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Newton's second law if mass changes

  1. Dec 24, 2011 #1

    Philip Wood

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    Are there any cases in newtonian physics where it is valid to apply Newton's second law in the form ƩF = m dv/dt + v dm/dt, in which dm/dt is non-zero?

    It is my belief that there are no such cases. For example, if one applies momentum conservation to a rocket in a field-free region, we obtain an equation which is consistent with ƩF = m dv/dt (that is ƩF = ma), but not with ƩF = m dv/dt + v dm/dt.

    Despite my scepicism, the original question is a genuine one.
     
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  3. Dec 24, 2011 #2
    You have to keep in mind all the components of your system. For example, if the mass of your rocket decreases with time and you're keeping into account only that mass, you must use: ƩF = m*dv/dt + dm/dt*v, if you're considering the system to be formed of the rocket plus the consumed fuel, than the mass of the system is constant and you have your ƩF = m*dv/dt and the momentum is conserved for the center of mass of the system.
     
  4. Dec 24, 2011 #3

    Philip Wood

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    I was taking m as the (changing) mass of the rocket itself, yet obtained a result (using the Principle of Conservation of momentum) which was inconsistent with F= mdv/dt + vdm/dt, but consistent with F= mdv/dt.

    Using the Pof C of M, I get w dm = m dv in which w is the velocity of the exhaust gases relative to the rocket (and is negative), m is the mass of the rocket itself, and v is its forward velocity. dm is negative.

    Thus w dm/dt = m dv/dt.

    But -w (-dm)/dt = w dm/dt is the rate of gain of backward momentum by the exhaust gases, so the forward force, F, on the rocket (from the gases) is w dm/dt.

    So, for the rocket, we have F = m dv/dt.
     
    Last edited: Dec 24, 2011
  5. Dec 24, 2011 #4

    D H

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    The answer is it depends on what you mean by "force" and it also depends on whether you think Newton's second law has any business being applied to a system of non-constant mass. There are some who argue that it doesn't. I'll ignore this latter concern.

    If you define force via F=dp/dt then yes, you get a "force" from [itex]\dot m v[/itex]. But now there's a big problem with this definition. Force is no longer frame invariant. If you define force via F=ma then there is no [itex]\dot m v[/itex] term. But now there's a big problem here as well. This definition creates problems with respect to the conservation laws. Pick your poison ...

    One way around this is to work in an inertial frame instantaneously co-moving with the system center of mass. Now F=dp/dt and F=ma are identical, just as they are for a system of constant mass.
     
  6. Dec 24, 2011 #5

    Philip Wood

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    Thank you, DH, for a very interesting post. When you say that force, defined as as dp/dt (with non-zero v dm/dt), is frame-dependent, does this remark apply if we consider only inertial frames?
     
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