I know that,
$$L=I\omega$$
where $L$ is the angular momentum vector, $I$ is the inertial tensor, and $\omega$ is the angular velocity.
Now here are my doubts :-
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Before I was taught the moment of inertia tensor concept, we were taught that moment of inertia is always calculated about an axis. However this tensor matrix seems to calculate it about a point. Am I right?
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Assuming that I am correct in my previous doubt and moment of inertia is calculated about a point (the same point about which $L$ is to be found),
is the formula
$$L_{0}=I_{0}\omega$$
correct even if $0$ is not the center of mass of the body?
Assuming that it is correct about any point, is it correct even if point $0$ is moving with any velocity and acceleration (say uniform circular motion)?
Best Answer
If rotation is restricted about a fixed axis, then there is a single mass moment of inertia component associated with this axis. It is defined as
$$\text{(angular momentum)} = \text{(mass moment of inertia)} \text{(rotation)}$$
$$ L_{\rm axis} = I_{\rm axis} \omega_{\rm axis} \tag{1}$$
The rotation vector is really just a direction and a magnitude. Any single MMOI component relates only the magnitude along a specified direction to the resulting angular momentum.
To generalize this problem, you can take all possible rotation directions and describe the resulting angular momentum vectors as a mass moment of inertia tensor. Mathematically this is a 3×3 matrix that transforms a 3×1 rotation vector into a 3×1 angular momentum vector
$$ \boldsymbol{L} = \mathbf{I}\, \boldsymbol{\omega} $$ $$ \pmatrix{L_x \\ L_y \\ L_z} = \begin{vmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{vmatrix} \pmatrix{\omega_x \\ \omega_y \\ \omega_z } \tag{2}$$
Now in the general sense, the rotation vector is not associated with a particular location. I mean the components of $\boldsymbol{\omega}$ do not change from point to point, like the components of linear velocity $\boldsymbol{v}$. So in the above equation rotation only defines direction and magnitude.
But angular momentum is defined at a point. Meaning, that if expressed at a different location the components change in a manner similar to velocities and torques:
$$ \begin{aligned} \boldsymbol{v}_A & = \boldsymbol{v}_B + \boldsymbol{r} \times \boldsymbol{\omega} & & \text{transformation of velocities} \\ \boldsymbol{L}_A & = \boldsymbol{L}_B + \boldsymbol{r} \times \boldsymbol{p} & & \text{transformation of angular momentum} \\ \boldsymbol{\tau}_A &= \boldsymbol{\tau}_B + \boldsymbol{r} \times \boldsymbol{F} & & \text{transformation of torque} \\ \end{aligned} \tag{3}$$
Here the vector $\boldsymbol{r}$ goes from $A \rightarrow B$, and $\boldsymbol{p}$ is linear momentum.
So the definition of mass moment of inertia needs to include locational information, making equation (2) incorrect unless the location is specified somehow.
$$ \boldsymbol{L}_A = \mathbf{I}_A \,\boldsymbol{\omega} \tag{4}$$
and transformation from one point to another is done using the parallel axis theorem
$$ \begin{aligned} \boldsymbol{L}_A & = \boldsymbol{L}_B + \boldsymbol{r} \times \boldsymbol{p} \\ & = \boldsymbol{L}_B + \boldsymbol{r} \times (m \boldsymbol{v}) \\ & = \mathbf{I}_B\, \boldsymbol{\omega} + \boldsymbol{r} \times (m \boldsymbol{\omega} \times \boldsymbol{r}) \\ \boldsymbol{L}_A &= \mathbf{I}_A\,\boldsymbol{\omega} \end{aligned} \tag{5}$$
where $$\mathbf{I}_A = \mathbf{I}_B - m [\boldsymbol{r}\times] [\boldsymbol{r} \times] $$
and $[\boldsymbol{r}\times] = \begin{vmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{vmatrix} $ is mathematical construct to represent cross products. It is called the cross product matrix operator.
In summary: