Question

A uniform ring having mass $m$, radius $R$, cross sectional area of the wire $A$ and Young's modulus $Y$ is rotating with an angular speed $\omega$ (where $\omega$ is small) about an axis passing through centre and perpendicular to plane of ring on a smooth horizontal surface. Which of the following options are correct?

• A. Tension in the wire is $\frac{mR{\omega }^2}{2\pi }$.
• B. Change in length of the wire is $\frac{m{R}^2{\omega }^2}{2A.Y}$.
• C. Change in radius of the ring is $\frac{m{R}^2{\omega }^2}{2\pi AY}$.
• D. Elastic potential energy stored is $\frac{1}{4\pi }\left(\frac{m^2{\omega }^4{R}^3}{AY}\right)$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A Tension in the wire is $\frac{mR{\omega }^2}{2\pi }$.
C Change in radius of the ring is $\frac{m{R}^2{\omega }^2}{2\pi AY}$.
D Elastic potential energy stored is $\frac{1}{4\pi }\left(\frac{m^2{\omega }^4{R}^3}{AY}\right)$.

Let T be the tension in the string.

$2Tsin\frac{d\theta }{2}=\frac{m}{2\pi R}\times R{\omega }^2.Rd\theta$

Using relation if x << 1, sin x=x
we get
$T=\frac{mR{\omega }^2}{2\pi }$

$Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$

$\frac{\Delta l}{l}=\frac{T}{Y.A}\Rightarrow \Delta l=\frac{T}{Y.A}\times l\Delta l=\frac{mR{\omega }^2}{2\pi }\times \frac{1}{Y.A}\times 2\pi R=\frac{m{R}^2{\omega }^2}{Y.A}$

$\frac{\Delta l}{l}=\frac{\Delta R}{R}=\frac{T}{Y.A}=\frac{mR{\omega }^2}{2\pi A.Y}\Rightarrow \Delta R=\frac{m{R}^2{\omega }^2}{2\pi A.Y}$

$U=\frac{1}{2}K{x}^2=\frac{1}{2}\left(\frac{Y.A}{l}\right)\times (\Delta l{)}^{2}U=\frac{1}{2}\frac{Y.A}{2\pi R}\times {\left(\frac{m{R}^2{\omega }^2}{Y.A}\right)}^2=\frac{1}{4\pi }\left(\frac{m^2{R}^3{\omega }^4}{Y.A}\right)$