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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 $0.065 \;cm ^{2} .$ Calculate the elongation of the wire when the mass is at the lowest point of its path.
$7.341 \times 10^{-5}\; m$
$3.985 \times 10^{-3}\; m$
$5.697 \times 10^{-5}\; m$
$1.539 \times 10^{-4}\; m$
Solution
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
$F=m g+m l \omega^{2}$
$=14.5 \times 9.8+14.5 \times 1 \times(2)^{2}=200.1 N$
Young's modulus $=\frac{\text { Stress }}{\text { Strain }}$
$Y=\frac{F}{A} \frac{l}{\Delta l}$
$\therefore \Delta I=\frac{F l}{A Y}$
Young's modulus for steel $=2 \times 10^{11} Pa$
$\therefore \Delta l=\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}}=1539.23 \times 10^{7}$
$=1.539 \times 10^{-4} m$
Hence, the elongation of the wire is $1.539 \times 10^{-4}\; m$
