at the message #183 ? it is the same but I replace the friction by an elastic attached between the needle and the red wall. It is more a problem of calculation here because the needle is at the position of the dot of contact.

What is the point of this?

If you want to use elastics just to measure the contact distances, to compute the slip distance from them, then it's just the same as the method I showed you:
https://www.physicsforums.com/threa...vector-of-friction.987415/page-3#post-6332661

If you want the energy of the elastic to match the energy that would be dissipated by friction, then first try to make that work for pure rolling, where the dissipated energy would be zero.

 

[JrK]

What you have in your original scenario is the opposite of rolling: Relative to the wall the rotation of the circle is in opposite direction to rolling. And that increases the slip distance.

You can capture all cases with this formula (rest frame of the wall, wall is horizontal, with the circle above it):

slip_distance = | right_translation_along_wall - CW_circle_rotation_angle * circle_radius |

In your scenario CW_circle_rotation_angle is negative (because the rotation is CCW), so that subtraction makes the slip_distance greater.

Yeah, I understood your method, I tried it and ok it works on the drawings for not composed examples. But I see the rolling when I move the objects in reality with my example. And even when I used the dot 'c' fixed on one end of the red arm I see the distance like d2 not d1. And when I use the method of the dot of contact fixed on the ground I have d2 not d1. Very odd.

What is the point of this?

Just to understand my mistake in the calculations. It could not help me to understand why I see the rolling in the example with the friction.

 

[A.T.]

But I see the rolling when I move the objects in reality with my example.

When both objects are moving, it's much harder to deduce the relative motion accurately. That's why you should transform into the rest frame of one of the objects.

 

[JrK]

Yes, very difficult !

I drew for a small angle of rotation, I drew at left the start and ending position. The circle moves in horizontal translation and the red wall rotate around A0. I drew at right your method:

An enlargement:

How I measure the distance of friction ? There is no "green" distance but there is an angle for the circle.

 

[A.T.]

I drew for a small angle of rotation,

Why? To make it even more difficult? Draw a bigger difference, if you want to work graphically.

BTW: If you don't want to transform the entire image to the rest frame of the wall, just transform the initial contact point twice, for each body separately to their final locations. Then in the final image, for each body draw the contact areas, from the transformed initial contact to the final contact.

 

[JrK]

Ok, all is fine, I understood my mistake in the drawing and a mistake in the integral. So even with a small angle the friction is well equal to d1. Great !

 

[A.T.]

So even with a small angle the friction is well equal to d1.

What do mean by "even with a small angle"? I think you have the logic of checking geometric formulas backwards: If a formula works for larger angles, then it definitely also works for small / infinitesimal angles. It's the other way around, where you have to be careful, because small angle approximation can fail for larger angles.

That's why your images with tiny displacements are so useless: It hard to see anything, easy to make errors, and even if you get it right you cannot automatically generalize it.