Newton’s second law in its most basic form states that for a single point mass, \(\dot{\textbf p}=\textbf F\) in any inertial frame. Combining this with Newton’s third law (antisymmetry of forces \(\textbf F_{i\to j}=-\textbf F_{j\to i}\), similar to many other antisymmetric phenomena in physics such as relative velocities \(\textbf v_{ij}=-\textbf v_{ji}\)) gives the most useful system formulation of Newton’s second law \(\dot{\textbf P}=\textbf F^{\text{ext}}\) in any inertial frame where \(\textbf P:=\sum_{i=1}^N\textbf p_i\) is the total momentum of a system of \(N\) point masses.
For the single-particle form \(\dot{\textbf p}=\textbf F\) of Newton’s second law, crossing both sides with the particle’s position \(\textbf x\) (relative to any origin, stationary, moving, accelerating, etc. as seen in the inertial frame) gives \(\dot{\textbf L}=\boldsymbol{\tau}\) in any inertial frame (this is thanks to a combination of two facts, namely: the cross product is a derivation, and it is non-degenerate, I wonder if there is a theory which generalizes this observation?) a rotational analog of Newton’s second law. Thus, naturally one might hope that an analog \(\textbf L=\boldsymbol{\tau}^{\text{ext}}\) should hold at the system level too, with \(\textbf L=\sum_{i=1}^N\textbf L_i\) and \(\boldsymbol{\tau}^{\text{ext}}=\sum_{i=1}^N\boldsymbol{\tau}^{\text{ext}}_i\). This identity is indeed true provided that the angular momenta \(\textbf L_i\) and external torques \(\boldsymbol{\tau}^{\text{ext}}_i\) are all measured in an inertial frame relative to a point \(\tilde{\textbf X}\) moving parallel at all times to the center of mass \(\textbf X\) of the system. The proof is a direct computation:
$$\dot{\textbf L}=\sum_{i=1}^N\dot{\textbf L}_i=\sum_{i=1}^N(\dot{\textbf x}_i-\dot{\tilde{\textbf X}})\times\textbf p_i+(\textbf x_i-\tilde{\textbf X})\times\textbf F_i$$
Now, the idea is to decompose the forces on each particle as \(\textbf F_i=\textbf F^{\text{ext}}_i+\sum_{j=1}^N\textbf F_{j\to i}\) with the non-degenerate \(\textbf F_{i\to i}=\textbf 0\). This finally gives:
$$\dot{\textbf L}=-\dot{\tilde{\textbf X}}\times\textbf P+\boldsymbol{\tau}^{\text{int}}+\boldsymbol{\tau}^{\text{ext}}$$
where \(\boldsymbol{\tau}^{\text{int}}=\sum_{1\leq i<j\leq N}(\textbf x_i-\textbf x_j)\times\textbf F_{j\to i}\) are \(N\choose 2\)\(=\frac{N(N-1)}{2}\) internal, reference-independent couples (and one would normally use one’s intuition to think about these rather than explicitly appealing to the formula) and \(\boldsymbol{\tau}^{\text{ext}}=\sum_{i=1}^N(\textbf x_i-\tilde{\textbf X})\times\textbf F^{\text{ext}}_i\). So it is clear that in order to have \(\dot{\textbf L}=\boldsymbol{\tau}^{\text{ext}}\) there are two obstacles in the way. Although actually, one of them is always zero, namely the internal couples \(\boldsymbol{\tau}^{\text{int}}=\textbf 0\). This follows from Galilean covariance (rather than from God, as my high school physics teacher told me); forces at a distance like gravity, electrostatics, masses pulled by springs or tension forces through ropes all have to be central (and thus conservative) so the cross product vanishes. On the other hand, the only time one can achieve an azimuthal/tangential/angular component of force is with friction, but then friction is not a force at a distance, but is rather a contact force and so the relative position \(\textbf x_i-\textbf x_j=\textbf 0\) at the point of transmission (the normal force is thus also zero because it simultaneously falls into both of those categories). I suspect any force of only position has to be like this, velocity-dependent forces like the magnetic force may be more subtle depending on Newtonian vs. relativistic frameworks, etc. To zero the remaining obstacle, since \(\textbf P=M\dot{\textbf X}\), it is both sufficient and necessary to have \(\tilde{\textbf X}||\textbf X\) at all times as claimed (e.g. a common, but certainly not the only choice is \(\tilde{\textbf X}:=\textbf X\)).
For instance, consider a hollow cylinder rolling without slipping down a ramp (couched in some inertial frame obviously). Take \(\tilde{\textbf X}\) to be the point of contact of the wheel with the ramp (clearly not associated with any physical particle). However, because \(\tilde{\textbf X}\) clearly moves parallel to \(\textbf X\) (indeed, it moves with the exact same velocity, just displaced from it), we know that \(\dot{\textbf L}=\boldsymbol{\tau}^{\text{ext}}\) relative to \(\tilde{\textbf X}\). Since this particular system of particles is in fact a rigid body, it is convenient to work with their common \(\omega\) and so \(L=I\omega\) (this step only works because of rolling without slipping) where by the parallel axis theorem \(I=2MR^2\). Now, one convenience of choosing \(\tilde{\textbf X}\) as the point of contact is that all contact forces acting at \(\tilde{\textbf X}\) by definition exert no torque about \(\tilde{\textbf X}\). So the only one is gravity. This yields \(\dot{\omega}=\frac{g\sin(\theta)}{2R}\). Since \(\omega\) is the same about any point (still don’t have very good intuition for this fact), one can do the derivation choosing \(\tilde{\textbf X}:=\textbf X\) instead, but it will be more lengthy since one has to more explicitly invoke the rolling without slipping constraint and Newton’s second law in its translation form to eliminate the static friction enforcing the rolling without slipping. The result is the same however, giving credence to the method.
My first year Cambridge physics supervisor also showed me a cute way to think about the situation which is mathematically equivalent to the above but conceptually somewhat different. The thing is that Newton’s second law refers only to a vector addition of force vectors, with no regards for where the point of application of those forces is (clearly if one is only interested in the center of mass motion, it doesn’t matter!). The torques are the devices that are sensitive to the point of application. So the idea is to basically draw the free body diagram of a system. Then, any forces that are not acting at the center of mass, draw a fictitious copy of the force acting at the center of mass but then to balance this out draw its negation also acting at the center of mass (if it acted elsewhere it wouldn’t affect the CoM motion but definitely the rotational dynamics!). Do this for all forces not acting at the CoM. The final result is that the torque about the CoM is completely comprised of reference-independent couples which dramatically clarifies the situation. Meanwhile, there is a unbalanced force left acting on the center of mass, which, intuitively, dictates its motion by Newton’s second law (actually, all this discussion holds equally well with CoM replaced by \(\tilde{\textbf X}\)). Hopefully this discussion will help me go back to learn rigid body dynamics more carefully and understand better.