Problem in understanding angular momentum of a rigid body

  • #1
MatinSAR
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Homework Statement
Please see below.
Relevant Equations
##L_i= \sum_j I_{ij} \omega_{j}##
Hello. I am reading Classical dynamics of particles and systems(Book by Stephen Thornton), I have problem in understanding the coordinate system they choose to define angular momentum for a rigid body. At the beginning of the chapter 11 they say:
1716899333445.png

They use 2 coordinate systems to describe motion of a rigid body. One is the fixed coordinate system. In this system we can use Newton's law of motion without any problem. The second coordinate system is fixed with respect to the body. In the inertial reference frame, the rigid body and the second coordinate system move in sync, causing them to remain stationary relative to each other. So if a rigid body consist of ##n## particles with mass of ##m_{\alpha}## in this system the particle ##\alpha## doesn't move. I can say that it doesn't have momentum in this system because it is fixed with respect to it.
Following that, they demonstrate the process of deriving an equation for the kinetic energy of a rigid body.
1716900128269.png

If we make the origin of the body corrdinate system coinside with center of mass of the body, it results in:
1716900314233.png

Up to this point, everything is clear. The issue arises in the next section, where they introduce the definition of angular momentum for a rigid body.
1716900508532.png

I cannot understand this. If point O is fixed in the body coordinate system, it could be any point on the body because the body doesn't move in the body coordinate system. Therefore, the angular momentum of this rigid body with respect to a fixed point in the body coordinate system should be 0! I believe the angular momentum should be defined with respect to the fixed coordinate system rather than a fixed point in the body coordinate system. Am I wrong?

Any help would be appreciated.

Edit 1: I edited a typo.
 
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  • #2
MatinSAR said:
Homework Statement: Please see below.
Relevant Equations: ##L_i= \sum_j I_{ij} \omega_{j}##

I cannot understand this. If point O is fixed in the body coordinate system, it could be any point on the body because the body doesn't move in the body coordinate
Even if O is fixed, the rest of the body would rotate around O.
 
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  • #3
anuttarasammyak said:
Even if O is fixed, the rest of the body would rotate around O.
But I think If point O is fixed in body coordinate system it should rotate with body so the body should be at rest with respect to the point O and it shouldn't have momentum.
 
  • #4
MatinSAR said:
Any help would be appreciated.
I'm not clear about this myself, so won't attempt a direct answer. But maybe this will help.

Let me quote from this link: https://galileoandeinstein.phys.virginia.edu/7010/CM_27_Eulers_Equations.html
“..but we’re looking at the changing angular momentum, why isn’t the angular momentum just zero in a frame in which the body is at rest? And the angular velocity, too?“​
This seems to be the same issue that you are having. The above link attempts to explain.

Also see discussion here: https://physics.stackexchange.com/questions/32787/motion-in-the-body-fixed-frame
 
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  • #5
Steve4Physics said:
I'm not clear about this myself, so won't attempt a direct answer. But maybe this will help.

Let me quote from this link: https://galileoandeinstein.phys.virginia.edu/7010/CM_27_Eulers_Equations.html
“..but we’re looking at the changing angular momentum, why isn’t the angular momentum just zero in a frame in which the body is at rest? And the angular velocity, too?“​
This seems to be the same issue that you are having. The above link attempts to explain.

Also see discussion here: https://physics.stackexchange.com/questions/32787/motion-in-the-body-fixed-frame
Thank you for sharing the links. The first one is still unclear. I will check the second link soon
 
  • #6
Steve4Physics said:
I'm not clear about this myself, so won't attempt a direct answer. But maybe this will help.

Let me quote from this link: https://galileoandeinstein.phys.virginia.edu/7010/CM_27_Eulers_Equations.html
“..but we’re looking at the changing angular momentum, why isn’t the angular momentum just zero in a frame in which the body is at rest? And the angular velocity, too?“​
This seems to be the same issue that you are having. The above link attempts to explain.

