Moment of Inertia of a cube (arbitrary axis)

Nice problem. Consider a point on the line along the given direction and passing through the origin:

$\vec{r}_L = t\left(\hat{i} + 4 \ \hat{j} + 3 \ \hat{k}\right)$

Any point within the cube is:

$\vec{r}_P =x \ \hat{i} + y \ \hat{j} + z \ \hat{k}$

Vector joining these two points is:

$\vec{r}_D = \vec{r}_P - \vec{r}_L$

Since we consider perpendicular distances from the axis to compute moments of inertia, the vector $\vec{r}_D$ must be perpendicular to the line:

$\implies \vec{r}_D \cdot \left(\hat{i} + 4 \ \hat{j} + 3 \ \hat{k}\right)$

This gives:

$t = \frac{x}{26}+\frac{2\,y}{13}+\frac{3\,z}{26}$

So finally, the vector $\vec{r}_D$ in terms of $x$ , $y$ and $z$ is:

$\vec{r}_D= \left(\frac{2\,y}{13}-\frac{25\,x}{26}+\frac{3\,z}{26}\right) \ \hat{i} + \left(\frac{2\,x}{13}-\frac{5\,y}{13}+\frac{6\,z}{13}\right) \ \hat{j} + \left(\frac{3\,x}{26}+\frac{6\,y}{13}-\frac{17\,z}{26}\right) \ \hat{k}$

A volume mass element is:

$dM = \frac{1}{0.2 \times 0.2 \times 0.2} \ dx \ dy \ dz$ $\implies dM = 125 \ dx \ dy \ dz$

The moment of inertia is therefore:

$I = \int_{V} \lvert \vec{r}_D \rvert^2 dM$ $I = 125\int_{-0.1}^{0.1} \int_{-0.1}^{0.1} \int_{-0.1}^{0.1} \lvert \vec{r}_D \rvert^2 \ dx \ dy \ dz$

Leaving out further evaluations, the result comes out to be:

$I = \frac{1}{150} \approx 0.006667$

Perform a triple integral over the cube interior. Determine the distance from each infinitesimal mass to the axis using the formula at this Wolfram Mathworld link . The formula for $d^2$ in terms of cross products near the end is particularly useful. The moment comes out to $\approx 0.006666$

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import math

Num = 400

M = 1.0
S = 0.2

V = S**3.0
rho = M/V

dx = S/Num
dy = dx
dz = dx

dV = dx*dy*dz
dm = rho*dV

#####################################

# Two points used in distance calcs

x1 = 0.0
y1 = 0.0
z1 = 0.0

x2 = 1.0
y2 = 4.0
z2 = 3.0

#####################################

I = 0.0

x = -S/2.0

while x <= S/2.0:

    y = -S/2.0

    while y <= S/2.0:

        z = -S/2.0

        while z <= S/2.0:

            v1x = x2 - x1
            v1y = y2 - y1
            v1z = z2 - z1

            v2x = x1 - x
            v2y = y1 - y
            v2z = z1 - z

            v1sq = v1x**2.0 + v1y**2.0 + v1z**2.0

            crossx = v1y*v2z - v1z*v2y
            crossy = -(v1x*v2z - v1z*v2x)
            crossz = v1x*v2y - v1y*v2x

            dsq = (crossx**2.0 + crossy**2.0 + crossz**2.0)/v1sq

            dI = dm*dsq

            I = I + dI

            z = z + dz

        y = y + dy

    x = x + dx

#####################################

print Num
print I

#>>> 
#100
#0.00666653846154
#>>> ================================ RESTART ================================
#>>> 
#200
#0.00666663461539
#>>>
#>>> 
#400
#0.00666665865385
#>>> 

Let the position vector of a point in the cube be $r = [x, y, z]^T$ , and let the unit vector of the direction of the axis be $n = \frac{1}{\sqrt{26}} [1, 4, 3]$ .

Then the required moment of inertia $I$ is defined by

$I = \rho \displaystyle \int_V | u |^2 dV$

where $\rho$ is the density of the cube, $\rho = \dfrac{m}{V}$ , and $u$ is the vector perpendicular to $n$ from the point $r$ . Using the standard projection formula, we obtain,

$u =\ (I - n n^T ) r$

Hence,

$I = \rho \displaystyle \int_V r^T (I - n n^T)(I - n n^T) r dV = \rho \int_V r^T (I - n n^T) r dV$

Further, this becomes,

$I = \rho \displaystyle \int_V r^T r - r^T n n^T r dV = \rho \int_V r^T r - n^T r r^T n dV = \rho \int_V r^T r dV - \rho n^T \int_V r r^T dV n$

Now $r^T r = x^2 + y^2 + z^2$

and

$r r^T = \begin{bmatrix} x^2 && xy && xz \\ xy && y^2 && yz \\ xz && yz && z^2 \end{bmatrix}$

From symmetry (about the origin), we have

$\displaystyle \int_V x^2 dV = \int_V y^2 dV = \int_V z^2 dV = \dfrac{1}{12} V a^2 = I_0$

where a is the side length of the cube. The result in the last equation is straight forward to compute. In addition, we have, also from symmetry,

$\displaystyle \int_V xy dV = \int_V xz dV = \int_V yz dV = 0$

Thus the moment integral boils down to,

$I = \rho ( 3 I_0 - I_0 n^T n ) = 2 \rho I_0 = \dfrac{1}{6} m a^2$

Substituting $m = 1$ and $a = 0.2$ , gives $I = \frac{0.04}{6} \approx \boxed{0.00666}$