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### Author Topic: Sum of torque  (Read 94924 times)

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #30 on: October 29, 2014, 08:49:03 PM »
Like that, there is a torque ?

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #30 on: October 29, 2014, 08:49:03 PM »

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #31 on: October 29, 2014, 09:19:15 PM »
If it's possible to create a torque with external objects on disk, the energy is won. So, I think it's possible to change the disk, one part with mass, another without mass, and to place point at the good position for have only a torque on it.

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #32 on: October 30, 2014, 08:30:10 AM »
If I take the disk in a double rotation w1 and w2, with w1>w2. It's possible to add a translation to all the system for have to point P1 that move in one direction. Other point P2 friction with black stem, this add forces F1 to disk and F2 to stem. Shock with external purple object for the point P1. P2 friction only, not shock.

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #33 on: October 30, 2014, 09:06:33 AM »
With velocities like that :

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #34 on: October 30, 2014, 10:05:59 AM »
With this ? this need a translation in addition for have forces like that.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #34 on: October 30, 2014, 10:05:59 AM »

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #35 on: October 30, 2014, 10:59:55 AM »
Vleft=R1w2−(R1−d)w1
Vright=R2w2−(R2−d)w1 With:
R1=10
R2=4
w1=10
w2=2
d=4
I find:
Vleft=-40
Vright=8

Use ring to be sure the formula is good: 1/2md²w1²+1/2mr²(w1−w2)²

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #36 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

#### Free Energy | searching for free energy and discussing free energy

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

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #37 on: October 30, 2014, 02:09:42 PM »
I think like that it's correct, someone can verify ?

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #38 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.

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #39 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.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #39 on: October 31, 2014, 08:55:37 AM »

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #40 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 »

#### EOW

• Sr. Member
• Posts: 361
##### Re: Sum of torque
« Reply #41 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 »

#### rc4

• Full Member
• Posts: 147
##### Re: Sum of torque
« Reply #42 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

#### rc4

• Full Member
• Posts: 147
##### Re: Sum of torque
« Reply #43 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

#### rc4

• Full Member
• Posts: 147
##### Re: Sum of torque
« Reply #44 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

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #44 on: November 06, 2014, 09:22:43 AM »