- #106
A.T.
Science Advisor
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Why?JrK said:... I would like to use the dot of contact...
Why?JrK said:... I would like to use the dot of contact...
Because I would like to make the analogy with the example with the friction to understand perfectly the first example.A.T. said:Why?
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.JrK said: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.
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 said:Because I would like to make the analogy with the example with the friction to understand perfectly the first example.
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 ?A.T. said:So again, why do you insist on focusing on it?
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 said:If you think of the dot of contact as rigid body, its atoms have moved in the y-direction.
JrK said:You think here with the needle, true ?
Isn't focusing on something irrelevant the mistake?JrK said:I know the movement of the dot of contact is irrelevant but I would like to understand my 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.A.T. said:Isn't focusing on something irrelevant the mistake?
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.Stephen Tashi said: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
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.JrK said: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.
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.JrK said: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).
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 said:Won't work.
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.JrK said: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.
Yes, the circle moves in horizontal translation like I specified.jbriggs444 said:It would anchor the circle to a fixed horizontal trajectory.
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 said:Yes, the circle moves in horizontal translation like I specified.
I don't understand the words "shear force" but I counted the energy to move in translation the circle : force by distance.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?
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?JrK said:I hope, I could say that only when I will understand the mistake with the needle.
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:The circle doesn't rotate around itself.
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.JrK said:I don't understand the words "shear force" but I counted the energy to move in translation the circle : force by distance.
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.jbriggs444 said:The normal force from the wall does work on the circle as it translates horizontally.
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. 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.
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.
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.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%).
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. 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.
Not the "distance between the two surfaces", but the slip distance. Mathematically, the integral of the relative velocity at the contact location over time.JrK said:That doesn't mean the distance increases between the two surfaces ...
"Slip" is the relative motion of the material parts in contact.JrK said:First, I would like to precise the word "slip",
Draw the wall horizontally. We only care about the relative motion here, not how the wall is oriented in your original frame.JrK said:I try to build the drawings,
Your images are really not very helpful.JrK said:I have a difference enough to measured:
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.JrK said:... let me draw it like it is on the first drawing, please.
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):A.T. said:the wall is in the same position at both time points, not in different ones like you draw it.
slip_distance = circle_displacement = 0A.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
The circle is displacing relative to the wall. The circle displacement is non-zero.JrK said:slip_distance = circle_displacement = 0
slip_distance = rotation_angle * circle_radius = d2
slip_distance = circle_displacement + rotation_angle * circle_radius = d2
I drew the drawings AT asked.jbriggs444 said:The circle is displacing relative to the wall. The circle displacement is non-zero.
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:JrK said:I drew the drawings AT asked.
The relevant phrases: "translating circle" and "the wall is fixed in this frame".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.
A.T. said:Draw the wall horizontally.
I translate the circle, I don't rotate it. I let AT reply if it is not what he asked.jbriggs444 said:The relevant phrase: "translating circle".