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Author Topic: Sum of torque  (Read 86057 times)

Offline EOW

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Sum of torque
« on: October 12, 2014, 11:36:02 AM »
Blue liquid can turn around red point. I suppose red disk like a mass M, this mass attrack liquid like gravity can do in water on Earth. I suppose pressure linear with distance like on Earth. There is buoyancy in liquid. I compute sum of torque :

With:

R1 = 1
R2 = 2

In this case L1 = 1 and L2 = sqrt( 2²-1² ) = sqrt(3)

T1:

integrate from 0 to 1 of (1+x)*(1-x) this give 0.666

T2:

integrate from 0 to sqrt(3) of x*(sqrt(3)-x)/sqrt(3) this give 1/2

Sum of torque is not 0, the system turn alone and give energy I think. Gravity can be change by electrostatic or magnetism forces, small balls can replace liquid.

« Last Edit: October 12, 2014, 06:23:51 PM by EOW »

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Sum of torque
« on: October 12, 2014, 11:36:02 AM »

Offline EOW

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Re: Sum of torque
« Reply #1 on: October 14, 2014, 04:23:23 PM »
Oops, there is an error, the integration for T2, the sum is zero.


It's possible to use 3 disks like second image showing, they turn at different rotationnal velocity X and Y but in the same direction without friction. Red disks don't turn around itself. At time t, friction is on, this accelerate one disk and decellerate others but if inertia are differents work from torques are different, so the sum of energy is not constant (take in account friction = energy = temperature).

The Work for torque = Torque * angle of rotation ( W=T*Theta )

Theta = 1/2 * a * t² with a the rotationnal acceleration

Second law of Newton : Torque = I * a with I the inertia

Angle of rotation = 1/2 * T/I * t²

Work of torque = 1/2 * T²/I * t²

W1 = 1/2 * T²/I1 * t² with I1 the inertia of grey disk
W2 = -1/2 * T²/I2 * t² with I2 the inertia of red disks

It's logical that W1 is not equal to W2 (in value) if I1 is not equal to I2, no ?

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Offline EOW

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Re: Sum of torque
« Reply #2 on: October 15, 2014, 04:51:59 PM »
Even with linear velocity there is a problem. E=1/2mV². Two objects move at +V1 and +V2. At a moment Object1 give +F to Object2, and Object2 give -F to Object1 without friction. Object1 move faster than Object2. The sum of energy is:

S=1/2m1(V1+F/m1*t)²+1/2m2(V2-F/m2*t)² - (1/2m1V1²+1/2m2V2² )

With data:

m1=10
m2=10
F=100
V1=10
V2=15
t=1

S=500 J
the sum of energy is positive


« Last Edit: October 15, 2014, 08:39:37 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #3 on: October 20, 2014, 06:58:00 PM »
3 disks (no gear just friction). Before start, blue disk, red disk and grey disks turn without friction. Note, red disk turn at w2 around itself and at w1 around blue point. A t=0, friction1 and friction2 are ON. This give a torque to grey and blue disks, energy of blue and grey disks increase. Red disk increase its rotational velocity around blue point (w1 increase). Rotationnal velocity of red disk around its center of gravity slow down SO the energy increase too (energy is 1/2I(w1-w2). All energies increase here. And there is energy from friction too. Friction 1 and 2 are choose for have a torque on red disk and only a torque. For this, friction must change all along the distance. It's just a technical problem. Note this works only fex second, until w2 is stable.


Offline EOW

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Re: Sum of torque
« Reply #4 on: October 21, 2014, 07:54:23 AM »
Note about w2, w3 and w1. absolute value of w2 must be greater than 0. Absolute value of w2 must be greater than (abs( ((R+r)/r)*w1) +abs(w3)  ). R=radius of big disk. r=radius of small disk. This works only for a transitional regime due to these conditions.

 I think big disk (disk that turn at w1) must be lower than small disk (disk that have w2) for have a difference of energy (w1-w2) increasing when w2 increase like image shows

better with 2 diameters for  disk 2 (second image). Remember energy come from red/yellow disk, it gives energy to grey disk and to blue disk. The goal is to increase energy of red/yellow disk in the same time.

