Question

2. A 15 kg mass fastened to the end of a steel wire of un-

Answer 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