Question

A uniform rod of length $l$ is acted upon by a force $F$ in a gravity free region, as shown in the figure. If the area of cross-section of the rod is $A$ and its Young's modulus is $Y$, then the elastic potential energy stored in the rod due to elongation is

• A. $U=\frac{F^{2}l}{6AY}$
• B. $U=\frac{F^{2}l}{3AY}$
• C. $U=\frac{F^2{l}^2}{4A^{2}Y}$
• D. $U=\frac{F^2{l}^2}{2AY}$

Answer 1

The correct option is A $U=\frac{F^{2}l}{6AY}$


The acceleration of the rod is
$a=\frac{F}{m}$
So, the tension in the rod at a distance $x$ from the free end is given by,
$T\left(x\right)=m_x\times a$
$m_x=\lambda \times x$, where $\lambda =\frac{m}{l}$
$\Rightarrow T\left(x\right)=\frac{mx}{l}\times \frac{F}{m}=\frac{Fx}{l}$
The longitudinal stress developed inside the rod at a distance $x$ from the free end is
$\sigma =\frac{T\left(x\right)}{A}=\frac{Fx}{lA}$
The strain energy density (energy per unit volume) at distance $x$ is
$\frac{dU}{dV}=\frac{1}{2}\frac{{\sigma }^2}{Y}=\frac{1}{2}\frac{F^2{x}^2}{Y{A}^2{l}^2}$
[where $dV=Adx$ is the volume of a small part of the rod]
$\Rightarrow \frac{dU}{Adx}=\frac{1}{2}\frac{F^2{x}^2}{Y{A}^2{l}^2}$

So the energy stored in the rod is found by integrating
$dU=\frac{1}{2}\frac{F^2}{Y{A}^2{l}^2}A{x}^{2}dx$
$\Rightarrow U=\frac{1}{2}\frac{F^2}{YA{l}^2}{\int }_0^l{x}^{2}dx=\frac{F^2}{2AY{l}^2}\times \frac{l^3}{3}$
$\therefore U=\frac{F^{2}l}{6AY}$