# 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.

• It's not true that just because all the forces are perpendicular to the velocity that angular momentum is conserved. Take for example a particle in uniform circular motion. Then flip the force vector. It now spins the other way, and the force was perpendicular to velocity for all times. (I am also not totally sure what question you are asking.) – Pricklebush Tickletush Aug 29 '13 at 17:46
• @AlecS You are misreading me, or possibly I was not clear (where ?). I am not saying that angular momentum is preserved when forces are orthogonal. All I am saying is that the magnitude of the linear momentum is preserved. So, this tells you that when you flip the force, you flip also the angular momentum, but it keep the same magnitude (of course that has to be compensated by whatever structure is providing the forces, as total angular momentum is conserved). My question is in the last 2 paragraphs. – babou Aug 29 '13 at 19:56
• Well, that's not true either. You can change an object's velocity arbitrarily with a perpendicular force. And the last two paragraphs refer to "the mathematical formulation of 'this'" or why "this" is not considered in practice... it is not clear what "this" is. – Pricklebush Tickletush Aug 29 '13 at 20:14
• @AlecS Yes, you can change velocity with an orthogonal force, but I am talking of speed along a known trajectory, possibly enforced by whatever means. If the trajectory is determined, acceleration orthogonal to that trajectory will not change speed along the trajectory. In other words, knowledge of the speed and the trajectory determines the acceleration, and the acceleration is orthogonal to the trajectory if the speed is constant. And the speed remains constant if the acceleration remains orthogonal to the trajectory. – babou Aug 29 '13 at 21:12
• @AlecS Sorry, the example was obviously wrong (I must be tired). I removed it. A phenomenon that increases speed is hardly an example of speed conservation. Thanks – babou Aug 29 '13 at 22:10

If you know about vectors then all you need is Euler's laws of motion for a rigid body.

1. For linear motion there is \begin{aligned} \vec{p} &= m\, \vec{v}_{\rm cm} \\ \sum \vec{F} &= m\, \vec{a}_{\rm cm} \end{aligned}
2. 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}
3. 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.

• Thanks. I actually meant something quite different about a different look at problems when trajectories are known. I found a proof. – babou Aug 31 '13 at 21:50

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).