Question

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm² . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer 1

Given

• Mass: \( m = 14.5 \,\text{kg} \)
• Unstretched length of wire (radius): \( L = r = 1.0 \,\text{m} \)
• Angular velocity at bottom: \( \omega = 2 \,\text{rev/s} = 4\pi \,\text{rad/s} \)
• Cross-sectional area: \( A = 0.065 \,\text{cm}^{2} = 0.065 \times 10^{-4} \,\text{m}^{2} = 6.5 \times 10^{-6} \,\text{m}^{2} \)
• Young’s modulus for steel: \( Y = 2 \times 10^{11} \,\text{N m}^{-2} \)
• Acceleration due to gravity: \( g = 9.8 \,\text{m s}^{-2} \)

1. Forces at the lowest point

At the lowest point, tension in the wire provides centripetal force and supports the weight:

\( T = mg + m \omega^{2} r \)

Compute each term:

\( mg = 14.5 \times 9.8 = 142.1 \,\text{N} \) \( \omega = 4\pi \Rightarrow \omega^{2} = 16\pi^{2} \approx 157.9 \) \( m \omega^{2} r = 14.5 \times 157.9 \approx 2290 \,\text{N} \)

\( T \approx 142.1 + 2290 \approx 2432 \,\text{N} \)

2. Elongation of the wire

Using Young’s modulus:

\( Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{T/A}{\Delta L / L} \Rightarrow \Delta L = \dfrac{T L}{Y A} \)

\( \Delta L = \dfrac{2432 \times 1.0}{(2 \times 10^{11})(6.5 \times 10^{-6})} \) \( = \dfrac{2432}{1.3 \times 10^{6}} \approx 1.87 \times 10^{-3} \,\text{m} \)

Elongation of the wire at the lowest point: \( \Delta L \approx 1.87 \times 10^{-3} \,\text{m} = 1.87 \,\text{mm} \).