What is the orientation of the vector of friction?

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The discussion centers on determining the correct orientation of the friction vector between a translating circle and a rotating wall. The friction vector is defined to be parallel to the wall and opposes the relative sliding motion. Participants explore the implications of the wall's rotation on the energy dynamics, noting that the energy required to move the circle exceeds the energy recovered from friction due to slipping. The conversation emphasizes the need to consider the changing normal force and the geometry of the system as the wall rotates. Ultimately, the key takeaway is that the friction force is constant in magnitude but its effectiveness varies with the relative motion of the two objects involved.
  • #121
jbriggs444 said:
So the cylinder supplies not only a horizontal force along its axis but also a shear force at right angles to its axis, right?
I don't understand the words "shear force" but I counted the energy to move in translation the circle : force by distance.
 
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  • #122
JrK said:
I hope, I could say that only when I will understand the mistake with the needle.
The needle is a proxy for the contact location, the movement of which is irrelvant for friction work. Thats the mistake with the needle. What is there more to understand?
 
  • #123
JrK said:
The circle doesn't rotate around itself.
Then the circle does rotate in the restframe of the wall, which allows you to see the relative motion better. Here the circle slides along the wall with a backspin. The backspin increases the slip distance, compared to pure slip with a non-rotating wheel.
 
  • #124
JrK said:
I don't understand the words "shear force" but I counted the energy to move in translation the circle : force by distance.
A "shear" is a force at right angles to the direction of a beam. This distinguishes it from a force in the direction of the beam which would be a tension or a compression.

Although the circle only moves horizontally, the vertical force of the cylinder makes possible a normal force from the wall. The normal force from the wall does work on the circle as it translates horizontally.
 
  • #125
jbriggs444 said:
The normal force from the wall does work on the circle as it translates horizontally.
It is odd. Anyway, you could compare the example with the needle+elastic and the example with the friction: it must be the same. If you say the up force works on one example it must work on the other. And think there is the force F3 on the circle.

I can, for example, use a fixed wall at top:

t10.png

Like that I don't take in account the vertical force.
A.T. said:
Then the circle does rotate in the restframe of the wall, which allows you to see the relative motion better. Here the circle slides along the wall with a backspin. The backspin increases the slip distance, compared to pure slip with a non-rotating wheel.
Maybe it is what I called 'a slip' in the example of the friction (I'm not sure to understand your message at 100%). In the example with the needle + elastic, like there is no friction, I measure only the length of the elastic.
 
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  • #126
A.T. said:
Then the circle does rotate in the restframe of the wall, which allows you to see the relative motion better. Here the circle slides along the wall with a backspin. The backspin increases the slip distance, compared to pure slip with a non-rotating wheel.

JrK said:
Maybe it is what I called 'a slip' in the example of the friction (I'm not sure to understand your message at 100%).
To analyze relative motion you have to transform to the rest frame of one of the bodies: In your frame the circle moves right, while the wall rotates CW. In the rest frame of the wall the circle still moves right along the wall, but also rotates CCW. That is the relative motion of the bodies, that determines the slip distance, which determines work by friction.

That slip distance for a circle spinning backwards while it slides forward is greater than for a non-spinning slipping circle, because there is more relative motion at the contact:

Non-spinning slipping wheel:
slip_distance = wheel_displacment

Back-spinning slipping wheel:
slip_distance = wheel_displacment + back_spin_angle * radius
 
  • #127
A.T. said:
To analyze relative motion you have to transform to the rest frame of one of the bodies: In your frame the circle moves right, while the wall rotates CW. In the rest frame of the wall the circle still moves right along the wall, but also rotates CCW. That is the relative motion of the bodies, that determines the slip distance, which determines work by friction.
That doesn't mean the distance increases between the two surfaces because the dot of contact changes it position between the two surfaces. You can rotate a wall around a fixed circle without any friction: look at the dot of contact in the same time is important.
 
