The maths solution
All the volumes are constant. I didn't draw the gaskets between the wheel and the black shape but there are (no gas escapes). I suppose the friction at 0 to study the sum of energy. The radius of the wheel is at 1.
If I'm looking at points A and B, one is moving more at right than one another. I calculated this with a big precision, like that it's possible to decrease the angle [tex]\alpha[/tex]
I don't need an energy to move in translation the black shape. So, the sum of energy must be at 0 too for the wheel, but for find the work of the circle I must integrate
[tex]\int_{pi/2}^{pi/2+da}(x-sin(x)) sin(x) dx[/tex]
[tex]\int_{pi/2}^{pi/2+da} (1-cos(x)) cos(x) dx[/tex]
[tex]\int_{3pi/2}^{3pi/2+da} (x-sin(x)) sin(x) dx[/tex]
[tex]\int_{3pi/2}^{3pi/2+da} (1 - cos(x)) cos(x) dx[/tex]
[tex]\int_{0}^{x}(x-sin(x))sin(x)dx=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))[/tex]
[tex]\int_{0}^{x}(1-cos(x))cos(x)dx=-x/2.0+sin(x)-1/4.0*sin(2.0*x)[/tex]
The result is 0.5 for an angle of 0.1
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xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNUMERICAL SOLUTIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
[quote]from mpmath import *
mp.dps=200; mp.pretty=True
da=0.000000001
a=pi/2
x1=a-sin(a)
y1=1-cos(a)
x2=a+da-sin(a+da)
y2=1-cos(a+da)
x3=a+pi-sin(a+pi)
y3=1-cos(a+pi)
x4=a+pi+da-sin(a+pi+da)
y4=1-cos(a+pi+da)
w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1-(y4-y3))*cos(a+da)/2
print w1
print w2
print "dx=", x2-x1-(x4-x3)
print "dy=", y2-y1-(y4-y3)
There is a difference of distance:
dx=1.00000000000000012447984e-18
dy=2.00000000000000012422984e-9
But, I calculated the mean works with the worst case:
w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1+(y4-y3))*cos(a+da)/2 ***** I need to calculate the integrale
print w1
print w2
And the results of the works are:
w1= 0.00000000000000000100000000000000012397984
w2=-0.00000000000000000100000000000000012422984
I can reduce the angle there is always a difference in the axis x.
I calculated for A and B like I drawn but it's possible to change the angle of the black shape, this could change the sign of the result.
from mpmath import *
mp.dps=200; mp.pretty=True
da=0.000000001
a=pi/2
def rectangles(f,a,b,n) :
h=(b-a)/float(n)
z=0
for i in range(n) :
z=z+f(a+i*h)
return h*z
def sx(x):
return (x-sin(x)) *sin(x)
def cx(x):
return (1-cos(x)) *cos(x)
def sxp(x):
return (x+pi - sin(x+pi)) *sin(x+pi)
def cxp(x):
return (1 - cos(x+pi)) *cos(x+pi)
m1=rectangles(sx,a,a+da,10000)
m2=rectangles(cx,a,a+da,10000)
m3=rectangles(sxp,a,a+da,10000)
m4=rectangles(cxp,a,a+da,10000)
print m1
print m2
print m3
print m4
print m1-m2+m3-m4
The result is:
Sum of works = 0.00000000514
But I'm not sure about my integration
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xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxMATH INTEGRATIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Here all math integrations:
x=pi/2.0
p1=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p2=-x/2+sin(x)-1/4*sin(2*x)
x=pi/2.0+da
p3=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p4=-x/2+sin(x)-1/4*sin(2*x)
x=pi/2.0
p5=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p6=1/2*(-x-2*sin(x)-sin(x)*cos(x))
x=pi/2.0+da
p7=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p8=1/2*(-x-2*sin(x)-sin(x)*cos(x))
The result is 0.00000000514
The difference is small because the angle is small, if I take [tex]\alpha[/tex] at 0.1 rd the result is 0.5
I change the math integrals and I obtain this :
x=pi/2.0
p1=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p2=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=pi/2.0+da
p3=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p4=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=3.0*pi/2.0
p5=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p6=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=3.0*pi/2.0+da
p7=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p8=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
print p1
print p2
print p3
print p4
print p5
print p6
print p7
print p8
z1=-(p3-p1)+p4-p2
z2=p6-p5+p7-p8
print abs(z1)-abs(z2)
It is the same result. And if I take the angle à pi the sum is 0.