Classical Systems With Variable Mass And Other Geometric Systems:

[arildno]

Classical Systems With Variable Mass And Other Geometric Systems

1. INTRODUCTION
1a) The material particle/system
The fundamental object in classical mechanics is the material particle.
A material particle has a position and a mass, and can be subject to forces.
A material system is a physical system consisting of the SAME set of material particles over time.
1b) Classical laws valid for the material particle:
MASS CONSERVATION
The classical material particle has the same mass throughout time, that is, the material particle is subject to the law of mass conservation.
Clearly, since a material system consists of the same material particles throughout time, the total mass of the material system can't change with time either.
NEWTON'S LAWS OF MOTION:
The material particle can be subject to forces, and its acceleration $$\vec{a}$$ is related to the forces $$\vec{F}$$ it is subject to through Newton's 2.law of motion:
$$\vec{F}=m\vec{a}$$
Since the material particle is presumed to have constant mass, a classically equivalent form of Newton's 2.law is:
$$\vec{F}=\frac{d\vec{p}}{dt},\vec{p}=m\vec{v},\vec{a}=\frac{d\vec{v}}{dt}$$
where $$\vec{v}$$ is the particle's velocity, and $$\vec{p}$$ the particle's momentum.
1c) Galilean invariance
Note in particular that since the value of the particle's mass is presumed independent of different observers' relations to each other (i.e, mass is seen as an "absolute", or invariant quantity), Newton's 2.law is a Galilean invariant in either of its formulations, since, by the law of mass conservation, $$\frac{dm}{dt}=0$$
2. GEOMETRIC PHYSICAL SYSTEMS
2a) MOTIVATION
It is an unnecessarily restrictive physics which only wants to deal with the behaviour of strictly material systems, since there are lots of other systems which might be of interest to study.
For example, consider a fluid flowing through a bent tube section.
It might be convenient to try and find a way to calculate what force the fluid exerts upon the tube section we're looking at throughout time, but because it is not the same fluid elements which is in contact with/in the vicinity of the tube section throughout time (the fluid flows through it and beyond), the isolated system of tube section+fluid contained in it is NOT a material system!
Also, a rocket moves forward by expelling exhaust backwards, so if we're interested in describing the motion of the rocket as a function of time, we're effectively wanting to describe the behaviour of a system which is continually losing mass through fuel ejection.
That is, the rocket+remaining fuel-system is not a material system.
Such naturally occurring systems are called geometric systems, and our aim is to formulate the proper form of the laws of motion for such systems.
To be continued..

 

[arildno]

Geometric systems continued

2b). THE COINCIDENT MATERIAL SYSTEM
Our trick will be to consider the dynamics of a material system instantaneously coincident with our geometric system, and through that, determine what laws a geometric system must obey.

The fancy way of doing this is by deriving what is known as Reynolds' transport theorem, we'll choose a more visual approach.

First, let us describe our geometric system:
Let S be a region in space, and let there exist material particles enclosed by S and outside of S.
S can be moving through space or undergo deformations, or it might be fixed and stationary.

However, we require that material particles can both enter and leave S, that is :
S is an arbitrarily chosen GEOMETRIC region, where in particular the boundary of S is not determined by whatever material particles happen to be there.

Let us consider S and its surroundings at the two times $$t$$ and $$t+\delta{t}$$
Now, material particles can be subdivided into 4 base types:
1) Those particles which were enclosed in S at time $$t$$ and which is also enclosed in S at time $$t+\delta{t}$$
We assign to that subset of particles a momentum $$\vec{p}_{E}(t)$$ at time "t", and momentum $$\vec{p}_{E}(t+\delta{t})$$ at time $$t+\delta{t}$$
2)Those particles which were enclosed in S at time $$t$$ and but which are NOT enclosed in S at time $$t+\delta{t}$$
We assign to that subset of particles a momentum $$\vec{p}_{-}(t)$$ at time "t", and momentum $$\vec{p}_{-}(t+\delta{t})$$ at time $$t+\delta{t}$$
3)Those particles which were NOT enclosed in S at time $$t$$ and but which ARE enclosed in S at time $$t+\delta{t}$$
We assign to that subset of particles a momentum $$\vec{p}_{+}(t)$$ at time "t", and momentum $$\vec{p}_{+}(t+\delta{t})$$ at time $$t+\delta{t}$$
4) Those particles which is outside of S at both times; these are ignored in the following.

