I am trying to determine the inertia tensor of a rectangular block. I will later be implementing this inertia tensor into the dynamics equation of a satellite body frame, which means the origin must lie on the block's center of mass. I've looked at examples for rectangular blocks and the only ones I could find have the origin of the coordinate frame on one edge of the block, and I could't find how to translate them into a center of mass based coordinate system.
Since the body is symmetric, obviously the non diagonal elements of the inertia tensor would be 0. Having said that, I calculated the diagonal elements having the frame at the edge of the block using :
$$I_{xx} = \frac{m}{3} (y^2+z^2) ; I_{yy} = \frac{m}{3} (x^2+z^2); I_{zz} = \frac{m}{3} (x^2+y^2)$$
which comes from the integral:
$$ I = \int_m (y^2+z^2)dm $$
So how I can I go about calculating these moments of inertia using the center of mass of the origin? Is it as simple as halving the distances? Or is there more to it?
Thank you in advance!
Best Answer
Method 1
By scaling. If the origin of co-ordinates $O$ is at one corner of the cuboid then you can make that point the centre of mass of a new cuboid with 8x the mass and 2x the dimensions.
Your cuboid extends from $O$ into one octant $(+x, +y, +z)$. Replicate that cuboid in the other 7 octants. The moments of inertia for this larger cuboid are 8x the previous values. The new mass is $M=8m$. The new dimensions are $x'=2x, y'=2y, z'=2z$. Therefore : $$I_{xx}'=8I_{xx}=\frac13(8m)(y^2+z^2)=\frac{1}{12}M(y'^2+z'^2) \text{ etc}$$
Method 2
Use the Parallel Axis Theorem. This relates the moment of inertia $I_O$ about any axis through the COM to the moment of inertia $I$ through any parallel axis a distance $d$ from the COM by $$I=I_O+md^2$$ Here we shift the axis from $Ox$ to a parallel axis through the centre of mass which is located at $(\frac12x, \frac12y, \frac12 z)$. The axis through the COM is a distance $d$ from axis $Ox$ where $$d^2=(\frac12y)^2+(\frac12z)^2=\frac14(y^2+z^2)$$ Therefore the new moments of inertia about the COM are $$I_{xx}'=I_{xx}-md^2=(\frac13-\frac14)m(y^2+z^2)=\frac{1}{12}m(y^2+z^2) \text{ etc}$$