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 $\omega$, find the small increase in its radius.

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

Answer 1

The correct option is B

$\frac{\rho {\omega }^2{R}^3}{Y}$
Consider an element $\text{PQ}$ of arc length $dl$. Let $T$ be the tension and $A$ be the area of cross section of wire so, mass of element,
$dm=\text{volume}\times \text{density}=\left(Adl\right)\rho$


The component of $T$, towards the centre provides the necessarry centripetal force.

$\therefore 2T\sin \left(\frac{\theta }{2}\right)=\left(dm\right)R{\omega }^2....\left(1\right)$

For small angles $\sin \frac{\theta }{2}=\frac{\theta }{2}=\frac{dl}{2R}$

Substituting in equation $\left(1\right)$, we have

$T\frac{dl}{R}=A\left(dl\right)\rho R{\omega }^2$

$\Rightarrow T=A\rho {\omega }^2{R}^2$

Let $\Delta R$ be the increase in radius,

Longitudinal strain,

$\frac{\Delta l}{l}=\frac{\Delta \left(2\pi R\right)}{2\pi R}=\frac{\Delta R}{R}$

Now, we know that, $Y=\frac{T/A}{\Delta R/R}$

$\Rightarrow \Delta R=\frac{TR}{AY}$

Substituting the value of $T$ we get,

$\Delta R=\frac{\left(A\rho {\omega }^2{R}^2\right)R}{AY}$

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

Hence, option (b) is correct answer.