Question

A circular ring of radius $R$ and mass $m$ made of a uniform wire of cross-sectional area $A$ is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. The breaking stress of the material of the ring is ${\sigma }_b$, If Young's modulus of the material of the ring is $Y$, then determine the increment in the radius $\Delta R$ of the ring.

Answer 1

Tension in the ring $F=\frac{mR{w}^2}{2\pi }$

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

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

From Hooke's law $Y=\frac{Stress}{Strain}=\frac{\sigma }{ϵ}$ where $\sigma =\frac{F}{A}$

$\therefore$ $Y=\frac{mR{w}^2/\left(2\pi A\right)}{\Delta R/R}$ $⟹\Delta R=\frac{m{R}^2{w}^2}{2\pi AY}$