What is the orientation of the vector of friction?

[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.

 

[JrK]

For me I don't have the right to do that:

Because I rotate the circle CCW and it is not true, the circle is moving in horizontal translation. For me, I need to do that:

Even your method gives the distance d1 (that represents the work needed to move the circle in translation), I don't understand why you do that. It is like to add a movement that doesn't exist in reality.

 

[A.T.]

I will think again with the example with the needle+elastic because I don't find the mistake.

Does your needle+elastic analogy work for trivial cases, like pure rolling? If not, there is no point in applying it to more complex cases.

 

[JrK]

Does your needle+elastic analogy work for trivial cases, like pure rolling? If not, there is no point in applying it to more complex cases.

Yes, it will be the same calculations, but like the distance doesn't increase for a pure rolling, the elastic doesn't increase its length and there is no energy needed to move the circle because the circle don't move. But yes, it is possible to think with a part of rolling. The advantage with the needle and the elastic: no friction ! I'm sure to increase the elastic of d2 (it is a mechanical constraint), the work needed to move the circle is represented by d1 and ok, I'm not sure about the forces F3 and F4.

 

[A.T.]

Does your needle+elastic analogy work for trivial cases, like pure rolling?

... the distance doesn't increase for a pure rolling, the elastic doesn't increase its length...

Really? Your elastic is fixed to a point on the wall and to the needle, which moves relative to the wall, along with the contact location. How can the elastic keep its length here?

 

[JrK]

Really? Your elastic is fixed to a point on the wall and to the needle, which moves relative to the wall, along with the contact location. How can the elastic keep its length here

No, you're right in a pure rolling, the distance of the elastic increases. It is another example, at start it wasn't a pure rolling but a mixed.

 

[A.T.]

No, you're right in a pure rolling, the distance of the elastic increases.

So the energy in the elastic has nothing to do with the energy dissipated by friction. It is a completely useless analogy, that fails even in the most trivial cases.

 

[JrK]

So the energy in the elastic has nothing to do with the energy dissipated by friction. It is a completely useless analogy, that fails even in the most trivial cases.

You're right, it is to find the mistake in that example I built at first to find the good length of friction. But maybe you could help me ? With the pure rolling it is easy, the circle is fixed to the ground and the wall rolling around the circle. The length of the elastic increases, where is the effort I give ? The torque on the wall is near 0 or with math it could be by limit is at 0.

 

[A.T.]

You're right, it is to find the mistake in that example I built at first to find the good length of friction. But maybe you could help me ? With the pure rolling it is easy, the circle is fixed to the ground and the wall rolling around the circle. The length of the elastic increases, where is the effort I give ? The torque on the wall is near 0 or with math it could be by limit is at 0.

The torque around which point?

 

[mcastillo356]

Summary:: It is a cinematic question of a basic movement of two shapes.

Hi,

It is my first message :) I hope you are all fine and safe in these difficult days !

I cannot find the good orientation of the vector of friction. A circle moves in translation to the right and in the same time the wall rotates around A0. A0 is fixed to the ground. There is always the contact between the circle and the wall. The circle doesn't rotate around itself, it is just a translation. There is a friction between the circle and the wall, I supposed the friction constant in value, not in orientation, for a small angle of rotation of the wall. At start, I thought the vector of friction has the same orientation than the wall but if I supposed that : the work from the translation of the circle is higher than the work from the friction. So, I think the orientation of the vector of friction is not like I think, how can I draw the good orientation of the vector of friction ? When I draw the length that the force moves along the wall I find a distance lower than I thought because there is a "slip" due to the modification of the angle of the wall and so the position of the dot of contact between the circle and the wall. So, is there a method to construct by drawings the orientation of the vector of friction ?

I drew 3 positions of the device with a small angle of rotation of the wall. And I drew an enlargement of the dot of contact:

View attachment 260742

Have a good day guys !

Friction is always opposite to movement. You've chosen the wrong frame of reference. I guess.

 

[JrK]

The torque around which point?

Difficult to name a fixed dot, I can use the dot of contact, I think the needle applies a torque on the wall relatively to the dot of contact, but the torque gives an energy and the elastic won an energy too, I don't see where is the negative torque or force here.

 

[A.T.]

Difficult to name a fixed dot, I can use the dot of contact,

That is a non-inertial reference point. Not really useful for energy calculations.

 

[JrK]

I saw the force from the needle: F1

The needle is very small: the size of one atom if I can. The circle is fixed.

At start, I would like to study my example not the pure rolling. But why not, in that case, the circle and the wall has tooth like gears, otherwise the red wall will escape. The tooth are useful for the red wall not for the needle.

 

[jbriggs444]

You're right, it is to find the mistake in that example I built at first to find the good length of friction. But maybe you could help me ? With the pure rolling it is easy, the circle is fixed to the ground and the wall rolling around the circle. The length of the elastic increases, where is the effort I give ? The torque on the wall is near 0 or with math it could be by limit is at 0.

