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 data,
Mass fastened to the end of a steel wire, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, $\omega$ = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, A = 0.065 cm² = 0.065 × 10^-4 m²

Let Δl is the elongation of the wire when the mass is at the lowest point of its path.

When the mass is whriled in a vertical circle, the total force on the mass is:
F = mg + ml$\omega^2$
⇒ F = (14.5 × 9.8) + (14.5× 1 × (12.56)²)
⇒ F = 2429.53 N
Young’s modulus = Stress/Strain
Y = (F/A)/(Δl/l)
so,  Δl = Fl/AY
Young’s modulus for steel = 2× 10¹¹ Pa
Δl = 2429.53× 1 / (0.065× 10^-4× 2× 10¹¹)
⇒ Δl = 1.87× 10^-3 m

Hence, the elongation of the wire is 1.87 × 10^-3 m.