A small body slides down a helical (spiral) groove of radius…

A small body slides down a helical (spiral) groove of radius $R$ and pitch $h$ (the drop per turn). The coefficient of friction between the body and the groove is $\mu$. With what steady speed does the body eventually slide down?

Explanation of Method

Steady speed in a helical groove.

Step 1: Slope of the groove

One turn unrolled is an incline of length $L = \sqrt{(2\pi R)^2 + h^2}$ and angle $\alpha$: $\tan\alpha = \frac{h}{2\pi R}$.

\[ \cos\alpha = \frac{2\pi R}{L}. \]

Step 2: The two parts of the normal force

The groove pushes on the body in two perpendicular directions: it carries the part of the weight across the groove, $mg\cos\alpha$, and it pushes the body horizontally toward the axis to keep it on the helix. The horizontal part of the speed is $v\cos\alpha$ on a circle of radius $R$:

\[ N_1 = mg\cos\alpha, \]

\[ N_2 = \frac{m v^2\cos^2\alpha}{R}, \]

\[ N = \sqrt{N_1^2 + N_2^2}. \]

Step 3: Steady speed

The speed stops growing when friction balances the pull along the groove: \[mg\sin\alpha = \mu N \;\Longrightarrow\; \frac{g^2\sin^2\alpha}{\mu^2} = g^2\cos^2\alpha + \frac{v^4\cos^4\alpha}{R^2}.\]

\[ v^2 = \frac{gR}{\cos\alpha}\sqrt{\frac{\tan^2\alpha}{\mu^2} - 1}. \]

In terms of $R$, $h$ and $\mu$: $v = \sqrt{\frac{g\,L\,\sqrt{h^2 - (2\pi\mu R)^2}}{4\pi^2\mu R}}$. with $L = \sqrt{4\pi^2R^2 + h^2}$. This requires $\tan\alpha \gt \mu$, i.e. $h \gt 2\pi\mu R$; otherwise the body stops. The body goes down at $v\sin\alpha$ vertically.

Check

At the threshold $\tan\alpha = \mu$: $v = 0$, as on an ordinary incline $\checkmark$

As the speed grows, the sideways push $N_2$ grows and brakes more; that is why a steady speed exists at all $\checkmark$

$h$ $2R$ $2\pi R$ $h$ $\alpha$ $\tan\alpha = \frac{h}{2\pi R}$ $\text{one turn unrolled}$

Left: the helical groove. Right: one turn unrolled into an incline of angle $\alpha$. Unlike a straight incline, the groove also has to push the body sideways toward the axis to keep it on the helix, and that extra push adds friction.

Result

\[\boxed{\,v = \sqrt{\frac{g\,L\,\sqrt{h^2 - (2\pi\mu R)^2}}{4\pi^2\mu R}}\,}\]

Remark

About the original solution. It used the incline condition $\tan\alpha = \mu$, under which the speed would stay at whatever value it had, and then an energy formula for one turn that does not give a steady speed. On a helix the groove must also supply the centripetal force, which adds friction and makes the steady speed depend on $R$. The formula above is the steady speed.