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 Sum of torques and sum of energy
Сообщение11.11.2016, 12:42 


05/09/16
27
It is a theoretical problem, the friction like the mass is very small.

The device:

Изображение

The device in another position:

Изображение

height of water : 0.1 m

pressure at top: 0 Pa
pressure at the middle: 0.0025 Pa
pressure at the bottom: 0.01 Pa

law for the pressure: $x^2$, from the bottom.

with:

$a=0.1$

$b=\sqrt{1-0.9^2}$


Moment of the magenta arm on the orange dot:

on y:

$\int_0^b (0.01-(1-\sqrt{1-x^2})^2)x = 6.41e-4 Nm$

on x:
$\int_0^a x(x^2-0.01) dx = 2.5e-5 Nm$

Moment of the black arm around the black dot:

$\int_0^a x^2 = 3.33e-4 Nm$

I give my python code:

Код:
from mpmath import *
mp.dps = 20

a=mpf(0.1)

def f1(x):
  return (0.01-(1-mp.sqrt(1-x**2))**2)*x

def f2(x):
  return  x*(x**2-mpf(0.01))

def f3(x):
  return x**2


i=mp.quad(f1,[0,mp.sqrt(1-0.9**2)])
j=mp.quad(f2,[0,a])
k=mp.quad(f3,[0,a])

print i,j,k
print mp.fabs(i)+mp.fabs(j)-mp.fabs(k)   




I think the sum of torques on the pink arm around the orange dot can't be the same than the torque of the black arm around the black dot because in this case the sum of energy is not constant if the black arm turns of a small angle $\delta$.

Have a good day

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 Re: Sum of torques and sum of energy
Сообщение13.11.2016, 11:16 


05/09/16
27
In my first message, I turn the device counterclockwise. But like that the energy recovered by the black arm is lower than the energy needed by the external device to turn the pink wall (the angle between the black arm and the pink arm is constant, the external device must do that). Even, I supposed there is no friction in theory, it is possible to lost an energy somewhere. So I turn the black arm clockwise, like that the energy recovered by the external device is higher than the energy needed by the black arm.

Like I calculated the moments on 2 different axes, this is not necessary to have the sum of moments at 0. But I have a problem with the sum of energy in that case.

I changed my formulas to have variables in the equations:

height of water : 0.1 m

pressure at top: $0 \cdot 0=0 Pa$
pressure at the middle: $0.05 \cdot 0.05=0.0025 Pa$
pressure at the bottom: $0.1 \cdot 0.1=0.01 Pa$

law for the pressure: $x^2$, from the bottom.

with:

$a=0.1$ the height of the water
$R=1$ the radius of the circle

$b=\sqrt{R^2-(R-a)^2}$ length of the pink part of circle in absisse


Moment of the magenta arm on the orange dot:

on y:

$\int_0^b (a^2-(1-\sqrt{R^2-x^2})^2)x = 6.41e-4 Nm$

on x:
$\int_0^a x(x^2-a^2) dx = 2.5e-5 Nm$

Moment of the black arm around the black dot:

$\int_0^a x^2 = 3.33e-4 Nm$

I calculated the moments only, and to have the energy I calculated with a very small angle $\delta$ but even like that I need to supposed the functions are continuous. The difference in the sum of torques is high (relatively) so if the functions are continuous, the energy needed by the black arm is lower than the energy recovered by the external device. I calculated the sum of energy to take in account all data and I come back when it's done.

At final, i think the calculations of my torques are correct, I just need a confirmation to go further, please.

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 Re: Sum of torques and sum of energy
Сообщение13.11.2016, 16:25 


05/09/16
27
I found my mistake, it is in the calculation of the pressure:

Moment of the magenta arm on the orange dot:

on y:

$\int_0^b (a-(R-\sqrt{R^2-x^2}))^2x$

on x:
$\int_0^a x(x-a)^2 dx $

Moment of the black arm around the black dot:

$\int_0^a x^2$

Like that it is near 0 but not exactly, with 1500 decimals digits I have always the same error: 6e-51 Nm maybe it is an error with my program I will verify that.

Wolfram detects the error:

Изображение

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