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 $\omega$, then -

${(}^{\prime }{A}^{\prime }$ is cross-section of ring$)$

• A. Tension produced in the ring, $T=A\rho {\omega }^2{R}^2$
• B. Tension produced in the ring, $T=A\rho {\omega }^{2}R$
• C. Small increase in its radius, $\Delta R=\frac{\rho {\omega }^2{R}^3}{Y}$
• D. Small increase in its radius, $\Delta R=\frac{\rho {\omega }^2{R}^2}{2Y}$

Answer 1

Answer: (A), (C)

Small increase in its radius, $\Delta R=\frac{\rho {\omega }^2{R}^3}{Y}$


Consider an element $PQ$ of length $dl$.

Let $T$ be the tension in the wire.

Now, mass of element $PQ$,

$\Rightarrow dm=A\left(dl\right)\rho$

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

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

For small angle,
$\sin \frac{\theta }{2}\approx \frac{\theta }{2}=\frac{\left(dl/R\right)}{2}$

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

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

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,
$Y=\frac{T/A}{\Delta R/R}$

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

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