Question

A uniform rod of length $L$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $\omega$. Find increase in length of the rod, if $\rho$ and $Y$ are the density and Young's modulus of the rod respectively,

• A. $\frac{\rho {\omega }^{3}Y}{4L^2}$
• B. $\frac{4\rho {\omega }^2{L}^3}{3Y}$
• C. $\frac{\rho {\omega }^3{L}^3}{3Y}$
• D. $\frac{\rho {\omega }^3{L}^3}{8Y}$

Answer 1

Answer: (C)

Given,
length of uniform $\mathrm{rod}=L$
angular speed $=\omega$
and density of rod $=\rho$ When rod is rotated in a horizontal plane, centrifugal force is responsible for increasing the length of rod, which is also equal to stress.
If $dl$ be the change in length of small element $dx$
due to rotation, by application of force $dF$.

Young's modulus,
$Y=\frac{\text{ stress }}{\text{ strain }}=\frac{dF/A}{dl/x}$
$\Rightarrow dl=\frac{dF\cdot x}{AY}$
$\Rightarrow dF=\frac{AYdl}{x}...(i)$
Where, $dF=$ centrifugal force due to mass of the
element of length, $dx$.
$dF=dm\cdot x{\omega }^2$
$dF=\frac{m}{L}\cdot xdx{\omega }^2...(ii)$
$\left[\because \text{ Change in mass }dm=\frac{m}{L}dx\right]$
From Eqs. (i) and (ii), we get
$\frac{AYdl}{x}=\frac{m}{L}\cdot xdx{\omega }^2$
$\Rightarrow dl=\frac{m{x}^2{\omega }^2}{LAY}\cdot dx$
$\Rightarrow$ Total change in length,
$l=\int dl={\int }_0^L\frac{m{x}^2{\omega }^2}{LAY}dx$
$=\frac{m{\omega }^2{L}^2}{3AY}=\frac{m}{AL}\cdot \frac{{\omega }^2{L}^3}{3Y}=\frac{\rho {\omega }^2{L}^3}{3Y}$