No. The torque on the wall is fixed. The limit is fixed.

As the position of the needle approaches the point of contact, the compression force on the needle increases without bound in order that the tangential force remain fixed. Meanwhile, the moment arm is decreasing toward zero. As a result, the torque approaches a fixed limit.

 

[JrK]

The torque on the wall is fixed. The limit is fixed.

Yes, you're right on the pure rolling example, and the elastic decreases its length.In the example with the circle in horizontal translation, I calculated the energy to move the circle and to rotate the wall with the program, it is the same (the sum is at 0), but the elastic win an energy d2F, and the circle needs the energy d1F, where I win an energy ? The example:

 

[A.T.]

So the energy in the elastic has nothing to do with the energy dissipated by friction. It is a completely useless analogy, that fails even in the most trivial cases.

You're right, it is to find the mistake in that example I built at first to find the good length of friction.

If you agree that your needle-elastic scenario is not modelling the friction of the original scenario in this thread, then you should start a new thread about your needle-elastic scenario, where you present it in a clear manner.

This thread is already way to long, and the original friction question has been settled.

 

[JrK]

the original friction question has been settled

If I have the right to add that case where I think your method doesn't work, if not the moderator can delete that message ? I imagine a circle that I blowed up (inflate), like a balloon but it is a circle. The radius of the circle increases more and more but the center of the circle is fixed to the ground. I drew the device at start and at end:

An enlargement to watch there is no friction:

I used your method (look at the dot of contact, they are at the same place):

Enlargement:

I replace the circle at the center just to see the magenta length:

Enlargement:

When I place the walls at the exact position, I need to count the movement of the magenta length. But there is no green length, the last drawing shows a difference of position of the dot of contact but look at the difference of length at the end of the walls, it is the same distance. I'm not sure at 100% there is no friction between the circle and the wall when I blow the circle, but the images seem to watch that.

The distance measure of friction is small but not exactly at 0, I measured: 0.03 but the distance calculated of the magenta length is 0.19, I can compare the lengths with the 2 following images at the same scale:

 

[A.T.]

... inflate ...

What does this have to do with your original scenario? It was rigid bodies only.

 

[JrK]

But the method is not universal ? it works for some examples not for others, so how to know if for an example it works or not.

 

[A.T.]

But the method is not universal ?

The universal method is integrating force dot relative velocity.

 

[A.T.]

For me I don't have the right to do that:

This is a trivial coordinate transformation to the rest frame of the wall.

 

[JrK]

I understood the problem (at least mine): the wall rotates around A0, but the wall doesn't rotate around the circle ! look the 2 positions:

If I place the circle at the same center:

And if the wall doesn't rotate around the circle, I need to correct its orientation just after rotate the group. So, the image :

is correct with your method.

 

[A.T.]

I need to correct its orientation just after rotate the group.

If you do this, then it's not a mere coordinate transformation. You are changing their relative orientation and thus the physical situation.

 

[JrK]

Are you agree that the wall doesn't rotate around the circle ?

 

[A.T.]

Are you agree that the wall doesn't rotate around the circle ?

Depends on what "rotate around a circle" means.

 

[JrK]

When you are fixed on the circle (you can see the wall), you see no modification of the orientation of the wall. You are fixed at the dot 'f' on the circle:

https://www.physicsforums.com/attachments/262602

When the circle moves in translation and when in the same time the wall rotates around A0, you see the same orientation of the wall.

 

[A.T.]

When you are fixed on the circle (you can see the wall), you see no modification of the orientation of the wall.

In your original scenario the wall changes orientation in the rest frame of the circle.

 

[JrK]

Yes, it is because I think with the dot of contact but I don't have the right...
Have you the equations of what I need to integrate to measure the length of friction ?

 

[A.T.]

Yes, it is because I think with the dot of contact ...

It has nothing to do with their contact point. The wall rotates while the circle doesn't, so the orientation of the wall relative to the circle changes.

 

[JrK]

The universal method is integrating force dot relative velocity.

Have the equations I need to integrate ?

 

[A.T.]

Have the equations I need to integrate ?

The 2nd equation in this section (v is the relative velocity of the materials in contact)
https://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

But in your specific scenario you can derive an analytical formula for the slip distance (sum of green and violet lines in post #179), and then multiply it by the constant force of friction.

 

[JrK]

I have a question about the direction of the graphical method. I built all the drawings with the final position back to the initial, why ? (at the beginning of the thread I done that, without think of the direction) the true direction is from the initial position to the final. If I drew from the initial to the final I have less than d2, it is worst: I have less than d2. With basic examples, it is easy and from initial to final or the reverse it is the same result, here not at all.

 

[A.T.]

With basic examples, it is easy and from initial to final or the reverse it is the same result, here not at all.

If you get different diagrams depending on what part you draw first, then that's more a problem with your consistency, rather than physics.