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Re: Sum of torque
« Reply #4 on: October 21, 2014, 07:54:23 AM »
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Offline EOW

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Re: Sum of torque
« Reply #5 on: October 21, 2014, 09:56:41 PM »
Magenta disk has a velocity around red point and around blue point too at start, this energy is gave. At t=0, 2 disks are like figure, with friction between 2 disks. Magenta disk increase rotationnal velocity of grey disk. Grey disk turn around blue point only. Magenta disk decrease its velocity around blue point but decrease its velocity around red point too. The sum of energy is not 0. Black arm is force to turn like grey disk.

Offline EOW

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Re: Sum of torque
« Reply #6 on: October 22, 2014, 09:14:38 AM »
A disk with radius R turn at w1 around red axis of rotation. A cylinder inside the disk of radius r turn at w2 around black axis of rotation. Black axis of rotation is supported by disk. At start, disk and cylinder are turning. Friction exist between cylinder and disk, this friction give torque F1/F2 and -F1/-F2. Like radius of disk is different of radius of cylinder torques not works the same value, because it's possible to have the same angle when choose good value w1 and w2. The works of torques can be positive with good value and friction add energy too.

F1/F2 works with a radius r on cylinder. -F1/-F2 works with a radius R on disk.

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Re: Sum of torque
« Reply #6 on: October 22, 2014, 09:14:38 AM »
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Offline EOW

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Re: Sum of torque
« Reply #7 on: October 22, 2014, 03:31:40 PM »
2 disks are turning together. Blue disk turn at w1 and yellow disk turn at w2. With w2<w1. A support not drawn aplly forces like image shows. torque on blue disk don't change energy because forces pass to axis of rotation. In the contrary, forces on yellow disk increase energy like (w1-w2)², like w1>w2, when w2 decrease, the energy increase. Friction add energy too. For maintain yellow disk at contact of black walls, this need to add a spring or something like that.

Offline EOW

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Re: Sum of torque
« Reply #8 on: October 23, 2014, 08:24:37 AM »
 Maybe like that ? N disks all around disk1. Each disk (except disk1) receive a torque that slow down rotationnal velocity of diskX, so the energy increase is W1 > Wx with X from 2 to N. And N the number of disks (except disk1). W1 is clockwise, Wx are anticlockwise. W1 don't decrease because forces pass through axis of rotation (blue point in center of grey disk1)

Friction create forces. Each disk except disk1 receive a torque, this torque decrease rotationnal velocity and increase energy because energy is a function of (W1-Wx)². W1 > Wx. Disk1 turn faster than others disks.
« Last Edit: October 23, 2014, 10:34:16 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #9 on: October 23, 2014, 10:43:08 PM »
At start the system is turning freely (no axis of rotation), the system turn around its center of gravity. w2 > w1 like that when w2 slow down, the energy increase, energy if a function of (w1-w2)².

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Re: Sum of torque
« Reply #9 on: October 23, 2014, 10:43:08 PM »
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Offline EOW

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Re: Sum of torque
« Reply #10 on: October 24, 2014, 10:27:52 PM »
All red disks will slow down, so like energy is (w1-w2)², energy increase. Except for 4 corners. But N can be great and there are always 4 corners. Before start, give energy for turn red disks around themselves and around green axis. After, friction is ON, this increase energy of the system.

Offline EOW

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Re: Sum of torque
« Reply #11 on: October 25, 2014, 08:16:36 AM »
With severals layers.

With one layer only: 4 corners give 4 torques but 4 disks at corners don't decrease their rotationnal velocity around their center of gravity.
With 2 layers only: the same for 4 next corners, previous torque are reported at this layer but 4 disks at layers one decrease rotationnal velocity =>energy is bigger
Wit 3 layers: 4 disks at corners of the second layer win energy

Etc.

