Physics Solution: Spring Extension Calculation
The problem involves a mass undergoing circular motion in a horizontal plane, where the centripetal force is provided by the tension in a spring.
Physics Principles Involved
- Centripetal Force: The force required to keep an object moving in a circular path. Formula: \$F_c = m \omega^2 r$
- Hooke's Law: The force exerted by a spring is proportional to its extension. Formula: \$T = k \Delta L$
Problem Analysis
Given:
- Mass, \$m = 0.5 \ kg$
- Natural length of spring, \$L_0 = 1 \ m$
- Spring constant, \$k = 800 \ N/m$
- Angular speed, \$\omega = 20 \ rad/s$
- Let the extension in the spring be \$\Delta L$.
- The radius of the circular path is the extended length: \$r = L_0 + \Delta L = 1 + \Delta L$
Calculation Steps
- The tension in the spring provides the centripetal force: \$T = F_c$.
- Substitute the formulas: \$k \Delta L = m \omega^2 r$.
- Substitute the expression for radius: \$k \Delta L = m \omega^2 (L_0 + \Delta L)$.
- Plug in the given values: \$800 \times \Delta L = 0.5 \times (20)^2 \times (1 + \Delta L)$.
- Simplify the equation: \$800 \Delta L = 0.5 \times 400 \times (1 + \Delta L)$ \$800 \Delta L = 200 \times (1 + \Delta L)$
- Solve for \$\Delta L$: \$800 \Delta L = 200 + 200 \Delta L$ \$800 \Delta L - 200 \Delta L = 200$ \$600 \Delta L = 200$ \$\Delta L = \frac{200}{600}$ \$\Delta L = \frac{1}{3} \ m$
Conclusion
The extension in the length of the spring is \$\frac{1}{3} \ m$.
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