Introduction to the Conical Pendulum
A mass attached to a string rotating in a horizontal circle is a classic physics problem known as a conical pendulum. Unlike a standard pendulum swinging back and forth, the mass here traces out a circle while the string forms the side of a cone.
Given Data
- Mass ($m$): $1 \text{ kg}$
- Length of string ($l$): $1 \text{ m}$
- Radius of the circle ($r$): $60 \text{ cm} = 0.6 \text{ m}$
- Acceleration due to gravity ($g$): $9.8 \text{ m/s}^2$
Solving for Tension ($T$)
First, we determine the vertical angle $\theta$ the string makes with the vertical axis. Using the triangle formed by the string ($l$) and the radius ($r$): $$\sin \theta = \frac{r}{l} = \frac{0.6}{1} = 0.6$$ Using trigonometry: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - 0.36} = 0.8$.
For a conical pendulum, the vertical component of tension balances weight, and the horizontal component provides centripetal force:
- $T \cos \theta = mg$
- $T \sin \theta = m \omega^2 r = \frac{mv^2}{r}$
From (1): $T = \frac{mg}{\cos \theta} = \frac{1 \times 9.8}{0.8} = 12.25 \text{ N}$.
Solving for Time Period ($T_{period}$)
The formula for the time period of a conical pendulum is derived from the centripetal force equations: $$T_{period} = 2\pi \sqrt{\frac{l \cos \theta}{g}}$$ Substituting our values: $$T_{period} = 2\pi \sqrt{\frac{1 \times 0.8}{9.8}} = 2\pi \sqrt{0.0816} \approx 2 \times 3.14159 \times 0.2857 \approx 1.79 \text{ s}$.
Summary
- Tension: The string must support both the weight and the centripetal acceleration, resulting in a tension of $12.25 \text{ N}$.
- Time Period: The duration for one full rotation is approximately $1.79 \text{ seconds}$.