Sum of torque

[EOW]

The torque from the liquid is not the same the torque from the solid, the wheel rotates and moves alone:

The law of gravity is 1/d² not 1/d.

[EOW]

Radius of the wheel = 0.5 m
Direction of gravity : vertical
Origin of gravitation: 0.5 m below the ground
Law of attraction: 1/d²

Torque on the solid:

     integrate from 0 to 0.5 of integrate from -pi/2 to pi/2 of x^2*cos(y)/(1+x*sin(y))² dx = 0.0986123 Nm

Torque on the walls from the liquid:

     integrate from 1.5 to 1 of (x-1)*(1/1.5-1/x) - integrate from 1 to 0.5 of (x-1)*(1/1.5-1/x) dx = 0.0112016 - 0.1098 = - 0.0986123 Nm

There is no torque on the wheel.

Now, if I attract from a point below the red dot at 0.5 m, the pressure on the walls from the liquid don't change but the torque on the liquid change

Like it is the same density for the liquid and the solid, the potential energy of gravity don't lost any energy when the wheel rotates.

[EOW]

Like the attraction come from a point (the green point), I need to give an energy in the center of the wheel but that energy is a potential energy that the wheel wins. The torque 0.0986123 - 0.083333 = 0.0152 Nm is the torque that can create an energy when the wheel rotates counterclockwise. Maybe it's possible to use a sphere.

[EOW]

Like that the cg of the semi disk solid don't work. There is negative torque on the wheel from the liquid on the walls but the work is :

( 0.1098 - 0.0112 ) x = 0.0986 x with x a small angle of rotation

There is an energy from the center of the wheel:

(0.1098 + 0.0112 ) = 0.121 x

The vertical force don't work and the cg of the semi solid disk can't work

[EOW]

I have the good values for the forces with the integrals I have the sum of forces at 0 (not with my programs).

The torque from the solid part is -0.252879+0.0675103+1.36114 = 1.5465087 Nm

The torque from the liquid is 1.38147-0.0277572 = 1.3537128 Nm

There is a difference.