Question

A $14.5kg$ mass, fastened to the end of a steel wire of unstretched length $1.0m$, is whirled in a vertical circle with an angular velocity of $2rev/s$ at the bottom of the circle. The cross-sectional area of the wire is $0.065c{m}^2$. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer 1

Step 1: Given Data

Mass of the body $m=14.5kg$

Length of the steel wire $l=1.0m$

Angular velocity $\omega =2rev/s$

The cross-sectional area of the wire $A=0.065c{m}^2$

We know that Young's modulus $Y_{steel}=2\times {10}^{11}N{m}^{-2}$

Step 2: Calculate the Force

The wire is fastened to a given mass and is rotating with the given angular velocity.

The forces acting on the wire are the weight of the mass and the centrifugal force due to the circular motion with a radius $l$.

Therefore, the total force assuming $g=9.8m/s^2$ is given as,

$F=mg+ml{\omega }^2$

$=14.5\left\{9.8+1\times {\left(2\times 2\mathrm{\pi }\right)}^2\right\}$

$=14.5\times 169.8$

$=2462N$

Step 3: Calculate the Elongation

We know that Young's modulus is given by

$Y=\frac{Stress}{Strain}$

where $Stress=\frac{F}{A}$, where $F$ is the force,

$Strain=\frac{∆l}{l}$

$\therefore Y=\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}$

$\Rightarrow Y=\frac{Fl}{A∆l}$

$\Rightarrow ∆l=\frac{Fl}{AY}$

Upon substituting the values we get,

$∆l=\frac{2462\times 1}{0.065\times {10}^{-4}\times 2\times {10}^{11}}$

$=\frac{2462}{65\times 2\times {10}^4}$

$=18.9\times {10}^{-4}m$

$=1.89mm$

Hence, the elongation of the wire when the mass is at the lowest point of its path is $1.89mm$.