My task is to compute the moment of inertia and the radius of gyration of a constant-density tetrahedron defined by $x,y,z\ge0$ and $\frac xa + \frac yb + \frac zc \leq 1$ about the x-axis.
I know that the moment of inertia, $I$, of an object is defined by: $I=\int_RdI$ over some region of integration R. Here, I'd let R be the volume of the tetrahedron. Then, $dI = r^2dM$ for mass element $dM=ρ(x,y,z)dV$ for some density constant density $ρ(x,y,z) = k$. So, we should have:
$I = \int_0^c \int_0^{b- \frac bcz} \int_0^{a- \frac aby – \frac acz} ρ(x,y,z)r^2dxdydz$
$r^2$ should represent the distance from any point in the tetrahedron to the x-axis, so we can (I think) let $r^2=y^2+z^2$. Hence:
$I = k\int_0^c \int_0^{b- \frac bcz} \int_0^{a- \frac aby – \frac acz}(y^2+z^2)dxdydz$.
From here, I would compute $I$ and then I would be able to calculate the radius of gyration, $r_g$, where $r_g=\sqrt{\frac IM}$.
This integral looks unnecessarily complicated and I'm not sure if I set it up correctly. Any idea where I might have gone wrong?
Best Answer
The set up
$$I = k\int_0^c dz \int_0^{b- \frac bcz} dx\int_0^{a- \frac aby - \frac acz}(y^2+z^2)\,dy$$
seems correct to me, the integration is maybe quite long but it shouldn't be complicated since we are dealing with polynomials integration.