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## Homework Statement

Hi

Calculate the forces F1, F2, F3. F1 is the force on the red wall. F2 is the force on the black center and F3 is the force from springs on the red center. Calculate the torques T1 from F1 and T2 from F2 and give the energy from these torques for and angle of 0.2 radian of the red arm.

The device:

The study is in 2 dimensions. The device is composed of a half disk where there are a lot of small balls. The balls have no mass and there is no friction. There are springs that attract balls with the law 1/d² to have the pressure on the walls like gravity can do with water. The springs have no mass. Each spring is attached from the red center to a ball. All the device turns around the red center. The center of the half disk turns too around the red center but it is the black arm that take the black center. Like the centers of rotation are not the same, the black arm must increase its length, I drawn a cylinder for that. The law of attraction of a spring is 1/d² with d the distance from the red center. Note, the springs never lost any potential energy because the balls keep their position in the half disk.

I show the device after 5° of rotation clockwise:

## Homework Equations

Torque. Force from pressure.

## The Attempt at a Solution

**The forces and the torques:**

I found the forces F1, F2 and F3 and I verified the sum is 0 so I think it's correct. I found the torques too but the sum of energy is not 0, so my error come from the forces or the torques:

The torque T2 I gave above is the torque around the red center not the green center, just for verify the sum of torques is well at 0 around the red center.

I needed to compute F3x and F3y due to a problem of accuracy so maybe I made a mistake, I'm really not sure about F3x and F3y.

**Torque T2 and the energy:**

__I noted :__

A: the angle of rotation of the red arm (clockwise), the black arm turns of A/2

##Fx=1/6 ##

##Fy=\int_{-\pi/2}^{\pi/2}( 0.5 - 1/\sqrt( 0.5*\cos(x) + (1.5 + 0.5*\sin(x) ) )*\sin(x)*0.5 dx = 0.182355##

##L=\sqrt( (1.5*\cos(2*x))^2+(1.5+1.5*\sin(2*x))^2)##

I need to have 2x inside trigo functions because when I integrate I do from 0 to A/2

##F=\sqrt( Fx^2 + Fy^2 )##

##B=\arctan( Fx / Fy )##

Torque T2:

##T2= F * L * \sin( pi/4 -x + B )##

Energy recover from black arm:

##\int_0^{A/2}( F * L * \sin( pi/4 -x + B ) dx##

Energy needed to increase the length of the black arm:

##\int_0^{A/2}( F * L ) )* \cos( pi/4 -x + B ) dx##

Energy needed from red arm:

##A*\int_0^1( ( 0.5 – 1/(2-x) ) * (x-2) ) dx##

##\int_0^A/2( F * L * ( \sin( pi/4 - x + B ) - \cos( pi/4 - x + B ) ) dx - A*\int_0^1( ( 0.5 – 1/(2-x) ) * (x-2) ) dx##

But the result is not 0, it is 0.05 - 0.0494171. The difference is small but it is not 0 because the integration take the function sin(x)-cos(x). So, I think I made a mistake about an angle. If you can help me not to find the result just to find the error please ?

Have a good day

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