Question

A thin ring of radius R is made up of a material of density $\rho$ and young's modulus Y. If the ring is rotated about its centre in its own plane with an angular velocity $\omega$ then the small increase in radius $\left(\Delta R\right)$ of the ring is

• A. $\frac{\rho {\omega }^2{R}^3}{Y}$
• B. $\frac{\rho {\omega }^2}{Y{R}^3}$
• C. $\frac{Y}{\rho {\omega }^2{R}^3}$
• D. None

Answer 1

Answer: (A)

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

Consider an element PQ of length dl. Let T be the tension and A the area of cross-section of the wire .

Mass of element $dm=volume\times density=A\left(dl\right)\times \rho$

The component of T, towards the centre provides the necessary centripetal force to portion PQ $F=2Tsin\left(\frac{d\theta }{2}\right)=\left(dm\right)R{\omega }^2$

For small angles $\frac{sin\left(d\theta \right)}{2}=\frac{d\theta }{2}=\frac{dl/R}{2}$

or $d\theta =\frac{d}{R}$

Substituting in Eq. (i), we have

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

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

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

$\Delta R=\frac{TR}{AY}=\frac{A\rho {\omega }^2{R}^2)R}{AY}$ or $\Delta R=\frac{\rho {\omega }^2{R}^3}{Y}$