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?

  • $\begingroup$ Hint: have a look to Steiner's theorem $\endgroup$
    – FGSUZ
    Jul 30, 2020 at 23:20

2 Answers 2


It looks like you are not adjusting the center of rotation after cutting. After cutting, you calculate the moment of inertia about the center of each of the cut pieces. Once you glue them back, the total moment of inertia is the sum of the two pieces, plus the inertia from rotating each of them off-center.


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.

With a moment of inertia tensor, you have to keep in mind where the origin is.


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