Question

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

$\frac{m{\omega }^2{L}^2}{3AY}$

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 therod at a distance $x$ from the axis of rotation will be due to the cetripetal force due to all elements between $x=x$ to $x=L$
$\therefore T={\int }_x^L\mu {\omega }^{2}xdx=\frac{\mu {\omega }^2}{2}[L^2-x^2]....\left(i\right)$

Let $dl$ be increase in length of the element. Then
$Y=\frac{T/A}{dl/dx}$

$dl=\frac{Tdx}{YA}=\frac{\mu {\omega }^2}{2YA}[L^2-x^2]dx$ [Using (i)]

So the total elongation of the whole rod is
$l={\int }_0^L\frac{\mu {\omega }^2}{2YA}[L^2-x^2]dx$

$=\frac{\mu {\omega }^2}{2YA}{[L^{2}x-\frac{x^3}{3}]}_0^L=\frac{1}{3}\frac{\mu {\omega }^2{L}^3}{YA}=\frac{1}{3}\frac{m{\omega }^2{L}^2}{YA}$