Is moment of inertia additive? If so, why doesn't adding two halves of solid box work?

I'm doing a project where I want to calculate the moment of inertia for some objects. I've broken the objects down into simple objects like cubes, spheres, cylinders, etc -- things I know the moment of inertia tensor for. My plan was to take these smaller parts and "glue them together." Since I know the inertia tensors for these simpler shapes and I thought moment of inertia was additive, I thought I could sum them and get the moment of inertia for the composite object. However, the following thought experiment has convinced me I'm mistaken.

Suppose I have a solid box (aka cuboid) with height $$h$$, width $$w$$, length $$\ell$$, and mass $$m$$. Now, suppose I cut this solid box in two, so the height is now $$\tfrac{1}{2}h$$ for each half box and similarly the mass for each is $$\tfrac{1}{2}m$$. The moment of inertia tensor for a solid box is

$$\mathbf{I}_{cube} = \begin{bmatrix} \frac{1}{12} m (h^2 + d^2) & 0 & 0\\ 0& \frac{1}{12} m (d^2 + w^2) & 0\\ 0& 0& \frac{1}{12} m (w^2 + h^2)\\ \end{bmatrix}$$

So for each half box, denoted by an index $$i\in \lbrace 1,2\rbrace$$, the tensor should be

$$\mathbf{I}_i = \begin{bmatrix} \frac{1}{12} \left(\frac{m}{2}\right) \left( \left(\frac{h}{2}\right)^2 + d^2\right) & 0 & 0\\ 0& \frac{1}{12} \left(\frac{m}{2}\right) (d^2 + w^2) & 0\\ 0& 0& \frac{1}{12} \left(\frac{m}{2}\right) \left(w^2 + \left(\frac{h}{2}\right)^2\right)\\ \end{bmatrix}$$

So now suppose I meld the two halves back together, thus restoring my initial cube. If moment of inertia is additive, then I should be able to sum $$\mathbf{I}_1$$ and $$\mathbf{I}_2$$ and get the original tensor of the cube. However, you can see this doesn't happen.

\begin{align} \mathbf{I}_1 + \mathbf{I}_2 &= 2 \begin{bmatrix} \frac{1}{12} \left(\frac{m}{2}\right) \left( \left(\frac{h}{2}\right)^2 + d^2\right) & 0 & 0\\ 0& \frac{1}{12} \left(\frac{m}{2}\right) (d^2 + w^2) & 0\\ 0& 0& \frac{1}{12} \left(\frac{m}{2}\right) \left(w^2 + \left(\frac{h}{2}\right)^2\right)\\ \end{bmatrix}\\&= \begin{bmatrix} \frac{1}{12} m\left( \left(\frac{h}{2}\right)^2 + d^2\right) & 0 & 0\\ 0& \frac{1}{12} m (d^2 + w^2) & 0\\ 0& 0& \frac{1}{12} m \left(w^2 + \left(\frac{h}{2}\right)^2\right)\\ \end{bmatrix}\\ &\ne \mathbf{I}_{cube} \end{align}\\

Why is this the case? In what sense is moment of inertia additive?

• Hint: have a look to Steiner's theorem Jul 30, 2020 at 23:20

Your $$I_1$$ is for a box that extends from $$-h/4$$ to $$+h/4$$, not one from $$0$$ to $$h/2$$. So it’s a “half-box”, but not the one you thought, and adding another one like it doesn’t give you the original box.