Sol:

Given:

Mass, m = 14.5kg

Length of the steel wire, l = 1.0m

Angular velocity, = 2rev/s

= 2 * 2π rad/s

= 12.56 rad/s

Cross-sectional area of the wire, A = 0.065\( cm^{2} \)

= \( 0.065 * 10^{-4} m^{2} \)

Let \(\Delta l \) 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 + \(m\omega _{1}^{2} l\)

= \( 14.5 * 9.8 + 14.5 * 1(12.56)^{2}\)

= 2429.53 N

Young’s modulus = \(\frac{Stress}{Strain}\)

Y = \(\frac{(F/A)}{(\Delta l / l)}\) \(\Delta l = Fl / AY\)

Young’s modulus for steel = \(2 * 10^{11}Pa\) \(\Delta l = 2429.53 * \frac{1}{0.065 * 10^{-4} * 2 * 10^{11}}\)

= \(1.87 * 10^{-3}m\)

Hence, the elongation of the wire is \(1.87 * 10^{-3}m\)

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