Question

A thin ring has a radius R, density $\rho$ and Young's modulus Y. The ring is rotated in its own plane about an axis passing through its centre with angular velocity $\omega$. Then the small increase in its radius is :

• A. $dR=\frac{\rho {\omega }^2{R}^3}{Y}$
• B. $dR=\frac{3\rho {\omega }^2{R}^3}{Y}$
• C. $dR=\frac{6\rho {\omega }^2{R}^3}{Y}$
• D. $dR=\frac{\rho {\omega }^2{R}^3}{2Y}$

Answer 1

Answer: (A)

$dR=\frac{\rho {\omega }^2{R}^3}{Y}$

$\begin{array}{c}\text{Mass of element}dm=\text{Area of crosssection}\times dl\times \rho \\ \text{Let a be area of cross section.}\end{array}$

$\begin{array}{cc}{F}_C=2T\sin \theta /2& \\ \text{As}\theta \text{is small,}\therefore \text{Sin(}\theta \text{/2)=}\theta \text{/2}& \\ \left(dm\right)R{\omega }^2=2T\frac{\theta }{2}& \end{array}$

$\begin{array}{cc}T& =\frac{dmR{\omega }^2}{\theta }(\because \theta =\frac{dL}{R})\\ T& =a{R}^2\rho {\omega }^2\end{array}$

$\begin{array}{c}\text{Longitudinal}\mathrm{strain}\frac{\Delta l}{l}=\frac{2\pi \Delta R}{2\pi R}=\frac{\Delta R}{R}\\ \frac{\Delta R}{R}=\frac{T/a}{Y}=\frac{R^2\rho {\omega }^2}{Y}\\ \Delta R=\frac{\rho {\omega }^2{R}^3}{Y}\end{array}$