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  4. Sum of energy not at 0 in that movement
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Sum of energy not at 0 in that movement

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Offline LB7 (OP)

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Sum of energy not at 0 in that movement
« on: 15/04/2020 09:56:56 »
A simple circle moves in translation to the right, the circle don't rotate around itself. I drew 3 positions of the circle. A red stem is always in contact with the circle. I drew 3 positions of the stem too. There is friction between the circle and the stem. The energy needed to move the circle is higher than the energy recovered by the friction, because the angle changes.


* him.png (105.67 kB . 699x562 - viewed 5459 times)


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ceux qui veulent me parler eh bien, faites le moi savoir avant 14h aujourd'hui de vive voix

« Last Edit: 05/07/2021 09:28:58 by LB7 »
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #1 on: 17/04/2020 03:17:26 »
I do not understand what you are trying to say.
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?
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #2 on: 17/04/2020 05:24:21 »
Quote from: Bobolink on 17/04/2020 03:17:26
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.

Quote from: Bobolink on 17/04/2020 03:17:26
What do you mean by 'circle' in physical terms for this scenario, a cylinder, maybe?
Yes, it is a cylinder for example.
« Last Edit: 17/04/2020 20:19:28 by LB7 »
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #3 on: 23/04/2020 10:39:30 »
I know it is difficult to measure the friction when two surfaces are in movement, so I used another method to prove I'm right: I used an elastic to prove the distance is well d2:


* vf3.png (66.01 kB . 1027x838 - viewed 4416 times)


* cdh.png (42.14 kB . 659x594 - viewed 4273 times)
« Last Edit: 23/04/2020 10:41:45 by LB7 »
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #4 on: 23/04/2020 14:50:09 »
Quote from: LB7 on 17/04/2020 05:24:21
I count that energy and I compare it with the energy recovered from the friction and there is a difference.
What is the difference?
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #5 on: 24/04/2020 11:41:27 »
I resume on the drawing:


* kns.png (188.81 kB . 1655x789 - viewed 4254 times)
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #6 on: 24/04/2020 12:14:04 »
So you don't know what the difference is?  Are just assuming there is a difference?
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #7 on: 24/04/2020 12:30:07 »
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. The sum of energy is not 0. For me it was logical with the friction (I posted here that solution before) but I never successed to prove it. Here, with the method of the elastic I prove the distance is d2.
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #8 on: 24/04/2020 13:22:39 »
I do not see where you proved anything.  It seems that you are only making claims.  Can you put numbers into your analysis showing the violation of the conservation of energy?
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #9 on: 24/04/2020 13:32:22 »
There are only 2 rigid objects and the friction (or the elastic). I move the circle in an horizontal translation, look at the angle, I need the energy d1*F with F the force of the friction (or the elastic). The difference of length won by the elastic is only d2*F. What else ? The rotation of the red arm doesn't need/give any energy. The circle doesn't rotate around itself. Where I can lost/win another energy ?
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #10 on: 24/04/2020 14:01:49 »
Quote from: LB7 on 24/04/2020 13:32:22
The rotation of the red arm doesn't need/give any energy
Really?  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.
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #11 on: 24/04/2020 14:22:22 »
Like I wrote #2: a pneumatic cylinder moves the circle at a constant velocity. I count the energy needed to move the circle in translation. I count the energy recovered from the heating by the friction (or the energy recovered from the elastic). That's all. Geometry, a length, etc. are proof.
« Last Edit: 24/04/2020 14:39:06 by LB7 »
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Offline Bobolink

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Re: Sum of energy not at 0 in that movement
« Reply #12 on: 24/04/2020 15:04:25 »
Quote from: LB7 on 24/04/2020 14:22:22
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.
Quote from: LB7 on 24/04/2020 14:22:22
I count the energy needed to move the circle in translation.
Great, please show the math how you counted it.
Quote from: LB7 on 24/04/2020 14:22:22
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.
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Offline LB7 (OP)

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Re: Sum of energy not at 0 in that movement
« Reply #13 on: 24/04/2020 15:11:05 »
What you call 'the calculations' ? the method ?

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 4205 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.


« Last Edit: 24/04/2020 15:17:19 by LB7 »
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Re: Sum of energy not at 0 in that movement
« Reply #14 on: 24/04/2020 16:23:15 »
Quote from: LB7 on 24/04/2020 15:11:05
What you call 'the calculations' ? the method ?
Are you telling me you do not know what calculation means?
Quote from: LB7 on 24/04/2020 15:11:05
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.
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Re: Sum of energy not at 0 in that movement
« Reply #15 on: 24/04/2020 16:52:43 »
I would like to know what you mean about calculations, numerical, ...

Here the forces:


* f1f2.png (23.89 kB . 377x265 - viewed 4077 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.

Quote from: Bobolink on 24/04/2020 16:23:15
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.
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.

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.
« Last Edit: 24/04/2020 17:04:33 by LB7 »
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Re: Sum of energy not at 0 in that movement
« Reply #16 on: 24/04/2020 18:02:27 »
I still do not see why you want to multiply the energy by cosine.
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.
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Re: Sum of energy not at 0 in that movement
« Reply #17 on: 24/04/2020 19:13:29 »
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.

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.

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Re: Sum of energy not at 0 in that movement
« Reply #18 on: 24/04/2020 20:03:21 »
Quote from: LB7 on 24/04/2020 19:13:29
I don't understand why there is a problem there.
Nor do I.

Why did you start the thread?
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Re: Sum of energy not at 0 in that movement
« Reply #19 on: 24/04/2020 20:12:43 »
To show how the law of the conservation of the energy is broken.


* bez.png (242.4 kB . 1646x772 - viewed 4021 times)
« Last Edit: 24/04/2020 21:00:18 by LB7 »
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