Conservation of linear momentum magnitude along a trajectory I was once criticized for "taking angular momentum as momentum going
in a circle". I was loosely trying to state, in classical mechanics,
that in using conservation of momentum, one can switch between linear
and angular momentum, in a problem when one is not concerned with
rotation of the body on itself, and even treat rotational
motion with linear momentum. I think that, in a very real sense,
angular momentum can also be seen as momentum going in a circle.  This is because circular motion can be seen as velocity going in a
circle, meaning of course that its direction changes to stay
tangential, even though there are other more abstract formulations or formalizations with a different dimentionality.
Actually I realize that this is hardly a physics problem, but rather
pure kinematics. Very simply because when considering only linear (not
necessarily straight) motion of a mass, one can simply factor out the
mass and deal with speed and acceleration, rather than momentum and
forces.
The idea is that the speed (momentum magnitude) of a body (mass) is
constant when all accelerations (forces) are orthogonal to the
trajectory, whatever the shape of that trajectory.
This does not seem too original.
It provides a very simple treatment of some single body (angular) momentum conservation
problem, but no one seem to ever use it.
It is particularly useful if one has to analyze strange trajectories,
for example imposed by rails.
Of course, it can be extended to the case of non-orthogonal
accelerations (forces) by projecting the acceleration (force) on the
trajectory tangent to get the speed (momentum magnitude) variation.
So I would like to know the proper mathematical formulation of this,
or a web reference where this is discussed and formulated
mathematically, especially in the case of non orthogonal forces. I
could not find it myself, but it may be a question of having the right
keywords.
I am also curious as to why this seems not much considered in
practice. I feel it gives beginners or amateurs a wrong perception of
momentum conservation laws which are much more interesting when used
to analyze interactions between parts of a system. Dynamics with a
single mass is hardly dynamics.
 A: If you know about vectors then all you need is Euler's laws of motion for a rigid body.


*

*For linear motion there is
$$ \begin{aligned} \vec{p} &= m\, \vec{v}_{\rm cm} 
\\ \sum \vec{F} &= m\, \vec{a}_{\rm cm} \end{aligned}$$

*For angular motion there is
$$ \begin{aligned} \vec{L}_{\rm cm} &= I\, \vec{\omega} 
\\ \sum \vec{M}_{\rm cm} &= I\, \vec{\alpha} + \vec{\omega}\times I \vec{\omega} \end{aligned} $$

*For any constraint a reaction force does no work, and hence $\vec{F} \cdot \vec{v} =0$


Now you can solve any problem in rigid body mechanics, without friction or contacts.
A: Not getting much interest or understanding for that question (it may
be poorly stated?), I am trying to answer it myself. I was hesitating to
do that because, not having any contact with that kind of mathematics for a
very long time, I have limited confidence in my own formal
competence. Comments and alternate answers are of course welcome
The basic idea is that we are considering problems where the
trajectory of a mass is know. For example it could be enforced by a
rail or a pipe system, or it is known for some other reason.
The point I am trying to make is that, when the trajectory is known,
whatever it is, then acceleration orthogonal to the motion is
irrelevant since it serves only to orient the velocity, and that is
already known from the trajectory. So speed can then be analyzed by
considering only tangential acceleration, and, in particular, it is
conserved when there is no tangential acceleration.
Formally, at time $t$, we have $\vec{a}=\vec{a_T}+\vec{a_C}$ where
$\vec{a_T}$ is the tangential acceleration, and $\vec{a_C}$ is the
centripetal acceleration (see figure from Wikipedia where I also found hints for this answer).

We define a unit vector $\vec{u_T}$ to orient the tangent to the
trajectory at time $t$ in the direction of the velocity $\vec{v}$:
$\;\,\vec{u_T}=\vec{v}/v$, where $v$ is the speed.
Let $a_T$ be the algebraic value of $\vec{a_T}$ on the oriented
tangent.
We thus have by definition of the projection: $a_T=\vec{a}.\vec{u_T}$
We now prove that $a_T=dv/dt$
Since $\vec{v}=v\vec{u_T}$, we have by differentiation
$\vec{a}=d\vec{v}/dt=(dv/dt)\vec{u_T}+v(d\vec{u_T}/dt)$
According to Frenet-Serret formulae, $d\vec{u_T}/dt=(v/r)\vec{u_C}$,
where $r$ is the current radius of curvature, and $\vec{u_C}$ is
orthogonal to the trajectory tangent.
Hence the tangential projection of $\vec{a}$ is the first term of the
sum, i.e.  $\vec{a_T}=(dv/dt)\vec{u_T}$.
Thus $a_T=\vec{a_T}.\vec{u_T}=(dv/dt)\vec{u_T}.\vec{u_T}=dv/dt$.
Which gives finally: $dv/dt=a_T=\vec{a}.\vec{u_T}$
If there is no tangential acceleration, then $dv/dt=0$, which implies
that the speed on the trajectory is constant.
If there is some acceleration on the trajectory, then the variation of
the speed between two points A and B is
$\int_{t_A}^{t_B}{\vec{a(t)}.\vec{v(t)}dt/v(t)}$, i.e., $\int_{t_A}^{t_B}{a_T(t)dt}$.
This result can simplify the analysis of some problems where the
trajectory is somehow fixed and known. The analysis can be based
exclusively on speed along the trajectory and its variation due to
tangential acceleration, assuming rotation of the mass on itself can
be neglected.
An exemple is the analysis of the motion of a rollercoaster into a
loop.
What I meant about angular momentum is that it can become implicit in
the analysis, since only speed (or linear momentum) is considered,
even when the trajectory is circular.
However, the last remark of my question is not quite accurate. This
does not have to be dynamics with a single mass, i.e., actually reduced
to kinematics. It is quite possible to consider two bodies bound to
the same trajectory, hitting each other and bouncing.
As said in the last paragraph of the question, I am still curious why
this is not used in simple problems (though while getting information for
this answer, I found that similar techniques are used for harder
problems ... but that is another story).
