Question 8 of 13advanced🔧 ApplyNumerical5 marks

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.

Correct Answer

The elongation of the wire is approximately 1.87×101.87 \times 10⁻³ m (or 1.87 mm).

Exercise: EXERCISES | Q: 8.11
For More Understanding

Explanation

This problem combines concepts of circular motion with elastic properties of materials. At the lowest point of the vertical circle, the tension in the wire must provide both the centripetal force for circular motion and support the weight of the mass. The elongation is then calculated using Young's modulus formula. The Young's modulus for steel (2.0×102.0 \times 10¹¹ N m⁻²) is taken from Example 8.1 in the context.

Solution Steps

  1. Step 1: Convert angular velocity to rad/s ω = 2 rev/s = 2×22 \times 2π rad/s = 4π rad/s

  2. Step 2: Calculate tension at the lowest point At the lowest point, tension T = Centripetal force + Weight T = mω²L + mg T = 14.5 × (4π)² × 1.0 + 14.5×9.814.5 \times 9.8 T = 14.5×1614.5 \times 16π² + 14.5×9.814.5 \times 9.8 T = 14.5 × (16π² + 9.8) T = 14.5 × (157.9 + 9.8) T = 14.5×167.714.5 \times 167.7 T ≈\approx 2432 N

  3. Step 3: Convert cross-sectional area to SI units A = 0.065 cm² = 0.065×100.065 \times 10⁻⁴ m² = 6.5×106.5 \times 10⁻⁶ m²

  4. Step 4: Calculate elongation using Young's modulus From the context, Young's modulus for steel Y = 2.0×102.0 \times 10¹¹ N/m² ΔL = TL/(AY) ΔL = 2432×1.02432 \times 1.0/(6.5×106.5 \times 10⁻⁶ × 2.0×102.0 \times 10¹¹) ΔL = 2432/(1.3×101.3 \times 10⁶) ΔL ≈\approx 1.87×101.87 \times 10⁻³ m

PreviousNext