Question

A thin ring of radius R is made of a material of density $\rho$ and Young’s modulus Y. If the ring is rotated about its centre in its own plane with angular velocity $\omega$ and the small increase in its radius is $\frac{\rho {\omega }^2{R}^3}{\lambda Y}$, then find $\lambda$___

Answer 1

$F_c=m{R}_c{\omega }^2$
$\Rightarrow 2T=\left(\rho A\pi R\right)\frac{2R}{\pi }{\omega }^2$
$\sigma =\frac{T}{A}=\rho {\omega }^2{R}^2$
Now,
$\frac{\Delta L}{L}=\frac{\sigma }{Y}$
$\frac{2\pi \Delta R}{2\pi R}=\frac{\rho {\omega }^2{R}^2}{Y}$
$\Delta R=\frac{\rho {\omega }^2{R}^3}{Y}$