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Author Topic: Sum of torque  (Read 102827 times)

Offline EOW

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Re: Sum of torque
« Reply #30 on: October 30, 2014, 01:33:36 PM »
I made a mistake R1 must be < R2, so value can be:

Vleft=-R1w2+(R1−d)w1
Vright=-R2w2+(R2−d)w1 With:
R1=0.5
R2=15
w1=10
w2=8
d=4
I find:
Vleft=-49
Vright=-20

Add a translation for have +30:

Vleft= -19
Vright=+10

CQFD

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Re: Sum of torque
« Reply #30 on: October 30, 2014, 01:33:36 PM »

Offline EOW

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Re: Sum of torque
« Reply #31 on: October 30, 2014, 02:09:42 PM »
I think like that it's correct, someone can verify ?

Offline EOW

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Re: Sum of torque
« Reply #32 on: October 30, 2014, 04:30:33 PM »
If I take a simple example. Trajectories are in doted lines. F1/F2 are created with a shock from external object. F11/F21 create a torque on ring, this increase the energy of disk. F12/F22 create a force to blue axis, this axis is free to move in space, so forces can't give a torque on axis but only move it, this give energy too. For simplify the problem, think with mass of blue axis like very high compared to the grey ring. I think like that the energy is not constant.

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Re: Sum of torque
« Reply #32 on: October 30, 2014, 04:30:33 PM »
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Offline EOW

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Re: Sum of torque
« Reply #33 on: October 31, 2014, 08:55:37 AM »
I forgot a 2 forces, so there is a torque on black arm, this decrease energy. But the force (F12+F22) on blue axis move all the system, this must increase a little its energy, no ? What's the difference between fixed blue axis and free ? Fixed blue axis F12+F22 don't works and free F12+F22 works.

Offline EOW

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Re: Sum of torque
« Reply #34 on: November 02, 2014, 07:57:29 AM »
It's seems it's not possible to have trajectories like I want. But it's possible to add N system in parallel. Each system give two energy E1 in heating and energy from disk E2. Sum for each disk is 2E1+E2. With N system this will give N(2E1+E2). I need to add energy at 2 last systems, need 2E1, so the sum of energy is N(2E1+E2)-2E1.

edit: for have friction, it's possible to place blue axis of rotation further (or closer), and change the angle, like angles can be shows between black and magenta lines. Add 3° at each new system. It's only for have "constact" during a time for have friction.

edit: in a last message I said it's not possible to find 2 points where horizontal velocity is negative, but it's possible with r > d (the mass is in the green point only, not at all the surface of the disk). Like that I can think like that: hits only one ball, and for have a torque I give the energy for this, like that I recover energy from ball but I lost energy I gave, BUT th energy of the disk increase !
« Last Edit: November 02, 2014, 12:27:49 PM by EOW »

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Re: Sum of torque
« Reply #34 on: November 02, 2014, 07:57:29 AM »
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Offline EOW

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Re: Sum of torque
« Reply #35 on: November 02, 2014, 07:36:08 PM »
A force F is needed for generate friction, so energy is needed for have friction. Force F is apply from an angle 'a1' to another angle 'a2'. It's not possible to have friction from 0 to 360° but only a part of 360°, it depend of the configuration.This force give energy to disks because F increase w1 of each disk. Friction don't decrease F, so all energy from F is giving to w1. Green arrows are forces from friction. Magenta arrows are forces that I need to give energy. 

Edit: it's possible to add forces F at bottom and at top, like I'm sure I don't lost energy from forces F.
« Last Edit: November 02, 2014, 09:42:21 PM by EOW »

Offline rc4

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Re: Sum of torque
« Reply #36 on: November 04, 2014, 05:52:53 AM »
Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Fdw3t with w3 the mean of w2

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforce=−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Fdw3t−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t

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Re: Sum of torque
« Reply #36 on: November 04, 2014, 05:52:53 AM »
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Offline rc4

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Re: Sum of torque
« Reply #37 on: November 05, 2014, 07:28:58 AM »
I made a mistake in my calculations:

Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Frw3t with w3 the mean of w2 **************************** I noted 'd' but it is 'r'

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforces=−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Frw3t−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t

Offline rc4

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Re: Sum of torque
« Reply #38 on: November 06, 2014, 09:22:43 AM »
I posted here :

http://physics.stackexchange.com/questions/143377/one-disk-ring-in-double-rotation-and-sum-of-energy

for have a reply, maybe if you up (+1) the question, more people will be interested about it


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Re: Sum of torque
« Reply #38 on: November 06, 2014, 09:22:43 AM »
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Offline EOW

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Re: Sum of torque
« Reply #39 on: November 09, 2014, 11:17:10 AM »
The problem with my last idea are the trajectories. I can't have green forces in this direction because points don't move like I thought. For that, I need to have a rotationnal velocity higher than w1. I use for this gears, that will increase rotationnal velocity. These addionnal gears has no mass (in theory) like that they don't lost energy.

