Centre of Mass of a Semicircular Ring

Result

The CM of a thin uniform semicircular ring (wire) of radius $R$ lies on the axis of symmetry at:

$y_{cm} = \frac{2R}{\pi} \approx 0.637R$

Derivation

This is the special case $\alpha = \pi/2$ of the circular arc formula $y_{cm} = \frac{R\sin\alpha}{\alpha}$:

$y_{cm} = \frac{R\sin(\pi/2)}{\pi/2} = \frac{R \cdot 1}{\pi/2} = \frac{2R}{\pi}$

Direct derivation

Place the semicircle with diameter along the $x$-axis, arch upward. By symmetry $x_{cm} = 0$.

A small arc element at angle $\theta$ from the positive $x$-axis (so $\theta$ goes from $0$ to $\pi$):

• Position: $(R\cos\theta, R\sin\theta)$
• Arc length: $R \, d\theta$
• Mass: $dm = \frac{M}{\pi R} \cdot R \, d\theta = \frac{M}{\pi} d\theta$

$y_{cm} = \frac{1}{M}\int_0^\pi R\sin\theta \cdot \frac{M}{\pi} \, d\theta = \frac{R}{\pi}\int_0^\pi \sin\theta \, d\theta = \frac{R}{\pi}[-\cos\theta]_0^\pi = \frac{R}{\pi}(1+1) = \frac{2R}{\pi}$

Physical interpretation

The CM at $\frac{2R}{\pi}$ is inside the semicircle — there is no wire there. This is not unusual: the CM of a curved body often lies in empty space.

The CM is about $63.7\%$ of the way from the centre to the rim, along the axis of symmetry.