Sum of torque

[EOW]

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.

[EOW]

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.

[EOW]

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.

[EOW]

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

[EOW]

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.