Understanding Centripetal Force in Horizontal Circular Motion
When an object of mass $m$ is whirled in a horizontal circle of radius $r$ at a constant speed $v$, it requires a centripetal force to maintain its circular path. This force is provided by the tension ($T$) in the string.
The Physics Principle
The centripetal force ($F_c$) is given by: $$F_c = \frac{mv^2}{r}$$
Since the tension $T$ must provide this centripetal force, we have $T = \frac{mv^2}{r}$. The string breaks when the tension exceeds the breaking strength of the material. In this problem, the maximum tension $T_{max} = 25\text{ N}$.
Step-by-Step Solution
1. Identify the given values:
- Breaking Tension ($T_{max}$) = $25\text{ N}$
- Mass ($m$) = $500\text{ g} = 0.5\text{ kg}$
- Radius ($r$) = $1\text{ m}$
2. Relate Velocity to Angular Velocity: The relation between linear velocity ($v$) and angular velocity ($\omega$) is $v = r\omega$. Substituting this into the force equation: $$T = m r \omega^2$$
3. Solve for angular velocity ($\omega$): $$\omega = \sqrt{\frac{T}{mr}} = \sqrt{\frac{25}{0.5 \times 1}} = \sqrt{50} \approx 7.071\text{ rad/s}$$
4. Convert to Revolutions Per Minute (RPM): We know $\omega$ in radians per second. To convert to RPM ($N$): $$N = \frac{\omega \times 60}{2\pi}$$ $$N = \frac{7.071 \times 60}{2 \times 3.14159} \approx \frac{424.26}{6.283} \approx 67.52\text{ rpm}$$
Thus, the greatest number of revolutions per minute the object can make without breaking the string is approximately 67.5 rpm.