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Q. A mass of 6.5kg in hanging from the end of a 60 cm long steel wire $ (Y=2\times {{10}^{11}}Pa) $ with area of corss-section 0.05 cm2. When it is revolving in a vertical circle it has an angular velocity of 2 revolutions per second, at the bottom of the circle. Approximate elongation of the wire (in metres) when the mass is at its lowest point of the trajectory is:

EAMCETEAMCET 2006Mechanical Properties of Solids

Solution:

Forces acting on the mass are the tension T and the weight W. At the lowest point, $ T-W=\frac{m{{v}^{2}}}{r} $ or $ T=W+mr{{\omega }^{2}} $ $ (\because v=r\omega ) $ $ =6.5\times 9.8+6.5\times 0.60\times {{(2)}^{2}} $ $ =63.7+15.6 $ $ =79.3\,N $ We have, $ Y=\frac{TL}{Al} $ $ \Rightarrow $ $ l=\frac{TL}{Ay} $ $ =\frac{79.3\times 0.60}{0.05\times {{10}^{-4}}\times 2\times {{10}^{11}}} $ $ =475.8\times {{10}^{-7}}m $ $ 4.75\times {{10}^{-5}}\,m $ $ \approx 4\times {{10}^{-5}}\,m $