About the angular momentum of the particle system relative to the center of mass reference

Question

I know its correct expression:$L=sum_{i}^{n}left(vec{r_i}times m_ivec{v_i}right)$
But the textbook thinks:$vec{F_i}=frac{d}{dt}left(m_ivec{v_i}right)$,But the speed here is in the center of mass reference frame, I don’t think this is the real force.
Summary given in the textbook:$tau_{i}^{ext}=frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ivec{v_i}right)=sum_{i}^{n}vec{r_i}timesvec{F_{i}^{ext}}$
I was confused, and finally I thought about why, but the textbook did not explain it. I don’t know if this is correct.
Below are my thoughts：
$sum_{i}^{n}vec{r_i}timesvec{F_{i}^{ext}}=sum_{i}^{n}vec{r_i}times m_i(vec{a_{cm}}+vec{{a_i}})=sum_{i}^{n}vec{r_i}times m_ivec{a_{i}}+left(sum_{i}^{n}vec{r_i}m_iright)times vec{a_{cm}}=sum_{i}^{n}vec{r_i}times m_ivec{a_{i}}$
Note that $sum_{i}^{n}vec{r_i}m_i=0$
$or$
$frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ivec{v_i}right)=frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ivec{v_i}+left(sum_{i}^{n}vec{r_i}m_iright)timesvec{v_{cm}}right)=frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ileft(vec{v_i}+vec{v_{cm}}right)right)=frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ivec{V_{i,groud}}right)=sum_{i}^{n}vec{r_i}timesvec{F_{i}^{ext}}$
Question：
I don't understand why the textbook is written like this,Relative to the speed in the center of mass reference frame, the speed is not the real speed,but it can become a real external force.
I think if the textbook uses this expression to be more clear,such as:$frac{d}{dt}left(sum_{i}^{n}vec{r_i}times m_ivec{v_i}right)=frac{d}{dt}left(sum_{i}^{n}{r_i}m_ileft({r_i}omegahat kright)right)=I_{cm}alphahat k$ (This is not absolutely correct, just a rigid body that rotates around the center of mass and has the same angular velocity)
The following is a clarification on the question：
If the ground is the reference frame, this can be close to the real force $F$.The velocity $v_i$ is relative to the center of mass reference frame,and the center of mass reference frame has a velocity $v_{cm}$ relative to the ground.So $v_i$ is not the speed relative to the ground, I cannot write it directly as $F$.The bottom is my explanation, but the textbook does not explain why this can be written directly as $F$.So I have doubts about this question.
Derivation given by textbook：
Their relationship is drawn from the picture

The following is about the angular momentum of the two parts

I can get such a relationship, but do they correspond to each other independently？

Part of the content is omitted here
  
There is no doubt that we have such a corresponding relationship, and I don't need to write the other relationship.
Another relationship:

Although I don’t need to write this step, because it can be concluded that the relationship of this step through the relationship of another step, they are undoubtedly corresponding.
But I have doubts about this step, what can be used to explain, can I eliminate the misunderstanding that I don’t know.
$frac{d}{dt}left(m_ivec{v_{cm,i}}right)neq vec{F_i}$ (I'm joking here)
Because even if I don’t deal with it by this way, another relationship can be derived from the relationship above. This is undoubtedly correct.
But I have also explained why this misunderstanding can be caused,If I look directly at it, I will probably reject this equal sign.
My explanation：
I can only explain to myself whether I have overlooked some of the rules of operation,although the symbols look similar.
What should I think，whether I have a wrong misunderstanding.Help!

pglpm · Accepted Answer

To be honest I'm not sure if I understand what you're asking or what your perplexities are. I'll make a couple of remarks about the balance of rotational momentum and give some references that examine it in detail and generality, hoping that this might help you.

If we consider one pointlike mass, the balance of rotational momentum with respect to the origin of the reference frame is a trivial consequence of the balance of translational momentum: we just need to vector-multiply the latter by $pmb{r}$, the position of the point-mass. The rotational momentum $pmb{H}$ of the point-mass is defined as $pmb{H}:=pmb{r}land(mdot{pmb{r}})$, where the dot denotes the time derivative, $dot{x} := frac{mathrm{d}x}{mathrm{d}t}$. Then, considering that $dot{pmb{r}}landdot{pmb{r}} = 0$, we easily find
$$dot{pmb{H}} = pmb{r}land pmb{F}  , $$
where $pmb{F}$ is the total of all forces acting on the point-mass.
But already in this trivial case it's important to emphasize that this balance is valid in any reference frame, even a non-inertial one, provided that the inertial force is included among all forces acting on the point-mass. (It's good to remember that non-inertial forces are so-called objective quantities, that is, they are the same in all frames, inertial and non-inertial; more about this below.)

