A uniform rod of mass m, length L, 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 ω 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

The correct option is mω²L²/3AY.

To find the total elongation of the rotating rod, we can calculate the tension at any section of the rod and then integrate the resulting strain along its entire length.

Let us consider a uniform rod of mass m, length L, and cross-sectional area A, rotating in a horizontal plane with a constant angular velocity $\omega$ about an axis passing through one of its ends (at $x=0$).

The tension $T\left(x\right)$ at a distance $x$ from the axis of rotation is due to the centripetal force acting on the part of the rod extending from $x$ to $L$. Let $dm$ be the mass of an infinitesimal element of length $dx\text{'}$ located at a distance $x\text{'}$ from the axis:

$dm=\frac{m}{L}dx\text{'}$

The tension $T\left(x\right)$ is given by integrating the centripetal force of all such elements from $x$ to $L$:

$T\left(x\right)={\int }_x^L\left(dm\right)·x\text{'}·{\omega }^2$

$T\left(x\right)={\int }_x^L\left(\frac{m}{L}dx\text{'}\right)x\text{'}{\omega }^2=\frac{m{\omega }^2}{L}{\int }_x^{L}x\text{'}dx\text{'}$

$T\left(x\right)=\frac{m{\omega }^2}{L}{\left[\frac{x{\text{'}}^2}{2}\right]}_x^L=\frac{m{\omega }^2}{2L}\left(L^2-x^2\right)$

Now, consider a small element of length $dx$ at distance $x$. The elongation $d\delta$ of this element under the tension $T\left(x\right)$ is given by Young's modulus formula:

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T\left(x\right)/A}{d\delta /dx}\Rightarrow d\delta =\frac{T\left(x\right)dx}{AY}$

Substituting the expression for tension $T\left(x\right)$:

$d\delta =\frac{m{\omega }^2}{2LAY}\left(L^2-x^2\right)dx$

Integrating this elongation from $x=0$ to $x=L$ to get the total elongation $\Delta L$:

$\Delta L={\int }_0^L\frac{m{\omega }^2}{2LAY}\left(L^2-x^2\right)dx$

$\Delta L=\frac{m{\omega }^2}{2LAY}{\left[L^{2}x-\frac{x^3}{3}\right]}_0^L$

$\Delta L=\frac{m{\omega }^2}{2LAY}\left(L^3-\frac{L^3}{3}\right)$

$\Delta L=\frac{m{\omega }^2}{2LAY}\left(\frac{2L^3}{3}\right)$

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