What is the orientation of the vector of friction?

  • #52
jbriggs444
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It is an indirect method, you supposed the direction of the force, there is no other work needed than the circle. I mean a direct physicist method.

What do you think about the difference of the distances (p1,p3)-(p2,p4) moved by the dot on the wall: the distance (p1,p3) and on the circle: the distance (p2,p4) ?
@A.T. has given you the direct method that any physicist would use to define the work done.

Your insistence on using labels like "p1" through "p4" without putting them on your drawings is annoying. I have no idea what you are talking about.
 
  • #53
JrK
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I didn't know it is the direct method. I thought it was possible to integrate the movement of the dot of contact if I take in account all the movements, I mean the movement of the wall and the movement of the circle. I thought it was a geometric method.

Your insistence on using labels like "p1" through "p4" without putting them on your drawings is annoying. I have no idea what you are talking about.
I insist because I would like to understand how to think well. At start, at the dot of contact: the dot of contact, p1 and p2 are at the same dot. The dot p1 is fixed on the wall. The dot p2 is fixed on the circle. After a time, the dot p1 is now at a new position I called it p3, and the dot p2 is at a new position I called it p4. I have two vectors: p1->p3 and p2->p4, I measure the difference of these 2 vectors, and each time I use that method with 2 movements I have well the energy needed to move the object equal at that difference. But here, if I use that method I find d2 not d1.
 
  • #54
jbriggs444
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I didn't know it is the direct method. I thought it was possible to integrate the movement of the dot of contact if I take in account all the movements, I mean the movement of the wall and the movement of the circle. I thought it was a geometric method.


I insist because I would like to understand how to think well. At start, at the dot of contact: the dot of contact, p1 and p2 are at the same dot. The dot p1 is fixed on the wall. The dot p2 is fixed on the circle. After a time, the dot p1 is now at a new position I called it p3, and the dot p2 is at a new position I called it p4. I have two vectors: p1->p3 and p2->p4, I measure the difference of these 2 vectors, and each time I use that method with 2 movements I have well the energy needed to move the object equal at that difference. But here, if I use that method I find d2 not d1.
So ##p_2## and ##p_4## are the before and after positions of a dot painted on the circle. The incremental work done on the circle is given by ##\vec{f_{\text{wc}}} \cdot (\vec{p_4}-\vec{p_2})##

Meanwhile, ##p_1## and ##p_3## are the before and after positions of a dot painted on the wall. The incremental work done on the wall is given by ##\vec{f_{\text{cw}}} \cdot (\vec{p_3}-\vec{p_1})##

The sum of the two is guaranteed to be negative and gives the amount of mechanical energy dissipated into thermal energy by the force of kinetic friction.

According to your setup, ##\vec{p_1}## = ##\vec{p_2}##. Both are equal to the initial position of the point of contact.

You do not have a ##p_x## variable for the after position of the point of contact. That is good since that position is irrelevant.
 
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  • #55
JrK
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I found a method to count the real difference I think. I use for that a theoretical elastic between the circle and the wall. I resumed my method in the following drawing:

jrd.png

I found the distance d2 not d1. Is it correct ?
 
  • #56
jbriggs444
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If you are going to the trouble of typing in an explanation, please type it into the forums instead of burying it in a graphic in a font that is all but unreadable. The purpose of this is twofold. First, it allows us to read what you have written. Second, it allows us to quote and respond to what you have written.
the graphic said:
An elastic is attached between the dot A and the dot B around the wall of the circle, always around the wall of the circle, I mean the length of the elastic at start is pi/4 * R with R the radius of the circle.

I suppose the force F of the elastic constant when its length changes, at least for the difference of length of that example.

The dot A is fixed on the circle. The dot B is fixed on the red wall. The dot B is always at the same position than the dot of contact between the circle and the red wall.
If the dot B is both fixed to the red wall and continuously at the point of contact, it follows that the wall is moving so as to keep one point always in contact with the circle. It would have been good to have pointed this out explicitly.

the graphic said:
So I need to change the position of dot B so the length of the elastic : the position of the dot becomes B'. Like I change the length of the elastic I recover an energy : the length of B - B'. For the drawing at final I win the energy of d2F
[...]
I think I am finally grasping what you are talking about. The stretching of the elastic measures the mechanical energy that would have been dissipated into thermal energy due to kinetic friction -- assuming that the tension in the elastic is at all times equal to what the force of kinetic friction would have been.

That is nice, but it has little to do with the work done by the wall on the circle.
 
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  • #57
JrK
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I like the drawings with all information on it but it is true it is better for others people to have the text inside the message.

