2. A 15 kg mass fastened to the end of a steel wire of un- stretched length 0.5 m is whirled in a vertical circle with an angular

0 thoughts on “2. A 15 kg mass fastened to the end of a steel wire of un- stretched length 0.5 m is whirled in a vertical circle with an angular”

1. Explanation:

   Elongation of the wire is:

   ΔL = F L₀ / (E A)

   where F is the force,

   L₀ is the initial length,

   E is Young’s modulus,

   and A is the cross sectional area.

   ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

   ΔL = T (1.25×10⁻⁶ m/N)

   T = (80,000 N/m) ΔL

   Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

   Sum of forces in the centripetal direction:

   ∑F = ma

   T − mg = mv²/r

   T − mg = mω²r

   T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

   T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

   Substitute:

   (80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

   (80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

   (797631.3 N/m) ΔL = 1331.35 N

   ΔL = 0.00167 m

   ΔL = 1.67 mm

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