Post by: LB7 on 15/04/2020 09:56:56
him.png (105.67 kB . 699x562 - viewed 5329 times) fjr.png (45.73 kB . 843x547 - viewed 5037 times)ceux qui veulent me parler eh bien, faites le moi savoir avant 14h aujourd'hui de vive voix
Post by: Bobolink on 17/04/2020 03:17:26
Why is the circle moving to the right if the force is to the left?
What do you mean by 'circle' in physical terms for this scenario, a cylinder, maybe?
Post by: LB7 on 17/04/2020 05:24:21
Why is the circle moving to the right if the force is to the left?For example, a pneumatic cylinder moves the circle to the right at a constant velocity. The circle cannot rotate around itself, it is a mechanical constraint. The pneumatic cylinder needs an energy to move the circle. I count that energy and I compare it with the energy recovered from the friction and there is a difference. I think the vector of friction is parallel of the red wall so I don't recover any energy from the rotation of the red wall. There is a slip because the angle 'a' changes and that slip lost a distance in the count of the energy from the friction. I measured the distance 'd1' and the distance 'd2' and for a small angle of rotation 'a' it is easy to compare these two energies. I can suppose the mass of the wall very low (to simplify the calculations) and to have the force of friction I can 'pince' the 2 walls with, for example, a spring. The spring is there to give the normal force. I can suppose the normal force passes always by the dot of contact. The length of the spring is always the same.
What do you mean by 'circle' in physical terms for this scenario, a cylinder, maybe?Yes, it is a cylinder for example.
Post by: LB7 on 23/04/2020 10:39:30
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Post by: Bobolink on 23/04/2020 14:50:09
I count that energy and I compare it with the energy recovered from the friction and there is a difference.What is the difference?
Post by: LB7 on 24/04/2020 11:41:27
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Post by: Bobolink on 24/04/2020 12:14:04
Post by: LB7 on 24/04/2020 12:30:07
Post by: Bobolink on 24/04/2020 13:22:39
Post by: LB7 on 24/04/2020 13:32:22
Post by: Bobolink on 24/04/2020 14:01:49
The rotation of the red arm doesn't need/give any energyReally? You can move a mass with no force? That seems like a violation of Newton's laws.
You are going to have to show the calculations of you want to prove anything. Pointing at a picture is not going to do it.
Post by: LB7 on 24/04/2020 14:22:22
Post by: Bobolink on 24/04/2020 15:04:25
Like I wrote #2: a pneumatic cylinder moves the circle at a constant velocity.Well that's good! I was confused by this comment: The rotation of the red arm doesn't need/give any energy.
I count the energy needed to move the circle in translation.Great, please show the math how you counted it.
I count the energy recovered from the heating by the friction (or the energy recovered from the elastic).Good, please show the math that you used to count that.
You have counted all the energies, all I am asking is for the calculations you used.
Post by: LB7 on 24/04/2020 15:11:05
Be careful, I changed the name of 'd1' in the first figure.Take in account only the last drawing. The distance moved by the circle is 'lg' not 'd1':
fve.png (59.04 kB . 711x605 - viewed 4138 times)For the circle:
I measure the distance moved by the circle and I multiply it by the force and multiply by the cosine of the angle of the wall relatively to the horizontal, note the distance moved by the circle multiply by the cosine of the angle is equal to d1. The energy given is lg*F*cos(a) = d1*F.
For the length of the elastic (because it is difficult to calculate the energy from the friction between two surfaces in movement):
I measure the distance I reduce the elastic, it is an energy I recovered: the length is d2. Like I suppose the force of the elastic constant. The energy recovered is d2*F.
Post by: Bobolink on 24/04/2020 16:23:15
What you call 'the calculations' ? the method ?Are you telling me you do not know what calculation means?
I measure the distance moved by the circle and I multiply it by the force and multiply by the cosine of the angle of the wall relatively to the horizontal, note the distance moved by the circle multiply by the cosine of the angle is equal to d1. The energy given is lg*F*cos(a) = d1*F.That does not look right to me. Work or energy is equal to force over a distance. So:
E = F * lg. Why did you multiply this times a Cosine? What is (a), there is no 'a' on your drawing.
