Question

A copper wire of negligible mass, 1 m length and cross-sectional area ${10}^{-6}{m}^2$ is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotating with an angular velocity of 20 rad/s. If the elongation in the wire is ${10}^{-3}$ m.
b. If for the same wire as stated above, the angular velocity is increased to 100 rad/s and the wire breaks down, find the breaking stress (in terms of $\times {10}^{10}N/m^2)$.

Answer 1

1
Sln :
The restoring force (F) produced in the wire provides the necessary centripetal force i.e.
F = ML ${\omega }^2$
(Since radius of circular path = length of wire = L)
We have, Young's modulus
$Y=\frac{FlA}{\Delta LlL}orF=YA.\frac{\Delta L}{L}$
Comparing Eqs. (i) and (ii), we get
$YA=\frac{\Delta L}{L}=mL{\omega }^{2}orY=\frac{m{L}^2{\omega }^2}{A\Delta L}$
Substituting given values, we get
$Y=400\times {10}^9=4\times {10}^{11}N/m^2$
The breaking force,
$F=mL{\omega }_{max}^2=\left(1kg\right)\left(1m\right)(100rad/s{)}^2={10}^4$ newton
Therefore, breaking stress = $\frac{Breakingforce}{Area}$
$=\frac{{10}^4}{10-6}={10}^{10}N/m^2$