For the cycle: give rotationnal velocity w1 and w2 (and kw2), this is for launch the system. And after let "live" the system like it is, no external motor. There is only friction between magenta/magenta disks. Forces are like that because kw2 > w1. Here I can take w1=10 clockwise, w2=-7 counterclockwise and kw2=21 clockwise.

First image: At start: all disks (or rings) are turning around blue axis at closckwise w1. All bigger green disks are turning at w2 counterclockwise around green axis. All smaller magenta disks are turning at kw2 clockwise with k =3 (in this example but k can be higher). Green disks have mass. Magenta disks haven't mass. Note there is no friction between magenta/green disks because it's gears (and in theroy I consider no friction here).

Second image: Look at forces. Friction generate F1 and F2 forces. Note there is heating => energy. F1 and F2 create F3, F4, F5, F6, F7, F8, F9, F10 like image shows and 2 additionnal -F3 -F4 not drawn. Each magenta disk receive a counterclockwise torque. They reduce their rotationnal velocity but note they have no mass. Each green disk receive a clockwise torque and increase their rotationnal velocity, and like they have mass, they increase kinetic energy. In the lab frame reference, the kinetic energy is 1/2md²w1²+1/4r²w2² with m the mass of the disk, d the lenght of the arm and r the radius of the disk. Fx-1 and Fx+1 show the forces come from another basis system when I repeat them.

Third image: repeat N systems

I define H the energy from one magenta/magenta friction. I define K the additionnal kinetic energy of one green disk. The sum of energy increases, we have (N-1)*H energy from heating . Add N*K kinetic energy from each green disk. Remove H for two last system, I need to give energy for give Fx-1 and Fx+1 at 2 terminal system. The sum of energy is (N-2)*H+NK.

Edit: like before, blue axes are fixed to the ground. Think with k very high (radius of magenta disks are very small), it's easy to see forces can be like I drawn but not all around the circle just like I drawn at start with arms horizontal.

Offline EOW

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Re: Sum of torque
« Reply #40 on: November 11, 2014, 09:56:02 AM »
The idea is to increase rotationnal velocity (in the lab frame reference) of disks with friction forces. Friction is energy and kinetics energy too. I try to choose good velocities, but event there is a problem, it's possible to chose one disk without mass if its rotationnal velocity decrease. For those disks that increase their rotationnal velocitiy set mass not at 0. I hope it's clear enough. Look at images, please.

Before $t=0$: I give rotationnal velocity $w_1$, $w_{2a}$, $w_{2b}$ and $w_3$, this is for launch the system, during this step I need to give energy. And after I let "live" the system like it is, no external motor. The sum of energy must be constant. I count all energies in the system, heating too. Green disk is turning around blue axis clockwise and around itself counterclockwise (green axis). Magenta disk is turning around blue axis clockwise and around itself clockwise (magenta axis) for the upper and counterclockwise around itself for the lower. If I want to guess no sliding between green and magenta disks, I need to have $w_3$ different of $w_3'$. There is only friction between magenta/magenta disks, but before $t=0$, I set $friction=0$ for launch the system. I need to set $|w_1| > |w_2|$. I can take, for example, $w_1=-10$, $w_{2a}=-7$, $w_{2b}=-5$ and $w_3=+9$ and $w_3'=-29$ (all angular velocities are labo frame reference). There is a relation between $w_{2x}$ and $w_3$ because there is no sliding between magenta and green disks.

Blue axes are fixed to the ground. Arms must turn together and gears (not drawn and without friction) force all arms to turn together at the same rotationnal velocity. If a torque is present on an arm it will be apply on others through gears. Green disks have mass. Magenta disks haven't mass.

I need to have friction between magenta disks AND in the good direction, for that I need to set $w_{2x}$ different for each green disk with $|w_1| > |w_2|$. Look the description and the image where I calculate velocities.

Note I study the sum of energy only in a transcient analysis. From $time=0$ to $time=t_x$ with $t_x$ very small.

At start:  Note there is no sliding between magenta/green disks because green disk force to turn magenta disk like gear can do, it's not a gear but consider contact magenta/green disks are like gears without friction (no heating dissipiation) and no sliding. Sure, magenta disk force green disk to turn like gear can do. But I guess no friction between magenta/green disks.

Look image N°1

Friction generate $F1$ and $F2$ forces. I noted all others forces I see. Note there is heating => energy between magenta/magenta disks. $F1$ and $F2$ create others forces. Each magenta disk receive a counterclockwise torque. They reduce their rotationnal velocity but they have no mass, so they don't decrease their kinetic energy. Each green disk receive a clockwise torque and increase their rotationnal velocity, and like they have mass, they increase kinetic energy. Image shows $Fx-1$ and $Fx+1$ forces, it comes from another basis system when I repeat them. Look image N°2

Another position: Look image N°3

To be sure trajectories are correct, I calculate linear velocities in this position: look image N°4

$P1$ and $P2$ are on each magenta disk, where there is friction. With $r'$ the radius of magenta disk. The linear velocity of $P1$ is $(d-r')(10)-r'(15)=-25r'$ and the linear velocity of $P2$ is $(d+r')(10)-r'(19)=-9r$, I counted positive the right direction. So, $P1$ move faster at left than $P2$ and forces can be like I drawn.

