Angular Momentum vs Moment of Inertia

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?

Answer

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.

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.

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.}