Find the centroid of a circular arc of radius R

Find the centroid of a circular arc of radius $R$ subtending a central angle $\alpha$.

Step 1: Set up

$x=R\cos\theta,\ y=R\sin\theta,\ \theta\in[-\alpha/2,\alpha/2]$; $ds=R\,d\theta$; total length $L=R\alpha$.

Step 2: Moment about the $y$-axis

\[M_y=\int_{-\alpha/2}^{\alpha/2}R\cos\theta\cdot R\,d\theta=R^2\big[\sin\theta\big]_{-\alpha/2}^{\alpha/2}=2R^2\sin\frac\alpha2\]

Step 3: Moment about the $x$-axis

By the odd symmetry of $\sin\theta$ over $[-\alpha/2,\alpha/2]$: $M_x=0$.

Step 4: Centroid

\[\bar x=\frac{M_y}L=\frac{2R^2\sin\frac\alpha2}{R\alpha}=R\cdot\frac{\sin(\alpha/2)}{\alpha/2},\qquad \bar y=0\]

\[\boxed{\left(R\cdot\frac{\sin(\alpha/2)}{\alpha/2},\,0\right)}\]

Step 5: Special cases

Semicircle ($\alpha=\pi$): $\bar x=\dfrac{2R}\pi\approx0.637R$, matching Problem 138. Quarter circle ($\alpha=\pi/2$): $\bar x=\dfrac{2\sqrt2R}\pi\approx0.900R$. As $\alpha\to0$: $\bar x\to R$ (the arc becomes indistinguishable from a straight chord at radius $R$).