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

Draw a free body diagram of the given situation
The situation can be depicted as in figure below:
Find the total tension force in the string
Given,

Mass, $m=14.5kg$
Length of the steel wire $l=1m$
Angular velocity, $\omega =2rev/s=2\times 2\pi =12.56rad/s$
Cross-sectional area of the wire,
$A=0.065c{m}^2=6.5\times {10}^{-6}{m}^2$

Let the elongation of the wire when the mass is at the lowest point of its path $=\Delta l$

When mass is whirled in vertical circle than at the lowest point of the circle the net force acting on the mass would be the sum of the downward force due to gravity and centrifugal force.
$T=mg+ml{\omega }^2$

Substituting the values, we get
$T=14.5\times 9.8+14.5\times 1\times (12.56{)}^2$
$=142.1+2287.4$
$T=2429.5N$

Find the elongation of the wire

As Young’s modulus is given by $\left(Y\right)=\frac{\text{stress}}{\text{Strain}}$

$\because \text{Stress}=\frac{T}{A},\text{Strain}=\frac{\Delta l}{l}$
$\because Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$
$\because \Delta l=\frac{Tl}{YA}$
where

$T=$ Tension force

$A=$ Area of cross section of steel wire

$l=$ Length of steel wire

$\Delta l=$ Elongation in steel wire

$\Delta l=\frac{Tl}{YA}$

$T=$ Tension force

$A=$ Area of cross section of steel wire

$l=$ Length of steel wire

$\Delta l=$ Elongation in steel wire

Substituting the values:

$\Delta l=\frac{2429.5\times 1}{2\times {10}^{11}\times 6.5\times {10}^{-6}}$

$\Delta l=1.87\times {10}^{-3}$