# What does it mean to take a mass moment of inertia about a single point?

Shows the following: How is it possible to define a mass moment of inertia about the origin which is a point and not an axis. What is the direction of rotation referred to?

What situation would this represent?

consider total angular momentum of a single particle as a vector, denoted by $$\vec{L}$$ , by definition $$\vec{L} \equiv \vec{r} \times \vec{P}$$ componentwise: \begin{aligned} L^i & =(r \times P)^i \\ & =\varepsilon^{i j k} r_j P_k \end{aligned} but $$\vec{P}=(\vec{\omega} \times \vec{r})m$$ , so its $$k$$th component is: $$P_k=(\varepsilon_{klm}\omega^l r^m)m$$ so we have: \begin{aligned} L^i & =\varepsilon^{i j k} \varepsilon_{k l m} r_j r^m \omega^l m \\ & =\left(\delta_l^i \delta_m^j-\delta_m^i \delta_l^j\right) r_j r^m \omega^l m \\ & =\left(r_j r^j \delta_l^i-r^i r_l\right) \omega^l m \\ & =\left(|r|^2 \delta_l^i-r^i r_l\right) m \omega^l \end{aligned} so it's natural to introduce a tensor(called inertia tensor) by: $$I_j^i=\left(|r|^2 \delta_j^i-r^i r_j\right) m$$ then we can express angular momentum in a tensorial form: $$L^i=I_j^i \omega^j$$
for continuous distribution, simply replace m by $$\rho d^3x$$ and do integral