Introduction
In physics, when an object moves in a circular path, it requires a centripetal force directed toward the center of the circle to maintain that path. In this problem, that force is provided by the tension in a string. Understanding the relationship between force, mass, radius, and frequency is key to solving circular motion problems.
The Physics Concept
For an object of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$, the required centripetal force $F_c$ is given by: $$F_c = m \cdot r \cdot \omega^2$$
Since the angular velocity $\omega = 2 \cdot \pi \cdot f$, where $f$ is the frequency in revolutions per second, we can substitute this into the formula: $$F_c = m \cdot r \cdot (2 \cdot \pi \cdot f)^2 = 4 \cdot \pi^2 \cdot m \cdot r \cdot f^2$$
Step-by-Step Solution
Given Data:
- Mass ($m$) = $0.5 \text{ kg}$
- Radius ($r$) = $1 \text{ m}$
- Maximum Tension ($T_{max}$) = $50 \text{ N}$
Formula:
Since the tension provides the centripetal force, $T_{max} = 4 \cdot \pi^2 \cdot m \cdot r \cdot f^2$
Calculation:
- Rearrange the formula to solve for frequency ($f$): $$f^2 = \frac{T_{max}}{4 \cdot \pi^2 \cdot m \cdot r}$$
- Substitute the values: $$f^2 = \frac{50}{4 \cdot (3.14159)^2 \cdot 0.5 \cdot 1}$$ $$f^2 = \frac{50}{2 \cdot 9.8696} \approx \frac{50}{19.739} \approx 2.533$$
- Find the square root: $$f = \sqrt{2.533} \approx 1.59 \text{ rev/s}$$
Conclusion
The greatest number of revolutions per second the object can sustain without breaking the string is approximately $1.59 \text{ Hz}$.