Question

A steel wire $1.5m$ long and of radius $1mm$ is attached with a load $3kg$ at one end the other end of the wire is fixed it is whirled in a vertical circle with a frequency $2Hz$. Find the elongation of the wire when the weight is at the lowest position:
($Y=2\times {10}^{11}N/m^2$ and $g=10m/s^2$)

• A. $1.77\times {10}^{-3}m$
• B. $7.17\times {10}^{-3}m$
• C. $3.17\times {10}^{-7}m$
• D. $1.37\times {10}^{-7}m$

Answer 1

The correct option is A $1.77\times {10}^{-3}m$

Forces at the lowest point are gravitational force and centrifugal force

$F=mg+m{\omega }^{2}r=mg+m{\omega }^{2}L$

$\Delta l=\frac{FL}{YA}=\frac{(mg+m{\omega }^{2}L)L}{Y\times \pi r^2}$

$\omega =2\pi f=2\pi \times 2=4\pi$

$\Delta l=\frac{(30+3\times 16{\pi }^2\times 1.5)\times 1.5}{2\times {10}^{11}\times \pi \times ({10}^{-3}{)}^2}=1.77\times {10}^{-3}m$