A 14.5 kg mass, fastened to the end of a steel wire of unstretched len

1.	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» L = 1 m W = 2 revolutions/s A = 0.065 cm² = 6.5 x 10^-6 m² F = mg + mlw² = 14.5 × 9.8 +14.5 × 1 × (2)² = 200.1 N ∆L = \(\frac{FI}{AY}\) [∵ Y = \(\frac{FI}{AΔI}]\) Young’s modulus of steel is 2 × 10¹¹Nm^-2 ∆L = \(\frac{200.1 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}}m\) = 1.539 × 10^-4m.

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»

L = 1 m

W = 2 revolutions/s

A = 0.065 cm² = 6.5 x 10^-6 m²

F = mg + mlw²

= 14.5 × 9.8 +14.5 × 1 × (2)²

= 200.1 N

∆L = \(\frac{FI}{AY}\)

[∵ Y = \(\frac{FI}{AΔI}]\)

Young’s modulus of steel is 2 × 10¹¹Nm^-2

∆L = \(\frac{200.1 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}}m\)

= 1.539 × 10^-4m.