This is not a question, but just a theory post.
Here we would like to calculate the position of center of mass of the circular arc subtending an angle φ at the center of the circle which it is a part of.
I made this graph on Desmos, here's the link to the graph: https://www.desmos.com/calculator/wwa9ofhgkb
for those who want the formula : Center of mass lies at a distance $R$$\biggr($ ${sin({{φ}\over 2})}\over {{φ}\over 2}$$\biggr) \ from\ the\ center.$
The proof is in the answer section
**Proof
Reference image:
Graph link: https://www.desmos.com/calculator/ufddnkhjum
At an angle $\theta$ from the x axis consider a small element of dm mass and length $dl$ = R•$d\theta$
The total arc length = R$φ$
Let M be the mass of the given circular arc
Linear mass density of the circular arc, $\lambda$ $=$ ${mass}\over {per\ unit\ length}$
$\lambda$ $=$ ${M}\over{Rφ}$
Mass = linear density×length
$dm$ $=$ $\lambda$×$dl$
$dm$ $=$ ${{M}\over{R φ}}×Rdθ$
Center of mass of this small element $r_{dm}$ = $R sinθ$
Center of mass of the total arc, $y_{com}$ $=$ ${\int dm.r_{dm}}\over{\int dm}$
$y_{com}$ $=$ ${{{MR}\over{φ}}\int_{{{\pi}\over{2}}-{φ\over2}}^{{{\pi}\over{2}}+{φ\over2}} sinθdθ}\over{{M}\over{φ}}\int_{{{\pi}\over{2}}-{φ\over2}}^{{{\pi}\over{2}}+{φ\over2}} dθ$
$y_{com}$ $=$ ${R\int_{{{\pi-φ}\over{2}}}^{{{\pi+φ}\over{2}}} sinθdθ}\over{\int_{{{\pi-φ}\over{2}}}^{{{\pi+φ}\over{2}}}} dθ$
$y_{com}$ $=$ R× ${-cos\theta \biggr |_{{\pi-φ}\over2}^{{\pi+φ}\over2}}\over{\theta \biggr |_{{\pi-φ}\over 2}^{\pi+φ\over 2}}$
$y_{com}$ $=$ R× ${2\ sin{φ\over 2} \over φ}$
$y_{com}$ $=$ R× ${sin{φ\over 2} \over {φ\over2}}$
For $x_{com}$: By symmetry $x_{com}$ will lie at the center
To prove it
$r_{dm}\ =\ R cos\theta$
$x_{com}$ $=$ ${R\int_{{{\pi-φ}\over{2}}}^{{{\pi+φ}\over{2}}} cosθdθ}\over{\int_{{{\pi-φ}\over{2}}}^{{{\pi+φ}\over{2}}}}dθ$
$x_{com}$ $=$ R× ${sin\theta \biggr |_{{\pi-φ}\over2}^{{\pi+φ}\over2}}\over{\theta \biggr |_{{\pi-φ}\over 2}^{\pi+φ\over 2}}$
$x_{com}$ $=$ R× ${cos{φ\over2}\ -\ cos{φ\over2}}\over{φ}$ $=\ 0$
Hence the center of mass of an arc is at a distance of R× ${sin{φ\over 2} \over {φ\over2}}$ from the center
Coordinates ≡ $\Biggr(0,{Rsin{φ\over2}\over{φ\over2}}\Biggr)$