Also see discussion here: https://physics.stackexchange.com/questions/32787/motion-in-the-body-fixed-frame
One of the answers from the second link that is somewhat logical:

I fussed about this as well. My resolution: for these calculations the fixed-body frame is not to be considered as co-moving with the body, but rather a non-rotating frame that instantaneously aligns with the body.
The Euler angles translate between the body and the space frames. The Euler angles are indeed functions of time, and the fixed-body frame is as well, but angular velocity and momentum are measured with respect to a fixed "snapshot" of the body frame at a particular time.

It's interesting that they didn't make this clear in their book. I hope to be able to ask this question to my professor because he uses this book as the main reference.

@Steve4Physics and @anuttarasammyak Thank you for your time.
 
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  • #7
Rigid Body Rotation.png
Here is what I think. The axis of rotation of a rigid body is a straight line consisting of all points that are fixed in space relative to a non-rotating inertial frame, while the rest of the points rotate with angular velocity ##\vec {\omega}## about it. In the picture on the right the rotation axis is the dotted line. The primed coordinate system is rotating with the body except for the origin O which is on the axis.

Point O, as all other points on th dotted line, is not rotating by definition. not only because it is on the axis, but also because, as a mathematical point, it has zero extent. You cannot tell if it is rotating or not. If you are an observer at point O and at rest with respect to it, you will see all the points in the rigid body (except those on the dotted line) move with respect to you. The only line that you see fixed on the rigid body is the dotted line, the axes will be rotating.

That's because you have to think of yourself as a "point" observer at rest on the dotted line. It is only human to think of ourselves as extended beings with points off-axis and that's where the fallacy understandably occurs, in my opinion.
 
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  • #8
@MatinSAR: this is a subtlety, but the answer is not too complicated. In brief, the angular momentum is measured with respect to the space-fixed frame, however it's just more convenient to express our results in terms of components in the body-fixed basis (since, most importantly, the components of the moment of inertia tensor in the body-fixed frame are constants!!)

In specific terms, imagine a rigid body rotating around a fixed point ##O##. The angular momentum measured about this point in the space-fixed frame is just\begin{align*}
\mathbf{L} &= \sum \mathbf{r} \times \mathbf{p} \\
&= \sum m \mathbf{r} \times (\boldsymbol{\omega} \times \mathbf{r}) \\
&= \sum m \left( |\mathbf{r}|^2 \boldsymbol{\omega} - (\mathbf{r} \cdot \boldsymbol{\omega}) \mathbf{r} \right)
\end{align*}So far, the angular momentum is still just that which is measured in the space-fixed frame. However, the key step is that we'll now just expand the vectors ##\boldsymbol{\omega}## and ##\mathbf{r}## in terms of the body-fixed basis, something like:\begin{align*}
\boldsymbol{\omega} = \bar{\omega}_i \bar{\mathbf{e}}_i \\
\mathbf{r} = \bar{x}_i \bar{\mathbf{e}}_i
\end{align*}If you plug this in, you end up with$$\mathbf{L} = \left[ \sum m \left( |\mathbf{r}|^2 \delta_{ij} - \bar{x}_i \bar{x}_j \right) \right] \bar{\omega}_j \bar{\mathbf{e}}_i := \bar{I}_{ij} \bar{\omega}_j \bar{\mathbf{e}}_i$$where ##\bar{I}_{ij}## is the moment of inerta tensor also expressed in the body-fixed coordinates.
 
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  • #9
kuruman said:
View attachment 346135Here is what I think. The axis of rotation of a rigid body is a straight line consisting of all points that are fixed in space relative to a non-rotating inertial frame, while the rest of the points rotate with angular velocity ##\vec {\omega}## about it. In the picture on the right the rotation axis is the dotted line. The primed coordinate system is rotating with the body except for the origin O which is on the axis.

