I'm trying to find the centre of mass of a $120°$ sector that doesn't intersect the center isosceles triangle (the intersection being $Ac$). E.g. the center of mass for the sum of the areas $2(A7 + B3)$ or expanded out $2(A1+A2+A3+B1+B2)$ (note that I've left out the area of $Ac$). I've now drawn a red outline around the area in question.
Image explanation
The first 3 lines of text in the picture are for my own reference and are in degrees. All other angles are in radians. I've used ** for exponent because I was using gnuplot for testing. Function names that repeat are for my benefit and just shows the transformation of the function. They mean the same thing. Also, because gnuplot overwrites the last one with the new one, only the last one is significant. Variable names starting with a capital are areas. Variable names starting with lowercase are lengths except for $sin3060$ which is a ratio and $ang$ which is an angle. Number's after an area variable name is used as a bit flag. I.e. A2 + A4 = A6 reads as Area 2 + Area 4 = area of both 2 and 4 which also be referenced as Area 6.
Note that I'm only wanting the centre of mass for the bottom sector. The top left and top right sectors are not relevant other than they have the same geometry as the bottom sector, just rotated $120°$ around the centre of the circle.
Reasons for the names I gave them:
A = area B = area and next letter after A (duh!) :D r = radius b = dunno, picked at random c = centre of mass Ap = area of pie slice Ac = area of triangle from centre Apg = area of pie slice (general) Acg = area of triangle from centre (general) hyp = hypotenuse ang = angle
For $Apg$ and $Acg$ I was thinking general because it works for any angle.
Actual question
So basically, I'm trying to find the value of $c$ (centre of mass = $c + {r \over 2} =$ distance from centre along line going down). The centre of mass lies somewhere on that line below the interface between $A1$ and $A2$ because of symmetry.
These are the following transformations that I tried to use, but when I get to the last line, I'm stuck as to where I should go. Does anyone have any suggestions?
A1(r, b) + B1(r, c) = B2(r, c)
Ac(r, 0) - Ac(r, b) + Ap(r ) - Ac(r, 0) - B2(r, c) = B2(r, c)
Ac(r, 0) - Ac(r, b) + Ap(r ) - Ac(r, 0) = 2 * B2(r, c)
- Ac(r, b) + Ap(r ) = 2 * B2(r, c)
- Ac(r, b) + Ap(r ) = 2 * (Apg(r, c) - Acg(r, c))
- Ac(r, b) + Ap(r ) = 2 * Apg(r, c) - 2 * Acg(r, c)
- ((r-b)**2)*sin3060/2 + pi*(r**2)/3/2 = 2 * (r**2)*ang(r, c)/2 - 2 * ((r/2 + c) / sin(pi/2 - ang(r, c)) * (r/2 + c) / 2)
- ((r-b)**2)*sin3060 + pi*(r**2)/3 = 2 * (r**2)*ang(r, c) - 2 * ((r/2 + c) / sin(pi/2 - ang(r, c)) * (r/2 + c) )
- ((r-b)**2)*sin3060 + pi*(r**2)/3 = 2 * (r**2)*ang(r, c) - 2 * ((r/2 + c)**2 / sin(pi/2 - ang(r, c)))
- ((r-b)**2)*sin3060 + pi*(r**2)/3 = (2 * (r**2)*ang(r, c) * sin(pi/2 - ang(r, c)) - 2 * (r/2 + c)**2) / sin(pi/2 - ang(r, c))
- ((r-b)**2)*sin3060 + pi*(r**2)/3 = (2 * (r**2)*asin(sqrt(r**2 - (r/2 + c)**2)/r) * sin(pi/2 - asin(sqrt(r**2 - (r/2 + c)**2)/r)) - 2 * (r/2 + c)**2) / sin(pi/2 - asin(sqrt(r**2 - (r/2 + c)**2)/r))
I've tried to keep the expansions in line from one transformation to the next, to make it clear what I've done.
Prettifying the final line comes out as:
$ - \left(r-b\right)^2 sin3060 + {\pi r^2 \over 3} = {2 r^2 asin\left({\sqrt{r^2 - \left(r/2 + c\right)^2} \over r}\right)sin\left({\pi \over 2} - asin\left({\sqrt{r^2 - \left(r/2 + c\right)^2} \over r}\right)\right)- 2 \left({r \over 2} + c\right)^2 \over sin\left({\pi \over 2} - asin\left({\sqrt{r^2 - \left({r \over 2} + c\right)^2} \over r}\right)\right)} $
or
$ - {\left(r-b\right)^2 \sqrt 3 \over 4} + {\pi r^2 \over 3} = {2 r^2 asin\left({\sqrt{r^2 - \left(r/2 + c\right)^2} \over r}\right)sin\left({\pi \over 2} - asin\left({\sqrt{r^2 - \left(r/2 + c\right)^2} \over r}\right)\right)- 2 \left({r \over 2} + c\right)^2 \over sin\left({\pi \over 2} - asin\left({\sqrt{r^2 - \left({r \over 2} + c\right)^2} \over r}\right)\right)} $
There is also another equation for $ang$ which I could have used instead, but I think that it is equally as messy.
Please note that my trig is pretty rusty. I've not used it extensively in about 10-20 years, so I'm probably missing something.
Since the figure is symmetric, the x coordinate lies on the axis of symmetry.
As for the y coordinate (or the distance from the center).
$$\frac {3}{\pi r^2} \left(\int_0^{r\cos 60^\circ} \int_{y\sin -60^\circ}^{y\sin 60^\circ} y \ dx \ dy + \int_{r\cos 60^\circ}^{r} \int_{-\sqrt {r^2-y^2}}^{\sqrt {r^2-y^2}} y \ dx \ dy\right)\\ \frac {3}{\pi r^2} \left(\int_0^{\frac {r}{2}} (\sqrt 3) y^2\ dy + \int_{\frac {r}{2}}^{r} 2y\sqrt {r^2-y^2} \ dy\right)\\ \frac {3}{\pi r^2}\left((\sqrt 3) (\frac {r^3}{24}) + \frac {2}{3}(\frac {3r^2}{4})^\frac {3}{2}\right)$$
$$\frac {3}{\pi r^2}\left( (\frac {r^3\sqrt 3}{24}) + \frac {2}{3}(\frac {3\sqrt 3 r^3}{8})\right)\\ \frac {3}{\pi r^2}\left( \frac {r^3\sqrt 3}{24} + \frac {r^3\sqrt 3 }{4}\right)$$
$$\frac {7r\sqrt 3}{24\pi}$$