E = F x d is the formula for the energy of the cylinder. 10N * .5m = 5j is a calculation.
At this point there is no where near enough information to learn anything else. We need more info, such as:
What is the weight of the cylinder.
What is the force applied to the cylinder
How fast is the bar moving.
What is the modulus of elasticity of the band.
If the cylinder pushes against the bar what is the resistance to movement of the bar.
Post by: LB7 on 24/04/2020 16:52:43
Here the forces:
f1f2.png (23.89 kB . 377x265 - viewed 4008 times)F1: the force on the wall
F2: the force on the circle
I need to multiply by the cos(a) with 'a' the angle from the horizontal to have the energy needed to move the circle, because it moves in horizontal translation.
We need more info, such as:It is possible to think with a very, very low mass (in theory the mass at 0) to forget it in the equations. Like the velocity of the circle is constant, the cylinder needs only to cancel the force F2*cos(a). I suppose all the materials perfects, especially the elastic. There is no friction between the elastic and the wall.
What is the weight of the cylinder.
What is the force applied to the cylinder
How fast is the bar moving.
What is the modulus of elasticity of the band.
If the cylinder pushes against the bar what is the resistance to movement of the bar.
To study the sum of energy it is easy: there the energy to move the circle and the energy recovered from the elastic that's all. The sum must be at 0.
Post by: Bobolink on 24/04/2020 18:02:27
This is just starting to seem like hand waving.
Calculate the initial energy that goes into the cylinder in joules.
Calculate the final energy of the cylinder, the energy absorbed by the elastic and the energy absorbed by the bar, all in joules. Then we can discuss your findings. At this point there is not really anything else to discuss.
Post by: LB7 on 24/04/2020 19:13:29
The force on the circle is at 45° relatively to the horizontal so I need to correct it with the cosine function. The movement of the circle is horizontal, the force is at 45°, the work is F*lg*cos(45°). I don't understand why there is a problem there.
Post by: Bored chemist on 24/04/2020 20:03:21
I don't understand why there is a problem there.Nor do I.
Why did you start the thread?
Post by: LB7 on 24/04/2020 20:12:43
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Post by: Bobolink on 24/04/2020 21:10:25
I calculate the energy in/out from start to end. I start at 45° and I stop at 45.001° for example. The conservation of the energy is always true, even for a small movement, even for a small time.One of the problems is that you are not very clear.
The force on the circle is at 45° relatively to the horizontal so I need to correct it with the cosine function. The movement of the circle is horizontal, the force is at 45°, the work is F*lg*cos(45°). I don't understand why there is a problem there.
So are you saying, that as the end of the red bar moves towards the lower right, a force is applied to the cylinder parallel to the bar and to the top right? And further the cylinder is not free to roll, however there is also no friction because the cylinder adheres to an elastic material which stretches to accommodate the movement of the cylinder?
Post by: Bobolink on 24/04/2020 21:30:22
You have one gray line that goes from the center of the circle to the left to point B.
When the circle is shifted to the right the new gray line is not parallel with the old gray line. If it was it would intersect point B'.
Post by: LB7 on 24/04/2020 21:54:41
No, there is no mistake in the drawing. I have the exact movement with the friction. The dot B is ALWAYS at the dot of contact between the circle and the red wall. At start, I studied that device without an elastic but with fricion between the circle and the red wall. Like it is difficult with the friction of 2 objects in movement, I found that example with the elastic. Here, there is no friction: I use the elastic to have the SAME forces than the friction at each time (There is no friction between the elastic and the wall. The circle cannot rotate around itself). Like that, I have the same energy needed to move the circle (compared to the energy with the friction). And I don't need/recover any energy from the rotation of the red wall around A0. At final, I need to measure the modification of the length of the elastic and it is well d2 not d1. It is easy because I have a dot fixed: A. You can compare the drawing with the friction, it is the same.
