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.
Answers
Answer :
Given data,
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
⇒ 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)
∴ 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× 10-3m.
Answer:
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young's modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m.