A thin uniform copper rod of length $l$ and mass m rotates uniformly with an angular velocity $\omega$ in a horizontal plane about a vertical axis passing through one of its ends. The elongation of the rod is $\frac{\rho {\omega }^a{l}^b}{cY}$. The value of $(a+b+c)$ is ________(Y is Young’s modulus of copper)

To investigate the variation of tension along its length, consider an element of thickness  $dx$  of the rod at a distance  $x$ from the axis of rotation. Let the tension in the rod at $x=x$ be equal to T and at $x=x+dx$  to  $T+dT$. Forces on the element  $dx$ are shown in the figure below. Application of Newton’s second law to the element gives.

$T-(T+dT)=\left(dm\right){\omega }^{2}x$

$-dT=\left(\frac{m}{l}dx\right).{\omega }^{2}x.$

To find out tension at $x=r$, integrate the above equation with limits $x=0$  to  $x=r$ and $T=T_0$  to $T=T$
 This gives.
$$\begin{gathered} -\underset{T_0}{\overset{T}{\int }}dT=\underset{0}{\overset{r}{\int }}\left(\frac{m}{l}dx\right){\omega }^{2}x \\ or T=T_0-\frac{m{\omega }^2}{l}.\frac{r^2}{2} \\ =\frac{1}{2}m{\omega }^{2}l-\frac{1}{2}m{\omega }^2\frac{r^2}{l} \\ =\frac{m{\omega }^{2}l}{2}(1-\frac{r^2}{l^2}) \end{gathered}$$

Note that the tension decreases from the axis of rotation to the end of rod. Different cross-sections of the rod are differently stressed.
 Let $d\epsilon$ be the elongation caused by the stress in the element $dx$. This elongation is released to the stress at the strip as
$$\begin{gathered} d\epsilon =\frac{m{\omega }^{2}l}{2AE}(1-\frac{x^2}{l^2})dx \\ \Rightarrow \underset{0}{\overset{\epsilon }{\int }}d\epsilon =\frac{m{\omega }^{2}l}{2AE}\underset{0}{\overset{1}{\int }}(1-\frac{x^2}{l^2})dx \\ or \epsilon =\frac{m{\omega }^{2}l}{3AE}=\frac{\rho {\omega }^2{l}^3}{3E} \end{gathered}$$