I know the mathematical formula of both about an axis and a point. But where exactly we use moment of Inertia about a point. For 2d objects like discs rotating in its own plane, the moment of Inertia we use is also of the axis passing through center of mass. If we dont use moment of Inertia about a point anywhere, then whats the point of defining it?


3 Answers 3


Inertia tensor

In mechanics, for general problems in 3-dimensional space, rotational inertia of a rigid body is represented by a tensor, namely the inertia tensor $\mathbb{I}_P$.

Inertia tensor is a second-order symmetric tensor, and it's defined w.r.t. a point $P$ as,

$$\begin{aligned} \mathbb{I}_P & := - \int_{\mathbf{r} \in V} \rho (\mathbf{r} - \mathbf{r}_P)_{\times} (\mathbf{r} - \mathbf{r}_P)_{\times} \\ & = \int_{\mathbf{r} \in V} \rho \left( |\mathbf{r} - \mathbf{r}_P|^2 \mathbb{I} - (\mathbf{r} - \mathbf{r}_P) \otimes (\mathbf{r} - \mathbf{r}_P) \right) \ . \end{aligned}$$

Even without focusing on the actual expression of the inertia tensor, it's worth stressing that inertia tensor is defined w.r.t. a point, "because" this object contains the information of the rotational inertia of the object w.r.t. every axis passing through point P.


I agree with you if you say that you usually use the center of mass $G$ as the reference point for the rotational dynamics because equations have simpler expressions (and are easier to remember).

But if you've always referred rotational inertia w.r.t. the center of mass $G$, it's always possible to relate inertia w.r.t. a point $P$ with inertia w.r.t. the center of mass $G$ of the body, just writing $$\mathbf{r} - \mathbf{r}_P = \mathbf{r} - \mathbf{r}_G + \mathbf{r}_G - \mathbf{r}_P \ ,$$

to get (after using properties of the center of mass to set some terms equal to zero in the derivation)

$$\mathbb{I}_P = \mathbb{I}_G - m (\mathbf{r}_G - \mathbf{r}_P)_{\times} (\mathbf{r}_G - \mathbf{r}_P)_{\times} \ .$$

  • $\begingroup$ Actually I am studying about Inertial tensor,thanks for answering, One more thing I want to ask is that, I learnt that every body with atleast one point on it at rest have 3 principle axes of rotation. My doubt is that what does this principle axes signify? I know for a simpler bodies like sphere, the angular momentum vector and angular velocity vector are always in same direction. But i got to know that in complex 3d shapes with angular velocity having 3 components and angular momntum havng 3 components, they need not be parallel. This is eating my brain. could you please help me with this? $\endgroup$ Commented Jun 22 at 17:02
  • $\begingroup$ I'll try later. So far, I can tell you that the 3 principle axes (passing through the center of mass) are 3 orthogonal directions for which rotational dynamics "decouples", i.e. what happens around one axis doesn't influence what happens around the other two. These directions are material (or Lagrangian) properties of the rigid body, i.e. they move/rotate with it. So far, if you're ok with this answer, upvote and accept if you think it's good enough, and open another question about the meaning of principle axes. This is the way to do here, to try to keep things clean $\endgroup$
    – basics
    Commented Jun 22 at 17:08
  • $\begingroup$ Thank you so much, As you said, I willl open a different question for each of my doubt. I want to upvote but unfortunately my account doesnt fulfill upvote requirements since i am new here. $\endgroup$ Commented Jun 22 at 17:11
  • $\begingroup$ No problem at all. You should add a brief bio in your profile to get a bunch of credits (+100?) that unlocks basic permissions here. Anyway it's balance and common sense: not too many questions in a single question, not further questions in comments if they need too much space to answer (and formatting), and so on $\endgroup$
    – basics
    Commented Jun 22 at 17:17
  • $\begingroup$ Here you can find some properties of inertia tensor basics.altervista.org/test/Physics/Me/inertia.html Hand-written notes about classical mechanics. Light-blue for links $\endgroup$
    – basics
    Commented Jun 22 at 17:21

As described above, the inertia tensor is 2nd moment of the mass distribution:

$$ I_{ij} = \langle ||r||^2\delta_{ij}-r_ir_j \rangle $$

and it contains 6 numbers that describe the rotational interia about any axis.

The 6 DoF are basically 3 moments ($I_1, I_2, I_3$) about the principle axises, and the 3 components that are the rotation, $R_{ij}$, required to diagonalize it:

$$ I'_{ij} = R_{ik}I_{kl}R_{lj} = {\rm diag}(I_1, I_2, I_3) $$

If you pick an arbitrary unit axis, $\hat n$, then the (scalar) moment about that axis is:

$$ I_{\hat n} = I_{ij}n_in_j $$

  • $\begingroup$ opposite sign in first expression of $I_{ij}$. As a check, in Cartesian coordinates $I_{xx} = \int_V \rho (y^2 + z^2)$. $\endgroup$
    – basics
    Commented Jun 22 at 17:50
  • $\begingroup$ @basics that's funny because I use my personal python tensor package and calculated for a unit mass at $r=(1,0,0)$ and got $I={\rm diag}(0,-1,-1)$ and just ignored the minus signs, and then forgot about them. $\endgroup$
    – JEB
    Commented Jun 22 at 21:05
  • 1
    $\begingroup$ okok. It seems ok now. We can also delete the comments here, I guess $\endgroup$
    – basics
    Commented Jun 22 at 21:16

They are the same thing. Rotation is always around an axis.

Often people people think in $2$ dimensions instead of $3$. They talk about a flat object rotating in a plane. The axis is perpendicular to the plane. They show the point where the axis passes through the plane.


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