Sum of torque

[EOW]

An arm of length d is turning clockwise at w1. A red disk of radius r is turning at w2. w1 and w2 velocities are in labo frame reference. I define w2′ the rotationnal velocity in the arm frame reference. I set w2′ for have the velocity at the origin of F1 at 0: w1(d+r)−rw2′=0, w2′=w1(d+r)/r. An external force F1 is applied during a very short time, like there is no movement, the work needed by the force F1 is 0. I define F the value of F1 or F2. Like I give F1 on disk, F2 appears on arm. The work from F2 on arm is Fdtw1. The work from F1 on disk is −Frtw2′=−Ftw1(d+r). The sum of energy is not 0.

[EOW]

The black arm is turning clockwise at w1. The blue axis is fixed to the ground. The grey cruz (grey arms) is turning at w2, with w2 < w1, so the grey cruz is turning around itself counterclockwise in the black arm reference. Each  red disk is turning at w3, with w3 < w2. For example, w1=10, w2=9 and w3=1. Friction from disk/disk gives black forces. The rotationnal velocity of each red disk on grey arm reference is w3' with w3'= w2-w3 = 8, so each red disk is turning counterclockwise around itself on the grey arm reference. So w3' can give forces like I drawn. These black forces give a torque on each disk and give a torque on the grey cruz. Each red disk increases its kinetic energy, because the torque is clockwise and w3 < w1. The black arm don't have a torque, so w1 is constant. Green forces reduces w2. BUT, the inertia of grey cruz can be like I want and don't depend only of the disks, I can add a big mass in the center of the cruz for example. Remember, torque=inertia*acceleration and kinetic energy is 1/2*inertia*w², so kinetic energy is 1/2w²/inertia. If inertia of cruz is very high the rotationnal velocity of the cruz don't decrease (imagine a mass very high), or very few in practise. If I increase the inertia of the cruz, I change w1 ? no. I change w3 ? no. I change the work from friction ? no. It is an independant parameter.

a/ w1 is constant
b/ I can choose the inertia of the grey cruz as high I want (in the center of the cruz)
c/ The heating is a positive energy
d/ The kinetic energy of each red disk increases
e/ I can reduce the lost of kinetic energy from grey cruz by adding a mass in the center of the cruz


Cycle:

1/ Friction is OFF. Launch the device with w1, w2, w3 with external motor
2/ No external motor or force
3/ Set friction ON
4/ Measure the sum of energy, heating too, for me the energy increases, the heating and the kinetic energy.

What do you think about that ?

[EOW]

For recover energy (if the heating is not the goal), just add 4 generators between red disks. Generator must be with few mass. Interface between red disk / rotor =  gear. I think I need to adjust forces from friction (or torque from generator) at each time for have no torque on w1. At inner radius, set friction higher and at outer set friction lower.

[EOW]

With one arm and one disk, the energy is not constant. Friction between disk/arm gives F1 and F2.

[EOW]

No friction, no gravity.

Case A
1/ turn counterclockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

Case B
1/ turn clockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

The energy needed for turn the ring around itself and with the arm is the same in case A and case B. In the contrary, the energy recover in case A is not the same than in case B.