That is nice, but it has little to do with the work done by the wall on the circle.
I have the same force on the circle and on the wall than the friction does during the movement. So the energy to move the circle is the same and the energy to rotate the wall is the same (the rotation of the wall doesn't need any energy nor give) with the friction or with the elastic. I need to change the length of the elastic step by step, it is theoretically but I can compare it to the energy recovered by the friction: it is the same. The advantage of the elastic allows to watch the distance d2. And for the friction of for the elastic the energy must the the same than the energy to move the circle.
 
  • #58
jbriggs444
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(the rotation of the wall doesn't need any energy nor give)
Wrong.
 
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  • #59
A.T.
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The advantage of the elastic allows to watch the distance d2. And for the friction of for the elastic the energy must the the same than the energy to move the circle.
Your elastic would stretch and store energy even if there was no slip, just pure rolling, which dissipates no energy in the sliding friction case. The two are in no way equivalent, and your d2 irrelevant for the energy dissipated by friction.
 
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  • #60
A.T.
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The stretching of the elastic measures the mechanical energy that would have been dissipated into thermal energy due to kinetic friction -- assuming that the tension in the elastic is at all times equal to what the force of kinetic friction would have been.
I don't think this is true. See my post #59.
 
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  • #61
jbriggs444
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There cannot be pure rolling given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.
the graphic said:
The dot B is always at the same position than the dot of contact between the circle and the red wall.
 
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  • #62
A.T.
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There cannot be pure rolling ...
I'm using pure rolling just to show that this elastic band method cannot be used to determine the energy dissipated by friction.

...given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.
I thought B is merely the contact location, constrained to be somewhere on the wall, but not necessarily fixed to the same physical wall position.

It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.
 
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  • #63
jbriggs444
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It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.
It is the OP's scenario. He specified it.
graphic said:
The dot A is fixed on the circle. The dot B is fixed on the red wall.
 
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  • #64
A.T.
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It is the OP's scenario. He specified it.
It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.
 
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  • #65
jbriggs444
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It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.
I agree that it does not make much sense. The only way I saw reconcile it, which indeed may not have been the intent, is to have a fixed point on the wall sliding around the circle and maintaining contact.

This interpretation certainly violates the constraint in the original post that the wall is fixed to a pivot on the ground.
 
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  • #66
JrK
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Thanks to both of all !! you are very very nice :)

It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.

No, it is not a problem of language but I think with a step of integration, I change the position of the dot B at each step of the integration, B is at start (of the step of the integration) at the dot of contact between the circle and the wall and only at that time, after, the dot B is farther because it is fixed on the red wall. The dot B is not in the continuous time at the dot of contact, it is not possible. But I can increase the number of steps.

I know where is my mistake with the elastic: I didn't count the difference of the potential energy stored in the elastic (the potential energy stored from start to end of the study) and there is one ! it is (d1-d2)*F. So, like I drew, the energy lost to move the circle is recover by the elastic.

Remember, I took the idea of the elastic because I would like to find a method to count the energy because I see the friction at d2*F not d1*F. With the dot A fixed on the circle it is not like the friction could do. So, I took an elastic for each step of integration. That elastic at start has a length of 0, the two ends of that elastic is at the dot of contact, one end fixed on the circle, the other end fixed on the red wall. For each step, the elastic will increase its length of d2 not d1, or at least it is like I see that.

I resumed all here:

vffd.png


I think like that I will understand the distance is not d2 but d1.

I copy paste the text in the image because it seems it is not possible to download the image at its full definition:

An elastic is attached between the dot A and the dot B around the wall of the circle, always around the wall of the circle. I suppose the force F of the elastic constant when its length changes, at least for the difference of length of that example. The dot A is fixed on the circle. The dot B is fixed on the red wall. At start the length of the elastic is 0 and at final it is d2 : at start the dots A and B coincide. At each step of the numerical integration, I take a new elastic. The two ends of each new elastic is at the dot of contact between the circle and the red wall, but one end is fixed on the circle the other end is fixed on the red wall. Each elastic increases its length of d2 not d1. The force F1 is the force from the elastic to the red wall at the final of one step. The force F2 is the force from the elastic to the circle at the final of one step. The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F = lg*cos(a)*F. I don’t need nor recover any energy from the rotation of the red wall.
 
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  • #67
A.T.
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The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F.
Then your elastic is not modelling friction correctly, which has to dissipate all the work done moving the circle in the original setup.
 
  • #68
Stephen Tashi
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It is very difficult to pose (not resolve) the movement by equations (math) ?
We won't know until we try!

