# 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 - Physics

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.

#### Solution 1

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 × 10-4 m2

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 = Stress/Strain

Y = (F/A)/trianglel = F/A  l/trianglel

:.trianglel = (Fl)/(AY)

Young’s modulus for steel = 2 × 1011 Pa

trianglel = (2429.53xx1)/(0.065xx10^(-4)xx2xx10^11)

=>trianglel = 1.87 xx 10^(-3) m

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

#### Solution 2

Here, m = 14.5 kg; l = r = 1 m; v = 2 rps; A = 0.065 x 10-4 m2 Total pulling force on mass, when it is at the lowest position of the vertical circle is F = mg + mr w2 = mg + mr 4,π2 v2

=14.5 xx 9.8 + 14.5 xx 1 xx 4 xx (22/7)^2 xx 2^2

=142.1 + 2291.6 = 2433.9 N

Y = F/A xx l/(trianglel)

or trianglel = (Fl)/(AY)= (2433.7xx1)/(0.065 xx 10^(-4)xx(2xx10^11)) = 1.87 xx 10^(-3) m = 1.87 m

Concept: Elastic Moduli - Young’s Modulus
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#### APPEARS IN

NCERT Class 11 Physics
Chapter 9 Mechanical Properties of Solids
Exercises | Q 11 | Page 244