- #176
A.T.
Science Advisor
- 12,308
- 3,492
Yes. Now:JrK said:
A.T. said:Then mark in different colors:
- the part of the wall that was in contact with the circle
- the part of the circle that was in contact with the wall
Yes. Now:JrK said:
A.T. said:Then mark in different colors:
- the part of the wall that was in contact with the circle
- the part of the circle that was in contact with the wall
Why does green go beyond the final contact point?JrK said:View attachment 261975green: wall in contact with the circle
violet: circle in contact with the wall
?
Yes.JrK said:
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:JrK said:...you forgot something very important: ...
#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;
}
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 said: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.
A.T. said: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.
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):JrK said: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.
A.T. said: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.
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 said:I used the method of the dot of contact fixed on the ground. I tested that method with another movements
This is not the method I explained to you (see post #179):JrK said:
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 said:The method you explained works only for some examples not all.
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. said: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.
So which other scenario do you want to analyze where my method fails?JrK said:It is what you said.
You don't need to transform for my method either. It just makes it easier to avoid the errors you keep making.JrK said:Could you think with my last method please ? it is easy and we don't need to transform the drawing.
This one for example:A.T. said:So for which other scenario do you want to analyze where my method fails?
- There are no markings on the wall, to see how it moves. Is there slippage or pure rolling?JrK said:This one for example:
A.T. said:- 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)
Then top left is not "my method", but a completely different scenario (pure sliding). Bottom left looks OK, so let's apply my method:JrK said:It is a pure rolling.
View attachment 262418
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.A.T. said:- 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.
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:A.T. said:Then top left is not "my method", but a completely different scenario (pure sliding).
JrK said:
A.T. said:Then top left is not "my method", but a completely different scenario (pure sliding).
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.JrK said: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 you transformed correctly and it gives the correct result of zero slip distance, for this pure rolling scenario.JrK said:Yes, bottom left is ok,
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 said:and if I apply the method, it gives that for my example:
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:A.T. said: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.
Yes, that old one is correct for your original scenario:JrK said: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/
JrK said:
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.JrK said:I saw a rolling where there was not.
What is the point of this?JrK said: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.
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.A.T. said: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.
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. said:What is the point of this?
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 said:But I see the rolling when I move the objects in reality with my example.
Why? To make it even more difficult? Draw a bigger difference, if you want to work graphically.JrK said:I drew for a small angle of rotation,
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.JrK said:So even with a small angle the friction is well equal to d1.