Newton's laws in variable mass systems

AI Thread Summary
The discussion centers on the application of Newton's laws to variable mass systems, particularly in the context of a spray can or rocket. Participants debate whether the force generated is constant across different inertial frames, with arguments focusing on the relationship between mass flow rate, exhaust velocity, and momentum. Key points include the assertion that force is frame-dependent when mass is not constant, and the distinction between definitions of force as F=dp/dt versus F=ma. The conversation highlights the complexities of applying classical mechanics to systems with changing mass and the implications for conservation laws. Ultimately, the dialogue emphasizes the need for careful consideration of reference frames and the definitions used in physics.
  • #51
Dickfore said:
Question: What form of Newton's Second Law is used in Hydrodynamics?

For a non-relativistic continuum (solid, fluid, etc) Newton's second law is simply the conservation of linear momentum expressed as:

http://en.wikipedia.org/wiki/Cauchy_momentum_equation
 
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  • #52
F=dp/dt=d(mv)/dt. With this definition of force, that F=m·dv/dt+v·dm/dt is a straightforward application of the chain rule. Dick: Does the chain rule not apply in physics?

As for this "camp":

Sure the chain rule work, as long as you use it properly.

Leibniz' rule of differentiation of an integral with moving boundaries is a fancy version of the chain rule, and that is the one to use here.
 
  • #53
arildno said:
This is a meaningless position.

Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever.

I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O.

I said the same thing:

Dickfore said:
YARLY! Imagine a collection of non-interacting balls traveling all uniformly relative to an inertial reference frame with a velocity V. If you mentally isolate a smaller and smaller subset of them, according to your formula, it seems there is a force acting on this subset. But, by definition, this force can not come from the neighboring points. Where does this force come from then?

but was ridiculed by D-H.
 
  • #54
afallingbomb said:
For a non-relativistic continuum (solid, fluid, etc) Newton's second law is simply the conservation of linear momentum expressed as:

http://en.wikipedia.org/wiki/Cauchy_momentum_equation

All i see on the lhs is mass per unit fluid volume element times acceleration of the said element.
 
  • #55
Dickfore said:
I said the same thing:



but was ridiculed by D-H.
Well, DH is dead wrong on this issue, and that is why he turns to ridicule, since he doesn't have any solid arguments.

Sorry that I didn't see that part of the previous discussion.
 
  • #56
arildno said:
D H said:
F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.
This is a meaningless position.

Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever.

I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O.
This is far from a meaningless position. Whether you want to call \dot m\,dv/dt a force or not is mostly semantics. Whatever you call it, it has the units of force and affects the equations of motion just like a force. The equations of motion will be correct. The only gotcha here is that that calling this quantity a force represents a break between Newton's laws and the conservation laws.

This is the point of view taken by many, if not all, high-fidelity missile and spacecraft simulations. Those simulations don't care about the ultra high-fidelity details that can only be divined by resorting to CFD models. Those CFD-based simulations are too slow to use as the basis for long-term simulations or Monte Carlo testing. All those high-fidelity sims care about is developing the temporal history of the vehicle's state (or vehicles' states in the case of a multi-vehicle simulation).

Calling this a meaningless position is just wrong. You just have to be aware of the limitations of this position. This is not my typical take on this issue; I prefer the F=dp/dt point of view. (Although this is the approach I use when all I care about are the equations of motion).
 
  • #57
And why should the momentum changing effect of volume expansion of your arbitrarily chosen control volume be called a "force"?

Why not call it by its proper name, namely..flux of momentum?

You might as well call a dog an apple, since both are organic compounds.


Classical mechanics, the mental framework we're in right now, is blind to other sources for change in momentum than forces acting upon particles contained within a system, along with whatever accretion of matter (carriers of momentum) to the system, for example through expansion of the volume of the geometric region we are talking about (for example having the interior region of ab expanding balloon as our control volume).

That other sources for momentum change exists as well is no good reason to make "force" into a flabby concept.
 
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  • #58
arildno said:
And why should the momentum changing effect of volume expansion of your arbitrarily chosen control volume be called a "force"?
Because its a very useful concept?

Or maybe, aerospace engineers thinking that the thrust produced by a jet or a rocket is a meaningful quantity and testable quantity is just plain goofy. NOT.
 
