Question

A thin ring of radius $r$ is made of density $\rho$ and Young's modulus Y. It is spun about its own plane, about an axis through its centre with angular velocity w. The amount (assumed small) by which its circumference increases is

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

Answer 1

The correct option is D $\frac{2\pi p{\omega }^2{r}^3}{Y}$

$\text{Taking small element of angle}2d\theta \text{.}$

$\begin{array}{c}\text{Taking Vertical forces on small element,}\\ \text{2T}\sin \left(d\theta \right)={\omega }^{2}rPA\cdot R\cdot 2d\theta \end{array}$

$\sin \left(d\theta \right)\cong \text{d}\theta \text{(as d}\theta \text{is very small)}$

$\begin{array}{cc}2T\cdot d\theta & ={\omega }^{2}rPA\cdot R2d\theta \\ T& ={\omega }^2{r}^{2}PA\end{array}$

$\begin{array}{c}\text{By Hookes law}\\ \text{elongation}\left(\Delta l\right)=\frac{Fl}{AY}\\ \Delta l=\frac{{\omega }^2{r}^{2}PA\cdot 2\pi r}{AY}\\ \Delta l=\frac{2\pi P{\omega }^2{r}^3}{Y}\end{array}$