Question

A uniform ring of radius R and made up of a wire ofcross-sectional radius r is rotated about its axis witha frcquency f If density of the wire is p and Young's modulus is Y. Find the fractional change in radiusof the ring .

Answer 1

Let $T$ be the tension in the ring.

From figure, net radial component of the tension $T^{\prime }=2Tsin\left(d\theta \right)$

As $d\theta$ is very small, thus $sin\left(d\theta \right)\approx d\theta$

$⟹$ $T^{\prime }=2Td\theta$

Mass of the ring $m=\rho \times \left(\pi r^2\right)\left(2\pi R\right)$

Mass of the small element of the ring $dm=\frac{m}{2\pi }\left(2d\theta \right)=\frac{md\theta }{\pi }$

$\therefore$ $dm=\frac{2{\pi }^2{r}^{2}R\rho d\theta }{\pi }=2\pi r^{2}R\rho d\theta$

Also $T^{\prime }=\left(dm\right)R{w}^2$ where $w=2\pi f$

$\therefore$ $2T\left(d\theta \right)=2\pi r^{2}R\rho \left(d\theta \right)\times R(2\pi f{)}^2$ $⟹T=4{\pi }^3{R}^2{r}^2\rho f^2$

Elongation in the length of the ring $\delta =2\pi (R+\Delta R)-2\pi R=2\pi \Delta R$

$\therefore$ Strain in the ring $ϵ=\frac{\delta }{2\pi R}=\frac{\Delta R}{R}$

From Hooke's law $Y=\frac{T}{Aϵ}$ where $A=\pi r^2$

$\therefore$ $Y=\frac{4{\pi }^3{R}^2{r}^2{f}^2\rho }{\pi r^2\times \Delta R/R}$ $⟹\Delta R=\frac{4{\pi }^2{R}^3{f}^2\rho }{Y}$