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  • #59
Hi arildno, I remember seeing you present your definition of force before, but the textbooks I have used tend to agree with DH's definition. Do you have a supporting reference, or is this a personal approach?
 
  • #60
DaleSpam said:
DH's definition.
Actually, I gave three:
  1. F=dp/dt (e.g., Marion, Goldstein (just double-checked; Goldstein is rather insistent that F=dp/dt is the proper form)),
  2. F=ma (e.g., Halliday&Resnick plus several engineering texts, with the mdot·v term denoted as a force)
  3. F=dp/dt=ma is a potato/patato issue because Newton's laws apply only to constant mass point masses.

I generally prefer (1) F=dp/dt because of the connection with the conservation laws. I do use (2) when the driving concern is the equations of motion.

BTW, it is rather well-known that F=dp/dt is frame-dependent. I don't get what all the hub-bub is about.
 
  • #61
When dealing with systems with variable mass, one needs to consider not only the "reactive forces", but also the shifting of the center of mass of a system due to its mass redistribution. Let us find an equation of motion for the center of mass of an open and not isolated system A.

In order that Newton's Laws be applicable, we must consider both the closure and isolation of this system A + B by identifying all the particles that flow out and into the system during a time interval \Delta t as well as all particles that exert forces on the particles.

The acceleration of the center of mass of the system is given by:

<br /> \mathbf{a}^{(A)}_{\mathrm{C M}}(t) = \frac{d \mathbf{v}^{(A)}_{\mathrm{C M}}(t)}{dt}<br />

where the velocity of the center of mass is by definition given by:

<br /> \mathbf{v}^{(A)}_{\mathrm{C M}}(t) \equiv \frac{\mathbf{P}^{(A)}(t)}{M^{(A)}(t)}<br />

where:

<br /> \mathbf{P}^{(A)}(t) = \sum_{a \in A(t)}{\mathbf{p}_{a}(t)}<br />

is the total momentum of the particles in the system A at the instant t and:

<br /> M^{(A)}(t) = \sum_{a \in A(t)}{m_{a}}<br />

is their total mass. As was pointed out by D-H's derivation in post #37, these quantities can be time dependent in two different ways: Either because the quantities that enter the sum change with time, or because the set over which we sum changes itself, or both, of course.

According to the rules of Calculus, we may write:

<br /> \mathbf{a}^{(a)}_{\mathrm{C M}})(t) = \frac{1}{M^{(A)}(t)} \, \frac{d \mathbf{P}^{(A)}(t)}{d t} - \frac{\mathbf{P}^{(A)}(t)}{[M^{(A)}(t)]^{2}} \, \frac{d M^{(A)}(t)}{d t}<br />

or:

<br /> M^{(A)}(t) \, \mathbf{a}^{(A)}_{\mathrm{C M}}(t) = \frac{d \mathbf{P}^{(A)}(t)}{d t} - \frac{\mathbf{P}^{(A)}(t)}{M^{(A)}(t)} \, \frac{d M^{(A)}(t)}{d t} = \frac{d \mathbf{P}^{(A)}(t)}{d t} - \mathbf{V}^{(A)}_{\mathrm{C M}}(t) \, \frac{d M^{(A)}(t)}{d t}<br />

Let us find the derivatives on the rhs of this equality. In doing so, we will adopt the following notation. Let the index a enumerate the particles that were inside the system A at time t; let the index i enumerate the particles that exited system A into system B and let j enumerate the particles that entered from B to A during the time interval \Delta t. Then we have:

<br /> \mathbf{P}^{(A)}(t + \Delta t) = \sum_{a}{\mathbf{p}_{a}(t + \Delta t)} - \sum_{i}{\mathbf{p}_{i}(t + \Delta t)} + \sum_{j}{\mathbf{p}_{j}(t + \Delta t)}<br />

We will keep quantities up to order O(\Delta t) in the above sum. It is important to realize that the number of particles that enter or exit (and, therefore, both the mass and momentum they carry with them) is a quantity of the order O(\Delta t). That is why we can exchange the argument from t + \Delta t to t in the last two sums. Similarly, we can write:

<br /> p_{a}(t + \Delta t) = p_{a}(t) + \Delta t \, \sum_{b \in A + B, b \neq a}{\mathbf{F}_{b a}(t)} + o(\Delta t) = p_{a}(t) + \Delta t \, \left[\sum_{a&#039; \in A, a&#039; \neq a}{\mathbf{F}_{a&#039; a}(t)} + \sum_{b \in B}{\mathbf{F}_{b a}(t)}\right] + o(\Delta t)<br />

