Pretty sure that this question has already been answered in this site, but I cannot find it. Anyway, here's the question: What is the difference between angular momentum and moment of inertia?

  • $\begingroup$ I won't be surprised if this question is marked as duplicate, but I just need a link to the question. $\endgroup$ Jan 25, 2019 at 9:15

3 Answers 3


Angular momentum is the "moment of momentum", meaning it gives us an idea of how far is the linear momentum vector applied at. Torques involve the moment arm of a force, and angular momentum involves the moment arm of momentum.

Particle Mechanics

Take a single particle moving in a straight line (in the absence of external forces). It has mass $m_i$, it is located at vector $\boldsymbol{r}_i$ with velocity vector $\boldsymbol{v}_i$. This sets us up for the following definitions

  • Linear Momentum of particle, $$ \boldsymbol{p}_i = m_i \boldsymbol{v}_i $$
  • Angular Momentum about the origin, $$ \boldsymbol{L}_i = \boldsymbol{r}_i \times \boldsymbol{p}_i $$

The above is sufficient to recover the location of the path line, at least the point on the path of the particle closest to the origin.

$$ \boldsymbol{r}_{\rm path} = \frac{ \boldsymbol{p}_i \times \boldsymbol{L}_i }{ \| \boldsymbol{p}_i \|^2} $$

You can easily prove this if you show that $\boldsymbol{L}_i = \boldsymbol{r}_{\rm path} \times \boldsymbol{p}$, which you do with vector triple product identity $\boldsymbol a\times( \boldsymbol b \times \boldsymbol c) = \boldsymbol b (\boldsymbol a \cdot \boldsymbol c) - \boldsymbol c ( \boldsymbol a \cdot \boldsymbol b)$.

So in summary, Angular momentum describes the (permendicular) distance where linear momentum acts. The conservation of angular momentum law means that not only linear momentum as a vector is conserved, but also the location of this vector (or the line in space the vector acts through) is conserved.

Rigid Body Mechanics

When you extend the above to multiple particles clumped together as a rigid body the concept of moment of inertia arises. First off, Charles's theorem state that for all the distances to be maintained, each particle can only move with a combination of translation and rotation (with vector $\boldsymbol{\omega}$) about a common axis. Commonly the center of mass is used as a reference point, and so the motion of each particle in the body is restricted to $\boldsymbol{v}_i = \boldsymbol{v}_{\rm com} + \boldsymbol{\omega} \times \boldsymbol{r}_i$.

Commonly the motion is decomposed to the translation of the center of mass, and the rotation about the center of mass. This yields the following relationships

  • Linear Momentum of rigid body, $$\boldsymbol{p} = \sum \limits_i m_i \boldsymbol{v}_i = m \boldsymbol{v}_{\rm com}$$

  • Angular Momentum about center of mass, $$\boldsymbol{L}_{\rm com} = \sum \limits_i \boldsymbol{r}_i \times m_i \boldsymbol{v}_i = \sum \limits_i m_i \boldsymbol{r}_i \times ( \boldsymbol{\omega} \times \boldsymbol{r}_i ) $$

Mass moment of inertia

In order to understand angular momentum of a rigid body rotating about the center of mass better, it is common to separate the geometry parts from the motion parts

$$ \boldsymbol{L}_{\rm com} = \underbrace{ \mathrm{I}_{\rm com} }_{\text{geometry}} \;\;\underbrace{ \boldsymbol{\omega}}_{\text{motion}} $$ where

$$ \mathrm{I}_{\rm com} = \sum_i (-m_i [\boldsymbol{r}_i \times][\boldsymbol{r}_i \times]) = \sum_i m_i \left| \matrix{ y^2+z^2 & - x y & - x z \\ - x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2} \right| $$

This is the mass moment of inertia tensor. It is the rotational equivalent to mass, since $\boldsymbol{p} = m \boldsymbol{v}_{\rm com}$ and $\boldsymbol{L}_{\rm com} = \mathrm{I}_{\rm com} \boldsymbol{\omega}$ have a similar form.

So, mass moment of inertia describes how far away from the rotation axis is the mass distributed at. It conveys the geometry information of angular momentum. So if a known mass moment of inertia about an axis $I$ is described by $I = m r ^2$, it means the geometry of the problem is similar to that of a mass ring with radius $r$.

More details in this similar answer.

