5 Engineering Example 2

5.1 Radioactive decay

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 $R=10^{23}\left(0.1-z\right)e^{-t∕1000}$ . 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

$N={\int \; <!--nolimits\; -->}_{x=0}^{0.1}{\int \; <!--nolimits\; -->}_{y=0}^{0.1}{\int \; <!--nolimits\; -->}_{z=0}^{0.1}{\int \; <!--nolimits\; -->}_{t=0}^{100}10^{23}\left(0.1-z\right)e^{-t∕1000}dtdzdydx$

which may be evaluated as follows

Thus the number of decays is approximately equal to $4.75\times 10^{20}$

Task!

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.

Answer

Answer

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.

Answer

Task!

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:

   Answer

   The integrations over the cube are of the form ${\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}dV$ .
   
   The mass $M$ is given by

   $$\begin{array}{rcll}M& =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^2\rho dzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}kxdzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^{2}2kxdydx={\int \; <!--nolimits\; -->}_{x=0}^{2}4kxdx={\left[2k{x}^2\right]}_0^2=8k& \end{array}$$
2. Now find the position of the centre of mass of the cube:

   Answer

   The $x$ -coordinate of the centre of mass will be given by $\frac{\int \; <!--nolimits\; -->\rho xdV}{M}$ where the numerator is given by

   $$\begin{array}{rcll}\int \; <!--nolimits\; -->\rho xdV& =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^2\rho xdzdydx={\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}k{x}^{2}dzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^{2}2k{x}^{2}dydx={\int \; <!--nolimits\; -->}_{x=0}^{2}4k{x}^{2}dx={\left[\frac{4}{3}k{x}^3\right]}_0^2=\frac{32}{3}k& \end{array}$$

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

   The $y$ -coordinate of the centre of mass is given by $\frac{\int \; <!--nolimits\; -->\rho ydV}{M}$ where the numerator is given by

   $$\begin{array}{rcll}\int \; <!--nolimits\; -->\rho ydV& =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^2\rho ydzdydx={\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}kxydzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^{2}2kxydydx={\int \; <!--nolimits\; -->}_{x=0}^{2}4kxdx={\left[2k{x}^2\right]}_0^2=8k& \end{array}$$

   So $\bar{y}=\frac{8k}{8k}=1$ .
   
   By symmetry (the density depends only on $x$ ), $\bar{z}=\bar{y}=1$ .

   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:

   Answer

   The moment of inertia about the $x$ -axis is given by $I_x={\int \; <!--nolimits\; -->}_V\rho \left(y^2+z^2\right)dV$ (page 58). In this case,

   $$\begin{array}{rcll}{I}_x& =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}kx\left(y^2+z^2\right)dzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\left[kx\left(y^{2}z+\frac{1}{3}{z}^3\right)\right]}_{z=0}^{2}dydx={\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^{2}kx\left(2y^2+\frac{8}{3}\right)dydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\left[kx\left(\frac{2}{3}{y}^3+\frac{8}{3}y\right)\right]}_{y=0}^{2}dx={\int \; <!--nolimits\; -->}_{x=0}^{2}kx\left(\frac{32}{3}\right)dx& \\ & =& {\left[\frac{16}{3}k{x}^2\right]}_0^2=\frac{64}{3}k=\frac{8}{3}M& \end{array}$$

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

   Similarly, the moment of inertia about the $y$ -axis is given by $I_y={\int \; <!--nolimits\; -->}_V\rho \left(x^2+z^2\right)dV$ i.e.
   
   

   $$\begin{array}{rcll}{I}_y& =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}kx\left(x^2+z^2\right)dzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{2}k\left(x^3+x{z}^2\right)dzdydx={\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2{\left[k\left(x^{3}z+\frac{1}{3}x{z}^3\right)\right]}_{z=0}^{2}dydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^2\left(k\left(2x^3+\frac{8}{3}x\right)\right)dydx={\int \; <!--nolimits\; -->}_{x=0}^2\left(k\left(4x^3+\frac{16}{3}x\right)\right)dx& \\ & =& {\left[k\left(x^4+\frac{8}{3}{x}^2\right)\right]}_{x=0}^2=k\left(16+\frac{32}{3}\right)=\frac{80}{3}k=\frac{10}{3}M& \end{array}$$

   By symmetry, $I_z=I_y=\frac{10}{3}M$ .