The MATERIAL system which contains particles of type 1),2),3) is called M

The S-momentum:
At every time, there is an amount of momentum enclosed in S called $$\vec{p}_{S}$$, and from the description above, we have:
$$\vec{p}_{S}(t)=\vec{p}_{E}(t)+\vec{p}_{-}(t) (1)$$
$$\vec{p}_{S}(t+\delta{t})=\vec{p}_{E}(t+\delta{t})+\vec{p}_{+}(t+\delta{t}) (2)$$

The M-momentum
The amount of momentum in M is called $$\vec{p}_{M}$$, and from the description above, we have:
$$\vec{p}_{M}(t)=\vec{p}_{E}(t)+\vec{p}_{-}(t)+\vec{p}_{+}(t) (3)$$
$$\vec{p}_{M}(t+\delta{t})=\vec{p}_{E}(t+\delta{t})+\vec{p}_{-}(t+\delta{t})+\vec{p}_{+}(t+\delta{t})(4)$$

Now, M is a material system, and thus, we know the rate of change of its momentum is related to the external forces acting upon it in $$\vec{F}=\frac{d\vec{p}_{M}}{dt}$$

What we're interested in, is the appropriate relation which holds for the rate of change of momentum enclosed in S, that is, $$\frac{d\vec{p}_{S}}{dt}$$
But, by subtracting eq. (1) from (2), adding zero on the right-hand side in an intelligent manner, dividing with $$\delta{t}$$ and using (3)+(4), we gain:
$$\frac{\vec{p}_{S}(t+\delta{t})-\vec{p}_{S}(t)}{\delta{t}}=\frac{\vec{p}_{M}(t+\delta{t})-\vec{p}_{M}(t)}{\delta{t}}-\frac{\vec{p}_{-}(t+\delta{t})-\vec{p}_{+}(t)}{\delta{t}}$$

Now, the left hand side will go to $$\frac{d\vec{p}_{S}}{dt}$$ as $$\delta{t}\to0$$, whereas the first term on the right-hand-side will go to $$\frac{d\vec{p}_{M}}{dt}$$, and hence, by invoking Newton 2-law, $$\vec{F}$$

We'll focus on the second term on the right hand-side:
Simplify particles of type 2) as a "single" continously leaving particle, whose momentum may be written as:
$$\vec{p}_{-}(t+\delta{t})=\delta{m}_{-}\vec{v}^{(-)}$$
where $$\delta{m}_{-}$$ is its mass, and $$\vec{v}^{(-)}$$ is the velocity it has right AFTER it has left S.
Make the analogous rewriting for the particle of type 3):
$$\vec{p}_{+}(t)=\delta{m}_{+}\vec{v}^{(+)}$$
where $$\delta{m}_{+}$$ is its mass, and $$\vec{v}^{(+)}$$ is the velocity it has right BEFORE it has entered S.
Thus, by setting $$\frac{\delta{m}_{-}}{\delta{t}}=\dot{m}_{-},\frac{\delta{m}_{+}}{\delta{t}}=\dot{m}_{+}$$
we gain, in the limit when rearranging:
$$\frac{d\vec{p}_{S}}{dt}+\dot{m}_{-}\vec{v}^{(-)}-\dot{m}_{+}\vec{v}^{(+)}=\vec{F}(5)$$
This is the law we were after!
Interpretations follow in the next post..

 

[arildno]

