The uniform solid block in Fig 10-38has mass 0.172kg and edge lengths a = 3.5cm, b = 8.4cm, and c = 1.4cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

Masses and coordinates of four particles are

By using the concept of moment of inertia we can calculate the moment of inertia about its axis. For the slab, if the axis is not passing through the center, then we can use the parallel axis theorem to calculate the moment of inertia.

1. Fortheslab when axis is passing through its center then, ${\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{=}\frac{\mathbf{M}}{\mathbf{12}}\left(a^2+b^2\right)$
2. Whentheaxis is not passing through its center, then according totheparallel axis theorem,
   $\mathit{l}\mathbf{=}{\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{+}\mathit{m}{\mathit{h}}^{\mathbf{2}}$
   

For solving parallel axis theorem, wecalculate the perpendicular distance from the axis passing through its center and corner.

$h=\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}$

According to the parallel axis theorem,

localid="1661141288241" $\begin{array}{rcl}l& =& l_{com}+M{h}^2\\ & =& \frac{M}{12}\left(a^2+b^2\right)+M\left[{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2\right]\\ & =& \frac{M}{12}\left(a^2+b^2\right)+M\left[{\left(a\right)}^2+{\left(b\right)}^2\right]\\ l& =& \frac{M}{3}\left(a^2+b^2\right)\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}l& =& \frac{0.172kg}{3}\left[{\left(3.5\times {10}^{-2}m\right)}^2+{\left(8.4\times {10}^{-2}m\right)}^2\right]\\ & =& 4.7\times {10}^{-4}kg.m^2\end{array}$

Therefore, rotational inertia about an axis through one corner and perpendicular to the large faces is, $4.7\times {10}^{-4}kg.m^2$.