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

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 $dT$ $=dm\omega^{2}x$ $=\mu\omega^{2}x dx$
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 $x=L$
$\therefore \quad$ $T=\int\limits_{x}^{L} \mu \omega^{2}x \,dx$ $=\frac{\mu\omega^{2}}{2}\left[L^{2}-x^{2}\right]$ $\quad\ldots\left(i\right)$
Let $dl$ be increase in length of the element. Then
$Y$ $=\frac{T /A}{dl / dx}$
$dl$ $=\frac{Tdx}{YA}$ $=\frac{\mu\omega^{2}}{2 YA}$ $\left[L^{2}-x^{2}\right]dx$ $\quad\left[Using\left(i\right)\right]$
So the total elongation of the whole rod is
$l$ $=\int\limits_{o}^{L} \frac{\mu\omega^{2}}{2YA}$ $\left[L^{2}-x^{2}\right]dx$
$=\frac{\mu\omega^{2}}{2 YA}$ $\left[L^{2}x-\frac{x^{3}}{3}\right]_{o}^{L}$ $=\frac{1}{3}\frac{\mu\omega^{2} L^{3}}{YA}$ $=\frac{1}{3}\frac{m\omega^{2}L^{2}}{YA}$