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Newton's laws in variable mass systems

  1. Aug 28, 2010 #1

    cjl

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    Thread split from [thread=424724]this thread[/thread].

    Not true.

    The spray can is generating a constant force, not a constant power. As a result, the work done by the spray can depends on the distance it travels, and the power generated depends on the rate at which it is traveling. The farther the spray can travels, the more work is done.
     
    Last edited by a moderator: Aug 31, 2010
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  3. Aug 29, 2010 #2

    D H

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    Re: Zero-G reaction forces

    Careful there, cjl! You are mixing reference frames here. The spray can is generating a constant force in the inertial frame instantaneously co-moving with the person holding the spray can. It is not generating a constant force in all inertial frames.
     
  4. Aug 29, 2010 #3

    cjl

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    Re: Zero-G reaction forces

    Yes it is. In all frames, the massflow from the spray can is the same, and the velocity difference between the spray can and the spray is the same. Therefore, in all frames, the force is the same.
     
  5. Aug 29, 2010 #4

    D H

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    Re: Zero-G reaction forces

    No, it's not. The force is d/dt(mv), and this a frame-dependent quantity when the mass is not constant (which it most certainly is not in the case of a rocket, or a spray can).
     
  6. Aug 29, 2010 #5
    Re: Zero-G reaction forces

    In non-relativistic mechanics it is not.
     
  7. Aug 29, 2010 #6

    D H

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    Re: Zero-G reaction forces

    Sure it is. The definition of force is F=dp/dt, not F=ma. In the case of a variable mass object the force acting on the object is a frame-dependent quantity.
     
  8. Aug 29, 2010 #7

    cjl

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    Re: Zero-G reaction forces

    Nope.

    If the massflow from the can is constant, and the can's exhaust velocity is constant, then the force is constant. The acceleration will be variable if the can's mass is changing, but the force will not be variable.

    (Specifically, using your definition, dp/dt = dm/dt*v. If dm/dt is constant, and the exhaust velocity is constant, then dp/dt is constant. So, if dp/dt is constant, then by your own definition, F is constant).
     
  9. Aug 30, 2010 #8

    D H

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    Re: Zero-G reaction forces

    Wrong again.

    Now you are not only mixing frames but you are mixing quantities! Your v here is the velocity of the exhaust relative to the vehicle (or in this case, the velocity of the spray relative to DaveC).

    Do again without mixing things up. To avoid mixing things up, use
    • [itex]m[/itex] is the (time-varying) mass of the vehicle, including the yet-unburnt fuel.
    • [itex]v[/itex] is the velocity of the vehicle from the perspective of some inertial frame.
    • [itex]p[/itex] is the momentum of the vehicle from this frame, [itex]p=mv[/itex].
    • [itex]u[/itex] is the velocity of the exhaust relative to the vehicle.
    • [itex]\dot m[/itex] is the mass flow rate.

    F=dp/dt is the definition of force, and it is very much a frame dependent quantity here.
     
  10. Aug 30, 2010 #9
    Re: Zero-G reaction forces

    Please state the transformation law for the force components from one Galilean frame of reference to another.
     
  11. Aug 30, 2010 #10

    cjl

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    Re: Zero-G reaction forces

    Oh, I'm not the one mixing things up here.

    Vehicle mass is irrelevant, so m vanishes.
    Vehicle velocity is irrelevant, so v vanishes
    Vehicle momentum is irrelevant, since both terms it is dependent on are irrelevant.
    Exhaust velocity matters, so u is the first term out of your list that actually shows up.
    Exhaust flow rate matters, so [itex]\dot m[/itex] is the only other item on your list that actually shows up.

    This is easily demonstrable by the fact that from any frame, the vehicle's exhaust will have a constant speed relative to the vehicle (at the instant that it is emitted). Since the fuel was originally traveling with the vehicle, this relative velocity u is also equal to the [tex]\Delta[/tex]V of the fuel. The fuel mass flow rate [itex]\dot m[/itex] is then used to figure out the fuel's [itex]\dot p[/itex]. This is fairly easily done through the equation [itex]\dot p[/itex] = [itex]\dot m[/itex]*[tex]\Delta[/tex]V. Since [tex]\Delta[/tex]V is constant (as we defined exhaust velocity to be constant), and [itex]\dot m[/itex] is constant (as we defined massflow to be constant), the exhaust's [itex]\dot p[/itex] will always be the same, as viewed from any frame. Since conservation of momentum holds, the vehicle's [itex]\dot p[/itex] must be equal to -[itex]\dot p[/itex] for the fuel. Since the fuel's [itex]\dot p[/itex] is constant, the vehicle's [itex]\dot p[/itex] is also constant, and therefore F is constant.


