Conical Pendulum

The situation

A mass $m$ is attached to a string of length $L$. The mass moves in a horizontal circle while the string traces out a cone. The string makes angle $\theta$ with the vertical. This is a conical pendulum.

The radius of the circle: $r = L\sin\theta$

The height of the cone: $h = L\cos\theta$

Forces on the mass

• Weight $mg$ downward
• Tension $T_{string}$ along the string toward the pivot (upward and inward)

The mass moves in a horizontal circle — no vertical acceleration.

Resolve tension:

• Vertical: $T_{string}\cos\theta$ (upward)
• Horizontal: $T_{string}\sin\theta$ (centripetal, toward axis)

Derivation

Vertical equilibrium:

$T_{string}\cos\theta = mg \quad \text{...(1)}$

$T_{string} = \frac{mg}{\cos\theta}$

Horizontal — centripetal:

$T_{string}\sin\theta = \frac{mv^2}{r} = mr\omega^2$

$T_{string}\sin\theta = mL\sin\theta \cdot \omega^2 \quad \text{...(2)}$

Divide (2) by (1):

$\tan\theta = \frac{L\sin\theta \cdot \omega^2}{g}$

$\frac{\sin\theta}{\cos\theta} = \frac{L\sin\theta \cdot \omega^2}{g}$

Cancel $\sin\theta$ (assuming $\theta \neq 0$):

$\frac{1}{\cos\theta} = \frac{L\omega^2}{g}$

$\omega^2 = \frac{g}{L\cos\theta}$

$\boxed{\omega = \sqrt{\frac{g}{L\cos\theta}}}$

Period:

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$

$\boxed{T = 2\pi\sqrt{\frac{L\cos\theta}{g}}}$

Comparison with simple pendulum

A simple pendulum has period $T = 2\pi\sqrt{\frac{L}{g}}$.

The conical pendulum has period $T = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}}$

where $h = L\cos\theta$ is the vertical height of the cone.

The period depends on the vertical height of the cone, not the string length. As $\theta \to 0$ (nearly vertical), $h \to L$ and the period approaches the simple pendulum result.

How $\theta$ changes with $\omega$

$\cos\theta = \frac{g}{L\omega^2}$

• Spin faster (larger $\omega$) → $\cos\theta$ smaller → $\theta$ larger → cone opens wider
• Spin slower → cone closes, $\theta$ decreases
• At very high $\omega$: $\theta \to 90°$, string becomes nearly horizontal

Tension in the string

From equation (1):

$T_{string} = \frac{mg}{\cos\theta}$

Since $\cos\theta \leq 1$: tension is always at least $mg$, and increases as $\theta$ increases (faster spinning).