Moment of inertia easy-med

I learned that Wolfram Mathematica returns the moment of inertia in units of the fifth power of length, which when multiplied by the density gives the SI form of the moment of inertia: kg*meter ${}^2$ . That fact is present, but not immediately apparent, in Guy Fox's solution. It took me a while to learn how it was present. The integration across $z$ provides one and the integration across $r$ provides the other four.

$\text{cm}=\frac{1}{100}$

$\text{density}=2700\frac{\text{kg}}{m^3}$

$\text{wheel}=\text{Region}\left[\text{RegionUnion}\left[\text{RegionDifference}\left[\text{Cylinder}\left[\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 5\ \text{cm} \\ \end{array} \right),45\ \text{cm}\right],\text{Cylinder}\left[\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 5\ \text{cm} \\ \end{array} \right),30\ \text{cm}\right]\right], \\ \text{RegionDifference}\left[\text{Cylinder}\left[\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 5\ \text{cm} \\ \end{array} \right),12\ \text{cm}\right],\text{Cylinder}\left[\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 5\ \text{cm} \\ \end{array} \right),4\ \text{cm}\right]\right]\right]\right]$

$\text{inertia}=\text{density}\times \left(\text{MomentOfInertia}[\text{wheel},\{0,0,0\},\{0,0,1\}]\ \text{m}^5\right) \Rightarrow \frac{17879967\pi}{8000000}\text{kg}\,\text{m}^2 \approx 7.02144662170349\,\text{kg}\,\text{m}^2$