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Motion with non-constant mass

  1. Aug 7, 2010 #1
    I'm having trouble understanding motion with non-constant mass.
    Specifically, if (dm/dt) is positive when the mass of the system is increased, I find that:
    [tex]\vec{F} = \frac{dm}{dt}\vec{u} + \frac{d\vec{v}}{dt} m [/tex]
    Where u is the velocity of the mass leaving/entering the system relative to the system and dm/dt is the rate of change in the mass of the system.

    If I set F = 0, in a situation where there are no external forces, I get:
    [tex]\frac{d\vec{v}}{dt}m = -\frac{dm}{dt}\vec{u}[/tex]

    But if dm/dt is negative, doesn't this mean the acceleration due to mass leaving the system dv/dt is in the same direction as u, the velocity at which the mass is leaving the system?

    Edit: Sorry for posting this in the wrong forum. I saw the warning, but I assumed it meant not posting specific homework questions, rather than general queries.
    Last edited: Aug 7, 2010
  2. jcsd
  3. Aug 7, 2010 #2
    when you say 'u' is the velocity of the mass leaving/entering the system, do you mean its the RATE that it enters/leaves or it is the velocity at which it leaves? If its the 'velocity' at which is leaves, then the force isn't 0.
  4. Aug 7, 2010 #3
    It depends on your derivation. (dm/dt)u can be seen as the thrust force which is applied by the mass leaving the system on the system, or it can be seen as the effects of conservation of momentum, as mass leaves the system.

    At any rate, F in this case signifies external forces, while (dm/dt)u signifies the effects of mass leaving the system. As you can see, the system is accelerated.

    So, when I say u is the velocity, I do mean it's the velocity.
  5. Aug 7, 2010 #4
    I don't quite follow, if you start from the top
    F = \frac{d}{dt}m\vec{v} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt}
    The first term is standard, and the second term is conservation of momentum, and simply results from the initial mass 'magically' growing or shrinking, it does not include any kind of thrust.

    If you're losing mass non-magically, and instead its 'radiating' off at some velocity, then you have an external force in addition, no?
  6. Aug 8, 2010 #5
    You can derive the equation for motion with non-constant mass using two ways. One of these ways involves the equations:
    [tex]P(t+dt)=(m+dm)(\vec{v}+d\vec{v}) - dm(\vec{v}+\vec{u})[/tex]
    [tex]P(t + dt) - P(t) = F*dt[/tex]

    The other involves newton's 2nd law:
    [tex]\vec{F} = m\vec{a}[/tex]
    Where m depends on time. Now, at time t the momentum of the leaving mass:
    In respect to the system. However,
    Meaning, after some time dt the system grants some mass dm velocity u. As such:
    [tex]\vec{P}(t+dt) - \vec{P}(t) = \vec{F}_{thrust}dt[/tex]
    Which means:
    [tex]dm*\vec{u} = \vec{F}_{thrust}dt[/tex]
    [tex]\vec{F}_{thrust} = \frac{dm}{dt}\vec{u}[/tex]
    Since the system applies this force on the mass, the mass must apply the same force, in the opposite direction on the system (newton's third law). Thus:
    [tex]\vec{F}_{other} - \vec{F}_{thrust} = m\vec{a}[/tex]
    But in this case, we've chosen dm/dt to be the change in the mass 'leaving' the system, so we would have to switch the sign:
    [tex]\vec{F}_{other} + \frac{dm}{dt}\vec{u} = m\vec{a}[/tex]

    It doesn't matter what the thrust is caused by; it could be a guy kicking balls off a truck, or it could be pressurized gas. As you see, the mass does not 'magically' grow or shrink; there is a very real force which grants the leaving mass the velocity u. This is also true in conservation of momentum; we simply don't 'see' this force directly.
    Last edited: Aug 8, 2010
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