What is the moment of inertia of a cube with uniform density about any edge?

Answer 1

Think of it as the difference in moment of inertias for two solid cubes. Calculate the moment of inertia of a solid cube with dimensions equal to the inner dimensions of your hollow cube. Then calculate the moment of inertia of a solid cube with dimensions equal to the outer dimensions of your hollow cube. Subtract the moment of inertia of the inner dimensions from the moment of inertia of the outer dimensions to get the moment of inertia of what's left. Same concept applies to finding the area of a thin-walled circle. Outer area - inner area = total area. Outer moment of inertia - inner moment of inertia = total moment of inertia.

This approach won't work however if you're considering hollow shell - a cube with walls of zero thickness.

If the axis of rotation goes through the cube center, perpendicular to one of its walls, first calculate moment of inertia of the wall that the axis passes through (let's call it Ia).

For all equations below d equals surface density(mass per unit of area) and a is length of cube's side.

Ia= d * a4 / 6

Then you have to calculate moments of inertia of four walls parallel to the axis.

This will be Ib=4 * Iwall=4*d*a4/3.

Total moment of the shell will be then:

I=2*Ia+Ib=1.5*d*a4.

If the axis is through the center and ┴ one face, I = (m/6)*[a² - (a-t)²], or

I = (m/6)(2at - t²) for any value of t, however small.

Source: CRC Std Math Tables

Answer 2

The parallel axis theorem says, I = Io+md^2. We know Io of a cube (side length a) with an axis parallel to a side and through the centroid is 1/6*m*a^2. The distance d to the axis is a/2*sqrt(2). This yields the moment of inertia along the axis of 2/3*m*a^2.

Answer 3

if side length is a then (2/3)ma^2

Answer 4

Ma