  • #128
JrK said:
That doesn't mean the distance increases between the two surfaces ...
Not the "distance between the two surfaces", but the slip distance. Mathematically, the integral of the relative velocity at the contact location over time.

It's very easy to visualize. Draw yourself some pictures. The wall is fixed in this frame, and only the circle moves over it:

1) Non-rotating, translating circle:
slip_distance = circle_displacement

2) Rotating, non-translating circle:
slip_distance = rotation_angle * circle_radius

3) CCW-rotating, right-translating circle:
slip_distance = circle_displacement + rotation_angle * circle_radius

Note that 3) is just a combination of 1) and 2), and the relative velocities at contact (slip) are simply added up.
 
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  • #129
Thanks for the method, I built something like that before but I didn't know how to do the intermediary steps to have the good distance of friction. First, I would like to precise the word "slip", I'm focus on the friction (for the example with the friction), so for me, the word I used at start is a slip in the friction, I mean less distance in friction. I see for you, the slip is the distance of friction. It depends on what we are focused. For me, the distance d1-d2 is a lack of friction. Just to precise.

I try to build the drawings, but I have a problem in the steps in need to do because I need to built with some steps in the drawings to have the cases you described.

So, for me, I start with the drawing with the start and the end positions:

h1.png

I glue the circle and the red wall together of the end position and I rotate to have the angle of the wall at start, I move them to have the wall at the end position at the same place than the start position, I have that:

h2.png

What I need to do now ? I move the circle alone in translation (x and y) to be at the position of the circle at start ? And I don't rotate it ? In that case the distance measured is d2.

h3.png
If I rotate the circle after repositionning it like before, I have d1:

g4.png


If I don't repositionnate the circle before rotating, I have d1-d2:

h4.png
 
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  • #130
JrK said:
First, I would like to precise the word "slip",
"Slip" is the relative motion of the material parts in contact.
"Slip distance" is the time integral of the relative speed of the material parts in contact.

JrK said:
I try to build the drawings,
Draw the wall horizontally. We only care about the relative motion here, not how the wall is oriented in your original frame.

Choose two time points with more displacement between them, so you see the slip distances better.

Then draw the path of the contact point on each body:
- a line on the wall (same length as circle_displacement )
- an arc on the circle (rotation_angle * circle_radius)
The total slip length the the sum of them.
 
  • #131
I have a difference enough to measured:

gy.png
Your calculation doesn't take in account the position of the dot of contact relatively to the material. The friction between 2 surfaces depends of the position on the dot of contact too (not the trajectory), example: a wall rotate around a circle without any slip, so no friction, but with your calculations there is one. The distance of friction is well d2, because the dot of contact changes its position on the circle of a length of d1-d2.
 

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  • #132
JrK said:
I have a difference enough to measured:
Your images are really not very helpful.

Draw the wall horizontally (its fixed in that frame). Draw the entire circle in two positions at least 1/4 radii apart and at least 30° rotated (backspin).
 
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  • #133
It is difficult for me to think with the wall in horizontal position, so if it is not so difficult for you, let me draw it like it is on the first drawing, please.

I draw with a bigger angle of rotation, here I glue the wall and the circle of the end position, after I rotate the 2 objects to have the angle of the wall like the start angle, after I dettached the circle and I move it in translation x+y, and after I rotate the circle of the angle, but I'm not sure it is what you want:

s1.png


Here, with the translation and the rotation, note there is the start position, the end position and the re-positionning position:

s2.png


The dot '1' is fixed on the wall. The dot '2' is fixed on the circle.

The drawing with the translation of the circle but not the rotation:

s4.png


Here, without the translation but with the rotation:
s5.png


the last with the length le1=le2:

s6.png
 
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  • #134
JrK said:
... let me draw it like it is on the first drawing, please.
It's simply wrong for the rest frame of wall: In that frame, the wall is in the same position at both time points, not in different ones like you draw it.

Again, the relative motion between the contact materials is what matters. For example the motion of the circle material in the rest frame of the wall. You either accept this, or there is no point in further discussion, which is just repetitive now.
 
