Suppose that we have a helical wire
\begin{equation*}
\vr(t) = \big(x(t)\,,\,y(t)\,,\,z(t)\big)
=\big(a\cos t\,,\,a\sin t\,,\, bt\big)\qquad 0\le t\le 2\pi
\end{equation*}
and that this wire has constant mass density \(\rho\text{.}\) Let’s find the centre of mass of the wire. Recall that the centre of mass is \(\big(\bar x,\bar y,\bar z)\) with, for example, \(\bar x\) being the weighted average
\begin{equation*}
\bar x = \frac{\int x\rho \dee{s}}{\int \rho\dee{s}}
= \frac{\int x \dee{s}}{\int \dee{s}}
\qquad\text{(since $\rho$ is constant)}
\end{equation*}
of \(x\) over the wire. Similarly \(\bar y = \frac{\int y \dee{s}}{\int \dee{s}}\) and \(\bar z = \frac{\int z \dee{s}}{\int \dee{s}}
\text{.}\) For the given curve
\begin{align*}
\big(x(t)\,,\,y(t)\,,\,z(t)\big) &=\big(a\cos t\,,\,a\sin t\,,\, bt\big)\\
\big(x'(t)\,,\,y'(t)\,,\,z'(t)\big) &=\big(-a\sin t\,,\,a\cos t\,,\, b\big)\\
\diff{s}{t}(t) &=\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\\
&=\sqrt{a^2\sin^2t+a^2\cos^2t+b^2}\\
&=\sqrt{a^2+b^2}
\end{align*}
so that
\begin{align*}
\bar x
&= \frac{\int x \dee{s}}{\int \dee{s}}
= \frac{\int_0^{2\pi} x(t) \sqrt{a^2+b^2}\,\dee{t}}
{\int_0^{2\pi} \sqrt{a^2+b^2}\,\dee{t}}
= \frac{\int_0^{2\pi} a\cos(t) \,\dee{t}}{2\pi}
=0\\
\bar y
&= \frac{\int y \dee{s}}{\int \dee{s}}
= \frac{\int_0^{2\pi} y(t) \sqrt{a^2+b^2}\,\dee{t}}
{\int_0^{2\pi} \sqrt{a^2+b^2}\,\dee{t}}
= \frac{\int_0^{2\pi} a\sin(t) \,\dee{t}}{2\pi}
=0\\
\bar z
&= \frac{\int z \dee{s}}{\int \dee{s}}
= \frac{\int_0^{2\pi} z(t) \sqrt{a^2+b^2}\,\dee{t}}
{\int_0^{2\pi} \sqrt{a^2+b^2}\,\dee{t}}
= \frac{\int_0^{2\pi} bt \,\dee{t}}{2\pi}
=\frac{b}{2\pi}\Big[\frac{t^2}{2}\Big]_0^{2\pi}
=b\pi
\end{align*}
So the centre of mass is right on the axis of the helix, half way up, which makes perfect sense.
