Moment of inertia rotational mechanics

Question

Moment of inertia of cube about body diagnaol is $ma^2 /6$. Moment of inertia of cube about any axis passing through centre of mass is same only. Is this result correct? If yes how do we prove that $MI$ of cube about any axis passing through center of mass is $ma^2/6$?

Answer 1

If the cube is rotating about an any of the three axes passing through the center of oposing faces then symmetry shows that the resulting angular momentum vector is parallel to the angular velocity vector. Thus those there directions are eigenvectors of the inertia tensor. Now, again, by symemtry all three of the eigenvalues are equal so in the basis using these three axes the intertia tensor is proprtional to the identity matrix. If a cartesian tensor is proportional to the identity matrix in one orthogonal frame it is numerically the same matrix in any orthogonal frame. Thus a cube has the same rotational inertia properties as a sphere. In particlular it's moments of inertia about any axis are all the same.

Answer 2

Consider a general direction vector (in longitude/latitude form)

$$\vec{n} = \pmatrix{ \cos \varphi \cos\psi \\ \sin\psi \\ \sin\varphi \cos\psi } $$ and the axis aligned mass moment of inertia tensor $$I = \pmatrix{ \frac{m}{6} a^2 & & \\ & \frac{m}{6} a^2 & \\ & & \frac{m}{6} a^2 } $$

The mass moment of inertia value felt about the direction $\vec{n}$ is

$$ j = \vec{n}^\intercal I\, \vec{n} $$

Doing the linear algebra and using some trigonometric identities you find that

$$ \begin{multline} j = \frac{m}{6} a^2 \cos^2 \varphi \cos^2 \psi + \frac{m}{6} a^2 \sin^2 \psi + \frac{m}{6} a^2 \sin^2 \varphi \cos^2 \psi = \\ = \frac{m}{6} a^2 \left(\cos^2\varphi+\sin^2\varphi \right) \cos^2 \psi + \frac{m}{6} a^2 \sin^2 \psi = \frac{m}{6} a^2 \end{multline} $$

So regardless of the direction, the MMOI is $\boxed{ j =\frac{m}{6} a^2 }$.