Question

A 6 kg weight is fastened to the end of a steel wire of unstretched length 60 cm. It is whirled in a vertical~~ circle and has an angular velocity of 2 rev/s at the bottom of the circle. The area of cross-section of the. i 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 $=2\times {10}^{11}$ N/m^{2} .

Answer 1

Given : $L=60cm=0.6m$ $f=2rev/s$ $A=0.05c{m}^2=5\times {10}^{-6}{m}^2$ $m=6kg$

Angular velocity at the lowest point $w=2\pi f=2\pi \left(2\right)=12.56$ rad/s

Let the tension in the wire when the weight is at the lowest point be $T$

$\therefore$ $T-mg=mL{w}^2$

OR $T=mg+mL{w}^2=6\left(9.8\right)+6\left(0.6\right)(12.56{)}^2=626.7N$

Using $Y=\frac{TL}{A\Delta L}$ $⟹\Delta L=\frac{TL}{AY}$

$\therefore$ $\Delta L=\frac{626.7\times 0.6}{2\times {10}^{11}\times 5\times {10}^{-6}}=3.8\times {10}^{-4}m$