I Axial angular momentum calculation

[cianfa72]

TL;DR Summary

About the definition of axial angular momentum and its calculation using moment of inertia tensor

Hi, I'd ask for clarification about some topics on angular momentum calculation.

Consider a system of particles moving w.r.t. a given inertial frame $\mathcal A$. Picked a point $P$ in $\mathcal A$ one defines the angular momentum vector $\vec L$ w.r.t. the reference point/pole P.

Furthermore one defines the axial angular momentum $L_a$ as the component along $a$ of the system's angular momentum $\vec L$ calculated w.r.t. a pole $P$ on the axis $a$. Such a notion is well defined since it doesn't depend on the particular pole $P$ chosen on the $a$ axis to do the calculation.

Next the notion of system's moment of inertia tensor $I$ is introduced that allows to calculate the axial angular momentum $L_a$ for the system supposed to rotate "as a rigid body" about any given $a$ axis with angular velocity $\omega$.

 

[cianfa72]

Btw, from a mathematical viewpoint, angular momentum is a covector, i.e. a linear map from $\mathbb R^3$ as vector space to the field $\mathbb R$. Therefore the moment of inertia tensor $\mathbf I$ (shorted as inertia tensor) is actually a (0,2)-rank tensor that "eats" a vector (angular velocity $\vec \omega$) and returns a covector (angular momentum $\mathbf L$).

Then when it comes to matrix representation, fixed a basis in $\mathbb R^3$ and the associated dual basis in $\mathbb R^{3*}$, the inertia tensor is represented in such a basis by a symmetric matrix $[{I}]$.

Should the moment of inertia $\mathbf I$ be associated in some way with the system's CoM?

 

[A.T.]

Should the moment of inertia $\mathbf I$ be associated in some way with the system's CoM?

it's often useful to express $\mathbf I$ wrt to the CoM, but you can use any other point.

 

[cianfa72]

it's often useful to express $\mathbf I$ wrt to the CoM, but you can use any other point.

Ok, however I'm confused from the following.

Let's say the origin $O$ of reference frame coincides with the rigid body's CoM. Suppose the body isn't symmetric about the frame $z$-axis.

Then we ask for the angular momentum $\mathbf L$ w.r.t the pole $O$ of the body when rotating with constant angular velocity $\vec \omega$ along the $z$-axis. Such an angular momentum vector $\mathbf L$ (or better covector) isn't directed along $z$ but basically rotates about it.

On the other hand, by definition of inertia tensor $\mathbf I$, we have $$\mathbf L = \mathbf I \vec \omega$$ Does this mean that inertia tensor is (or can be) function of the body's "absolute position" w.r.t. the given reference frame?

 

[A.T.]

Ok, however I'm confused from the following.

I'm not sure If I understand what you are confused about, but this might help:
https://en.wikipedia.org/wiki/Parallel_axis_theorem

 

[TSny]

Ok, however I'm confused from the following.

Let's say the origin $O$ of reference frame coincides with the rigid body's CoM. Suppose the body isn't symmetric about the frame $z$-axis.

Then we ask for the angular momentum $\mathbf L$ w.r.t the pole $O$ of the body when rotating with constant angular velocity $\vec \omega$ along the $z$-axis. Such an angular momentum vector $\mathbf L$ (or better covector) isn't directed along $z$ but basically rotates about it.

On the other hand, by definition of inertia tensor $\mathbf I$, we have $$\mathbf L = \mathbf I \vec \omega$$ Does this mean that inertia tensor is (or can be) function of the body's "absolute position" w.r.t. the given reference frame?

For a given coordinate system, the components $I_{xx}, I_{xy},$ etc. of the inertia tensor depend on the orientation of the rigid body relative to the coordinate axes. So, if the body is rotating such that the mass distribution of the body changes relative to the coordinate axes, at least some of the tensor components will change with time.

However, if you choose axes that are fixed to the body (so the axes rotate with the body), then the components of the inertia tensor relative to these axes will be time independent.

For example, consider a "tilted dumbbell" rotating about the z-axis with constant angular velocity $\boldsymbol{\omega}$ as shown:

$\boldsymbol L$ is the angular momentum of the dumbbell and $\boldsymbol L$ precesses around the z-axis. If the xyz axes are fixed in the lab, then all the inertia tensor components relative to these axes will change with time (except for $I_{zz}$).

But, if we fix the blue coordinate axes to be the dumbbell so that the coordinate system rotates around the z-axis with the dumbbell, all of the inertia tensor components relative to this body-fixed coordinate system will be time independent.

 

[cianfa72]

But, if we fix the blue coordinate axes to be the dumbbell so that the coordinate system rotates around the z-axis with the dumbbell

I don't understand what you mean by "fix the blue coordinates axes to be the dumbbell" ?

What I got is that the "new" coordinate system is supposed to rotate around the z-axis "anchored" with the dumbbell.

all of the inertia tensor components relative to this body-fixed coordinate system will be time independent.

Yes, definitely.

 

[cianfa72]

So, fix a cartesian coordinate system in the lab with origin in the dumbbell's CoM. Use $(\theta,\varphi)$ spherical coordinates to describe the orientation of the dumbbell (system) relative to the coordinate axes. Then the inertia tensor $\mathbf I$ will depend on $(\theta, \varphi)$, hence its matrix representative in the given coordinate system will have entries $I_{i,j}$ function of $(\theta, \varphi)$ as well.

In the end, let me say, the inertia tensor behaves like a (0,2)-rank tensor field since it depends on the system configuration space point (charted for instance by $(\theta, \varphi)$ spherical coordinates).

 

[TSny]

I don't understand what you mean by "fix the blue coordinates axes to be the dumbbell" ?

What I got is that the "new" coordinate system is supposed to rotate around the z-axis "anchored" with the dumbbell.

Yes, it is common in the study of rigid body dynamics to use "body-fixed" axes. These axes rotate with the body. So, the components of the body's inertia tensor relative to these axes do not change with time.

In the example I used, you could choose the blue coordinate system to rotate with the dumbbell as body-fixed axes. Often, however, the orientation of the body fixed axes is chosen to make the off-diagonal components of the inertia tensor equal to zero. This is always possible. For the dumbbell, this occurs if you choose one of the body fixed axes to be along the axis of the dumbbell.

 

[cianfa72]

As general question: does the notion of inertia tensor also extend to non-rigid systems?

 

[ergospherical]

It might make sense to think in those terms if the deformation timescale is quite long compared to the timescale of the dynamics. Maybe, to give an arbitrary example, you are studying the rotational dynamics of a galactic disk, whose density profile is changing slowly with time. Otherwise, you are moving into the realm of continuum or fluid mechanics.

 

[TSny]

As general question: does the notion of inertia tensor also extend to non-rigid systems?

For any system of particles at some instant, you could calculate $I_{xx}, I_{xy},$ etc for some coordinate system. But, I don't know of any use for doing this unless the system is a rigid body. I'm only familiar with using the inertia tensor for rigid bodies.

 

[cianfa72]

For any system of particles at some instant, you could calculate $I_{xx}, I_{xy},$ etc for some coordinate system.

Ok yes, one can calculate such matrix entries at some time $t$ in a given coordinate system using the definition of inertia tensor.

However it is useless since it is no longer true that the angular momentum $\mathbf L$ w.r.t. the system's CoM of the system rotating with angular velocity $\vec \omega$ about the CoM at time $t$ is given by $$\mathbf L = \mathbf I \vec \omega$$

But, I don't know of any use for doing this unless the system is a rigid body. I'm only familiar with using the inertia tensor for rigid bodies.

Ok, definitely.