So, with: $|w_{2a}| > |w_{2b}| > |w_{2c}| > |w_{2d}|$ it's ok.


I need to give $Fc1$ and $Fc2$ forces for have friction between magenta disks. These forces don't work. Look Image N°5

I repeat $N$ systems like that last image

I define $H$ the energy from one magenta/magenta friction. I define $K$ the additionnal kinetic energy of one green disk. I define N the number of basic system {magenta disk + green disk}. The sum of energy increases, we have $(N-1)*H$ energy from heating . Add N*K kinetic energy from green disks. Remove $H$ for two last system: I need to give energy for give $Fx-1$ and $Fx+1$ at 2 terminal disks. The additionnal energy is $(N-2)*H+NK$, it's not 0.

Edit: with friction between magenta and green disks it could be easier to find good velocities. Maybe with no friction between magenta/magenta disks. I will calculate this. If necessary change radius of disks, magenta disks can be with different radius; The same for green disks.


« Last Edit: November 11, 2014, 12:01:51 PM by EOW »

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Re: Sum of torque
« Reply #40 on: November 11, 2014, 09:56:02 AM »
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Offline EOW

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Re: Sum of torque
« Reply #41 on: November 11, 2014, 02:12:01 PM »
Like image shows. F5 and F6 forces (and others) increase w1. There is heating. 4 red disks decrease kinetics energy but 3 black disks increase theirs and there is heating. |w1|>|w2|

Offline EOW

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Re: Sum of torque
« Reply #42 on: November 12, 2014, 12:19:54 AM »
With 'R' radius of purple disk.  With 'r' the radius of grey disk. 'F' the force from friction. 't' the time. A grey disk is rotating at w1 around red axis and at w2. Gears are turning too. I need to give kinetic energy for that. After, I let "live" the disk and gears. There is on friction between purple disk and grey disk, no between gears. No external motor. I let the device like it is and I count energy. The device works only few seconds. Purple disk is fixed to the ground. Friction generate forces F1 and F2, energy goes to heating. w2>(R+r)w1/r, with labo frame reference.

Gears are turning clockwise and counterclockwise. I guess between gear/grey_disk it's like a gear: no sliding, no heating dissipation.

Cycle:

1/ Friction is OFF. Launch grey disk and gears at w1 and w2
2/ Set friction ON
3/ Measure the sum of energy, heating too (H) !

H= | F(Rw1-r(w2-w1))t |

The sum of energy must be constant, but it decreases in this example. So with gears I can have good rotationnal velocities and forces. Like gears has mass (like disk), when grey disk decelerates, gears want to keep their rotationnal velocity due to inertia, this give forces. I can set forces like I want, for example F/2 for first gear and F/4 for second gear because forces depends of the inertia of each gear and it can be like I want. Here the delta energy is H+Ft(-rw2+Rw1+1/2rw1). I need to have w2>(R+r)w1/r for have F1 and F2 in these directions. So with the limit case w2=(R+r)w1/r the sum of energy is -1/2Frtw1+H. In the limit case H=0, the sum is not 0, it is < 0. If w2=(R+r)w1/r+x, the sum of energy is -Ftx-1/2Frtw1+ Ft ((Rw1-r(w1(R+r)/r+x)-w1)t)=-1/2Frtw1, it's not 0.

Calculations :

Interface PurpleDisk/GreyDisk: ( -1/2rw2+3/2(R+r)w1 )Ft
Interface GreyDisk/FirstGear: ( -1/4rw2-3/4(R+3r)w1) Ft
Interface FirstGear/SecondGear: ( -1/4rw2+1/4(R+5r)w1) Ft

Sum is (-rw2+Rw1+1/2rw1) Ft

« Last Edit: November 12, 2014, 11:56:04 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #43 on: November 12, 2014, 01:07:00 PM »
My last case, the energy increases of -rw2-1/2rw1, because I forgot one torque. Like H is FRtw2 it's not possible to have 0.

Now, grey disk rotate around itself counterclockwise. Forces are like image shows. The energy from torques is Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1

This result don't depend of w1, but for friction w2 is a parameter, H = Ft((R+r)w1-rw2), look below I verified this calculation

The additionnal energy is Ft((R+r)w1-rw2)-FRtw1= Ftr(w1-w2) and w2 < w1, the energy is positive
« Last Edit: November 13, 2014, 12:20:44 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #44 on: November 12, 2014, 08:09:05 PM »
I'm not sure about my forces and my sum of energy but in the first image, grey disk don't around itself. The energy from heating H=+FRtw1, and torques give the energy +Frtw1-F(R+r)rw1, the sum is 0.

In the second image, grey disk turns around itself counterclockwise. The energy from heating is Ft((R+r)w1-rw2). The energy from torque is +Frtw2-F(R+r)rw1, here the sum is 0, like that I'm sure of the calculation of the heating H for the last message where I find a sum different of 0.
« Last Edit: November 12, 2014, 11:51:24 PM by EOW »

 

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