Question

A steel cylindrical rod of length $l$ and radius $r$ is suspended by its end from the ceiling. Find the elastic deformation energy $U$ of the rod.

• A. $U=\frac{1}{6}\pi r^2{\rho }^2{g}^2\frac{l^3}{E}$
• B. $U=\frac{5}{6}\pi r^2{\rho }^2{g}^2\frac{l^3}{E}$
  
• C. $U=\frac{1}{6}\pi r^2{\rho }^2{g}^2\frac{2l^3}{E}$
  
• D. $U=\frac{5}{6}\pi r^2{\rho }^2{g}^2\frac{2l^3}{E}$
  

Answer 1

The correct option is A $U=\frac{1}{6}\pi r^2{\rho }^2{g}^2\frac{l^3}{E}$

Let's consider,

$g=$ acceleration due to gravity

$l=$ length of the rod

$r=$ radius of the road

$x=$ change in length

$\rho =$ density

$E=$ modulus

From the given figure,

$Stress=\frac{\rho .\pi r^2(l-x)g}{\pi r^2}=\rho g(l-x)$

$Strain,\frac{d\Delta x}{dx}=\frac{Stress}{E}=\frac{\rho g(l-x)}{E}$

The elastic deformation energy is given by

$dU=\frac{1}{2}\times stress\times strain\times volume$

$dU=\frac{1}{2}\times \rho g(l-x)\times \frac{\rho g(l-x)}{E}\times \pi r^{2}dx$

${\int }_0^{U}dU=\frac{\pi r^2{\rho }^2{g}^2}{2E}{\int }_0^l(l-x{)}^{2}dx$

$U=\frac{\pi r^2{\rho }^2{g}^2{l}^3}{6E}$

The correct option is A.