Solution
Reasoning:
As the cord unwinds from the rod, the ball describes an expanding horizontal circle while slowly descending. The problem states the turns are closely spaced and at the same level, allowing us to approximate the motion at any instant as a Conical Pendulum of length $l$ (where $l$ is the instantaneous length of the unwound cord).
1. Force Analysis (Conical Pendulum):
Let $\theta$ be the angle the cord makes with the vertical.
Vertical equilibrium: $$ T \cos \theta = mg \quad \dots(1) $$
Horizontal centripetal force provided by tension: $$ T \sin \theta = \frac{mv^2}{r} $$ Where radius $r = l \sin \theta$. $$ T \sin \theta = \frac{mv^2}{l \sin \theta} \quad \dots(2) $$
Dividing (2) by (1): $$ \tan \theta = \frac{v^2}{gl \sin \theta} $$ $$ v^2 = gl \sin \theta \tan \theta \quad \dots(3) $$
2. Work-Energy Theorem:
The ball starts from rest (assumed approx zero initial kinetic energy as it starts unwinding). The work done by gravity as the ball descends is converted into kinetic energy.
Vertical distance descended: $h = l \cos \theta$.
$$ W_{\text{gravity}} = \Delta K $$
$$ mgh = \frac{1}{2} mv^2 $$
$$ mg (l \cos \theta) = \frac{1}{2} m v^2 $$
$$ v^2 = 2gl \cos \theta \quad \dots(4) $$
3. Solving for $\theta$:
Equating (3) and (4):
$$ gl \sin \theta \tan \theta = 2gl \cos \theta $$
$$ \sin \theta \left( \frac{\sin \theta}{\cos \theta} \right) = 2 \cos \theta $$
$$ \sin^2 \theta = 2 \cos^2 \theta $$
$$ \tan^2 \theta = 2 $$
$$ \tan \theta = \sqrt{2} $$
Answer: $\theta = \tan^{-1}(\sqrt{2})$