1. Understand the geometry

A uniform semicircular wire of radius $R$ lies in the $x$-$y$ plane.

• Its centre is at the origin.
• The line joining its ends is the $x$-axis.

So the semicircular arc is symmetric about the $y$-axis, and hence the centre of mass must lie on the $y$-axis.

Therefore, $x_{\text{CM}}=0$ and we only need to find $y_{\text{CM}}$.

2. Use the centre of mass formula for a wire

For a uniform wire, $y_{\text{CM}}=\frac{1}{L}\int y\,dl$ where $L$ is the total length of the wire.

For a semicircular wire of radius $R$, $L=\pi R$

3. Parameterize the semicircle

Take the upper semicircle: $x=R\cos\theta, \qquad y=R\sin\theta, \qquad 0\le \theta \le \pi$

The small arc length element is $dl=R\,d\theta$

4. Compute $y_{\text{CM}}$

Substitute into the formula: $y_{\text{CM}}=\frac{1}{\pi R}\int_0^{\pi} (R\sin\theta)(R\,d\theta)$

$y_{\text{CM}}=\frac{R^2}{\pi R}\int_0^{\pi} \sin\theta\,d\theta$

$y_{\text{CM}}=\frac{R}{\pi}\left[-\cos\theta\right]_0^{\pi}$

$y_{\text{CM}}=\frac{R}{\pi}\left[ -\cos\pi + \cos 0 \right]$

Since $\cos\pi=-1$ and $\cos 0=1$, $y_{\text{CM}}=\frac{R}{\pi}(1+1)=\frac{2R}{\pi}$

5. Compare with the given form

Given position of centre of mass: $\left(0,\frac{xR}{\pi}\right)$

We found: $\left(0,\frac{2R}{\pi}\right)$

Thus, $x=2$ So, $|x|=2$

6. Comparison with stored answer

Stored correct answer: $2$

Our derived answer matches it.