The position of the centre of mass of a uniform semi-circular wire of radius $$R$$ placed in $$x-y$$ plane with its centre at the origin and the line joining its ends as $$x$$-axis is given by $$\left(0, \frac{xR}{\pi}\right)$$. Then, the value of $$|x|$$ is ___.

Consider a uniform semicircular wire of radius $$R$$ lying in the $$x$$-$$y$$ plane with its centre at the origin and its diameter along the $$x$$-axis. A small element of the wire at angle $$\theta$$ (measured from the positive $$x$$-axis) has coordinates $$(R\cos\theta,\, R\sin\theta)$$ and arc length $$dl = R\,d\theta$$, where $$\theta$$ ranges from $$0$$ to $$\pi$$ (upper semicircle).

By symmetry, the $$x$$-coordinate of the centre of mass is zero. For the $$y$$-coordinate:

$$y_{\text{cm}} = \frac{\displaystyle\int_0^{\pi} (R\sin\theta)\, R\,d\theta}{\displaystyle\int_0^{\pi} R\,d\theta} = \frac{R^2 \displaystyle\int_0^{\pi} \sin\theta\, d\theta}{R\pi}$$

$$\int_0^{\pi} \sin\theta\, d\theta = [-\cos\theta]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$$

$$y_{\text{cm}} = \frac{R^2 \times 2}{R\pi} = \frac{2R}{\pi}$$

The position of the centre of mass is $$\left(0,\, \dfrac{2R}{\pi}\right)$$. Comparing with the given form $$\left(0,\, \dfrac{xR}{\pi}\right)$$, we get $$x = 2$$, so $$|x| = 2$$.