Question

A $6$-kg weight is fastened to the end of a steel wire of un-stretched length $60$cm. It is whirled in a vertical circle and has an angular velocity of $2$ revolution per second at the bottom of the circle. The area of cross-section of the wire is $0.05c{m}^2$. Calculate the elongation of the wire when the weight is at the lowest point of the path. Young's modulus of steel $=2\times {10}^{11}$ Pa.

Answer 1

At the lowest position, $T-mg=m{\omega }^{2}l$

or $T=mg+m{\omega }^2(l_0+\Delta l)$

Stress $=Y\times$ Strain $=Y\frac{\Delta l}{l_0}$

$T=A\times$ stress $=A\times Y\left(\Delta l/l_0\right)$

or, $A\times Y\left(\Delta l/l_0\right)=mg+m{\omega }^2(l_0+\Delta l)$

or, $\Delta l=\frac{mg+m{\omega }^2{l}_0}{\left[\right(AY/l_0)-m{\omega }^2]}=3.77\times {10}^{-4}$m.