The particle moves in a circle in the x-y plane while the observer moves along the z-axis. Relative to the observer, the particle traces a helical path. Let the radius of the circle be $R$ and angular velocity be $\omega$. The relative vertical velocity is $v_z$.

The pitch $h$ is the vertical distance covered in one time period ($T = 2\pi/\omega$). $$ h = v_z T = v_z \frac{2\pi}{\omega} \implies v_z = \frac{h\omega}{2\pi} $$

Position vector: $\vec{r} = R \cos(\omega t)\hat{i} + R \sin(\omega t)\hat{j} – v_z t \hat{k}$ (Observer sees particle moving down/away).

Velocity vector ($\vec{v}$): $$ \vec{v} = \frac{d\vec{r}}{dt} = -R\omega \sin(\omega t)\hat{i} + R\omega \cos(\omega t)\hat{j} – v_z \hat{k} $$ $$ |\vec{v}| = \sqrt{ (R\omega)^2 + v_z^2 } = \text{constant} $$

Acceleration vector ($\vec{a}$): $$ \vec{a} = \frac{d\vec{v}}{dt} = -R\omega^2 \cos(\omega t)\hat{i} – R\omega^2 \sin(\omega t)\hat{j} $$ $$ |\vec{a}| = R\omega^2 $$

The radius of curvature $\rho$ is defined as: $$ \rho = \frac{|\vec{v}|^2}{a_n} $$ Since the speed $|\vec{v}|$ is constant, the tangential acceleration $a_t = 0$. Therefore, the total acceleration is purely normal ($a_n = |\vec{a}|$).

$$ \rho = \frac{R^2\omega^2 + v_z^2}{R\omega^2} = R + \frac{v_z^2}{R\omega^2} $$ Substituting $v_z = \frac{h\omega}{2\pi}$: $$ \rho = R + \frac{1}{R\omega^2} \left( \frac{h\omega}{2\pi} \right)^2 $$ $$ \rho = R + \frac{h^2 \omega^2}{4\pi^2 R \omega^2} = R + \frac{h^2}{4\pi^2 R} $$