### 5 Engineering Example 2

Introduction

A cube of an impure radioactive ore is of side 10 cm. The number of radioactive decays taking place per cubic metre per second is given by $R=1{0}^{23}\left(0.1-z\right){e}^{-t∕1000}$ . The dependence on time represents a half-life of $693$ seconds while the dependence on the vertical coordinate $z$ represents some gravitational stratification. The value $z=0$ represents the bottom of the cube and $z=0.1$ represents the top of the cube. (Note that the dimensions are in metres so 10 cm becomes 0.1 m.)

What is the total number of decays taking place over the cube in the 100 seconds between $t=0$ and $t=100$ ?

##### Solution

The total number of decays is given by the quadruple integral

$\phantom{\rule{2em}{0ex}}N={\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\int }_{z=0}^{0.1}{\int }_{t=0}^{100}1{0}^{23}\left(0.1-z\right){e}^{-t∕1000}dtdzdydx$

which may be evaluated as follows

$\begin{array}{rcll}N& =& {\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\int }_{z=0}^{0.1}{\int }_{t=0}^{100}1{0}^{23}\left(0.1-z\right){e}^{-t∕1000}dtdzdydx& \text{}\\ & =& {\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\int }_{z=0}^{0.1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-1000×1{0}^{23}\left(0.1-z\right){e}^{-t∕1000}\right]}_{t=0}^{100}\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\int }_{z=0}^{0.1}\left[1{0}^{26}\left(0.1-z\right)\left(1-{e}^{-0.1}\right)\right]\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\int }_{z=0}^{0.1}\left[9.5×1{0}^{24}\left(0.1-z\right)\right]\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& 9.5×1{0}^{24}{\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(0.1z-0.5{z}^{2}\right)\right]}_{z=0}^{0.1}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& 9.5×1{0}^{24}{\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}\left[0.005\right]\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& 0.005×9.5×1{0}^{24}{\int }_{x=0}^{0.1}{\int }_{y=0}^{0.1}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& 0.005×9.5×1{0}^{24}×0.1×0.1=4.75×1{0}^{20}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\end{array}$

Thus the number of decays is approximately equal to $4.75×1{0}^{20}$

For the solid prism shown below (the subject of the Task on page 50) find

1. the coordinates of the centre of mass
2. the moment of inertia about the $x$ - , $y$ - and $z$ -axes.

The $x$ , $y$ and $z$ coordinates of the centre of mass of a solid of constant density are given on page 55 by

$\phantom{\rule{2em}{0ex}}\stackrel{¯}{x}=\frac{\int x\phantom{\rule{0.3em}{0ex}}dV}{\int dV}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\stackrel{¯}{y}=\frac{\int y\phantom{\rule{0.3em}{0ex}}dV}{\int dV}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\stackrel{¯}{z}=\frac{\int z\phantom{\rule{0.3em}{0ex}}dV}{\int dV}$

For the triangular prism, the task on page 50 showed that the denominator $\int dV$ has value $1.5$ . The numerator of the expression for $\stackrel{¯}{x}$ is given by

$\begin{array}{rcll}\int x\phantom{\rule{0.3em}{0ex}}dV& =& {\int }_{z=0}^{1}{\int }_{y=0}^{1-z}{\int }_{x=0}^{3}x\phantom{\rule{0.3em}{0ex}}dxdydz={\int }_{z=0}^{1}{\int }_{y=0}^{1-z}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{3}\phantom{\rule{0.3em}{0ex}}dydz={\int }_{z=0}^{1}{\int }_{y=0}^{1-z}\frac{9}{2}\phantom{\rule{0.3em}{0ex}}dydz& \text{}\\ & =& {\int }_{z=0}^{1}{\left[\frac{9}{2}y\right]}_{y=0}^{1-z}\phantom{\rule{0.3em}{0ex}}dz={\int }_{z=0}^{1}\left(\frac{9}{2}\left(1-z\right)-0\right)\phantom{\rule{0.3em}{0ex}}dz={\int }_{z=0}^{1}\left(\frac{9}{2}-\frac{9}{2}z\right)\phantom{\rule{0.3em}{0ex}}dz& \text{}\\ & =& {\left[\frac{9}{2}z-\frac{9}{4}{z}^{2}\right]}_{0}^{1}=\frac{9}{2}-\frac{9}{4}-\left(0-0\right)=\frac{9}{4}=2.25& \text{}\end{array}$

So, $\stackrel{¯}{x}=\frac{2.25}{1.5}=1.5$ . By similar integration it can be shown that $ȳ=\frac{1}{3},\phantom{\rule{1em}{0ex}}\stackrel{̄}{z}=\frac{1}{3}$ .

The moment of inertia about the $x-$ axis, ${I}_{x}$ is given by ${I}_{x}=\rho {\int }_{V}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dV$ which for the solid under consideration is given by

$\begin{array}{rcll}{I}_{x}=\rho {\int }_{x=0}^{3}{\int }_{y=0}^{1}{\int }_{z=0}^{1-y}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx& =& \rho {\int }_{x=0}^{3}{\int }_{y=0}^{1}\left({y}^{2}-{y}^{3}+\frac{{\left(1-y\right)}^{3}}{3}\right)\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& \rho {\int }_{x=0}^{3}\frac{1}{6}\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{2}\rho & \text{}\end{array}$

Now, the mass $M$ of the solid is given by $M=\rho ×Volume=\frac{3}{2}\rho$ (where the volume had been calculated in a previous example) so

$\phantom{\rule{2em}{0ex}}{I}_{x}=\frac{1}{2}\rho =\frac{1}{2}\rho ×\frac{M}{\frac{3}{2}\rho }=\frac{1}{3}M$

