Elasticity
11.0 Potential Energy in a Stretched Wire
11.0 Potential Energy in a Stretched Wire
As shown in the Fig. E. 23. , the stress developed in the rod is, $$Stress = \frac{F}{A}$$ And the strain in the rod, $$Strain = \frac{{\Delta l}}{l}$$
Also we know,$$\begin{equation} \begin{aligned} Y = \frac{{Stress}}{{Strain}} \\ Strain = \frac{{Stress}}{Y} \\ \frac{{\Delta l}}{l} = \frac{F}{{AY}} \\ \Delta l = \frac{{Fl}}{{AY}} \\ k = \frac{{AY}}{l} \\\end{aligned} \end{equation} $$
Also, volume of the rod is given by, $$Volume = A \times l$$
Now, the potential energy stored in the spring is given by,
$$\begin{equation} \begin{aligned} U = \frac{1}{2}k{\left( {\Delta l} \right)^2} \\ U = \frac{1}{2}\left( {\frac{{AY}}{l}} \right){\left( {\frac{{Fl}}{{AY}}} \right)^2} \\ U = \frac{1}{2} \times \left( {\frac{F}{A}} \right) \times \left( {\frac{F}{{AY}}} \right) \times \left( {Al} \right) \\ U = \frac{1}{2} \times Stress \times Strain \times Volume \\ \mu = \frac{1}{2} \times Stress \times Strain \\\end{aligned} \end{equation} $$
where, $\mu $=Strain Energy Density
Example 7. Find the elastic potential energy stored in the uniform bronze rod of mass $m$, cross-section area $A$, length $l$ hanging vertically from the ceiling of the wall.
Take the young’s modulus of bronze = $Y$.
Solution: Uniform bronze rod having mass $m$, area of cross-section $A$ and length $L$.
$\lambda = \frac{m}{l},{\text{mass per unit length}}$
Take a small element of $dx$ length at a distance $x$ from $A$ as shown in Fig. E. 24.
Let the extension produced in small element of $dx$ length be $\Delta x$. This extension is produced due to the weight of $CB$
Length of $CB\ =\ (l-x)$
Mass of $CB\ =\ \lambda (l - x)$
Weight of $CB\ =\ \lambda (l - x)g$
From the FBD of the small segment $\Delta x$,
$$\begin{equation} \begin{aligned} Stress = \frac{{\lambda \left( {l - x} \right)g}}{A} \\ Strain = \frac{{\Delta x}}{{dx}} \\\end{aligned} \end{equation} $$So from Hooke’s law, $$\begin{equation} \begin{aligned} Strain = \frac{{Stress}}{Y} \\ \frac{{\Delta x}}{{dx}} = \frac{{\lambda \left( {l - x} \right)g}}{{AY}} \\ \Delta x = \frac{{\lambda g}}{{AY}}\left( {l - x} \right)dx \\\end{aligned} \end{equation} $$
Integration both sides with proper limits we get, $$\begin{equation} \begin{aligned} \mathop \smallint \limits_0^{\Delta l} \Delta x = \frac{{\lambda g}}{{AY}}\mathop \smallint \limits_0^l \left( {l - x} \right)dx \\ \Delta l = \frac{{\lambda g}}{{AY}}\left[ {lx - \frac{{{x^2}}}{2}} \right]_0^l \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} \Delta l = \frac{{\lambda g{l^2}}}{{2AY}} \\ \Delta l = \frac{{mgl}}{{2AY}}\left( {As,\lambda = \frac{m}{l}} \right) \\\end{aligned} \end{equation} $$
We know that, $$k = \frac{{AY}}{l}$$
So, the elastic potential energy is given by,
$$\begin{equation} \begin{aligned} U = \frac{1}{2}k{\left( {\Delta l} \right)^2} \\ U = \frac{1}{2}\left( {\frac{{AY}}{l}} \right){\left( {\frac{{mgl}}{{2AY}}} \right)^2} \\ U = \frac{1}{8}\frac{{{{\left( {mg} \right)}^2}l}}{{AY}} \\\end{aligned} \end{equation} $$