3) Interpretations

a) GALILEAN INVARIANCE
We will show that the left-hand side of (5) is, indeed, a Galilean invariant (and thus, that the right-hand side is invariant as well)
Now, if $$m_{S}(t)$$ is the mass contained in S at time t, and $$\vec{v}_{C.M}(t)$$ is the velocity of the center of mass in S, we clearly have:
$$\vec{p}_{S}(t)=m_{S}(t)\vec{v}_{C.M}$$
Furthermore, the rate of change of the mass contained in S must be influx of mass (in the form of new particles) minus outflux of mass (in the form of "old" particles):
$$\frac{dm_{S}}{dt}=\dot{m}_{+}-\dot{m}_{-}$$
whereby (5) can be rewritten as:
$$m_{S}\vec{a}_{C.M}+\dot{m}_{-}(\vec{v}^{(-)}-\vec{v}_{C.M})-\dot{m}_{+}(\vec{v}^{(+)}-\vec{v}_{C.M})=\vec{F}(6)$$
But, since the mass fluxes can only depend on the RELATIVE velocities of the particles and S, this is clearly an invariant feature, and thus, the whole left-hand side is a Galilean invariant.
b) THE MASS FLUX AND THE N-PARTICLE SYSTEM:
Now, $$\dot{m}_{-}$$ is the positive outflux rate of mass (of those particles which actually left S), whereas $$\dot{m}_{+}$$ is the positive influx rate of mass (of those particles which actually entered S.)

By introducing the MASS FLUX $$\dot{m}$$ to be outflux when positive, and negative when influx, and remembering that we are always to use the velocity a particle has when OUTSIDE S (whether "before" or "after"), we see that for a system where N particles either leaves or enters S, we have, from eq. (5):
$$\frac{d\vec{p}_{S}}{dt}+\sum_{i=1}^{N}\dot{m}_{i}\vec{v}_{i}=\vec{F}(7)$$
where $$\dot{m}_{i},\vec{v}_{i}$$ are the appropriate values for the i's particle's mass flux and velocity.
c) TRANSITION TO REYNOLDS' TRANSPORT THEOREM; THE MOMENTUM FLUX
Assume that a geometric boundary point on S has velocity $$\vec{v}_{A}$$, whereas a COINCIDENT material point (with density $$\rho$$) has velocity $$\vec{v}$$
Let the local area on the surface of S be dA, with outwards normal vector $$\vec{n}$$.
Then, during a time interval $$\delta{t}$$, a material particle with mass $$\rho(\vec{v}-\vec{v}_{A})\cdot\vec{n}dA\delta{t}$$ leaves S
(If the quantity is negative, it means that the particle enters)
The infinitesemal mass flux is therefore $$\rho(\vec{v}-\vec{v}_{A})\cdot\vec{n}dA$$
Thus, taking (7) to the limit of infinitely many particles, we gain, in integral form:
$$\frac{d}{dt}\int_{S}\rho\vec{v}dS+\int_{A}\rho\vec{v}(\vec{v}-\vec{v}_{A})\cdot\vec{n}dA=\vec{F}(8)$$
where A is the surface of S.
The left-hand side is what you get out of Reynolds' transport theorem, and we see that the surface integral can be interpreted as THE FLUX OF MOMENTUM THROUGH A, i.e., the momentum flux.
d)NOTE ON MATERIAL SYSTEMS:
For material systems, material particles enclosed cannot leave or enter the system's boundary, that is $$\vec{v}=\vec{v}_{A}$$ at all points on the boundary of the material system M.
Thus, we regain the Newton's 2. law of motion for the material system, since the momentum flux term vanishes identically:
$$\frac{d}{dt}\int_{M}\rho\vec{v}dM=\vec{F}$$
dM is the infinitesemal volume element of M.
e) NOTE ON $$\vec{F}$$
Note that forces among particles of types 1), 2), 3) are INTERNAL forces in M, and is, hence not included in $$\vec{F}$$

Examples of use follows in the next posts..

 

[arildno]

The rocket equation

A rocket gains its forward thrust by ejecting fuel backwards.
We will let our geometric system R be a region in space which include the rocket+remaining fuel-system, but does not include any of the ejected fuel.
We assume that the rocket is moving through vacuum, and is not subject to any external forces, other than pseudo-forces if we describe the motion in a non-inertial reference frame.
We have basically two choices of reference frames:
An inertial reference frame and a non-inertial reference frame.

1. The inertial reference frame description:
At time "t", let the rocket+remaining fuel be wholly contained within a region R, whereas no ejected fuel is contained in R.
The rocket and remaining fuel has velocity $$\vec{v}_{R}$$ with respect to I, and the fuel just ejected out of R has velocity $$\vec{v}_{f}$$ with respect to I.
R itself moves along with the rocket and remaining fuel, so that the mass flux out of R equals the loss of the rocket's mass, that is the mass flux rate is $$\dot{m}_{-}=-\frac{dm_{R}}{dt}>0$$
where $$m_{R}(t)$$ is the mass of the rocket+remaining fuel.