Years ago, my intuition sees that difference of length but I didn't think at the elastic to measure the distance. You can make the experience to see the trajectories. Here, I have 2 different examples, where the sum of energy is the same, not equal to 0, the example with the elastic is a direct measure.
Post by: Bobolink on 24/04/2020 22:22:28
Years ago, my intuition sees that difference of length but I didn't think at the elastic to measure the distance. You can make the experience to see the trajectories. Here, I have 2 different examples, where the sum of energy is the same, not equal to 0, the example with the elastic is a direct measure.Your intuition was wrong.
No, there is no mistake in the drawing. I have the exact movement with the friction. The dot B is ALWAYS at the dot of contact between the circle and the red wall.OK
To have the sum of energy at 0 (the energy needed to move the circle equal to the energy from the heating or the elastic) the distance must be d1. But I found d2.Of course the distance isn't d1! The elastic is holding the cylinder back! The 'missing' energy has been converted from kinetic energy to potential energy in the elastic.
Post by: LB7 on 25/04/2020 05:23:09
The 'missing' energy has been converted from kinetic energy to potential energy in the elastic.I see, the problem comes from the fixed dot, I can't use a fixed dot, so attach the elastic between the circle and the red wall and I change the elastic at each step:
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Post by: Bobolink on 25/04/2020 17:33:09
I see, the problem comes from the fixed dot, I can't use a fixed dot, so attach the elastic between the circle and the red wall and I change the elastic at each step:Like I said the 'missing' energy is in the elastic. You guys always crack me up, you think that if you make the scenario complicated enough then some law or other of physics will no longer apply. You can easily make a scenario I can't figure out, that doesn't mean the laws of physics don't apply it will just mean I can't figure it out.
Post by: evan_au on 25/04/2020 22:46:31
Quote
energy recovered from the frictionFriction tends to produce heat, which is the lowest form of energy.
It's really hard to recover useful energy from heat (unless it is really hot).
Next time, try generating a more useful form of energy, like electricity.
Lookup: https://en.wikipedia.org/wiki/Thermodynamics
This also tells you that you can't generate energy "for free"
Post by: LB7 on 25/04/2020 23:49:13
I come back with the method with the elastic, to have the distance 'd2' I need to place the dot A on a needle, and the needle is between the circle and the red wall:
Post by: Bobolink on 26/04/2020 02:50:09
What you really really need to do is stop wasting your time trying to beat the conservation of energy.
Post by: LB7 on 26/04/2020 07:36:27
I wrote a program:
vfz.png (136.69 kB . 1126x838 - viewed 3734 times)I used the notation in my first drawing. But I don't know if the calculations are correct. But I have well the energy recover from the friction lower than the energy needed to move the circle.
Post by: Bored chemist on 26/04/2020 09:16:49
But I don't know if the calculations are correct.If they tell you that energy is not conserved, then they are wrong.
Post by: LB7 on 26/04/2020 12:11:46
The method I used with the needle: it is just to prove my intuition is right. But it is worst than I think: the method of the needle is an amplificator of the energy, it destroys more energy. More the diameter of the needle is small more the forces aside the walls is bigger and like the distance moved by the red wall is lower than the distance of the circle: I lost more and more energy !
Post by: Bobolink on 26/04/2020 14:12:07
But it is worst than I think: the method of the needle is an amplificator of the energy, it destroys more energy.So if you have a certain amount of energy in a closed system some of the energy can just disappear? If you could do the opposite and make energy appear out of nowhere, you would be a rich person. Good luck.
Post by: LB7 on 26/04/2020 14:29:00
Post by: LB7 on 02/05/2020 16:32:22
fr3.png (63.98 kB . 752x698 - viewed 3237 times)I control in position the circle and the wall, I need 2 external devices, but I count all the energies I give or recover. There is always the contact between the 2 walls.
