Question

A uniform rod of mass $m$, length $L$ and area of cross-section $A$ is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity $\omega$ in a horizontal plane. If $Y$ is the Youngs modulus of the material of rod, the increase in its length due to rotation of rod is

• A. $\frac{m{\omega }^2{L}^2}{AY}$
• B. $\frac{m{\omega }^2{L}^2}{2AY}$
• C. $\frac{m{\omega }^2{L}^2}{3AY}$
• D. $\frac{2m{\omega }^2{L}^2}{AY}$

Answer 1

Answer: (C)

Consider a small element of length $dx$ at a distance $x$ from the axis of rotation.
Mass of the element, $dm$ $=\frac{m}{L}dx$ $=\mu dx$
where $\mu$ $=\frac{m}{L}$
The centripetal force acting on the element is $dT$ $=dm{\omega }^{2}x$ $=\mu {\omega }^{2}xdx$
As this force is provided by tension in the rod (due to elasticity), so the tension in the rod at a distance $x$ from the axis of rotation will be due to the centripetal force due to all elements between $x=x$ to $x=L$
$\therefore$ $T=\underset{x}{\overset{L}{\int }}\mu {\omega }^{2}xdx$ $=\frac{\mu {\omega }^2}{2}\left[L^2-x^2\right]$ $\dots \left(i\right)$
Let $dl$ be increase in length of the element. Then
$Y$ $=\frac{T/A}{dl/dx}$
$dl$ $=\frac{Tdx}{YA}$ $=\frac{\mu {\omega }^2}{2YA}$ $\left[L^2-x^2\right]dx$ $\left[Using\left(i\right)\right]$
So the total elongation of the whole rod is
$l$ $=\underset{o}{\overset{L}{\int }}\frac{\mu {\omega }^2}{2YA}$ $\left[L^2-x^2\right]dx$
$=\frac{\mu {\omega }^2}{2YA}$ ${\left[L^{2}x-\frac{x^3}{3}\right]}_o^L$ $=\frac{1}{3}\frac{\mu {\omega }^2{L}^3}{YA}$ $=\frac{1}{3}\frac{m{\omega }^2{L}^2}{YA}$