There are no external forces acting, I is an inertial frame, and $$\dot{m}_{+}=0$$; thus (5) is written as:
$$\frac{d}{dt}(m_{R}(t)\vec{v}_{R})+\dot{m}_{-}\vec{v}_{f}=\vec{0}$$
By identifying $$\vec{v}_{f}-\vec{v}_{R}=\vec{v}_{e}$$ as the ejection velocity of the exhaust gas, we may rewrite the above equation as:
$$m_{R}\vec{a}_{R}=\frac{dm_{R}}{dt}\vec{v}_{e}$$
This is the rocket equation, and since $$\frac{dm_{R}}{dt}<0$$
we see that the rocket accelerates in the direction opposite of the ejection velocity, as it should.

2. The non-inertial reference frame description:
The simplest choice of reference frame and R is, however, to let R be as in 1, but choose the NON-inertial reference frame in which R, the rocket, and the remaining fuel is at rest.
In this reference frame, I', a pseudo-force equal to $$-m_{R}(t)\vec{a}_{R}$$ acts on the matter enclosed in R.
The velocity of the ejected fluid is therefore $$\vec{v}_{e}=\vec{v}_{f}-\vec{v}_{R}$$ where $$\vec{v}_{f},\vec{v}_{R}$$ are the velocities with respect to the inertial frame I, under 1.
Since the matter enclosed in R is at rest, $$\frac{d\vec{p}_{R}}{dt}=\vec{0}$$ so (5) is written as:
$$\vec{0}+\dot{m}_{-}\vec{v}_{e}=-m_{R}\vec{a}_{R}$$
which agrees with the expression in 1.

 

[arildno]

Sand falling onto a moving conveyor belt

Let us consider a conveyor belt moving with constant velocity $$\vec{v}_{c}=V\vec{i}$$
Sand falls onto the conveyor belt with a velocity $$\vec{v}_{S}=V_{x,S}\vec{i}+V_{y,S}\vec{j}$$ measured relative to the inertial frame in which $$\vec{v}_{c}$$ is as above.
We assume that the sand is instantaneously accelerated to the conveyor belt once it hits it.

Let our system consist of the conveyor belt plus already co-moving sand.
The mass of the system is $$m_{c}(t)$$

What net force must act upon the conveyor belt in order for $$\vec{v}_{c}$$ to remain constant?

Now, the mass influx rate equals our system's mass gain, i.e $$\dot{m}_{+}=\frac{dm_{c}}{dt}$$
Thus, (5) can be written:
$$\frac{d}{dt}(m_{c}(t)\vec{v}_{c})-\frac{dm_{c}}{dt}\vec{v}_{S}=\vec{F}$$
That is,
$$\vec{F}=\frac{dm_{c}}{dt}(\vec{v}_{c}-\vec{v}_{S})$$

Note in particular, that if we have $$V_{x,S}=V$$, only a net normal force is required to act upon the conveyor belt in order to keep its velocity constant.

 

[e(ho0n3]

I've never seen this kind of treatment on the subject so forgive my misunderstandings below.

c) TRANSITION TO REYNOLDS' TRANSPORT THEOREM; THE MOMENTUM FLUX
Assume that a geometric boundary point on S has velocity $$\vec{v}_{A}$$, whereas a COINCIDENT material point (with density $$\rho$$) has velocity $$\vec{v}$$
Let the local area on the surface of S be dA, with outwards normal vector $$\vec{n}$$.

COINCIDENT as in located in the same stop?

Then, during a time interval $$\delta{t}$$, a material particle with mass $$\rho(\vec{v}-\vec{v}_{A})\cdot\vec{n}dA\delta{t}$$ leaves S
(If the quantity is negative, it means that the particle enters)

Hold on here. How did you get that expression for the mass of the material particle?

 

[arildno]

COINCIDENT as in located in the same stop?