Similarly, the moment of inertia about the $y-$ axis, ${I}_{y}$ is given by ${I}_{y}=\rho {\int }_{V}\left({x}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dV$ which for the solid under consideration is given by

$\begin{array}{rcll}{I}_{x}=\rho {\int }_{x=0}^{3}{\int }_{y=0}^{1}{\int }_{z=0}^{1-y}\left({x}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx& =& \rho {\int }_{x=0}^{3}{\int }_{y=0}^{1}\left({x}^{2}\left(1-y\right)+\frac{{\left(1-y\right)}^{3}}{3}\right)\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& \rho {\int }_{x=0}^{3}\left(\frac{1}{2}{x}^{2}+\frac{1}{12}\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{19}{4}\rho & \text{}\end{array}$

and so   $\phantom{\rule{1em}{0ex}}{I}_{y}=\frac{19}{4}\rho =\frac{19}{4}\rho ×\frac{M}{\frac{3}{2}\rho }=\frac{19}{6}M$ .  Finally, by symmetry, ${I}_{z}={I}_{y}=\frac{19}{6}M$ .

##### Exercise

For the solid shown below (the subject of the Task on page 47) find the centre of mass and the moment of inertia about the $x$ -, $y$ - and $z$ -axes.

$\left(\stackrel{̄}{x},ȳ,\stackrel{̄}{z}\right)=\left(0.75,1.6,3\right)$ ${I}_{x}=15.66M$ ${I}_{y}=12.8M$ ${I}_{z}=4.46M$

A cube of side 2 is made of laminated material so that, with the origin at one corner, the density of the material is $kx$ .

1. First find the mass $M$ of the cube:

The integrations over the cube are of the form ${\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}dV$ .

The mass $M$ is given by

$\begin{array}{rcll}M& =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}\rho \phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}kx\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}2kx\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{2}4kx\phantom{\rule{0.3em}{0ex}}dx={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2k{x}^{2}\right]}_{0}^{2}=8k\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
2. Now find the position of the centre of mass of the cube:

The $x$ -coordinate of the centre of mass will be given by $\frac{\int \rho x\phantom{\rule{0.3em}{0ex}}dV}{M}$ where the numerator is given by

$\begin{array}{rcll}\int \rho xdV& =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}\rho x\phantom{\rule{0.3em}{0ex}}dzdydx={\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}k{x}^{2}\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}2k{x}^{2}\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{2}4k{x}^{2}\phantom{\rule{0.3em}{0ex}}dx={\left[\frac{4}{3}k{x}^{3}\right]}_{0}^{2}=\frac{32}{3}k\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

So $\stackrel{¯}{x}=\frac{\frac{32}{3}k}{8k}=\frac{4}{3}$ .

The $y$ -coordinate of the centre of mass is given by $\frac{\int \rho y\phantom{\rule{0.3em}{0ex}}dV}{M}$ where the numerator is given by

$\begin{array}{rcll}\int \rho ydV& =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}\rho y\phantom{\rule{0.3em}{0ex}}dzdydx={\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}kxy\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}2kxy\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{2}4kx\phantom{\rule{0.3em}{0ex}}dx={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2k{x}^{2}\right]}_{0}^{2}=8k\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

So $\stackrel{¯}{y}=\frac{8k}{8k}=1$ .

By symmetry (the density depends only on $x$ ), $\stackrel{¯}{z}=\stackrel{¯}{y}=1$ .

The coordinates of the centre of mass are ( $\frac{4}{3},1,1$ ).

3. Finally find the moments of inertia about the $x$ -, $y$ - and $z$ -axes:

The moment of inertia about the $x$ -axis is given by ${I}_{x}={\int }_{V}\rho \left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dV$ (page 58). In this case,

$\begin{array}{rcll}{I}_{x}& =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}kx\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\left[kx\left({y}^{2}z+\frac{1}{3}{z}^{3}\right)\right]}_{z=0}^{2}\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{2}{\int }_{y=0}^{2}kx\left(2{y}^{2}+\frac{8}{3}\right)\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{2}{\left[kx\left(\frac{2}{3}{y}^{3}+\frac{8}{3}y\right)\right]}_{y=0}^{2}\phantom{\rule{0.3em}{0ex}}dx={\int }_{x=0}^{2}kx\left(\frac{32}{3}\right)\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[\frac{16}{3}k{x}^{2}\right]}_{0}^{2}=\frac{64}{3}k=\frac{8}{3}M\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

where the last step involves substituting that the mass $M=8k$ .

Similarly, the moment of inertia about the $y$ -axis is given by ${I}_{y}={\int }_{V}\rho \left({x}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dV$ i.e.

$\begin{array}{rcll}{I}_{y}& =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}kx\left({x}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}{\int }_{z=0}^{2}k\left({x}^{3}+x{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx={\int }_{x=0}^{2}{\int }_{y=0}^{2}{\left[k\left({x}^{3}z+\frac{1}{3}x{z}^{3}\right)\right]}_{z=0}^{2}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{2}{\int }_{y=0}^{2}\left(k\left(2{x}^{3}+\frac{8}{3}x\right)\right)\phantom{\rule{0.3em}{0ex}}dydx={\int }_{x=0}^{2}\left(k\left(4{x}^{3}+\frac{16}{3}x\right)\right)\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[k\left({x}^{4}+\frac{8}{3}{x}^{2}\right)\right]}_{x=0}^{2}=k\left(16+\frac{32}{3}\right)=\frac{80}{3}k=\frac{10}{3}M\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

By symmetry, ${I}_{z}={I}_{y}=\frac{10}{3}M$ .