Center of mass of a circular arc I have an infinitesimally thin wire in the shape of a circular arc, with uniform mass distribution. I would like to know the location of the center of mass of the wire.
The arc subtends an angle of 120° (one third of a full circle), and the radius is 3.
From the symmetry, I know that the center of mass is between the center of the circle, and the midpoint of the circular arc, but I do not know how to calculate the distance from the center of the circle to the center of mass.
I'm interested in the geometric aspects of the problem, not so much on the physical aspects of it.
 A: Let me show the most general way to find the answer.
For a curve (or an infinitesimally thin wire with uniform density, i.e. uniform linear mass distribution), the center of mass is at the centroid of the curve.
In the general 2D case, the centroid of a parametric curve $\vec{s}(t) = \left ( x(t) , y(t) \right )$, $t_0 \le t \le t_1$ is at $( \hat{x} , \hat{y} )$,
$$\begin{cases}
\hat{x} = \frac{1}{L} \displaystyle \int_{t_0}^{t_1} x(t) \, \delta(t) \, dt \\
\hat{y} = \frac{1}{L} \displaystyle \int_{t_0}^{t_1} y(t) \, \delta(t) \, dt
\end{cases} \tag{1}\label{NA1}$$
where $\delta(t) \, dt$ is the arc length parameter at $t$,
$$\delta(t) \, dt = \sqrt{ \left( \frac{ d\, x(t) }{ d t } \right )^2 + \left( \frac{ d\, y(t) }{ d\, t} \right) ^2 } \, dt$$
and $L$ is the total length of the curve,
$$L = \int_{t_0}^{t_1} \delta(t) \, dt$$ 
In this particular case, we have a circular arc,
$$\begin{cases}
x(\theta) = r \cos(\theta) \\
y(\theta) = r \sin(\theta)
\end{cases}$$
and therefore
$$\delta(\theta) \, d\theta = \sqrt{ \left(-r \sin(\theta)\right)^2 + \left(r \cos(\theta)\right)^2 } \, d\theta = \sqrt{ r^2 \left( (\sin\theta)^2 + (\cos\theta)^2 \right) } \, d\theta = \sqrt{ r^2 } \, d\theta = r \, d\theta$$
The arc distends one third of a full circle, or 120°. If we put the center of the circle at origin, and the midpoint of the arc on the positive $y$ axis, then $\theta$ ranges from $90°-120°/2 = 30°$ to $90°+120°/2 = 150°$, i.e. from $\theta = \pi/6$ radians to $\theta = 5 \pi/6$ radians.
The length $L$ of the circular arc we already know from geometry; it is one third of the perimeter of the circle of radius $r$,
$$L = \frac{2 \pi r}{3}$$
Substituting these to $\eqref{NA1}$ we get
$$\begin{cases}
\hat{x} = \frac{3}{2 \pi r} \displaystyle\int_{\pi/6}^{5\pi/6} r \cos(\theta) \, r \, d\theta \\
\hat{y} = \frac{3}{2 \pi r} \displaystyle\int_{\pi/6}^{5\pi/6} r \sin(\theta) \, r \, d\theta
\end{cases}$$
which simplify to
$$\begin{cases}
\hat{x} = \frac{3 r}{2 \pi} \displaystyle\int_{\pi/6}^{5\pi/6} \cos(\theta) \, d\theta = \frac{3 r}{2 \pi} \left(\Bigl[-\sin\theta \Bigr]_{\pi/6}^{5\pi/6} \right) \\
\hat{y} = \frac{3 r}{2 \pi} \displaystyle\int_{\pi/6}^{5\pi/6} \sin(\theta) \, d\theta = \frac{3 r}{2 \pi} \left(\Bigl[\cos\theta \Bigr]_{\pi/6}^{5\pi/6} \right)
\end{cases}$$
Because $-\sin(\pi/6) - -sin(5\pi/6) = 0$, $\hat{x} = 0$. Which is completely expected, because we arranged the arc to be symmetric around the $y$ axis.
