Understanding the Conical Pendulum
A bob attached to a string and whirled in a horizontal circle forms a conical pendulum. In this system, two main forces act on the bob:
- Gravity ($mg$): Acting vertically downwards.
- Tension ($T$): Acting along the string, directed toward the point of suspension.
Resolving Forces
We resolve the tension $T$ into two components:
- Vertical component: $T \cos(\theta)$ balances the weight ($mg$).
- Horizontal component: $T \sin(\theta)$ provides the necessary centripetal force ($F_c = \frac{mv^2}{r}$).
Given Data
- Mass $m = 200 \text{ g} = 0.2 \text{ kg}$
- Radius $r = 50 \text{ cm} = 0.5 \text{ m}$
- Angle $\theta = 30^\circ$
- Gravity $g \approx 9.8 \text{ m/s}^2$
Step-by-Step Solution
1. Calculating Tension ($T$)
From vertical equilibrium: $T \cos(30^\circ) = mg$ $T = \frac{mg}{\cos(30^\circ)} = \frac{0.2 \times 9.8}{\frac{\sqrt{3}}{2}} \approx \frac{1.96}{0.866} \approx 2.26 \text{ N}$
2. Calculating Speed ($v$)
From horizontal centripetal force: $T \sin(30^\circ) = \frac{mv^2}{r}$ Substitute $T = \frac{mg}{\cos(30^\circ)}$: $\frac{mg}{\cos(30^\circ)} \sin(30^\circ) = \frac{mv^2}{r}$ $g \tan(30^\circ) = \frac{v^2}{r}$ $v^2 = rg \tan(30^\circ)$ $v = \sqrt{0.5 \times 9.8 \times \tan(30^\circ)} = \sqrt{4.9 \times 0.577} \approx \sqrt{2.827} \approx 1.68 \text{ m/s}$