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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. diagram

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.

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Adrian
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  • You have so many nice colours in your image, but you don't seem to use them to tell us what sector you're talking about. I don't understand what you mean by "a $120^\circ$ sector that doesn't intersect the center isosceles triangle" – it would help a lot if you just described it in terms of the coloured areas in the image before delving into the details of your calculations. – joriki Jan 21 '20 at 21:07
  • Hi @joriki, I'm talking about the bottom sector that contains areas Ac, A1, A2, A4, B1, and B2. I'll try and clarify the text more. – Adrian Jan 21 '20 at 21:09
  • @joriki, are the changes better? I didn't reference the colours, but did use the names of the areas. – Adrian Jan 21 '20 at 21:20
  • Everything green or brown? – Doug M Jan 21 '20 at 21:20
  • @DougM, I guess you can call that awful colour brown. :D So, everything that is green, brown and grey, except for the brightest green ($Ac$), and only in the bottom sector. The sector to the top right also has the same brown because it is geometrically the same, just rotated $120°$ from the centre of the circle. The extra information in that sector was because I ran out of room in the bottom sector, and it is equivalent in terms of geometry. – Adrian Jan 21 '20 at 21:23
  • @Adrian: The red outline made all the difference. – joriki Jan 21 '20 at 21:57

1 Answers1

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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).

$$\frac {3}{\pi r^2} \left(\int_0^{r\cos 60^\circ} \int_{y\sin -60^\circ}^{y\sin 60^\circ} y \ dx \ dy + \int_{r\cos 60^\circ}^{r} \int_{-\sqrt {r^2-y^2}}^{\sqrt {r^2-y^2}} y \ dx \ dy\right)\\ \frac {3}{\pi r^2} \left(\int_0^{\frac {r}{2}} (\sqrt 3) y^2\ dy + \int_{\frac {r}{2}}^{r} 2y\sqrt {r^2-y^2} \ dy\right)\\ \frac {3}{\pi r^2}\left((\sqrt 3) (\frac {r^3}{24}) + \frac {2}{3}(\frac {3r^2}{4})^\frac {3}{2}\right)$$

$$\frac {3}{\pi r^2}\left( (\frac {r^3\sqrt 3}{24}) + \frac {2}{3}(\frac {3\sqrt 3 r^3}{8})\right)\\ \frac {3}{\pi r^2}\left( \frac {r^3\sqrt 3}{24} + \frac {r^3\sqrt 3 }{4}\right)$$

$$\frac {7r\sqrt 3}{24\pi}$$

Doug M
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  • Wow. My calculus sucks too. :D How did you come up with that calculation? Can you explain what the different terms mean on your first line? – Adrian Jan 21 '20 at 21:48
  • $\frac {1}{A} \iint y \ dx\ dy$ is gives the center of mass. I broke it up two integrals for the triangular part and for the rest of the section of the circle. I could have done polar or something like that, to avoid breaking it up, but I didn't. – Doug M Jan 21 '20 at 21:51
  • You can get displayed equations by enclosing them in double instead of single dollar signs. That's especially relevant when you mix integral limits, fractions and roots. – joriki Jan 21 '20 at 21:56
  • What's the diff between single and double dollar signs @joriki? – Adrian Jan 21 '20 at 22:08
  • @DougM, I need to go back to my calc. I feel that I should know that formula. – Adrian Jan 21 '20 at 22:10
  • I am not 100% on this. I found a couple of mistakes, since corrected. There may be more hiding. – Doug M Jan 21 '20 at 22:11
  • @Adrian: There may be a misunderstanding here. The difference is what I already wrote above: that double dollar signs produce displayed equations and single dollar signs don't. – joriki Jan 21 '20 at 22:12
  • Also, when you were messing around with the mathjax, you made a bit of a mess. Why did it go centred? Also, unbalanced and extraneous parentheses. – Adrian Jan 21 '20 at 22:13
  • No, single dollar signs do @joriki. That is how I wrote mine in the Q. I think it is causing the equations to become centred though. – Adrian Jan 21 '20 at 22:13
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    the double dollar signs center it, make it bigger, stretch out the integral signs place the limits closer to the top and bottom of each. Any extraneous parenthesis are my errors in editing. – Doug M Jan 21 '20 at 22:14
  • The vertical stretching looks better, but the centring is awful. :D – Adrian Jan 21 '20 at 22:19
  • @Adrian: As Doug M says. Displayed equations (produced by double dollar signs) are for standing alone outside of text; inline equations (produced by single dollar signs) are for writing equations among text. You don't have any displayed equations in the question; your equations are cramped and a strain on the eyes because you're using a style that's intended for inline equations. – joriki Jan 21 '20 at 22:20
  • @joriki, I just tried to use the double dollar sign and it caused the equations to wrap, which I didn't like, so I kept it as is. – Adrian Jan 21 '20 at 22:22
  • I've not verified this result, but using calculus seems to be the better way to go than the way I was doing it. I'll look into it. Thanks. – Adrian Jan 21 '20 at 22:34