The diagrams you have presented ignore the invention of Cartesian coordinates. They lack representations for crucial dimensions such as the radius of the circle and dimensions that establish the relation of the line of movement of the center of the circle to the point where the wall pivots.

You are attempting to discuss the problem in an abstract geometric style. Archimedes and Newton were able to think about things this way, but the modern approach is to use cartesian coordinates!

For the sake of getting advice about modern methods, you should be specific about cartesian coordindates. For example:

The wall pivots at (0,0). The radius of the circle is r. The center of the circle moves along the line (x,y_1) where y_1 is constant. .

Draw a diagram showing the above information. Try to find the coordinates of the point of contact (x_c, y_c) as functions of x and the constants.

You can get one equation by setting the distance between the center of the circle and (x_c, y_c) equal to r. You can get another equation from the condition that the vector from (x_c, y_c) to the center of the circle is perpendicular to the vector from (0,0) to (x_c,y_c). The complication in getting a solution from these two equations is that there are two possible solutions for (x_c, y_c) since there are two lines from (0,0) tangent to the circle.

I think there are clever people on the forum that can solve the above problem since it is posed in the modern style. Considering the problem in coordinate-free way is only leading to confusion.
 
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  • #69
JrK
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I like a lot geometry !! Yes, a math method would be better, I tried with my program with a numerical integration (post #24), I used coordinates in the program but I don't know if my calculations are correct. I can explain the program if necessary, the program uses the first drawings: the dots, distances.

To come back to the method with the elastic, I need to place the dot A fixed not on the circle but on a needle, and the needle needs to be between the circle and the red wall. A very thin needle ! Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall. The dot B is fixed on the red wall. I need only one elastic.

I give the details:

gjk.png


fht.png
 
  • #70
Stephen Tashi
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@JrK, you're still avoiding cartesian coordinates.

I see no reason for the imaginary needle. You seem to think your coordinate free diagrams are helpful. So far, they haven't helped anybody.
 
  • #71
JrK
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In my program I used the name of the dot of the first drawing.
The (0,0) is on A0
The dot of contact between the circle and the red arm is 'i'
The center of the circle is the dot 'p'
The intersection between the horizontal line and the red arm is the dot 'q'
The height from A0 to the top of the circle is 'D'
The radius of the circle is 'R'
The length of the distance from A0 to 'i' is 'l' (small L)
I integrate from 'xinf' to 'xsup'


For the dots i,p,q, I added a letter for the 'x' or the 'y' and I added an information for take in account the step: the number '1' is the step in calculation and the '2' is the last calculation.

'wf' is the energy to move the circle in translation
'wd' is the energy recovered from the friction
'wp' is to verify the vector of the force of friction is like the orientation of the red wall, must be at 0

What I need to add in the calculations ? Maybe the calculation of wd is not like I calculate because I use the dot 'i' and all people here says it is not the movement of the dot of contact that gives the energy from the friction. Maybe there is a method (an old method used in 16th) but I didn't find on internet, the name of the method could help me to find it and to apply it.
 
  • #72
A.T.
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To come back to the method with the elastic,
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.

More general case:

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.
 
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  • #73
jbriggs444
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Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall.
So this "needle" is not like a physical pin that pierces the elastic at a particular point. It is merely an indicator (like a "needle" on a speedometer) marking the contact point.

Like any other position marker, this indicator has no physical significance. It does not "do" anything. It is merely a convenience for the purposes of accounting.

It seems that you would like to use this marker to separate the elastic into a "circle piece" and a "wall piece". The idea would be that the stretching or shrinking of the "circle piece" corresponds to the work done on the circle by the force of kinetic friction. However, this is not correct.

To gauge the work done on the circle, the relevant place on the elastic is the place (if any) where it is stationary. This is the place where the elastic material does not move past a stationary line at right angles to the contacting surfaces. This place will not, in general, be at the point of contact.

It is worth noting that this point on the elastic changes with your choice of reference frame. Similarly, the work done on the circle changes with your choice of reference frame. Neither energy nor work are invariant quantities.
 
  • #74
JrK
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The needle is a real object. It is there to be near the dot of contact, I would like to be at the exact position of the dot of contact but it is possible only in theory. I suppose all the objects rigid and without any friction. Remember I done that because I don't know another method to calculate that example with the friction between the objects. I cannot place the image in the message, a bug ?
 

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  • #75
A.T.
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Remember I done that because I don't know another method to calculate that example with the friction between the objects.
I gave you a method using integrals and I gave you a method using elastics. The needles seem needless here.
 
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