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

mass (m) = 14.5 Kg
Length of wire ( L) = 1 m
Area of cross section of wire (A) = 0.065 cm² = 6.5 × 10^-6 m²
Angular frequency (v) = 2 rev/s
So, angular velocity(w) = 2πv
= 2π(2) = 4π rad/s
Young's modulus for steel (Y) = 2 × 10¹¹ N/m²
See the attachment ,
At the lowest point of the vertical circle,
T - mg = mw²L
T = mg + mw²L
= 14.5 { 9.8 + (4π)²×1 }
= 14.5 { 9.8 + 16π²}
= 14.5 × 167.72 N
= 2431.94 N

Now,
Youngs modulus = stress/strain
Y= TL/A∆L
∆L = FL/AY
= 2431.94 × 1 /(6.5×10^-6)×(2×10¹¹)
= 1.87 × 10^-3 m
= 1.87 mm

Answer 2

Answer:

mass (m) = 14.5 Kg

Length of wire ( L) = 1 m

Area of cross section of wire (A) = 0.065 cm² = 6.5 × 10^-6 m²

Angular frequency (v) = 2 rev/s

So, angular velocity(w) = 2πv

= 2π(2) = 4π rad/s

Young's modulus for steel (Y) = 2 × 10¹¹ N/m²

See the attachment ,

At the lowest point of the vertical circle,

T - mg = mw²L

T = mg + mw²L

= 14.5 { 9.8 + (4π)²×1 }

= 14.5 { 9.8 + 16π²}

= 14.5 × 167.72 N

= 2431.94 N

Now,

Youngs modulus = stress/strain

Y= TL/A∆L

∆L = FL/AY

= 2431.94 × 1 /(6.5×10^-6)×(2×10¹¹)

= 1.87 × 10^-3 m

= 1.87 mm

Explanation: