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 it's Young's modulus is $Y$, then the elastic potential energy stored in the rod due to elongation is

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

Answer 1

Answer: (A)

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
$ T(x)=\frac{m x}{l} \times \frac{F}{m}=\frac{F x}{l} $
The longitudinal stress developed inside the rod at a distance $x$ from the free end is
$ \sigma=\frac{T(x)}{A}=\frac{F x}{l A} $
The strain energy density (energy per unit volume) is
$\frac{d U}{d V}=\frac{1}{2} \frac{\sigma^{2}}{Y}=\frac{1}{2} \frac{F^{2} x^{2}}{Y A^{2} l^{2}}$, where $d V=A d x$ is the volume of a small part of the rod
So the energy stored in the rod is
$ U=\frac{1}{2} F^{2} \int_{0}^{1} x^{2} d x $
$ U=\frac{F^{2} l}{6 Y} $