Collecting everything together, we may write:

<br /> \mathbf{P}^{(A)}(t + \Delta t) = \mathbf{P}^{(A)}(t) + \Delta t \, \left[\sum_{a, a&#039; \in A, a&#039; \neq a}{\mathbf{F}_{a&#039; a}(t)} + \sum_{a \in A, b \in B}{\mathbf{F}_{b a}(t)}\right] - \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \mathbf{v}_{i}(t)} + \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \mathbf{v}_{j}(t)} + o(\Delta t)<br />

<br /> \frac{d\mathbf{P}^{(A)}(t)}{d t} = \sum_{a, a&#039; \in A, a&#039; \neq a}{\mathbf{F}_{a&#039; a}(t)} + \sum_{a \in A, b \in B}{\mathbf{F}_{b a}(t)} - \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \, \mathbf{v}_{i}(t)} \right]} + \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \, \mathbf{v}_{j}(t)} \right]}<br />

The first term in this equation is exactly equal to zero due to Third Newton's Law (the summation is over the same system A and \mathbf{F}_{a a&#039;} = -\mathbf{F}_{a&#039; a}) The second term is the sum of all the instantaneous external forces that act on the system \sum{\mathbf{F}_{\mathrm{ext}}(t)}. The third and the fourth term are the outgoing and incoming momentum flux, respectively, these are frame dependent quantities (because velocity is a frame dependent quantity). Although expressed as limits, these are actually finite quantities (if the particle flux is finite).

The other derivative is:

<br /> M^{(A)}(t + \Delta t) = \sum_{a}{m_{a}} - \sum_{i}{m_{i}} + \sum_{j}{m_{j}} = M^{(A)}(t) - \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i}} + \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j}}<br />

<br /> \frac{d M^{(A)}(t)}{d t} = -\lim_{\Delta t \rightarrow 0}{\left \frac{1}{\Delta t} \, \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i}}\right]} + \lim_{\Delta t \rightarrow 0}{\left \frac{1}{\Delta t} \, \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j}}\right]}<br />

Again, both of these terms are finite.

Combining everything together, we can finally write:

<br /> M^{(A)}(t) \, \mathbf{a}^{(A)}(t) = \sum{\mathbf{F}_{\mathrm{ext}}}(t) - \mathbf{\Pi}_{\mathrm{out}}(t) + \mathbf{\Pi}_{\mathrm{in}}(t)<br />

where \mathbf{\Pi}_{\mathrm{in/out}}(t) is the incoming (outgoing) momentum flux in the center-of-mass reference frame (a frame independent quantity, because relative velocities are Galilean invariants):

<br /> \mathbf{\Pi}_{\mathrm{out}}(t) = \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \, \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \, \mathbf{u}_{i}}\right]}, \; \mathbf{u}_{i} = \mathbf{v}_{i} - \mathbf{v}^{(A)}_{\mathrm{C M}}<br />

<br /> \mathbf{\Pi}_{\mathrm{in}}(t) = \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \, \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \, \mathbf{u}_{j}}\right]}, \; \mathbf{u}_{j} = \mathbf{v}_{j} - \mathbf{v}^{(A)}_{\mathrm{C M}}<br />

By specifying the number of particles that leave (enter) the system in unit time \varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t) with a relative velocity in a unit velocity-space volume around \mathbf{u} and in unit interval around the mass m, we can express the above quantities as:

<br /> \mathbf{\Pi}_{\mathrm{out}/\mathrm{in}}(t) = \int{m \, \mathbf{u} \, \varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t) \, d^{3}u \, dm}<br />

Also, the mass rate of change is:

<br /> \frac{d M^{(A)}(t)}{d t} = -\mu_{\mathrm{out}}(t) + \mu_{\mathrm{in}(t)}<br />

where the incoming (outgoing) mass flux rate is given by:

<br /> \mu_{\mathrm{out}/\mathrm{in}}(t) = \int{m \varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t) \, d^{3}u \, dm}<br />
 
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  • #62
Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?
 
  • #63
pallidin said:
Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?

But, no one said it's a closed system.
 
  • #64
There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?
 
