Sum of momentum

[m441]

Hello,

I'm not fluently in english, but I gave a french link where I explained the device in french if it's necessary.

I would like to study this device:



The support (a torus, blue color) turns counterclockwise around the center of rotation C1. The center C1 is fixed to the ground.
A red object with compressible balls and springs turns counterclockwise around the center of rotation C2. The center C2 is fixed to the support. The red object can turn around C2, I drawn the red object above the torus but it's for understand there are 2 objects.
The torus pass throught the red object and balls pressure the right wall of the red object and a part of the "circle" of the torus.




In a part of the red object there are a lot a small balls. Each ball is attracked by a spring (there are a lot of springs). I suppose the balls small like the molecules of water in theory, just for simplify the study. Like that I can apply the law of pressure. The springs act like the gravity can do, it's just for have a pressure to the right wall (I need it).



So, I will have a pressure at right and forces from this pressure, like water can do in a recipient under gravity. The forces from the springs are more and more higher too but a little less because there is the presence of the torus.

Like this the device is very unstable. But I accelerate more and more the support for have : Angular_velocity_of_the_support = Angular_velocity_of_the_red_object. At each time, the support must have the exact angular velocity of the object, it's difficult to do in practise but it's possible and I would like to study the device like that.

Now, I study the sum of momentums and the sum of forces. My goal is to have a torque on the red object but cancel the torque on the support. And it's possible with good values of the force Fx.

The device with distances and datas:



I want to have the sum of forces on the red object in the direction of (C1,C2):



**************************************************************************Calculations*********************************************************************

R=6.5 external radius of the torus
c=3 side of red object where there is balls
d=4.34 distance C1C2 (x or y)

The distance d is obtain with the equation $cos(\alpha)-sin(\alpha)+c/R=0$, in this case $\alplha=64.05°$

I suppose the force of pressure is equal to the height of the balls, it's possible because I can choose the force of the spring. Each spring attrack with same force. I did not give the unity but if it's necessary, it's meter for the distance, Newton for the force, Nm for the momentum.

******************************************Springs**************************************
The start of integration (x):
$s1=d-c/2=2.84$

The middle of the integration (x):
$sm=d=4.34$

The end of the integation (x):
$s2=d+c/2=5.84$

The height:
$H=d+c/2=5.84$

Momentums:

Integrate1:

$\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx$

Integrate2:

$\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx$

Value: $-0.20+1.63$

Sum of forces:

$\frac{1}{c}\int_{s1}^{s2} (H-\sqrt(R^2-x^2)) dx$

Value: $1.15$

******************************************Side Wall**************************************
Momentums:

$\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx$

$\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx$

Value:$ -0.37+1.87$

Sum of forces:
Value: $c/2=1.5 $
******************************************Left Wall**************************************

I want to cancel of torque on the support, so I need $Fx: 1.5-1.15=0.35$

Momentums:

$Value: -0.35*4.34=-1.5 $

******************************************The Torus************************************

The part of the circle that is in contact with balls don't give a torque to the torus because it's a circle. With Fx the sum of forces on the red object is in the direction of (C1,C2), so there is no torque.


*****************************************Results****************************************

Like that the sum of momentum on the red object is :

$ -0.20+1.63-0.37+1.87-1.5=1.42$

There is no torque on the support. Now, if I accelerated the support like the object, I can keep constant the relative position of the object on the support. It's like the support and the object are fixed together, but there are not ! I need to give an energy for accelerate more and more the support but this energy can be recover later. In the contrary I don't gave any energy for accelerate the red object.

So the device increases its potential energy alone ? Where I'm wrong ?

If it's not clear, I'm asking the question in french here:

http://www.chercheursduvrai.fr/forum/in ... topic=2021

[Oleg Zubelevich]

i understand your english but i do not understand the mechanism

-- Сб авг 01, 2015 11:54:04 --

the red frame what is this? is it a box filled with balls?

[Pphantom]

Тема перемещена из форума «Физика» в форум «Помогите решить / разобраться (Ф)»

[m441]

Yes, the red frame, is a solid with balls inside it. The balls are only where I drawn them (top/right). There are only two different "objects": the support (the torus) and the red frame+balls+springs. Each spring is attached between a ball and a green point to the red frame.

I drawn 4 positions, to look how the device turns.



The red frame will accelerate, but I accelerate the support in the same time for keep constant the relative position of the red frame on the support.

To simplify the study I consider the mass of the balls and the mass of the springs like zero or it's possible to cancel the centrifugal force of each object with an arm in practise.

The red frame has a mass and its center of gravity is on C2.

[m441]

I made a mistake in my sum of forces and in the momentum but there is always a difference, it is small.

Now, if I change the center C2, the difference of sum of momentum is very high:



I will calculate this later.

[m441]

At C2 at start, the sum of momentum on the support at C1 is not exactly the same than the momentum on the object at C2. The difference is small but there is one.