Snake eyes!

We know that the moment of inertia of a cube is, $I= (ma^2)/6$ ,through an axis passing through its centre and parallel to two opposite edges. By the parallel axis theorem, the moment of inertia through one of the edges is $Ix=I+md^2$ , where d is the distance of the edge from the centre of the cube.Now, $d^2=(a^2)/2$ .Therefore $Ix=(ma^2)/6+(ma^2)/2=(2ma^2)/3$ .Putting the values of m and a from the given data,we get $Ix=20$ .

The Moment of Inertia of a cube rotating about an axis passing through two face square centers is m*(a^2)/6

The axis along the edge is a distance of d = a/(/sqrt{2})

By the Parallel Axis Theorem, the Moment of Inertia along the Edge is equal to: Moment of Inertia Along the Face center axis (m (a^2)/6) + M (d^2)

Since this is a perfect cube ( consist of 12 edges) and its mass is distributed, it means every edges get 2,5 grams ( 30 / 12 = 2.5 grams) so that every vertices has the same mass as the edges.

From the question above, an axis passes through one of the edges(it means, 2 vertices will be passed through too). By using the formula below we can estimate the cube's inertia:

I = m(R 1)^2 + m(R 2)^2 + ...... + m(R_12)^2

Since 2 vertices are in axis, their inertia won't be calculated ( R = 0 ) and another 2 vertices, their distance to axis will be diagonally or (R = √2 ) finally :

I = 2.5 ×4 + 4 × 2.5 = 20 kg m^2

The coordinate frame of reference is assumed to be attached at a corner of the cube. The axis of rotation is assumed to be the x- axis.

Consider a differential volume element at an arbitrary position ${x,y,z}$ . It's mass dm would be given by ${d}m= \rho *{d}x*{d}y*{d}z.$

The distance of this element from the axis of rotation is $r= \sqrt(y^2 + z^2)$

The moment of inertia of the cube about this axis is then

$I=\int\limits_0^1 \int\limits_0^1 \int\limits_0^1 \mathrm{r}^{2}\,\mathrm{d}m$

Also $\rho = m/V$ where V is the total volume of the cube and m is it's mass

There are several ways to solve this problem, and the answer is $I=2ma^2/3=20~\mbox{g/cm}^2$ . Consider an axis (let's call the value of moment inertial along it equal to $I_o$ ) that is parallel with one of the edges and passes through the center of the dice (the cube), using the theorem of parallel axes one can get the relation $I_o=I-ma^2/2$ . Now let's think about the cube with the same mass density of $2$ times bigger in the length scale: the mass is $8$ times heavier and each edges are $2$ times longer. The moment inertial along an axis that passes through the center of mass and parallel with one of the edges of that new big dice is just equal to $8I$ (one can divide it into $8$ cubes) and also equal to $32I_o$ (using dimensional analysis), so finally one gets:

$I=4I_o=4(I-ma^2/2) \Rightarrow I = 2ma^2/3 =20~\mbox{g/cm}^2$