R=Radius of first square layer (a side of inside square = 2R)
r= radius of red disk
F=basic force from friction
t=time

1/ With first layer only, there are:
  - 20 torques work 20*T*(w1-w2)t = 2*20*Fr*w2t = 40Fr(w1-w2)t
  - Friction 1 = 40Frw2t
  - 8 torques back (R-r)Fw1t, with R=6r for example, this give -48Frw1t+8Frw1t
  Sum = 40Fr(w1-w2)t + 40Frw2t -48Frw1t+8Frw1t = 0 so no energy from first layer if alone


2/ With second layer only, there are:
  - 28 torques work 28*T*(w1-w2)t
  - Friction 2
  - 8 torques back (R+2r-r)Fw1t
  Sum is 0 too like first layer

3/ With first AND second layers only, there are:
  - 20+28+8 torques work 52*T*(w1-w2)t = 2*56Fr(w1-w2)t so that add 16Frw1t-16Frw2t
  - Friction 1+Friction2 + 8Frw2t
  - 8 torques back (R-r)Fw1t + 8 torques back (R+2r-r)Fw1t
 
The difference is 16Frw1t-8Frw2t, the sum is not 0, like w1>w2, the energy is positive.

Forces on corners' s disk will receive more torque (4F), so rotationnal velocity at start must be higher for 4 corners's disk for have better efficiency.



« Last Edit: October 25, 2014, 03:45:30 PM by EOW »


Offline EOW

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Re: Sum of torque
« Reply #12 on: October 25, 2014, 03:58:37 PM »
The sum is 0:

1/ With first layer only, there are:
  - 24 torques work 24*T*(w1-w2)t = 2*24*Fr*(w1-w2)t = 48Fr(w1-w2)t
  - Friction 1 = 48Frw2t
  - 8 torques back (R)Fw1t, with R=6r for example, this give -48Frw1t
 

2/ With second layer only, there are:
  - 32 torques work
  - Friction 2 =  64Frw2t
  - 8 torques back (R+2r)Fw1t
  Sum is 0 too like first layer

3/ With first AND second layers only, there are:
  - 24+32+8 torques work 64*T*(w1-w2)t = 2*64Fr(w1-w2)t add 16Fr(w1-w2)t
  - Friction 1+Friction2 + 8Frw2t add 16Frw2t
  - 8 torques back (R+2r)Fw1t  + 8 torques back (R+2r)Fw1t add 16Frw1t

Sum = 0

------------------------------------ IF W2 LAYER_2 IS DIFFERENT OF W2 LAYER_1---------------------------------

Maybe if w2 layer 2 noted w2' is lower than w2 of layer 1 the sum is not 0. I will compute it. I suppose frictions are different for have the same force.

 - 24+32+8 torques, Frt( 24(w1-w2)+64(w1-w2')+16(w1-w2)), don't change if w2'=w2 because force is the same and torque is apply on first layer not the second
  - Friction 1+Friction2 + 2*8Frw2t add 8Fr(w2+w2')t here friction change because it depend of w2 and w2'
  - 8 torques back (R+2r)Fw1t + 8 torques back (R+2r)Fw1t , don't change if w2' not equal to w2 because force is the same

here the difference exist it is 8Fr(w2+w2')t that can be different of 8Fr(2w2)t

Difficult to build in practice because friction must change in dymamics, corners's disks must have a friction that change for have the same force even velocity is not the same.
« Last Edit: October 25, 2014, 07:37:23 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #13 on: October 26, 2014, 02:44:30 AM »
I compute work from torques:

    +24+32+8 torques on red disks = Frt(48(w1−w2)+64(w1−w′2)+16(w1−w2))
    Friction : +24Fr(w2+w2)t+32Fr(w′2+w′2)t+8Fr(w2+w′2)t
    -8 torques layer1 - 8 torque layer2 =−8(R+2r)Fw1t−8(R+2r)Fw1t=−8(6r+2r)Fw1t−64Frw1t=−128Frw1t

The sum of works from torque = 128Frtw1−64Frtw2−64Frtw′2+56Frtw2+72Frtw′2−128Frw1t

The sum = −8Frtw2+8Frtw′2=8Frt(w′2−w2) it's different of 0

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Offline EOW

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Re: Sum of torque
« Reply #14 on: October 26, 2014, 10:12:31 AM »
How disks can be for have different friction.

Before t=0, an external system give energy for rotate disks around themselves and around green axis. At t=0, the external system is OFF and friction is ON. The energy must be conserved but not.

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Re: Sum of torque
« Reply #14 on: October 26, 2014, 10:12:31 AM »

 

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