Point O, as all other points on th dotted line, is not rotating by definition. not only because it is on the axis, but also because, as a mathematical point, it has zero extent. You cannot tell if it is rotating or not. If you are an observer at point O and at rest with respect to it, you will see all the points in the rigid body (except those on the dotted line) move with respect to you. The only line that you see fixed on the rigid body is the dotted line, the axes will be rotating.

That's because you have to think of yourself as a "point" observer at rest on the dotted line. It is only human to think of ourselves as extended beings with points off-axis and that's where the fallacy understandably occurs, in my opinion.
Thank you for sharing your perspective. I find it similar with my own interpretation. When considering a moving rigid body, we can seprate its motion into two separate movements. The translational movement of point O and the rotation of other points around it. When the point is fixed within an inertial frame, it aligns with the situation you have already described.
ergospherical said:
@MatinSAR: this is a subtlety, but the answer is not too complicated. In brief, the angular momentum is measured with respect to the space-fixed frame, however it's just more convenient to express our results in terms of components in the body-fixed basis (since, most importantly, the components of the moment of inertia tensor in the body-fixed frame are constants!!)

In specific terms, imagine a rigid body rotating around a fixed point ##O##. The angular momentum measured about this point in the space-fixed frame is just\begin{align*}
\mathbf{L} &= \sum \mathbf{r} \times \mathbf{p} \\
&= \sum m \mathbf{r} \times (\boldsymbol{\omega} \times \mathbf{r}) \\
&= \sum m \left( |\mathbf{r}|^2 \boldsymbol{\omega} - (\mathbf{r} \cdot \boldsymbol{\omega}) \mathbf{r} \right)
\end{align*}So far, the angular momentum is still just that which is measured in the space-fixed frame. However, the key step is that we'll now just expand the vectors ##\boldsymbol{\omega}## and ##\mathbf{r}## in terms of the body-fixed basis, something like:\begin{align*}
\boldsymbol{\omega} = \bar{\omega}_i \bar{\mathbf{e}}_i \\
\mathbf{r} = \bar{x}_i \bar{\mathbf{e}}_i
\end{align*}If you plug this in, you end up with$$\mathbf{L} = \left[ \sum m \left( |\mathbf{r}|^2 \delta_{ij} - \bar{x}_i \bar{x}_j \right) \right] \bar{\omega}_j \bar{\mathbf{e}}_i := \bar{I}_{ij} \bar{\omega}_j \bar{\mathbf{e}}_i$$where ##\bar{I}_{ij}## is the moment of inerta tensor also expressed in the body-fixed coordinates.
That makes complete sense to me now. I appreciate your help and time. Thank you.
 
  • #10
@MatinSAR the real pain for problems with rigid body dynamics, e.g. different types of spinning tops, is figuring out how to translate your results back into the space-fixed frame. Because we usually do most of the work using body-fixed components where the inertia tensor components are simple constants [for example Euler's equations of rigid body motion e.g. ##I_1 \dot{\omega}_1 + (I_3 - I_2) \omega_2 \omega_3 = \tau_1## refer to body-fixed components -- except I've now omitted the "bars" to conform to the standard convention].

After solving the Euler equation, you know the time evolution of the components ##\omega_i## (and therefore also ##L_i##) in the body-fixed basis. But somehow you need to figure out what this looks like in the space-fixed frame! It can sometimes be tricky. For a "free" (no torque) spinning top, for example, ##\mathbf{L}## is a conserved vector, so you can in a sense "invert" the problem and deduce how the body-fixed basis rotate around ##\mathbf{L}## in the space-fixed frame.

That might not make much sense if you're not used to it - have a look here to read some more (especially the worked examples)
https://www.damtp.cam.ac.uk/user/tong/dynamics/three.pdf
 
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  • #11
ergospherical said:
That might not make much sense if you're not used to it - have a look here to read some more (especially the worked examples)
https://www.damtp.cam.ac.uk/user/tong/dynamics/three.pdf
Thank you for sharing the PDF.





After discussing the problem with my professor and reviewing all the responses here, I feel confident that I have understood the concept. Thank you to everyone for their help.
 

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