    (Sorry for some of the strange formula appearances - I'm not that familiar with Latex)
     
  12. Aug 30, 2010 #11
    Re: Zero-G reaction forces

    "Vehicle mass is irrelevant"?
    I know of no such provision in any reputable contexts of physical laws.
     
  13. Aug 30, 2010 #12
    Re: Zero-G reaction forces

    "Vehicle momentum is irrelevant"
    Oh, please...
     
  14. Aug 30, 2010 #13
    Re: Zero-G reaction forces

    That inertia or momentum is irrelevant is preposterous.
    Those aspects exist in all mass-systems. Never been proven otherwise.
     
  15. Aug 30, 2010 #14

    D H

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    Re: Zero-G reaction forces

    Short answer: Suppose inertial frame B is moving at velocity V wrt inertial frame A. The force (using F=dp/dt) on some object with non-constant mass m in frame B is

    [tex]F_B = F_A + \dot mV[/tex]

    In short, force is not a galilean invariant.

    Long answer: Dealing with variable mass systems are a bit tricky. The choices are
    • Use F=dp/dt. This has the obvious advantage of being connected with the conservation laws. It has also the obvious disadvantage of not being a galilean invariant.

    • Use F=ma. This is a galilean invariant, but it is no longer connected with the conservation laws. Using F=ma will lead to all kinds of problems. It is not valid for computing work, for example.

    • Rewrite Newton's second law as [itex]F_{\text{ext}} = dp/dt - u\dot m[/itex]. For example, see Halliday and Resnick. (Did you sleep through freshman physics?)

      There is a slight problem here with work and momentum. This is essentially F=ma again.

    • Throw up your hands in disgust and claim that Newton's laws only apply to systems with constant mass. For example, see Plastino & Muzzio, Celestial Mechanics and Dynamical Astronomy, 53:3 (1992) http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html.


    Oh yes you are.

    'Nuff said. Well, almost 'nuff said.

    You are mixing frames. The exhaust velocity is only constant with respect to the vehicle, and the vehicle is accelerating. Arguing whether Newton's laws do or do not apply to objects with non-constant mass is, IMHO, being a bit too pedantic. Arguing whether Newton's laws do or do not apply in accelerating frames is a lot less pedantic. They don't -- at least not without the addition of fictitious forces.
     
    Last edited: Aug 30, 2010
  16. Aug 30, 2010 #15
    Re: Zero-G reaction forces

    This is wrong. Your invalid conclusion is a consequence of trying to use the definition F = dp/dt. This equation has Fnet on the left side and that it strictly holds for point particles. Mechanics teaches us that 2nd Newton's Law is not a definition of force. Forces are defined independently of it and depend on the nature of the bodies that interact and their relative position and velocity. For example, Newton's Law of universal Gravitation:

    [tex]
    \mathbf{F} = -G \, \frac{m_{1} \, m_{2}}{r^{2}_{1 2}} \, \hat{\mathbf{r}}_{1 2}
    [/tex]
     
  17. Aug 30, 2010 #16

    D H

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    Re: Zero-G reaction forces

    Good one! Tell me another. I have deadlines galore and humorous posts such as that do help.
     
  18. Aug 30, 2010 #17
    Re: Zero-G reaction forces

    If you think this is humorous, then your deadlines are probably related to some accounting project.
     
  19. Aug 30, 2010 #18

    cjl

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    Re: Zero-G reaction forces

    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.