  • #135
A.T. said:
the wall is in the same position at both time points, not in different ones like you draw it.
Be careful: I drew 3 positions ! The start position, the end position and the position with the transformations you asked, the position transformed is exactly at the start position because the wall doesn't move, for example the translation but not the rotation, look I moved a little the transformed position (grey color):

d1.png


Here the full drawing, to watch the wall doesn't move:

ff6.png


You can see there are 2 circles (grey and blue) because I attached to the circle the dotted lines to have the center.
 
  • #136
A.T. said:
1) Non-rotating, translating circle:
slip_distance = circle_displacement

2) Rotating, non-translating circle:
slip_distance = rotation_angle * circle_radius

3) CCW-rotating, right-translating circle:
slip_distance = circle_displacement + rotation_angle * circle_radius
slip_distance = circle_displacement = 0
slip_distance = rotation_angle * circle_radius = d2
slip_distance = circle_displacement + rotation_angle * circle_radius = d2
 
  • #137
JrK said:
slip_distance = circle_displacement = 0
slip_distance = rotation_angle * circle_radius = d2
slip_distance = circle_displacement + rotation_angle * circle_radius = d2
The circle is displacing relative to the wall. The circle displacement is non-zero.
 
  • #138
jbriggs444 said:
The circle is displacing relative to the wall. The circle displacement is non-zero.
I drew the drawings AT asked.
 
  • #139
JrK said:
I drew the drawings AT asked.
Let me look back and see. As I recall, he was discussing a combination of displacement and rotation and wanted to use the rest frame of the wall... Here we go: #128:
A.T. said:
It's very easy to visualize. Draw yourself some pictures. The wall is fixed in this frame, and only the circle moves over it:

1) Non-rotating, translating circle:
slip_distance = circle_displacement

2) Rotating, non-translating circle:
slip_distance = rotation_angle * circle_radius

3) CCW-rotating, right-translating circle:
slip_distance = circle_displacement + rotation_angle * circle_radius

Note that 3) is just a combination of 1) and 2), and the relative velocities at contact (slip) are simply added up.
The relevant phrases: "translating circle" and "the wall is fixed in this frame".
Here is is another quote.
A.T. said:
Draw the wall horizontally.
 
  • #140
jbriggs444 said:
The relevant phrase: "translating circle".
I translate the circle, I don't rotate it. I let AT reply if it is not what he asked.

Jbriggs444: have you find the mistake with the example with the needle+elastic ? because you asked the vertical force works, etc.
 
  • #141
JrK said:
I translate the circle, I don't rotate it. I let AT reply if it is not what he asked.
If you use the rest frame of the wall and mention a rotation angle, it must be the circle that rotates.

The wall does not rotate in its own rest frame. That's a tautology.
 
  • #142
jbriggs444 said:
If you use the rest frame of the wall and mention a rotation angle, it must be the circle that rotates.
I explained the method I used in the message #133. I glued, I rotate, I translate, etc. and I asked if it is correct or not. If not, explain the method I need to use, I mean by steps:

1/
2/
3/
etc.
 
  • #143
JrK said:
I explained the method I used in the message #133. I glued, I rotate, I translate, etc. and I asked if it is correct or not. If not, explain the method I need to use, I mean by steps:

1/
2/
3/
etc.
Well, let's check back. We've already reviewed #128. Three scenarios were under discussion.

1) Non-rotating, translating circle:

2) Rotating, non-translating circle:

3) CCW-rotating, right-translating circle:


This suggest that we would want three pictures. Each with a before and after circle. Now let us review #133.

The first picture shows a circle that is non-rotating and translating in the ground frame.

In the wall frame it is rotating and translating. The wall is not drawn horizontally as was requested.

So the first drawing is not what was asked.

The drawing that is purported to show rotation but not translation shows three circles and two walls. There is no rotation in the ground frame. It is far from clear whether there is translation in the wall frame.