  • $\begingroup$ Thank you so much! But can you please explain the mass moment of inertia tensor? I don't really get it. $\endgroup$ Jan 25, 2019 at 18:27
  • $\begingroup$ A tensor (or matrix) is multiple values combined into one quantity. Mass is a scalar quantity, but the mass moment of inertia is not. Push an object along various directions and you will feel the same mass regardless. Try to rotate a body about various directions and the inertial resistance you feel varies. This is expressed by the 3×3 symmetric inertia tensor $$\mathrm{I} = \left| \matrix{ I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} } \right| $$ $\endgroup$ Jan 25, 2019 at 21:06

Eventhough @ja72 contribuited with his answer analysing the physics of rotational physics, I would like to contribuite a bit in a more mathematical sense between the concept of moment of inertia and the mathematical object called tensor. Actually, the moment of inertial is the "freshman" encounter with tensors, because it's appearence is natural in Newtonian Physics.

It's well-known that the relationship between the angular momentum $\vec{L}$ of a particle and the angular velocity $\vec{\omega}$ is then a linear relationship:

$$ \vec{L} = \mathcal{I} \vec{\omega} \tag{1}$$

But this case is a particular case when the rotational system do not have the contribuition of moment of inertia from other directions. Physically speaking this system can rotate only around $z$-axis ($\hat{k}$ direction) or $y$-axis ($\hat{j}$ direction) or $x$-axis ($\hat{i}$ direction) and also rotating around a direction of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$

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But in a arbitrary rotational motion, the contribuition of the moment of inertial of other axis will be appreciable and then the linear relationship $(1)$ have no meaning.

Instead the relationship is performed by another kind of physical quantity called Inertia Tensor:

$$ \mathfrak{I} = \sum^{3}_{i=1}\sum^{3}_{i=1} I^{ij} \vec{e}_{i}\otimes\vec{e}_{j} \tag{2}$$

Where components $I^{ij}$, are called components of inertia tensor. These components have a matrix representation called Inertia matrix or (unfortunately) Inertia Tensor:

$$ I^{ij} = \sum^{3}_{i=1}\sum^{3}_{j=1} m[\delta^{ij} \mid\mid \vec{r}\mid\mid^{2} - r^{i}r^{j}] \tag{3}$$

The whole object $(2)$ is the real inertia tensor and tensors (in this case of rank 2) are objects which lies on a vector space called Tensor Product $V\otimes W$ (please keep this for all your academic life: the best (rapid) answer to the question "what is a tensor?", is "it's a element of tensor product".

Then, when we apply the tensor on angular velocity vectors, we get the vector called angular momentum:

$$ \vec{L} = \mathfrak{I}(\vec{\omega}) \tag{4}$$



In fact there's a minor technical (but crucial point) consideration on $(4)$: to perform a linear transformation like $(4)$, we have to define the object which have tensor components $I^{i}\hspace{0.3mm}_{j}$, i.e., an mixed tensor:

$$ \mathfrak{I} = \sum^{3}_{i=1}\sum^{3}_{j=1} I^{i}\hspace{0.3mm}_{j} \vec{e}_{i}\otimes\vec{e}^{j} \tag{5}$$

And then (in component form) we perfom a operation called contraction, which result the components of angular momentum vector:

$$L^{i} = \sum^{3}_{j=1}I^{i}\hspace{0.3mm}_{j} \omega^{j} \tag{6}$$

$$ * * * $$

Some references (For Tensors):

$[1]$ Steven Romam's Advanced Linear Algebra

$[2]$ Joshi's * Matrices and Tensors in Physics *

$[3]$ McMahon's Relativity Demystified

$[4]$ Lang's Linear Algebra


The moment of inertia is a measure of how a mass of a rotating body is distributed about the axis of rotation, and its symbol is $I$. The angular momentum is the mass times the cross product of the vectors $r$ and $v$, or the product of the moment of inertia and the angular velocity $\omega$. Its symbol is L.

  • 2
    $\begingroup$ Actually, Michael, the moment of inertia is a measure of how difficult it is to produce angular acceleration in a rotating body that is given a specified torque. Angular momentum is the rotating equivalent of linear momentum. For linear momentum, $p=mv$. The rotational equivalent of mass is $I$ and the rotational equivalent of velocity is $\omega$, so angular momentum is given by $L=I\omega$. $\endgroup$ Jan 25, 2019 at 19:23

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