Understanding the Physics of a Conical Pendulum
When an object (like a bob) is whirled in a horizontal circle while attached to a fixed point by a string, it forms a conical pendulum. Unlike a simple pendulum that swings back and forth, the bob here moves in a horizontal plane, tracing a circle while the string traces the surface of a cone.
The Problem Statement
We are given:
- Mass ($m$) = 200 g = 0.2 kg
- Radius of the circle ($r$) = 50 cm = 0.5 m
- Angle with the vertical ($\theta$) = $30^{\circ}$
- Acceleration due to gravity ($g$) = 9.8 m/s²
We need to find:
- The tension in the string ($T$)
- The speed of the bob ($v$)
Step-by-Step Solution
1. Identify the Forces
The forces acting on the bob are:
- Tension ($T$): Acting along the string, at an angle $\theta$ to the vertical.
- Weight ($mg$): Acting vertically downwards.
We can decompose the tension into two components:
- Vertical component ($T \cos \theta$): Supports the weight of the bob ($T \cos \theta = mg$).
- Horizontal component ($T \sin \theta$): Provides the necessary centripetal force for circular motion ($T \sin \theta = \frac{mv^2}{r}$).
2. Calculate Tension ($T$)
From the vertical equilibrium: $T \cos(30^{\circ}) = mg$ $T = \frac{mg}{\cos(30^{\circ})}$ $T = \frac{0.2 \times 9.8}{0.866} \approx 2.26 \text{ N}$
3. Calculate Speed ($v$)
From the horizontal centripetal force equation: $T \sin(30^{\circ}) = \frac{mv^2}{r}$
Divide the horizontal force equation by the vertical force equation to eliminate $T$ and $m$: $\tan(\theta) = \frac{v^2}{rg}$ $v^2 = rg \tan(\theta)$ $v = \sqrt{rg \tan(\theta)}$
Plugging in the values: $v = \sqrt{0.5 \times 9.8 \times \tan(30^{\circ})}$ $v = \sqrt{4.9 \times 0.577} \approx \sqrt{2.827} \approx 1.68 \text{ m/s}$
Summary of Results
- Tension in the string: 2.26 Newtons
- Speed of the bob: 1.68 m/s
This problem demonstrates the balance between gravity and centripetal force, showing how the angle of inclination effectively dictates both the required speed and the tension on the supporting cable.