Angular momentum during general motion I was reading David Morin's Introduction to Classical Mechanics. This is my first time with rotation so there are some doubts which I tried to google but cannot find an answer. One of the doubt is:


Let the position of the CM relative to a fixed origin be $ \mathbf{R}=(X, Y) $. And let the position of a given point relative to the $ \mathrm{CM} $ be $ \mathbf{r}^{\prime}=\left(x^{\prime}, y^{\prime}\right) $. Then the position of the given point relative to the fixed origin is $ \mathbf{r}=\mathbf{R}+\mathbf{r}^{\prime} $ (see Fig. 8.4). Let the velocity of the CM be $ \mathbf{V} $, and let the velocity relative to the $ \mathrm{CM} $ be $ \mathbf{v}^{\prime} $. Then $ \mathbf{v}=\mathbf{V}+\mathbf{v}^{\prime} $. Let the body rotate with angular speed $ \omega^{\prime} $ around the CM (around an instantaneous axis parallel to the $ z $ axis, so that the pancake remains in the $ x-y $ plane at all times). Then $ v^{\prime}=\omega^{\prime} r^{\prime} $.
Let's look at $ \mathbf{L} $ first. Let $ M $ be the total mass of the pancake. The angular momentum relative to the origin is $$ \begin{aligned}[b] \mathbf{L} &=\int \mathbf{r} \times \mathbf{v} d m \\ &=\int\left(\mathbf{R}+\mathbf{r}^{\prime}\right) \times\left(\mathbf{V}+\mathbf{v}^{\prime}\right) d m \\&=\int \mathbf{R} \times \mathbf{V} d m+\int \mathbf{r}^{\prime} \times \mathbf{v}^{\prime} d m \quad\text{(cross terms vanish; see below)}  \\&=M \mathbf{R} \times \mathbf{V}+\left(\int r^{\prime 2} \omega^{\prime} d m\right) \hat{\mathbf{z}} \\ &\equiv \mathbf{R} \times \mathbf{P}+\left(I_{z}^{\mathrm{CM}} \omega^{\prime}\right) \hat{\mathbf{z}} \end{aligned}\tag{8.9} $$ In going from the second to third line above, the cross terms, $ \int \mathbf{r}^{\prime} \times \mathbf{V} d m $ and $ \int \mathbf{R} \times \mathbf{v}^{\prime} d m $, vanish by definition of the $ \mathrm{CM} $, which says that $ \int \mathbf{r}^{\prime} d m=0 $ (See Eq. (5.58); basically, the position of the CM in the CM frame is zero.) This implies that $ \int \mathbf{v}^{\prime} d m=d\left(\int \mathbf{r}^{\prime} d m\right) / d t $ also equals zero. And since we can pull the constant vectors $ \mathbf{V} $ and $ \mathbf{R} $ out of the above integrals, we are therefore left with zero. The quantity $ I_{z}^{\mathrm{CM}} $ in the final result is the moment of inertia around an axis through the CM, parallel to the $ z $ axis. Equation (8.9) is a very nice result, and it is important enough to be called a theorem. In words, it says:

Here, Morin says that angular momentum of the body about origin is sum of the angular momentum of the CM of the body w.r.t. Origin and angular momentum of the body w.r.t. to the CM. So now, what if the angular momentum of CM is $\vec 0$ about the Origin. Then the sum will be equal to just angular momentum of the body w.r.t. to the CM. Doesn't this mean tha amgular momentum of the body about origin will be same as angular momentum of the body about its CM? I think I am not understanding the theory correctly.
 A: Angular momentum is a way of describing rotating objects.
If you see an object which does not rotate around its CM, but is rotating around something else, you will intuit that it has angular momentum.
If you see an object which does not rotate around something else, but spins on its own, you will also intuit that it has angular momentum.
These two forms (spin, and orbital) of angular momentum add up. And sometimes, one of the forms is zero, so that all the angular momentum comes from the other.
If the angular momentum of CM is 0 about the origin, then the total angular momentum will just be that of the body relative to the CM. In this situation, all of the angular momentum is from the body rotating around itself, it is spinning but not orbiting. You are understanding the theory correctly.
A: It sounds like you are understanding this correctly. If the angular momentum of the COM about the origin is zero (i.e., $M\mathbf{R}\times\mathbf{V} = \mathbf{0}$), then the angular momentum of the body about the origin is the same as its angular momentum about the COM. Consider the example of a disk rotating in two dimensions, with its COM moving radially outward from the origin.

Since $\mathbf{R}$ is parallel to $\mathbf{V}$, their cross product is zero, and the angular momentum about the origin is equal to the angular momentum about the COM, $\mathbf{L} = I\omega\,\hat{z}$, where $I$ is the moment of inertia about the COM.
A few comments on angular momentum that you might find helpful:

*

*Angular momentum is measured with respect to a particular frame of reference. That is, to measure $\mathbf{L}$, we need not only an origin, but also three axes. Since you have Morin's book, I would encourage you to read Section 9.1 (in particular, the paragraph at the end of page 375).

*A system's angular momentum is, by definition,
$$\mathbf{L} = \sum_{i}{\mathbf{r}_i\times\mathbf{p}_i}. \tag{1}$$
(In the case of a continuous mass distribution, this sum turns into an integral.) Eq. (1) does not always reduce to the familiar $I\boldsymbol{\omega}$. In fact, for rotations in three dimensions, $\mathbf{L}$ and $\boldsymbol{\omega}$ need not even point in the same direction!

*Any formula for $\mathbf{L}$ that follows logically from Eq. (1) is correct. Morin's Eq. (8.9), then, is unequivocally true, at least for a rigid pancake shaped object.

A: 
Doesn't this mean tha amgular momentum of the body about origin will
be same as angular momentum of the body about its CM?

Yes. From the picture, considering $\mathbf R$ constant, the velocity of CM is zero, and we can use $V$ to represent volume instead of velocity. The density is a function of the coordinates with respect to the CM:
$\mathbf L = \int{(\mathbf R + \mathbf r') \times d\mathbf p} = \int{(\mathbf R + \mathbf r') \times \mathbf v dm } = \int{(\mathbf R + \mathbf r') \times \mathbf v' \rho(x',y') dV } $
Taking the constant vector $\mathbf R$ out of the integral:
$\mathbf L = \mathbf R \times \int{ \mathbf v' \rho(x',y') dV } + \int{\mathbf r' \times \mathbf v' \rho(x',y') dV }$
For a rigid body, considering a constant rotation axis: $\mathbf v'= \omega \mathbf r'$, where $\omega$ is constant along the body.
$\int{ \mathbf v' \rho(x',y') dV } = \omega \int{  \mathbf r' \rho(x',y') dV }$
This integral is zero, because it is the definition of the coordinates of the CM, and the origin of them is the CM itself by hypothesis.
So, $\mathbf L = \int{\mathbf r' \times \mathbf v' \rho(x',y') dV }$ which is the angular momentum of the body with respect to the CM.