For a system of point-masses $m_i$ with position vectors $pmb{r}_i$ the situation becomes more interesting, and this is where the balance of rotational momentum appears as a law. Let's state this law for a general reference frame, inertial or non-inertial.
Suppose that on each point-mass $m_i$ acts the total external force $pmb{F}_i$, including the inertial force, and the forces $pmb{f}_{ik}$ from the other point-masses. The total rotational momentum of the system in this reference frame, with respect to its origin, is defined as $pmb{H}:=pmb{r}_iland(m_idot{pmb{r}_i})$.
The balance of rotational momentum states that
$$dot{pmb{H}} = sum_i pmb{r}_i land pmb{F}_i  ,$$
that is, the rate of change of the rotational momentum equals the total external torque. This equation holds in any frame, inertial or non-inertial (again, provided that we're including inertial forces).
The equation above in this case does not follow from the balance of translational momentum for this system. If we multiply the latter balances for each point-mass by $pmb{r}_i$, sum them up, and do some manipulations, we're left with an additional term $sum_{ik} (pmb{r}_i - pmb{r}_k) land pmb{f}_{ik}$. This term only vanishes if the mutual forces are central, that is, they are directed along the lines joining the respective point-masses (see Joos's reference below).
The balance of rotational momentum, taken as a law, therefore requires mutual forces to be central. In many physics textbooks the opposite point of view is often taken: they assume the mutual forces to be central, and the balance of rotational momentum then becomes a consequence of the balance of translational momentum plus this centrality assumption. You can choose the point of view that you prefer. When we go over to continuum mechanics, however, we have to take the balance of rotational momentum as primitive because of the appearance of contact forces; see the references below regarding this point.

Returning to your question, the point is that the law
$$frac{mathrm{d}}{mathrm{d}t}[pmb{r}_iland(m_idot{pmb{r}_i})] =
sum_i pmb{r}_i land pmb{F}_i $$
is valid in any frame, inertial and non-inertial, and therefore also in the centre-of-mass frame. We must only remember to include the inertial force on each point-mass. I suppose that "$F^text{ext}$" in your textbook includes that.
Additional note on objectivity of forces and on inertial force
If we consider two generic (rigid) frames in mutual motion, a position vector $pmb{r}'(t)$ in the second frame is related to the position vector $pmb{r}(t)$ in the first by
$$pmb{r}' = pmb{c}(t) + pmb{Q}(t)pmb{r}$$
where $pmb{c}(t)$ is a vector relating the origins of the two frames, and $pmb{Q}(t)$ an orthogonal rotation matrix relating the axes of the two frames. As seen from the $t$-dependence, this relation is also true for time-varying rotating frames.
Any (non-inertial) force is said to be objective because its expressions in any two frames are related by
$$pmb{F}' = pmb{Q}pmb{F}  .$$
In other words, observers in the two frames always agree about the magnitude of the force, $lvert pmb{F}'rvert = lvertpmb{F}rvert$ (owing to the norm-preserving property of orthogonal rotation matrices), and also on its direction with respect to masses in the system an to the fixed stars. For example, all observers agree that a Hookean force exerted by a spring is directed along the line joining the extremities of the spring; and all observers agree that the gravitational force between two masses is directed along the line joining the centres of mass of the two masses.
Inertial forces are the only exception. The most general form of the inertial force on a point-mass with position vector $pmb{r}$ is
$$pmb{F}_{text{inertial}} =
2 pmb{varOmega} (dot{pmb{r}} - dot{pmb{c}}) - (pmb{varOmega}^2- dot{pmb{varOmega}}) (pmb{r}-pmb{c}) + ddot{pmb{c}} ,$$
where $pmb{c}$ is the position vector of the origin of the frame with respect to any inertial frame, and $pmb{varOmega}$ is the angular velocity (expressed as a matrix) of the frame with respect to the inertial frame (or to the fixed stars). You see that the expression above includes the usual term $ddot{pmb{c}}$ coming from linear acceleration and the centrifugal, Coriolis, and Euler forces coming from rotation.
References
The general balance of rotational momentum, inertial forces, and the objectivity of forces are discussed in depth, for example, in

Truesdell, Toupin: The Classical Field Theories (Springer 1960), especially chapters B.III and D.I.

Truesdell: A First Course in Rational Continuum Mechanics. Vol. 1: General Concepts (2nd ed. Academic Press 1991), especially Chapter I.

Samohýl, Pekar: The Thermodynamics of Linear Fluids and Fluid Mixtures (Springer 2014), especially section 3.3; inertial forces are discussed in section 3.2 (note that this is a book on thermomechanics, not just thermodynamics).

The history about the law of balance of rotational momentum, especially with regard to Euler's discovery of its necessity as an independent law, is summarized in

Truesdell: Whence the law of moment of momentum? (1963/1968) (also available here)

A traditional presentation of this balance (but limited to inertial frames) is in

Joos: Theoretical Physics (3rd ed. Hafner 1958), especially sections 5.5 and 6.2 (older edition here).

R.W. Bird · Answer

The F found from the rate of change of the momentum of a particle is the resultant force acting on that particle. In a system of particles, that resultant consists of both external forces and some which are internal to the system. If you add all such forces (or torques) together, the ones that are internal will cancel out (action and reaction).

Answered by R.W. Bird on June 7, 2021

About the angular momentum of the particle system relative to the center of mass reference

2 Answers

Additional note on objectivity of forces and on inertial force

References

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