# What units is the moment of inertia for a point mass?

What are the units used in the integral equation for the point mass moment of inertia. Specifically what are the units of the double integral over a region R of $r^2· \rho· dA$, where $r$ stands for the radius and $\rho$ for the density. The integral is the standard definition of the moment of inertia that can then be used in the kinetic energy formula. I am struggling with the units , normally a double integral can represent a volume. Can anyone understand the units here?

• Volume is a triple integral, not a double one. – John Alexiou Aug 3 '17 at 16:32

Double integral is defined as the limit of a Riemann sum:

$$\iint\limits_R f(x,y)\,\mathrm{d}A=\lim_{N\to\infty}\sum\limits_{i=1}^N f(x_i,y_i)\,\Delta A_{i}.$$

As finding limit doesn't introduce any additional dimensionful constants in the expression under limit sign, the expression's units naturally become the units of the integral.

Thus the units are the same as of the integrand, including $\mathrm dA$. And since $\mathrm{d}A$ has the dimension of area, density is surface density, so its dimension is mass per area, and radius has the dimension of length, thus the units become

$$\mathrm{m}^2\cdot\frac{\mathrm{kg}}{\mathrm{m}^2}\cdot\mathrm{m}^2=\mathrm{kg}\cdot\mathrm{m}^2.$$

Normally it is $\rm kg\times m^2$. That is for every mass or body.

As you point out, they are the result of the integral $$\int_V r^2 · \rho· dV$$, where instead of the area, you usually have the volume.

Therefore, $$[r]^2·[\rho]·[V]=m^2·\frac{kg}{m^3}·m^3=kg·m^2$$

If, instead, you were taking into account the integral over the area, the units would also be $kg·m^2$: $$[r]^2·[\rho]·[A]=m^2·\frac{kg}{m^2}·m^2=kg·m^2$$ , where I suppose that the density is also well defined as 2D.

There are four layers of (triple) integrals that are needed to describe the constituent properties of a rigid body.

$$\begin{cases} V = \iiint {\rm d}V & :{\rm [L^3]} & & \mbox{volume} \\ m = \iiint \rho\, {\rm d}V & :{\rm \frac{[M]}{[L^3]}[L^3] = [M]} && \mbox{mass}\\ {\bf c} = \frac{1}{m} \iiint \rho\, {\bf r} \, {\rm d}V & : {\rm \frac{[M][L]}{[M]} } = [L] & & \mbox{center of mass}\\ I = \iiint \rho\, \| {\bf r} \|^2 , {\rm d}V & : {\rm [M][L^2]} & & \mbox{mass moment of inertia}^\star \end{cases}$$

$\star$ The MMOI calculation in tensor form is ${\bf I} = \int -[{\bf r}\times][{\bf r}\times]\,{\rm d}m$ where $[{\bf r}\times]$ is the 3×3 skew-symmetric matrix represnting a cross product. For the purposes of unit analysis the equation shown would suffice though. Just remember there is an additional $\rm [L^2]$ term inside the integral.