    (Specifically, F = [itex]\dot m[/itex]Ve for any rocket motor, in which [tex]\dot m[/tex] is the fuel mass flow rate, and Ve is the exhaust velocity)
     
    Last edited: Aug 30, 2010
  20. Aug 30, 2010 #19

    cjl

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    Re: Zero-G reaction forces

    Of course if you're in the vehicle frame, you have to add fictitious (inertial) forces, but that's somewhat irrelevant to my point here. The rocket motor (since that's effectively what we're talking about here) is a constant thrust device in any frame. If you ignore drag, and operate in an inertial frame, this means that the net force applied to an object by a rocket motor is constant (assuming the mass of the rocket motor is negligible). In a non-inertial frame, there will be other forces, yes, but the rocket motor is still generating the same force as it always was, even in a non-inertial frame.
     
  21. Aug 30, 2010 #20

    D H

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    Re: Zero-G reaction forces

    Actually, my deadlines are very closely related to what has become the topic of this thread. This stuff is my job. In other words, "Why, yes, I am a rocket scientist."

    The reason your post was humorous is because it is exactly contrary to the modern interpretation of Newton's second law. This modern interpretation is that Newton's second law is not truly a scientific law. Instead, Newton's second law is definitional. In particular, it defines the concepts of momentum and force.
     
  22. Aug 30, 2010 #21

    D H

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    Re: Zero-G reaction forces

    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 lets 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.
     
  23. Aug 30, 2010 #22

    Cleonis

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    Re: Zero-G reaction forces

    I think the interpretation you mention here may have been modern at some point in time, but it's been overtaken.
    (This is far away from the original topic of this thread. DH, if you prefer that this discussion is restarted in a thread of its own, please say so. )


    I will start with a general, sweeping statement. Then I will discuss a particular example, claiming that the lesson from that example is universally valid.

    All laws of physics have a dual character, in that they define the concepts they use, and make statements about these concepts.
    The upshot is that it's never a case of either/or. Newton's laws are definitions and true physical laws.


    Now discussion of a historical example:
    The definition of the concept of electric resistance is Ohm's law: R = V/I . There is no such thing as first defining the physics of electric resistance, and then proceed to discover Ohm's law R = V/I.

    There is no such thing as measuring electric resistance directly; the observables are current strength, I, and electromotive force, V. Also, in the early years of investigation of electrics there were a number of different ways in usage of how to gage electromotive force, and none of them was the same as the modern one. The modern way of gaging electromotive force is designed to be as linear as possible, but in the early days there was no way of knowing which of the methods was linear and which wasn't. However, it was noticed that with some definitions of electromotive force Ohm's law obtained, and with other definitions it didn't (or with much more deviation from the law). This had an influence on how the concept of electromotive force was defined: the scientists came to favor definitions of electromotive force for which Ohm's law obtained. In other words, the law was used to define the concept of electromotive force.

    So, is Ohm's law just circular reasoning? No, it isn't. Over time it became increasingly clear that a material's electric resistance can be predicted on the basis of its structural properties alone. The metal contains a large population of electrons that are so free that they can flow through the metal like a fluid or a gas.

    Historically, Ohm's law was intuited on the basis of very little evidence, and subsequently Ohm's law influenced the way that the concept of electromotive force was defined. Over time Ohm's law grew from strength to strength, gaining support in ways that were entirely independent from its first conception.

    Back to sweeping statements:
    This dual character applies for all laws of physics. Each law of physics serves both as law of physics and as operational definition of how the data must be organized so that the law obtains.

    No circular reasoning
    This does not mean the laws of physics are circular reasoning. The laws of physics are not circular reasoning: that should be obvious to anyone.
     
  24. Aug 30, 2010 #23
    Re: Zero-G reaction forces

    Actually, I don't care about the credentials of someone who posts over the Internet. Please give reference where this "modern definition of Newton's second law" is given.
     
  25. Aug 30, 2010 #24

    Cleonis

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    Re: Zero-G reaction forces

    I cannot speak for DH, but for later ideas I can give a source that dates to the 1970's. (This material has influenced me enormously.)

    In the book 'Gravitation' by Misner, Thorne and Wheeler.
    (paragraph 12.3):
    So ideas like this were in circulation in the 70's
     
  26. Aug 30, 2010 #25
    Re: Zero-G reaction forces

    That's all nice, but it is irrelevant to the discussion at hand.
     
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