Sum of torque

[EOW]

At start, I want the same angular velocity of the spiral1 and the spiral2, but like the diameter is not the same I need to increase the diameter of the tube. Increase the diameter cost nothing in theory. Like that the arm has a torque on it. The angular velocity of each spiral at start is -w.cos(a). The Spiral1 has a counterclockwise torque on it, so it will accelerate in the laboratory reference counterclockwise. The Spiral2 has a clocckwise torque on it so it will accelerate in the lab reference clockwise. In the arm reference, the Spiral1 accelerates from -w.cos(a) to X and the Spiral2 deccelerates from -w.cos(a) to Y, but radius of each spiral change in the same time and it's possible to adjust the thickness of the tube when it goes from the Spiral1 to the Spiral2.

This device must works with the angle a=0: axis of spiral are vertical.

I can use a full material for the tube with a low mass, it's easier to see the volume can be constant from Spiral1 to Spiral2.

[EOW]

Take the example in the image. There is no arm, just 2 pulleys and one tube. I want to have the same angular velocity for the Pulley1 and Pulley2 even the radius of the Pulley1 is twice than the Pulley2. The pulleys turn counterclockwise. There is pressure from a gas outside at P=100000Pa and inside the closed tube there is p=1000 Pa. The Pulley1 receives a counterclockse torque from F1, the Pulley2 receives a clockwise torque from the force F2. Like I want the same angular velocity if the section (surface) of the tube when it is on the Pulley1 is 2mm * 2 mm. I need to keep constant the surface of the material, it is 4 surfaces by 2*2 = 16 mm². Like the Pulley1 has a radius twice than the Pulley1 I need to move 16 in surface but the length of the arc is 2 times lower it's not 2 mm but 1 mm, so if I change only 2 surfaces, I need to have 2 surfaces at 2*1 and 2 surfaces at 6*1 = 16. So the front surface from F1 is 2*2=4 and from F2 it's 1*6=6. So the Force F2=1.5*F1. Or if I change 4 surfaces I will have 4 surfaces at 4mm*4mm.

The sum of the energy:

1/ The Pulley1 has a counterclockwise torque from F1 the energy is 2*R*F1*wt=8(P-p)Rwt, where (P-p) is the difference of the pressure
2/ The Pulley2 has a clockwise torque from F2 the energy is -R*F2*wt=-16(P-p)Rwt
3/ The volume of the tube increases like 8(P-p)Rwt
Note: move the wall don't need any energy because the force from pressure is perpendicular to the movement except where there are F1 and F2.

Even the sum of energy is constant do that:

Now place all this device on an arm that turns at w. The Pulley1 and the Pulley2 turn at -w in the reference arm. But look at the torque on the arm, it is positive. And the Pulley2 receives a clockwise torque but it turns at -w in the arm reference not in the lab reference ! so it decelerates in the arm reference (accelerates in the lab reference).

The sum can't be constant.

[EOW]

In the last example, the volume of the tube increase but it's possible to let it constant, for that I need to enter the tube inside itself. Like that repeat the cycle is easier.

I place the center of rotation of the device in the red center like that the force from pressure due to the conic section can't give a torque.

The pulley1 can support the tube in the upper side of the magenta line (third image) and the pulley2 can support the lower side of the magenta line. Like that there is a torque on the arm.

[EOW]

I change the direction of the force, like that the arm receives a clockwise torque. The volume of the tube is constant so I don't lost any potential energy. Each pulley will accelerate in the lab reference (decelerate in the arm reference). Like that the difference is the bigger pulley lost more energy more quickly (in the arm reference) but I don't gave any energy in fact

Example:

Radius of Pulley1 = 1000 mm
Radius of Pulley2 = 500 mm

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 7.4641 mm * 0.535898 mm = 4 mm² (resolv 2x+2y=4 and xy=4)

The tube don't lost any potential energy. The arm don't receive a torque.

Example2:

Radius of Pulley1 = 1000 mm
Radius of Pulley2 = 1000 mm (same radius)

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 0.472136 mm * 16.9492 mm = 8 mm² (resolv 2x+2y=8 and xy=8)

Here the volume of the tube increases and the arm receives a clockwise torque.

[EOW]

At start the squares don't turn around the blue center. The arm turns at w. The blue walls are supported by the squares so each square accelerates counterclockwise. The red center supports the red walls, in one turn of the squares, the torque could be at 0 on the arm, but not because the squares increase their angular velocities and there are less and less time for the torque. So the arm accelerates clockwise and the squares accelerate counterclockwise. Imagine with the thickness T near 0.

Outside the device there is a small pressure.
Inside the blue/red shape there is a bigger pressure from a gas for example.

The pressure at 100000Pa is only in the shape composed by 2 blue walls and 2 red walls (small volume because T is near 0). This shape gives the forces F1 to F4.

I give the Python code to test. With k=0 (no acceleration) there is no torque on the arm. But with k=0.2 for example there is a torque of 58 in one turn of the squares.

The squares turn together with the same angular velocity, they keep constant their relative position from each other.