  • #65
Let's get real here. Has a totally open system EVER been observed beyond mathematical construct in true reality?
Can't think of a single one...
 
  • #66
pallidin said:
There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?

Your body.
 
  • #67
pallidin said:
Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?
A rocket that in the ten minutes or so that it takes to get from the ground to insertion orbit spews out 90% of of its mass. The center of mass does shift, and this shift does indeed impact the dynamics. Ignoring that effect results in an error on the order of 10s (maybe 100?) of meters. (It's been a while since I attacked this problem.)

Even worse, at least as far as modeling is concerned, is the impact on rotation. For translational state, all that matters is the rate at which the center of mass is moving within the vehicle. The rotational analog of a shifting center of mass is a shifting inertia tensor. However, knowing the rate at which the inertia tensor is changing is not enough to solve the rotational problem. You get a nasty line integral; the path taken by the flowing fuel / exhaust gases inside the vehicle also come into play. Normally this effect is very tiny (a launch vehicle turns about 9 degrees/minute during the climb to insertion burn). There are some other circumstances where this effect can be significant.
 
  • #68
pallidin said:
There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?
Jets, rockets, turbines, ...
 
  • #69
Jets, rockets and turbines can not perform without an external environment of some sort, even if it's just gravity itself.
The external environment envelopes and defines the nature of the thrust.
Being thus intertwined, the system is NOT closed.
 
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  • #70
pallidin said:
Jets, rockets and turbines can not perform without an external environment of some sort, even if it's just gravity itself.
The external environment envelopes and defines the nature of the thrust.
Being thus intertwined, the system is NOT closed.

And that's what we said: the system is open.
 
  • #71
Open to what?
We have defined parameters that demands external forces are eminently and immutably existent.
How is that "closed" at all?
 
  • #72
Please chose one of the options below:

[1] Open mechanical systems exist;

[2] Open mechanical systems do not exist.

Read carefully before you chose.
 
  • #73
Dickfore said:
Please chose one of the options below:

[1] Open mechanical systems exist;

[2] Open mechanical systems do not exist.

Read carefully before you chose.

My answer, 2
Totally open mechanical systems are not physically possible.
Provide even one example of a totally open system.
 
  • #74
In classical physics and thermodynamics, systems can be isolated, closed, or open. An isolated system is truly isolated. It exchanges neither energy nor mass with its surroundings. The surroundings are completely irrelevant in the case of an isolated system. The surrounding environment is important for closed systems and open systems. A closed system exchanges energy but not mass with its surroundings. An open system exchanges energy and mass with its surroundings.

Isolated systems are nice for toy problems. The only true isolated system is the universe as a whole (maybe; I think the jury is still out even on that). Many systems are approximately isolated. The solar system, for example.

The solar system obviously is not truly isolated. It does interact gravitationally with nearby stars and the galaxies.The solar system is, in fact, an open system. Some of the solar wind does escape the solar system, and some of the interstellar medium crosses the heliopause into the solar system. So, how can the solar system be an open system and yet be accurately treated as an isolated system? Simple: The interactions are tiny. Those interactions can be safely ignored in many applications. Even the most accurate of the solar system ephemerides model do not model interactions with bodies outside of the solar system.

So the solar system illustrates one simple way to deal with the challenges presented by an open system: Pretend otherwise. A lot of physics involves knowing what simplifying assumptions could help with solving a problem and determining whether using those simplifying assumptions comprises the solution.

With a rocket ignoring the exhaust is not an option. The exhaust is an extremely important part of the dynamics of a rocket. (It is in a sense why the rocket goes.) The exhaust stream, however, is not important at all. What happens to the stuff in the exhaust stream once it is well past the shock boundary isn't going to impact what happens to the rocket one iota. Properly drawing the control surface that separates the system of interest from the surrounding environment is another one of those simplifying that physicists and engineers make.
 
  • #75
D H said:
An isolated system is truly isolated. It exchanges neither energy nor mass with its surroundings.

While true in a theoretical sense, it is not possible in the real world at all.
 
  • #76
Can it be shown, for example, that a totally isolated system can exist at all?
I want some evidence. Give me even a small piece to chew on.
Any takers?
 
  • #77
So what is the argument? DickFore has been talking about open systems. A rocket, for example. Reread the definitions of isolated, closed, and open systems, and then answer Dick's question in post #72.
 
  • #78
Jesus Christ! Will you choose a stance and stick to it?
 
