What is the orientation of the vector of friction?

[A.T.]

I count the energy stored inside the elastic.

If you mean a constant force elastic, that basically just measures the slipping distance relevant for friction work, then the correct method of attaching that elastic is described here:

https://www.physicsforums.com/threa...of-the-vector-of-friction.987415/post-6332661

It doesn't require any needles.

 

[jbriggs444]

We repeatedly see that the circle is described as being pushed by a cylinder and that we are concerned with the work done by the cylinder.

But we are never shown the attachment point of the cylinder, the anchor point for the cylinder nor any accounting for the work associated with the torque from the cylinder as it rotates with respect to the circle.

 

[Stephen Tashi]

I need a force that pulls,

You wish to have a force that impedes the motion of the point of contact.

We are dealing with rigid bodies, there is no property of rigid bodies that says such a force exists. If you want to assume such a force exists then you are adding information to the problem and changing it into a different problem.

You wish to say that work done on the point of contact is the same as work done on the circle.

Calculating work done on a tiny object that stays at the point of contact is not the same as calculating work done on the circle. The circle and the tiny object are not the same object.

 

[JrK]

with the torque from the cylinder as it rotates with respect to the circle.

No, the cylinder doesn't rotate, the circle moves in translation. You can glue the circle with the cylinder. The cylinder will move only in translation.

It doesn't require any needles.

I replied to you before, I used it at #55, that method doesn't use the dot of contact and I would like to use the dot of contact, and I need the needle for that. If I fixe the dot A on the circle I don't have the analogy with my first example with the friction.

You wish to have a force that impedes the motion of the point of contact.

I don't understand, the circle and the wall are controlled in position by external devices: cylinder and motor (for example). I need a force to simulate the force of friction because I see a slip with the friction and I could understand where is my mistake when I use an example that uses the dot of contact. The force of the elastic gives the force that the friction could do in the first example.

Thanks a lot for your patience !

 

[jbriggs444]

No, the cylinder doesn't rotate, the circle moves in translation. You can glue the circle with the cylinder. The cylinder will move only in translation.

Yes the cylinder does rotate. It remains parallel to the wall which is rotating.

If it does not rotate, it must translate. If it translates, then, since it is subject to a torque, that translation involves work.

 

[A.T.]

... I would like to use the dot of contact...

Why?

 

[JrK]

Why?

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

@jbriggs444: yes the cylinder needs a work ! I counted that work, it is the work to move the circle.

 

[Stephen Tashi]

because I see a slip with the friction and I could understand where is my mistake when I use an example that uses the dot of contact.

To repeat: calculating work done on the dot of contact is a different problem than calculating work done on the circle. If you think of dot of contact as a rigid body, it is not the same rigid body as the circle.


Let's try this: Suppose the point of contact is initially at (2, 15) and there is an atom $A$ of the circle at (2,15). Later atom $A$ has moved to (2+3, 15) = (5,15) and the point of contact has moved to (2+3,27) = (5,27). Atom $A$ has not moved in the y-direction. No other atom of the circle has moved in the y-direction. If you think of the dot of contact as rigid body, its atoms have moved in the y-direction. So the dot of contact and the circle are different rigid bodies.

 

[A.T.]

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

But in the case with friction the movement of the contact point is irrelevant for the work. So again, why do you insist on focusing on it?

 

[JrK]

So again, why do you insist on focusing on it?

Again, to make the exact analogy with the example with the friction. I know the movement of the dot of contact is irrelevant but I would like to understand my mistake. And, for you: why you don't want to use the needle and the elastic ?

If you think of the dot of contact as rigid body, its atoms have moved in the y-direction.

You think here with the needle, true ? if yes, I'm agree the needle moves in 'y' like my drawings showing. (I supposed x: horizontal and y:vertical).

 

[Stephen Tashi]

You think here with the needle, true ?

Yes, the needle moves in the y-direction. The needle and the circle are not the same object. Hence the work done on the circle and the work done on the needle are different numbers.

 

[A.T.]

I know the movement of the dot of contact is irrelevant but I would like to understand my mistake.

Isn't focusing on something irrelevant the mistake?

 

[JrK]

Isn't focusing on something irrelevant the mistake?

I hope, I could say that only when I will understand the mistake with the needle. The needle + elastic are not complex to study and the advantage: it is geometry, I can measure the difference of potential energy from the length of the elastic.

