Question

Find the moment of inertia through face diagonal of a cube. Edge length of a cube is $a$ units.

Answer 1

From the figure, Moment of Inertia of a square plate can be given as

$I=2\times$ (Moment of Inertia of the triangular plate)

We know, Moment of Inertia of the triangular plate is given as

$I_{tri}=\frac{1}{6}M{R}^2$

$\therefore I=2\times I_{tri}$

$I=2\times \left(\frac{1}{6}\frac{M}{2}{r}^2\right)$

From the figure,

$r=\sqrt{a^2-{\left(\frac{\sqrt{2}a}{2}\right)}^2}=\frac{a}{\sqrt{2}}$

$I=2\times \left(\frac{1}{6}\frac{M}{2}{\left(\frac{a}{\sqrt{2}}\right)}^2\right)$

$I=\frac{M{a}^2}{12}$

Now, the cube can be considered to be made up of square plates of $dm$ mass stacked upon each other till height $a$.

So, the moment of inertia of the cube about the diagonal is given by:

${\int }_0^{I}dI={\int }_0^M\frac{a^{2}dm}{12}$

$I=\frac{M{a}^2}{12}$

Hence, the answer is $\frac{1}{12}M{a}^2$.