The motion of the particle can be decomposed into two components: a uniform circular motion in the plane perpendicular to the cylinder's axis, and a uniform linear motion along the axis.

Let $\vec{v}$ be the velocity of the particle, with constant speed $v=|\vec{v}|$. Let $\vec{v}_t$ be the velocity component tangential to the circular path of the projection, and $\vec{v}_z$ be the axial velocity component. Their magnitudes are $v_t$ and $v_z$ respectively. By the Pythagorean theorem, the speeds are related by:

$$v^2 = v_t^2 + v_z^2$$

The projection of the particle's motion onto the plane perpendicular to the axis is a uniform circular motion of radius $R$ and period $T$. The speed of this projected motion is $v_t$. It can be calculated as the circumference of the circle divided by the period:

$$v_t = \frac{2\pi R}{T}$$

Since the axial velocity $\vec{v}_z$ is constant, the acceleration of the particle is entirely due to the change in the direction of the tangential velocity $\vec{v}_t$. This is the centripetal acceleration of the projected circular motion. It points towards the cylinder's axis and has a magnitude:

$$a = \frac{v_t^2}{R}$$

For any motion at a constant speed $v$ along a path with a local radius of curvature $\rho$, the magnitude of the acceleration is given by:

$$a = \frac{v^2}{\rho}$$

Equating the two expressions for the acceleration magnitude gives a relationship between the total speed $v$ and the tangential speed $v_t$:

$$\frac{v^2}{\rho} = \frac{v_t^2}{R} \implies v^2 = v_t^2 \frac{\rho}{R}$$

Now, we substitute this expression for $v^2$ into the velocity decomposition equation:

$$v_t^2 \frac{\rho}{R} = v_t^2 + v_z^2$$

We can now solve for the axial speed $v_z$:

$$v_z^2 = v_t^2 \frac{\rho}{R} - v_t^2 = v_t^2 \left(\frac{\rho}{R} - 1\right) = v_t^2 \frac{\rho - R}{R}$$ $$v_z = v_t \sqrt{\frac{\rho - R}{R}}$$

Finally, substituting the expression for $v_t$ in terms of $R$ and $T$:

$$v_z = \left(\frac{2\pi R}{T}\right) \sqrt{\frac{\rho - R}{R}} = \frac{2\pi R}{T} \frac{\sqrt{\rho - R}}{\sqrt{R}}$$ $$v_z = \frac{2\pi}{T} \sqrt{R(\rho - R)}$$