Undergrad Properties of angular momentum

[cianfa72]

TL;DR

About the properties of angular momentum calculated w.r.t. a fixed point P in the rest frame of the system's center of mass (CoM).

A very basic question about the properties of angular momentum for a system of particles.

In a reference frame (inertial or otherwise) fix a point P and calculate the angular momentum $\vec L_p$ w.r.t. it. Then take the direction $\vec a$ and project $\vec L_p$ along it. Then the following result holds $$\frac {dL_a} {dt} = \tau_a$$ where $L_a$ and $\tau_a$ are the projection of the angular momentum $\vec L_p$ and the total external torque $\vec \tau_p$ about P on the $\vec a$ direction, respectively.

Next, pick the rest frame of the system's center of mass (CoM). It has the property that the angular momentum $\vec L_p$ calculated about any fixed point P in this frame returns the same result. Note that is a kinematic fact, therefore it holds regardless of whether the system's CoM rest frame is inertial or is not.

Of course, in the latter case, the contribution to the total external torque $\vec \tau$ abount point P includes the torque due to the inertial/fictitious forces that appear to "act" on the system's particles.

Did I understand it correctly ? Thanks.

 

[wrobel]

I can just explain what one actually should understand regarding this stuff.
Assume we have a system of particles $A_1,\ldots, A_N$ with masses $m_1,\ldots,m_N$ respectively.
Let $Sxyz$ be a coordinate frame with the origin at the center of mass $S$ and the coordinate vectors $\boldsymbol e_x,\boldsymbol e_y,\boldsymbol e_z$ do not change their directions relative an inertial frame for all the time. Let $\boldsymbol\rho_1,\ldots,\boldsymbol \rho_N$ be the position vectors of the points $A_1,\ldots,A_N$ relative the frame $Sxyz$.
Then by definition put $\boldsymbol K_*=\sum_{k=1}^Nm_k\boldsymbol \rho_k\times \boldsymbol {\dot\rho}_k.$

Theorem. $$\frac{d\boldsymbol K_*}{dt}=\sum_{k=1}^N\boldsymbol\rho_k\times \boldsymbol F_k.$$
Here $\boldsymbol F_k$ is an external force applied to $A_k$.

If we have an inertial frame $OXYZ$ then
$$\boldsymbol K_O=\sum_{k=1}^Nm_k\boldsymbol r_k\times \boldsymbol {\dot r}_k,\quad \boldsymbol K_O=m\boldsymbol {OS}\times\frac{d}{dt}\boldsymbol {OS}+\boldsymbol K_*.$$
Here $\boldsymbol r_k=\boldsymbol{OA}_k,\quad m=m_1+\ldots+m_N$
Then following formula holds as well
$$\frac{d}{dt}\boldsymbol K_O=\sum_{k=1}^N\boldsymbol r_k\times \boldsymbol F_k.$$
We can consider projections of these laws on fixed axes. Let $\boldsymbol e$ be a fixed unit vector. Then for example we can write down a trivial consequence from above:
$$\frac{d(\boldsymbol K_*,\boldsymbol e)}{dt}=\Big(\boldsymbol e,\sum_{k=1}^N\boldsymbol\rho_k\times \boldsymbol F_k\Big).$$

 

[cianfa72]

The following formula holds as well
$$\frac{d}{dt}\boldsymbol K_O=\sum_{k=1}^N\boldsymbol r_k\times \boldsymbol F_k.$$

Suppose $OXYZ$ isn't inertial. Then your external $\boldsymbol F_k$ also includes the inertial force that appears acting on the particle $A_k$ in that (non inertial) frame (as implied by Newton's 2nd law written in that frame).

 

[cianfa72]

BTW, the point I'd like to be sure is what I asked in the OP.

Namely, in the context of Newtonian physics, at a given universal time $t$, the system's angular momentum $\boldsymbol L_p$ evaluated w.r.t. any point P in the rest frame of system's CoM is independent from the chosen point P. No requirement of inertial/non inertial applies to such frame.

This way in the rest frame of the system's CoM makes sense to define the system's angular momentum.

 

[jbriggs444]

Namely, in the context of Newtonian physics, at a given universal time $t$, the system's angular momentum $\boldsymbol L_p$ evaluated w.r.t. any point P in the rest frame of system's CoM is independent from the chosen point P. No requirement of inertial/non inertial applies to such frame.

I want to be sure that I understand the setup precisely.

When you speak of "the" rest frame of the system's CoM, I believe that you are implicitly requiring that this frame be non-rotating.

So we have this non-rotating CoM rest frame. The system may be under some non-zero net force. The CoM may be jiggling around as a result.

Now we choose a point P. This point has fixed coordinates relative to the CoM rest frame. If the frame jiggles, point P jiggles with it.

You ask whether the total angular momentum of the system then depends on which point we choose to be P.

Clearly not.

 

[cianfa72]

When you speak of "the" rest frame of the system's CoM, I believe that you are implicitly requiring that this frame be non-rotating.

I mean the frame in which the system's CoM stays always at rest (zero total $\vec P$ w.r.t. it). Basically it shares the system's CoM velocity. If the system is under some non-zero external net (real) force then such a system's CoM rest frame will be non inertial.

So we have this non-rotating CoM rest frame. The system may be under some non-zero net force. The CoM may be jiggling around as a result.

You mean that under some non-zero net (real) force, the system's CoM may be jiggling around from the viewpoint of an inertial frame.

Now we choose a point P. This point has fixed coordinates relative to the CoM rest frame. If the frame jiggles, point P jiggles with it.

You ask whether the total angular momentum of the system then depends on which point we choose to be P.

Clearly not.

You mean pick a point P with fixed coordinates w.r.t. the CoM rest frame above. Then the system total angular momentum calculated at any given time $t$ doesn't depend on which point we choose to be P (even though its value may change with time).

 

[jbriggs444]

I mean the frame in which the system's CoM stays always at rest (zero total $\vec P$ w.r.t. it). Basically it shares the system's CoM velocity. If the system is under some non-zero external net (real) force then such a system's CoM rest frame will be non inertial. What do you mean with non-rotating ?

By contrast to a rotating frame of reference. It is not spinning.

We may take it as an article of faith that any "frame" we speak of will be at least [Born] rigid.