Centre of Mass of a Circular Arc

Setup

A thin uniform wire bent into a circular arc of radius $R$. The arc subtends a total angle $2\alpha$ at the centre, symmetric about the $y$-axis. We find the CM along the $y$-axis (by symmetry, $x_{cm} = 0$).

Derivation

Consider a small element of the arc at angle $\theta$ from the $y$-axis (measured from the axis of symmetry).

The element subtends angle $d\theta$ and has length $R \, d\theta$.

Mass element: $dm = \lambda R \, d\theta$ where $\lambda = M/(2\alpha R)$ is the linear mass density.

The $y$-coordinate of this element: $y = R\cos\theta$

$y_{cm} = \frac{1}{M}\int_{-\alpha}^{\alpha} R\cos\theta \cdot \lambda R \, d\theta$

$= \frac{\lambda R^2}{M}\int_{-\alpha}^{\alpha} \cos\theta \, d\theta = \frac{\lambda R^2}{M} \cdot 2\sin\alpha$

Substituting $\lambda = \frac{M}{2\alpha R}$:

$y_{cm} = \frac{M}{2\alpha R} \cdot \frac{R^2}{M} \cdot 2\sin\alpha = \frac{R\sin\alpha}{\alpha}$

$\boxed{y_{cm} = \frac{R\sin\alpha}{\alpha}}$

Special cases

Semicircular arc ($2\alpha = \pi$, so $\alpha = \pi/2$):

$y_{cm} = \frac{R\sin(\pi/2)}{\pi/2} = \frac{R}{\pi/2} = \frac{2R}{\pi} \approx 0.637R$

Full circle ($2\alpha = 2\pi$, so $\alpha = \pi$):

$y_{cm} = \frac{R\sin\pi}{\pi} = 0$

CM at the centre — as expected for a complete ring.

Very small arc ($\alpha \to 0$, using $\sin\alpha \approx \alpha$):

$y_{cm} \to \frac{R\alpha}{\alpha} = R$

A very small arc is essentially a point at distance $R$ from centre — CM is at $R$. ✓

The CM is inside the arc, not on it

For a semicircular arc, $y_{cm} = \frac{2R}{\pi} \approx 0.637R < R$.

The CM lies inside the arc, not on the wire itself. This is typical for curved bodies — the CM may be in empty space.