The wall is not drawn horizontally as was requested

So the second drawing is not what was asked.

There is no drawing purporting to show both rotation and translation.

So there is no third drawing.
 
  • #144
Here:

f5.png


The start position is the wall at 45° with the blue circle. The wall is fixed at 45° (the wall at 30° is the end position, I need it to built the drawing). The circle in GREY color at left is translating not rotating. Are you agree ?

And the distance asked by AT is 0 in the case 1/

The second case is correct too, if you consider the wall fixed at 45° (I drew also the end position): rotating not translating. and the distance is well the distance I measured from start: d2.
 
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  • #145
JrK said:
The start position is the wall at 45° with the blue circle. The wall is fixed at 45° (the wall at 30° is the end position, I need it to built the drawing). The circle in GREY color at left is translating not rotating. Are you agree ?
As I understand it, you built the drawing by starting with the BLUE circle. However, that circle is just a reference from which you constructed the GREEN circle -- by translating right relative to the ground frame. And from which you constructed the GRAY circle -- by rotating 15 degrees counterclockwise.

The drawing is intended to depict the GRAY circle as the starting point and the GREEN circle as the ending point.

Yes, I agree that from a wall-relative point of view, this amounts to a translation without a rotation. I fail to understand the point of drawing two walls when the drawing is supposed to depict a wall-relative point of view.
JrK said:
The second case is correct too, if you consider the wall fixed at 45° (I drew also the end position): rotating not translating. and the distance is well the distance I measured from start: d2.
There is no second case in the quoted post.
 
  • #146
JrK said:

The grey circle is wrong. The transformation to the rest frame of the wall is a pure rotation around the pivot bottom left, such that the walls align. And the green circle needs to be transformed in the very same way. So the green circle ends up much further up-right along wall, not where you drew it. The distance of point 2 to the pivot doesn't change in the transformation.

This really trivial:
- Draw both time points in your frame
- Rotate both entire images around the pivot so their walls align (and ideally are horizontal)
 
  • #147
@jbriggs444: I don't represent the drawings like standards physicists do and I don't use the standard notations too, so I think it is more difficult to understand.

The second image, rotating not translating (start in blue, end in green, transformation in grey):

hee.png


@AT: I don't understand: I choose the wall fixed at 45°, I drew the start position and the end position (arbitrary). After, I glued the circle and the wall of the end position and I rotate it of 15°, after I dettached the wall and the circle and I translate the circle to be at the same position than the circle of the start position, it is not that I need to do ?
 
  • #148
JrK said:
@jbriggs444: I don't represent the drawings like standards physicists do and I don't use the standard notations too, so I think it is more difficult to understand.
Yes, your drawings are more difficult to understand than the drawings that are being requested.

For instance, a drawing containing notations "le1" and "le2" without any verbiage discussing what "le1" or "le2" denote might as well contain neither notation. And indeed, if it is not translating then "le1" and "le2" are completely devoid of significance.

A drawing that is supposed to use the rest frame of a wall but in which a "fixed" wall takes two different orientations -- that simply boggles the mind.
 
  • #149
JrK said:

Now the grey circle has the right position, but wrong orientation. In the original image it was the other way around.
 
  • #150
jbriggs444 said:
For instance, a drawing containing notations "le1" and "le2" without any verbiage discussing what "le1" or "le2" denote might as well contain neither notation. And indeed, if it is not translating then "le1" and "le2" are completely devoid of significance.
Yes, you're right I don't explained, le1=le2 : I noted that. What is le1 ? it is d2. And what is le2 ? it is the distance of the friction (the slip that AT spoke). So, le1=le2 said the distance of friction is well the distance I thought.

A.T. said:
Now the grey circle has the right position, but wrong orientation. In the original image it was the other way around.
You spoke about what image ? the image of the message # 147 is the second case: no translating, rotating.

The image for the translating, no rotating is the image in the message #144. It is the same dot, I'm sorry about that but yes, the distance in the first case is 0. Or I made a mistake in what you asked.
 

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