# Angular momentum around CM

1. Dec 24, 2006

### P3X-018

When considering a system of particles with respect to a fixed point O, then the total angular momentum around that point can be writtin as

$$\textbf{L}_O = \sum_i \textbf{r}_i\times m_i\textbf{v}_i$$

However in my book they add and substract the vektor $\sum_i \textbf{R}_{CM}\times m_i\textbf{v}_i$, which gives,

$$\textbf{L}_O = \sum_i (\textbf{r}_i-\textbf{R}_{CM})\times m_i\textbf{v}_i + \sum_i \textbf{R}_{CM}\times m_i\textbf{v}_i$$

It is then stated that the first term in the above equation, resembles the angular momentum around CM, and the second term "the angular momentum of CM", i.e.

$$\textbf{L}_O = \textbf{L}_{CM} + \textbf{R}_{CM}\times \textbf{P}$$

I understand the second term, but the first term seems a little confusing. How can the 1st term be considered as the angular momentum around CM? Indeed the vector $\textbf{r}_i -\textbf{R}_{CM}$ points from CM to the particle i, but how about the velocity vector term $\textbf{v}_i$ of particle i in $\textbf{L}_{CM}$, doesn't that need to be the velocity of the particle i relative to the CM, before it can be considered as the angular momentum around CM?

Last edited: Dec 24, 2006
2. Dec 24, 2006

### StatMechGuy

The idea behind this separation is to say that you can break the angular momentum into two parts. Let's use the earth for our example.

The first part, L_cm, comes from the body orbiting around its pole. The second part comes from the earth orbitting around the sun. It's nice because the second part means that you can completely separate out the dynamics of most problems into "body rotating around itself" and then "point mass moving around".

3. Dec 25, 2006

### P3X-018

I understand the idea behind it. But when you calculate the angular momentum around some point in an inertial frame, then you need the velocity with respect to that frame, right? I.e. the position vector from the origin of the coordinatsystem to the particle cross the velocity of the particle with respect to that point, correct?
But inorder to calculate the angular momentum around CM wouldn't need the velocity of the particle with respect to CM? Because $\textbf{v}_i$ is with respect to the point O. If CM was moving the the velocity with respect to CM wouldn't be $\textbf{v}_i$, which is just the velocity with respect to O.
So have I misunderstood something, and hence my assumptions are incorrect?

4. Dec 25, 2006

### Meir Achuz

Write $$R=\sum_i m_i r_i/M,$$ and
$$v_i(cm)=v_i-V$$. Plug this into your second equation to derive the fact that the first term is indeed what your last sentence requires.

Last edited: Dec 25, 2006
5. Dec 25, 2006

### P3X-018

Now that I think about, then the torque around the CM for the particle system, is

$$\mathbold{\tau}_{CM} = \sum_i (\textbf{r}_i-\textbf{R}_{CM})\times \textbf{f}_i$$

And the time derivative of the angular momentum around CM is

$$\frac{\mathrm{d}\textbf{L}_{CM}}{\mathrm{d}t} = \sum_i( \textbf{v}_i - \textbf{v}_{CM} ) \times \textbf{p}_i + \sum_i(\textbf{r}_i-\textbf{R}_{CM})\times \textbf{f}_i$$

But inorder for the angular momentum theorem to hold, the first term of the above equation most be $\mathbold{0}$, which means that the momentum $\textbf{p}_i$ most be parallel to the velocity of the particle in the CM-frame, $\textbf{v}_i - \textbf{v}_{CM}$. Therefore $\textbf{p}_i$ most be the momentum of particle in the CM-frame.

However, in the derivation in my first post, the velocity $\textbf{v}_i$ is with respect to the frame of O, so $m_i\textbf{v}_i$ can't be the momentum of particle in the CM-frame, and hence the expression

$$\sum_i (\textbf{r}_i-\textbf{R}_{CM})\times m_i\textbf{v}_i$$

can't be called the angular momentum around the CM. Or where do I go wrong?

Last edited: Dec 25, 2006
6. Dec 25, 2006

### Staff: Mentor

I agree that they went out of their way to make this derivation obscure! Try this: Substitute:

$$\textbf{v}_i = \textbf{v'}_i + \textbf{v}_{CM}$$

Where the primed velocities are with respect to the center of mass frame. Crank it out and you'll find that the terms with the velocity of the CM will cancel.