Because $\cos(\pi / 6) - \cos(5\pi / 6) = \sqrt{3}/2 - -\sqrt{3}/2 = \sqrt{3}$,
$$\hat{y} = \frac{3 r}{2 \pi} \sqrt{3} = \frac{3 \sqrt{3}}{2 \pi} r$$
In the case of $r = 3$, $$\hat{y} = \frac{9 \sqrt{3}}{2 \pi} \approx 2.48$$
This is in perfect agreement with King Tut's answer.
A: General formula  of distance of com arc from centre of generating circle is $  \frac{\sin(\theta /2)}{\theta/2}R$ where $\theta$ is angle that arc subtends on centre of circle.  This result is proved using integration.
Using this you get answer as $ \frac{3\sqrt{3}}{2\pi}R$
A: Let me show you the easiest way to calculate the centroid of a line of uniform mass. This can be expressed most easily in the complex plane as follows,
$$C=\frac{\int z~ds}{\int ds}=\frac{1}{L} \int z~ds=\frac{\int z~|\dot z|~d\theta}{\int |\dot z|~d\theta}$$
where $C$ is complex centroid, $L$ is the arc length, $\dot z$ is the derivative of $z$ w.r.t $\theta$. For a circular arc of radius r we then have
$$
z=re^{i\theta}\\
\dot z=ire^{i\theta}\\
|\dot z|=r
$$
then
$$
\begin{align}
C
&=\frac{r^2\int_{\theta1}^{\theta2}e^{i\theta}~d\theta}{r\int_{\theta1}^{\theta2}~d\theta}\\
&=r\frac{-i\left(e^{i\theta_2}-e^{i\theta_1} \right)}{\theta_2-\theta_1}
\end{align}
$$
Specializing to your case, with $\theta\in[0,2\pi/3]$ we find
$$C=r\frac{\sqrt{3}/2+i\cdot 3/2}{2\pi/3}$$
and, the radial distance to the centroid is given by
$$|C|=r\frac{3\sqrt{3}}{2\pi}$$
as you already know. As for this not being on the curve, of course it cannot be. Imagine that the curve is on a massless plane. Where would suspend the plane from a string so that it did not tilt?
A: Here is what I have tried, assuming that the mass distribution along the length of the arc is uniform.
$$\lambda=\frac{dm}{ds}=\frac{M}{L}$$
$$s=r\theta$$
Position your arc so that it is symmetrical about the $y$-axis and centered at the origin. Set the center of mass at $(h,k)$.
Due to symmetry about the $y$-axis, $h=0$.
The angle the arc subtends is $2\pi/3$, so let’s integrate along $[\pi/2-\pi/3,\pi/2+\pi/3]$ which equals $[\pi/6,5\pi/6]$.
By definition of center of mass,
$$\begin{align}
k &= \frac1M \int y\,dm \\
&= \frac1M \int r\sin(\theta)\, \lambda\,ds \\
&= \frac1M \int r\sin(\theta)\, \lambda r\,d\theta \\
&= \frac{r^2}M \int \lambda\sin(\theta)\,d\theta \\
&= \frac{r^2}{M}\bigl( -\lambda\cos\theta\bigr)\Bigr|^{5\pi/6}_{\pi/6}\\
&= \frac{r^2\lambda}{M}\left[ -\cos\left(\frac{5\pi}6\right) + \cos\left(\frac\pi6\right)\right] \\
&= \frac{r^2\lambda}{M} \left( +\frac{\sqrt3}2 + \frac{\sqrt3}2\right)\\
&= \frac{r^2\lambda\sqrt 3}M=\frac{r^2M\sqrt 3}{LM} = \frac{r^2\sqrt3}{2\pi r/3}\\
&= \frac{3r\sqrt3}{2\pi}\\
&= \frac{9\sqrt3}{2\pi}
\end{align}$$
where $M$ is the mass of the section of arc in question.
Big thanks to David K for spotting the error I had mentioned earlier.