The needle and the circle are not the same object. Hence the work done on the circle and the work done on the needle are different numbers

Yes, the circle and the needle are not the same object. The work done by the circle is d1*F. There is no work from the needle because the sum of forces on the needle is 0 at each time, the elastic pulls the needle but the walls of the circle and the red wall stop the needle. The other energy is the potential energy stored inside the elastic. And maybe the energy from the forces F3 and F4, I'm not sure about my calculation.

 

[jbriggs444]

Because I would like to make the analogy with the example with the friction to understand perfectly the first example.

@jbriggs444: yes the cylinder needs a work ! I counted that work, it is the work to move the circle.

Then it's not a cylinder. It's an arbitrary external force plus an arbitrary external torque. A cylinder has two attachment points, one at the business end where it attaches to the load and one at the opposite end where it is anchored.

As to your utilization of a needle as a proxy for the contact point, you need to account for the fact that in order to do its job, the needle must exert a non-zero force on the circle parallel to the wall. In order to do this, the needle cannot be exactly at the contact point. It must always be displaced some distance to the side. Further, no matter how small the distance, in order to do its job, the torque produced by the needle on the circle must be a constant.

 

[JrK]

Yes, it is a standard cylinder, fixed on the ground, it pushes the circle:

You can glue the circle to the cylinder (I mean the part of the cylinder that moves to the right).

 

[jbriggs444]

Yes, it is a standard cylinder, fixed on the ground, it pushes the circle:

View attachment 261874

You can glue the circle to the cylinder (I mean the part of the cylinder that moves to the right).

Won't work. You are not accounting for the resulting non-zero net torque on the circle. At least not if we are using friction.

But if we are using the needle and elastic and and some sort of sliding arrangement to clamp the circle to the cylinder and if the flat surface where the cylinder touches the wheel is greased... those are all central forces. No net torque. Yes, that could work.

 

[JrK]

Won't work.

What do you mean ? I can glue the circle on the cylinder (the part that moves). The circle doesn't rotate around itself. I don't want the circle rotate around itself like I specified.

 

[jbriggs444]

What do you mean ? I can glue the circle on the cylinder (the part that moves). The circle doesn't rotate around itself. I don't want the circle rotate around itself like I specified.

If you glue the circle to the cylinder then the other end of the cylinder could ride up and down a vertical greased wall and thereby provide torque to the circle. That could work.

A cylinder anchored to the ground and glued to the cylinder could not work. It would anchor the circle to a fixed horizontal trajectory -- which had not been my understanding of the problem.

 

[JrK]

It would anchor the circle to a fixed horizontal trajectory.

Yes, the circle moves in horizontal translation like I specified.

 

[jbriggs444]

Yes, the circle moves in horizontal translation like I specified.

So the cylinder supplies not only a horizontal force along its axis but also a shear force at right angles to its axis, right?

 

[JrK]

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.

 

[A.T.]

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?

 

[A.T.]

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.

 

[jbriggs444]

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.

 

[JrK]

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:

Like that I don't take in account the vertical force.

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.

 

[A.T.]

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%).

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

 

[JrK]

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.

 

[A.T.]

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.

 

[JrK]

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:

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:

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.

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

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

 

[A.T.]

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.

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.

 

[JrK]

I have a difference enough to measured:

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.

 

[A.T.]

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).

 

[JrK]

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:

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

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:

Here, without the translation but with the rotation:

the last with the length le1=le2:

 

[A.T.]

... 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.

 

[JrK]

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):

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

You can see there are 2 circles (grey and blue) because I attached to the circle the dotted lines to have the center.

 

[JrK]

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

 

[jbriggs444]

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.

 

[JrK]

The circle is displacing relative to the wall. The circle displacement is non-zero.

I drew the drawings AT asked.

 

[jbriggs444]

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:

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.

Draw the wall horizontally.

 

[JrK]

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.

 

[jbriggs444]

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.

 

[JrK]

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.

 

[jbriggs444]

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.

 

[JrK]

Here:

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.

 

[jbriggs444]

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.

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.

 

[A.T.]

Here:

View attachment 261948

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)

 

[JrK]

@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):

@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 ?

 

[jbriggs444]

@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.

 

[A.T.]

View attachment 261949

Now the grey circle has the right position, but wrong orientation. In the original image it was the other way around.

 

[JrK]

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.

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.