Question

A string of area of cross-section \(4mm^2\) and length \(0.5 m\) is connected with a rigid body of mass \(2 kg\). The body is rotated in a vertical circular path of radius \(0.5 m\). The body acquires a speed of \(5 m/s\) at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ______\(×10^{–5}\). 

(use young’s modulus \(10^{11} N/m^2\) and \(g = 10 m/s^2\))

Answer 1

Given : \(A = 4\,mm^2 = 4 \times 10^{-6}\,m^2\), \(m = 2\,kg\), \(r = 0.5\,m\), \(v = 5\,m/s\)

Young’s modulus \(Y = 10^{11}\,N/m^2\), \(g = 10\,m/s^2\)

Tension at the bottom of vertical circle

\(\begin{array}{l} T - mg = \frac{mv^2}{r} \end{array}\)

\(\begin{array}{l} T = mg + \frac{mv^2}{r} \end{array}\)

\(\begin{array}{l} T = 2\times10 + \frac{2\times 5^2}{0.5} \end{array}\)

\(\begin{array}{l} T = 20 + \frac{50}{0.5} = 20 + 100 = 120\,N \end{array}\)

Strain is given by

\(\begin{array}{l} \text{Strain} = \frac{\text{Stress}}{Y} = \frac{T/A}{Y} \end{array}\)

\(\begin{array}{l} = \frac{120}{4 \times 10^{-6} \times 10^{11}} \end{array}\)

\(\begin{array}{l} = \frac{120}{4 \times 10^{5}} = 3 \times 10^{-4} \end{array}\)

\(\begin{array}{l} = 30 \times 10^{-5} \end{array}\)

Hence, strain = 30 × 10⁻⁵