Precisely.
The region we're looking at is purely a geometric quantity which can be assigned a position and extent in space, relative to some reference frame.
As an example, let S denote the following moving set of points:
$$S(t)=\{(x,y)\in\Re^{2}|Ut\leq{x}\leq{Ut+1},0\leq{y}\leq1\}$$
where U is a constant.
S is therefore a moving unit square.
If a stationary material point P is located at (2,0.5), P will be outside of S up to time$$t=\frac{1}{U}$$ when it will be on the boundary; in the interval $$\frac{1}{U}\leq{t}\leq\frac{2}{U}$$ P will be inside S, and then it gets out of S again for $$t>\frac{2}{U}$$
With a material point, I mean a particle which has been assigned a mass (or density), and which can be subject to forces.
A geometric point has no mass, and is not subject to forces, but may have a well-defined position (function) with respect to a reference frame.

As for the expression of for mass:
The velocity by which a particle distances itself from a surface is
$$(\vec{v}_{p}-\vec{v}_{sur})\cdot\vec{n}$$
where $$\vec{v}_{p},\vec{v_{sur}},\vec{n}$$ are the particle velocity, surface velocity, and normal to the surface.
That is, the particle's RELATIVE normal velocity to the surface.
If the particle's relative velocity to the surface had been purely tangential, that is the same thing that the particle doesn't LEAVE the surface..
All right?

Now, I think it's great that you ask questions, ehoon, and I will do my best to answer them to your satisfaction.
But:
Could you consider PM'ing me instead, or post it in the discussion thread sticky in General Physics?
Since we're testing out if there is interest to making various FAQ sticky's, or stickys which deal with commonly occurring themes, and my thread here might if it generates enough support be one of those, it would be easier for future readers if the argument in the sticky is uninterrupted.
(I plan to write quite a few more posts in this thread..)

 

[arildno]

The external forces approach

Now, before moving onwards with another example, I'll describe another manner in which to derive the rocket equation.
Consider the MATERIAL system M which consist of the rocket hull and SOME of its remaining fuel mass, so that its mass, $$m_{M}$$ equals the mass of the rocket+all fuel remaining-system R at some time $$t^{*}$$
That is:
$$m_{M}=m_{R}(t^{*})$$
$$m_{R}(t)$$ is the mass of system R as a function of time.
Thus, up to time $$t^{*}$$ system M is certainly subject to EXTERNAL forces, namely those generated by that system R ejects other fuel pieces than those fuel pieces contained in system M.
That is, for $$t<t^{*}$$ Newton's 2.law of motion for M reads:
$$\vec{F}=m_{M}\vec{a}_{M}=m_{M}\vec{a}_{R}$$
where $$\vec{F}$$ is some force, and M is co-moving with R (hence, equal accelerations).
Taking this equation to the limit $$t\to{t}^{*}$$ yields:
$$\vec{F}^{*}=m_{R}(t^{*})\vec{a}_{R}$$
Our aim is now to estimate $$\vec{F}^{*}$$

At time $$t^{*}$$, system R consists of system M plus some fuel of infinitesemal mass $$\delta{m}$$ which sits on the OUTER boundary of M.
In the time interval $$\delta{t}$$ M imparts an impulse $$-\vec{F}^{*}\delta{t}$$ on that fluid mass, so that it experiences a momentum change $$\delta{m}(\vec{v}_{f}-\vec{v}_{R})=\delta{m}\vec{v}_{e}$$
where $$\vec{v}_{f}$$ is the ejected fuel's velocity relative to the inertial frame in which R (and M) has velocity $$\vec{v}_{R}$$
$$\vec{v}_{e}$$ is the exhaust velocity of the fuel.
Thus, we gain:
$$\vec{F}^{*}=-\frac{\delta{m}}{\delta{t}}\vec{v}_{e}=\frac{dm_{R}}{dt}\vec{v}_{e}$$
since the positive outflux rate of mass equals the negative rate of change of system R's mass.
Hence, we gain:
$$\frac{dm_{R}}{dt}\vec{v}_{e}=m_{R}\vec{a}_{R}$$
and since M and $$t^{*}$$ were arbitrary, this equation holds for all t, and is the rocket equation.