  • #79
Dickfore said:
Jesus Christ! Will you choose a stance and stick to it?

What? There IS NO SUCH THING as a closed system(excepting the universe of course)
That's my stance from the beginning if you didn't notice.
Hello?
 
  • #80
No, your stance has been that an open system doesn't exist. That is what has Dick and I confused.

That said, there certainly are systems that are for all practical purposes closed or isolated. If the effects of some external interaction are orders of magnitude smaller than the effects from state uncertainties, measurement errors, etc., or orders of magnitude smaller than any physically meaning result, who cares? If all external interactions are dwarfed by uncertainties, errors, and significance, the system is for all practical purposes isolated.
 
  • #81
Can ANYONE show me even ONE pure example of an open-system?
No.
That's my point, and highly relevant to physics.
 
  • #82
?

Will you take a stance, please? And do stop using words like "totally open" and "pure example of an open system". That doesn't make sense.
 
  • #83
Well, then, allow me to rephrase.
Is there any experimental evidence which demonstrates that systems are totally isolated from another? Any at all? Please give evidence.
 
  • #84
I think your confusion is that you do not understand that open is the opposite of isolated. Reread the definitions in post #74, and if what I wrote doesn't make sense, use google.

To answer your question, yes, except for systems that are isolated by virtue of the expansion of space, all systems interact with one another to some extent. That said, no physicist is going to care about the gravitational effect of some extremely remote star on their Earthbound gravity experiment. The interaction is so incredibly tiny that it can be utterly ignored. It doesn't exist as far as the experiment is concerned.

Modeling a system as being isolated or closed is a useful abstraction -- when that abstraction is applicable.
 
  • #85
D H said:
?

Will you take a stance, please? And do stop using words like "totally open" and "pure example of an open system". That doesn't make sense.

Oh, what is wrong with "totally open"?
That simply means open without boundaries.
And "pure example" is only asking for the truth, that's all.
 
  • #86
D H said:
Modeling a system as being isolated or closed is a useful abstraction -- when that abstraction is applicable.

Upon which standard do you define the abstraction? Not easy, my friend.
 
  • #87
pallidin said:
Oh, what is wrong with "totally open"?
That simply means open without boundaries.
What's wrong with your terminology is that it is in complete disagreement with the standard definition of an open system. A system without boundaries cannot be open. Think of your house. Open the doors and windows and air flows into and out of the house. The house with the open windows and doors is an open system. Close all the doors, close all the windows, seal all the cracks and the air flow pretty much stops. That's a closed system (approximately). It's still not an isolated system because heat flows between the inside and outside the house through the closed windows, the closed doors, the walls, the floor, and the ceiling.
 
  • #89
I will retire for the evening and consider your thoughts.
Thank you.
 
  • #90
Dickfore said:
lol, what ever you say. I don't want to quarrel and get a ban. If someone reads this thread and knows Physics, they will draw a conclusion for themselves.

A ridiculous thread. cjl and Dickfore clearly have lot stronger understanding of basic mechanics than D H :rolleyes:

In the beginning cjl was trying to explain this:

cjl said:
Rocket motors do not change in thrust just because you attach them to a different vehicle. The force they generate is purely a function of their massflow and their exhaust velocity. Hence, vehicle mass and velocity are irrelevant.

D H disagrees. According to him the thrust must depend on the vehicle into which the rocket is attached. He justifies this by defending the importance of the concept of thrust.

D H said:
Or maybe, aerospace engineers thinking that the thrust produced by a jet or a rocket is a meaningful quantity and testable quantity is just plain goofy. NOT.

Makes no sense! The thrust produced by a rocket must not exist, because it must exist?
 
  • #91


D H said:
cjl said:
(Specifically, F = \dot mVe for any rocket motor, in which \dot m is the fuel mass flow rate, and Ve is the exhaust velocity)
That is correct only if you toss the definition of force as F=dp/dt and use m·dv/dt=Fext+u·dm/dt in its place, where u is relative velocity of expelled material. Defining Freaction≡u·dm/dt let's one simply use F=ma, even for a system with non-constant mass. This form is admittedly very useful as the basis for the equations of motion of a rocket. It is however absolutely useless for computing things like work precisely because it throws out the connection with the conservation laws.
It is this \dot mVe term that this entire silly imbroglio is all about. Is it a force or is it something that happens to have units of force? The answer lies in how one interprets Newton's second law. Is force defined by F=dp/dt or F=ma? Different textbooks do use different definitions, and the two definitions definitely are not the same in the case of a variable mass system.
 
