According to Wikipedia to prove $I=I_{cm}+md^2$:

$I_\mathrm{cm} = \int (x^2 + y^2) \, dm.$ $$I = \int \left[(x + d)^2 + y^2\right] \, dm$$

Expanding the brackets yields

$$I = \int (x^2 + y^2) \, dm + d^2 \int dm + 2d\int x\, dm.$$ The first term is $I_{cm}$ and the second term becomes $md^2$. The integral in the final term is the $x$-coordinate of the centre of mass, which is zero by construction.

I'm confused as to why the third term is 0, probably because I don't understand how the $dm$ mass axis of integration is defined.

  • 1
    $\begingroup$ If it helps, you can think of it as $\int x dm=\int \int \int x \rho(x,y,z)dx dy dz $ $\endgroup$
    – user126422
    Jun 11, 2017 at 15:51

2 Answers 2


It's zero by definition of the centre of mass.

We define the centre of mass as the unique point $\vec{r}_{CM}$ where $\int dm (\vec{r}-\vec{r}_{CM}) = 0$.

Then we define the moment of inertia about an axis passing through the point $O$ with position vector $\vec{r}_O$ as:

$I_O = \int dm (\vec{r} - \vec{r}_O)^2$

In this expression we add and subtract $\vec{r}_{CM}$ to get:

$I_O = \int dm (\vec{r} - \vec{r}_{CM} + \vec{r}_{CM} - \vec{r}_O)^2$

Relabelling $\vec{r}_{CM} - \vec{r}_O = \vec{d}$ gives $I_O = I_{CM} + Md^2 + 2\vec{d} \cdot \int dm (\vec{r}-\vec{r}_{CM})$

The last integral is the one you had represented as $\int dm x$ Presumably your $x$ is defined relative to the centre of mass, I've just made this explicit by having everything defined in terms of vectors with an arbitrary origin. The last integral vanishes by comparison to the definition of centre of mass at the beginning.

P.S. all the integrals with respect to mass are defined as $\int dm = \int dV \rho $ where both integrals are taken over the body in question and $\rho$ is the mass density in the volume considered. (Or alternatively, they're defined over all space, and the fact that $\rho$ vanishes when you're not in the body causes those contributions to vanish).


Here's a much simpler explanation without vectors. Let the new axis (i.e., the axis that's at a distance $d$ from the CoM) pass through the origin. In this frame, let the CoM be at $x$. We simply need to evaluate the integral for the moment of intertia in this frame.

$I = \int_{-r}^{r}(x+d)^2dm\\ I = \int_{-r}^{r}(x^2)dm + \int_{-r}^{r}(d^2)dm + 2d\int_{-r}^{r}(x)dm$

As you can see, the third term goes to zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.