Now, the external forces approach brings along an important insight:
Consider the conveyor belt example where the belt moves with constant horizontal velocity V, and the sand falls upon it strictly vertically, i.e, with 0 horizontal velocity.
In order for this to be possible, we've seen that the horizontal component of our force equation must be:
$$F=\frac{dm}{dt}V$$

Now, note that this superficially can be written as:
$$F=\frac{dm}{dt}V=\frac{d}{dt}mV=\frac{dp}{dt}$$
where p is the (horizontal) momentum in the conveyor belt+attached sand system.
Thus, it is at times said that the reason why we need an external horizontal force acting upon our system, is that the mass of the system increases.

This is totally wrong, and displays a complete misunderstanding of the issues involved!
The reason why the conveyor belt PROPER needs to be subject to an external force F, is that it also is subject to the force from the incoming particles, which have been accelerated by the conveyor belt up to the conveyor belt's own velocity.
That is in this case, the external forces approach will derive the (horizontal) equation:
$$F-\frac{dm}{dt}V=0$$

If, initially, the incoming particles had had velocity V, their attachment to the conveyor belt would not have changed their horizontal momentum at all, only their vertical momentum.
Thus, in that case, only a VERTICAL external force is needed on the conveyor belt in order to keep its velocity constant.

 

[arildno]

Flow through a tube

While the two variable-mass examples have been shown in the last post to be amenable to an alternative approach which might be regarded as somewhat simpler than the geometric system approach, the folllowing example is decidedly simpler to analyze properly when using a geometric system approach.

Consider a U-tube at rest with cross sectional area A, and let a fluid of density $$\rho$$ enter the tube at one open end with velocity $$U\vec{i}$$ and leave the tube at the other end with velocity $$-U\vec{i}$$
We let U be constant across the cross section, and for simplicity, we let the fluid pressure at both entrances be set to 0.
The fluid's velocity field is considered stationary.

Let V be the volume enclosed by the surface S consisting of the tube wall T and the two sections $$A_{1}, A_{2}$$
By neglecting the pressure forces from the fluid outside V along with the gravity force, the only force experienced by the fluid enclosed in V is that from the tube, $$-\vec{R}$$
$$\vec{R}$$ is therefore the force exerted by the fluid on the tube.

Recalling Reynolds' transport theorem, we have:
$$\frac{d}{dt}\int_{V}\rho\vec{v}dV+\int_{S}\rho\vec{v}(\vec{v}-\vec{v}_{S})\cdot\vec{n}dS=-\vec{R}, \vec{v}\mid_{A_{1}}=U\vec{i},\vec{v}\mid_{A_{2}}=-U\vec{i}$$
Since S is fixed, $$\vec{v}_{S}=\vec{0}$$, and since fluid cannot flow through the wall, the non-zero contributions of S is: $$\int_{A_{1}}U\vec{i}U\vec{i}\cdot(-\vec{i})dA_{1}+ \int_{A_{2}}(-U\vec{i})(-U\vec{i}\cdot(-\vec{i})dA_{2}=-2\rho{A}U^{2}\vec{i}$$

Since volume V is fixed in space, we have: $$\frac{d}{dt}\int_{V}\rho\vec{v}dV=\int_{V}\frac{\partial}{\partial{t}}(\rho\vec{v})dV=\vec{0}$$
since the velocity field was considered stationary.

Thus, we gain a very the simple expression for the force $$\vec{R}$$ acted upon the tube by the fluid:
$$\vec{R}=2\rho{A}U^{2}\vec{i}$$

Now, the great thing about this approach is that we may calculate the net force on the tube irrespective of the tube's actual geometry, and the fluid's complicated velocity field within it!

All we need to know, is :
1) The velocity field at the inlet and outlet zones
2) That the velocity field inside the tube is stationary
3) The pressure forces acting on the inlet&outlet plus volume forces (say gravity)

The pressure forces will often be small compared to the momentum flux term, and the gravity term is easy to include.

Thus, if we call the momentum flux term $$\vec{M}$$, and the easily calculated forces $$\vec{F}$$, the reaction force on the tube satisfies the simple equation:
$$\vec{R}=\vec{F}-\vec{M}$$

This is a well-known formula for any engineer; flow details are less important than their overall effect.
If the flow in the tube is turbulent, we would need to tweak the formula a bit, since the flow can't be regarded as strictly stationary anymore.