  • #92
It is a force. It comes from the ejected particles who act on the reamining part of the rocket according to 3rd Newton's Law.
 
  • #93
Fine. Is this force the same in all frames?

Answering yes puts you in the F=ma camp.
Answering no puts you in the F=dp/dt camp.
 
  • #94
D H said:
Fine. Is this force the same in all frames?

Answering yes puts you in the F=ma camp.
Answering no puts you in the F=dp/dt camp.

Forces are Galilean invariants. This one is too. To see this, notice that both \mathbf{u} and \dot{m} are Galilean invariants. There are no camps. Your transformation formula for the force was incorrect.
 
  • #95
Then tell that to Marion, and Goldstein (or the authors who have taken over for them) and others as well.

And please do find the flaw in that derivation. HINT: it is not my derivation.
 
  • #96
The "third camp", the one you said retreated from the discussion, was right. I guess they saw that they were dealing with obtuse opponents, so there was no point for them to propagate the discussion.

You (and I) derived something which is not Second Newton's Law. It simply gives the acceleration of the center of mass of a system with variable mass in terms of the external forces (real ones) acting on the particles that happen to consist the system at that instant and the momentum flux (according to the CM frame) that goes out and into the system. The equation happens to be of the form:

<br /> m(t) \, \mathbf{a}(t) = \sum{\mathbf{F}_{\mathrm{ext}}(t)} - \mathbf{\Pi}_{\mathrm{out}}(t) + \mathbf{\Pi}_{\mathrm{in}}(t) <br />

However, this is NOT Second Newton's Law! It is an equation of motion. I don't even think Newton wrote down an equation of motion for a system with variable mass. I think this equation is derived by Meshchersky in the beginning of the XX century.

Furthermore, this equation tells us about the motion of the CM of the variable mass system. As an extreme example, a variable mass systme is a two-body system where one of the particles exits the (geometric) boundaries. It is obvious that the above equation is of little use in that case. The case where this equation is useful is where it makes sense to approximate the momentum flux $\mathbf{\Pi}$ as a continuous function of time and the object you are considering has a much bigger mass than the mass it exchanges. Also, it is convenient if the geometric boundaries of the object are actually rigid walls and the motion of the object is then that of a rigid body (with a variable mass fuel compartment). This is an abstraction of a 'rocket'.

As you mentioned very well, this equation is not sufficient to predict the motion of such a 'rocket'. One still needs the equation of motion for the rotation angles in relation to the external torques. As you noted, the problem is quite complicated in realistic situations.
 
  • #97
DH's position is just silly.
It is unsurprising that he employs the logical fallacy ad authoritam

To take a case where it is utterly nonsensical to describe the momentum flux as a force, we can look at a fluid moving at constant (horizontal) velocity U, and choosing as our control volume that at t=0 starts out as a line segment of length 1, broadening into a rectangle, where one side remains stationary at the initial position of the line segment, the other vertical side moving with (horizontal) velocity V.
Letting d be the density of the fluid, the momentum containe in our control volume is simply U*V*d*t, with a rate of change U*V*d

In this case, the rate of change of momentum within our control volume cannot be ascribed as the effect of an acting force.
Forces act solely upon material particles, not upon arbitrarily chosen spatial regions*. The rate of change of moementum is solely due to flux of momentum, i.e, a quantity that has the same units as force, but is still wholly distinct from force.


Note:
Some use "momentum flux" to designate what I'd call "momentum flux density", i.e, the rate of momentum transfer per unit area(by means of momentum-carrying particles leaving the control volume).


*Remember that in the classical world time&space are dynamically inactive quantities, merely the empty box within which dynamics and the play of forces occur.
 
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  • #98
My view might be silly, but it is the one most widely supported in literature.

Oh, and it is also useful.
 
  • #99
Can a rocket, as opposed to the rocket + exhaust, be modeled with a Lagrangian?
 
  • #100
If it is going through a non-conservative medium (e.g., the atmosphere), no. If it has a control system that turns thrusters on and off, throttles them, gimbals them, or activates some other effector to keep the vehicle under control, no.

In short, the answer is no.
 
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