Question

A solid copper cylinder of length $l=65cm$ is placed vertically on a horizontal surface and subjected to a vertical compressive force, $F=1000N$ directed downward and distributed uniformly over the end face. If the resulting change of the volume of the cylinder is $\frac{(x+0.84)}{10}m{m}^3$, find the value of $x$.
Young's modulus for copper is $1.2\times {10}^{11}$ and Poisson's ratio for copper is 0.33

Answer 1

Given,
$\mu$ is the Poisson's ratio.
For copper $\mu =0.33$
Young's modulus for copper:$E=1.2\times {10}^{11}$
Volume of a solid cylinder
$V=\pi r^{2}l$
So, $\frac{\Delta V}{V}=\frac{\pi 2r\Delta rl}{\pi r^{2}l}+\frac{\pi r^2\Delta l}{\pi r^{2}l}=\frac{2\Delta r}{r}+\frac{\Delta l}{l}$ (1)
But longitudinal strain $\frac{\Delta l}{l}$ and accompanying lateral strain $\frac{\Delta r}{r}$ are related as
$\frac{\Delta r}{r}=-\mu \frac{\Delta l}{l}$ (2)
Using (2) in (1), we get,
$\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2\mu )$ (3)
But $\frac{\Delta l}{l}=\frac{-\frac{F}{\pi r^2}}{E}$
(Because the increment in the length of cylinder $\Delta l$ is negative)
So, $\frac{\Delta V}{V}=\frac{-F}{\pi r^{2}E}(1-2\mu )$
Thus, $\Delta V=\frac{-Fl}{E}(1-2\mu )$
The negative sign means that the volume of the cylinder has decreased.
Given,
$\Delta V=\frac{x}{10}m{m}^3=\frac{x}{10}\times {10}^{-9}{m}^3$
$F=100N$
$l=65\times {10}^{-2}m$
$\Rightarrow \frac{x}{10}\times {10}^{-9}{m}^3=\frac{100\times 65\times {10}^{-2}}{1.2\times {10}^{11}}(1-2\times 0.33)$
$\Rightarrow 10x=\frac{65\times 0.34}{1.2}=18.4$
$\Rightarrow x=1.84$