To have the friction between the walls I consider a spring presses the two walls at the dot of contact, only at that dot:
hhg3.png (34.19 kB . 357x340 - viewed 3136 times)With the wall fixed at 45° (reference) I translate the circle but not rotate it. The dot '1' is fixed on the wall and the dot '2' is fixed on the circle. I just move a little to the right the grey circle because it is at the same place than the blue circle. I have:
f5.png (47.06 kB . 903x683 - viewed 3259 times)Now, I rotate and don't translate the wall (the wall is always fixed at 45°, it is the reference):
fds.png (41.83 kB . 941x814 - viewed 3242 times)At final the distance moved is the sum of the two last: it is d2 not d1 in the global drawing.
The circle need the energy d1*F but the friction lost d2*F. With F the force of the friction.
That device destroys the energy but it is possible to use the example with the needle and the elastic for example to create it.
Post by: LB7 on 08/05/2020 05:30:17
1/ Before start: place a fixed dot on each object at the dot of contact.
2/ The dot of contact is fixed to the ground: fixed in position but also fixed in orientation.
3/ After end: measure the distance from the dot of contact to the fixed dot along the path of the object1: the line or the circle. Measure the distance moved by the dot of contact from start to end from the end of the dot of contact: be careful, the dot of contact is fixed ALSO in orientation.
4/ The algebric sum is the distance of the 2 distances.
For the following example :
vzd.png (115.64 kB . 1091x837 - viewed 3161 times)The energy to move the circle is lg*cos(44.5°)*F = d1*F. The distance of friction is d2, so the energy from heating is d2*F. The energy is not conserved. With a line and a circle ... I like !!!!
Post by: Bobolink on 08/05/2020 11:32:05
The energy to move the circle is lg*cos(44.5°)*F = d1*F. The distance of friction is d2, so the energy from heating is d2*F. The energy is not conserved. With a line and a circle ... I like !!!!That is wrong energy was conserved. If there was no friction then d1 would equal d2. But there was friction so d2 is shorter than d1. So the friction contribution is d1- d2, not d2.
The only way to prove your idea is to do the math, your drawings are not enough.
Post by: LB7 on 08/05/2020 15:27:20
The only way to prove your idea is to do the math, your drawings are not enough.Logic, geometry, measure, etc. it is math. Elementary math (primary school), for an elementary example. Even I think before math, geometry, measure, logic existed, but math had taken all of theses matters for it.
If there was no friction then d1 would equal d2. But there was friction so d2 is shorter than d1. So the friction contribution is d1- d2, not d2.The energy to move the circle is d1*F, so I need to have d1*F from the friction but I have d2*F.
Post by: LB7 on 13/05/2020 13:44:17
1/ draw the start and final position
2/ group the all objects in the final position
3/ rotate the group of the final position back to have the same orientation for one object (the wall for example)
4/ rotate one object forward (the circle for example) of : the relative rotation of one object to the other less the difference of the orientation of the dot of contact
5/ measure the distance of friction.
Post by: LB7 on 14/05/2020 05:08:08
Post by: LB7 on 29/05/2020 15:52:13
Post by: LB7 on 08/06/2020 15:04:06
1/ draw the objects at the initial and final positions
2/ group the objects at the final position and rotate back to have the same orientation of the initial position
3/
3.1/ place A0 the common dot, measure the distance moved by the dot of contact
3.2/ place the center of the circle the common dot, measure the distance moved by the dot of contact
4/ calculate the algebric sum: this is the distance of friction
Post by: LB7 on 09/06/2020 10:39:23
1/ the red arm is fixed, I move the circle along the red arm
2/ I glue the circle and the red wall, I rotate that group
3/ the red arm is fixed, I rotate the circle
in that case, the sum of energy is conserved. So the combination of 2 movement in the same time changes the result.
When I write the method:
1/ draw the start and final position
2/ group the all objects in the final position
3/ rotate the group of the final position back to have the same orientation for one object (the wall for example)
4/ rotate one object forward (the circle for example) of : the relative rotation of one object to the other less the difference of the orientation of the dot of contact
5/ measure the distance of friction.
I need to precise, "the difference of the orientation of the dot of contact" is measured after the dot 3/, it is not the true orientation, it is the orientation after rotate back the group.