Solution
Geometric Analysis of the Helix
We can analyze the motion by “unrolling” the cylindrical surface of the helix into a flat plane. In this view, the path of the bead becomes the hypotenuse of a right-angled triangle where the base is the circumference of the cylinder ($2\pi R$) and the height is the pitch ($h$).
Let $\alpha$ be the angle of inclination of the helix with the horizontal. $$ \sin \alpha = \frac{h}{\sqrt{(2\pi R)^2 + h^2}} $$ $$ \cos \alpha = \frac{2\pi R}{\sqrt{(2\pi R)^2 + h^2}} $$ For brevity, let $L = \sqrt{4\pi^2 R^2 + h^2}$ be the length of one turn. Thus $\sin \alpha = h/L$.
(a) Time to Descend Height H
The motion along the wire is equivalent to sliding down a smooth inclined plane of angle $\alpha$. The acceleration along the wire is $a = g \sin \alpha$.
The distance $S$ traveled along the wire corresponds to a vertical drop $H$. From geometry: $$ S = \frac{H}{\sin \alpha} $$
Using the kinematic equation $S = \frac{1}{2} a t^2$ (starting from rest): $$ \frac{H}{\sin \alpha} = \frac{1}{2} (g \sin \alpha) t^2 $$ $$ t^2 = \frac{2H}{g \sin^2 \alpha} \implies t = \frac{1}{\sin \alpha} \sqrt{\frac{2H}{g}} $$
Substitute $\sin \alpha = h / \sqrt{4\pi^2 R^2 + h^2}$: $$ t = \frac{\sqrt{4\pi^2 R^2 + h^2}}{h} \sqrt{\frac{2H}{g}} $$
(b) Force Exerted by the Wire
The normal force $\vec{N}$ exerted by the wire has two orthogonal components:
- $N_{\perp}$: Component balancing the normal component of gravity (perpendicular to the path on the “unrolled” plane).
- $N_{rad}$: Component providing the centripetal force required for circular motion (radial direction).
1. Perpendicular Component ($N_{\perp}$):
From the inclined plane analogy:
$$ N_{\perp} = mg \cos \alpha $$
2. Radial/Centripetal Component ($N_{rad}$):
The bead moves in a helical path which projects to a circle of radius $R$. The horizontal component of velocity is $v_x = v \cos \alpha$.
$$ N_{rad} = \frac{m (v_x)^2}{R} = \frac{m (v \cos \alpha)^2}{R} $$
We find the speed $v$ using conservation of energy after descending height $H$: $$ v = \sqrt{2gH} $$ So, $N_{rad} = \frac{m (2gH) \cos^2 \alpha}{R} = \frac{2mgH \cos^2 \alpha}{R}$.
Total Normal Force ($N$): $$ N = \sqrt{N_{\perp}^2 + N_{rad}^2} $$ $$ N = \sqrt{(mg \cos \alpha)^2 + \left( \frac{2mgH \cos^2 \alpha}{R} \right)^2} $$ $$ N = mg \cos \alpha \sqrt{1 + \left( \frac{2H \cos \alpha}{R} \right)^2} $$
Substitute $\cos \alpha = \frac{2\pi R}{L}$ where $L = \sqrt{4\pi^2 R^2 + h^2}$: $$ N = mg \frac{2\pi R}{L} \sqrt{1 + \frac{4H^2}{R^2} \frac{4\pi^2 R^2}{L^2}} $$ $$ N = \frac{2\pi mg R}{L} \sqrt{\frac{L^2 + 16\pi^2 H^2}{L^2}} $$ $$ N = \frac{2\pi mg R}{L^2} \sqrt{L^2 + 16\pi^2 H^2} $$
Substituting $L^2 = 4\pi^2 R^2 + h^2$: