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Variable-mass systems

  1. Dec 26, 2011 #1
    Newton's second law gives that

    [tex]\sum\mathbf{F} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} [/tex]

    In a system where mass varies with time, m(t), one would simply think that this would lead to

    [tex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/tex]

    Yet everywhere online I see that this is not the case, (especially here, page 228) says that this is not the case. In fact, it is said that

    [tex]\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}[/tex]

    where [itex]\mathbf{u}[/itex] is the velocity of dm, the mass entering or leaving the main body. What then is wrong with [itex]\sum\mathbf{F} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}[/itex]?

    Happy holidays!
  2. jcsd
  3. Dec 26, 2011 #2
    I took a quick look at the book, and my immediate guess is that they have absorbed a negative sign into either the velocity or the dm/dt. In the case of a rocket, for example, dm/dt is negative, because mass is leaving the rocket and thus the mass of the rocket is decreasing. Also be careful with the velocities of the rocket and propellant. Draw out a simple force diagram with a rocket and its propellant.
  4. Dec 27, 2011 #3

    D H

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    You won't see F=dp/dt in Newton's Principia. You won't see F=ma, either. Newton's Principia is pretty much calculus free.

    There's a big problem with using ƩF=dp/dt as the definition of force. This makes force a frame dependent quantity for variable mass systems. This problem vanishes if one uses ƩF=ma. This introduces a new problem. Now the connection with the conservation laws is broken. There are two ways around this. One is to rewrite Newton's second law as ƩFext+u*dm/dt=ma. Another is to define thrust (u*dm/dt) as one of the external forces acting on the system, from which one gets ƩF=ma once again.

    One last way around this problem is to use an inertial frame that is instantaneously co-moving with the center of mass of the variable mass system. This makes the v*dm/dt term that arises in ƩF=dp/dt vanish. In this frame, ƩF=dp/dt is equivalent to ƩF=ma, but the connection with the conservation laws is clearly maintained if one uses ƩF=dp/dt.
    Last edited: Dec 27, 2011
  5. Dec 27, 2011 #4
    Thanks DH. Just a couple of questions. First of all, what do you mean by the conservation laws, and how does this relate to the form of Newton's second law?

    Also, I noticed that in many derivations of [itex]\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}[/itex] the author suddenly says that the ejected mass has velocity [itex] \mathbf{v} + d\mathbf{v} - \mathbf{u} [/itex] relative to the ground. However, if the ejected mass is launched with a relative velocity [itex] \mathbf{u}[/itex] to the main body, then is not its velocity relative to the ground simply [itex] \mathbf{v} + d\mathbf{v} + \mathbf{u} [/itex]? See here for example.
  6. Dec 27, 2011 #5

    D H

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    You'll see people using v-u as opposed to v+u when they reduce the rocket problem to a 1D problem and treat velocity as a scalar as opposed to a vector. Let v is the velocity of the rocket with respect to the ground, positive upward, and u be the speed (positive) at which the exhaust leaves the rocket relative to the rocket. This means the velocity of the exhaust relative to ground is v-u. Treating everything as a vector leads to v+u.

    Regarding the v+u versus v+dv+u imbroglio, that's a silly argument based on a typical physicist abuse of mathematics. If you want to be pedantically correct, the momentum of the exhaust cloud is given by the integral
    [tex]\mathbf p_e(t) = \int_{t_0}^t \dot m_e(\tau)(\mathbf v_r(\tau)+\mathbf u(\tau))\,d\tau[/tex]
    Differentiating with respect to time via the Leibniz integral rule yields
    [tex]\dot{\mathbf p}_e(t) = \dot m_e(t)(\mathbf v_r(t)+\mathbf u(t))[/tex]

    One last point with regard to the article cited in post #4. Regardless of the author's opinion, there is an ambiguity with regard to the projectile velocity. Someone buying such a system had better read the specs and read how the spec values were tested. Those specs and the test reports are inevitably classified. A university professor without proper clearance knows naught of the topic on which he is writing.
    Last edited: Dec 27, 2011
  7. Dec 27, 2011 #6
    Interesting! Yet I noticed that the derivation using F = dp/dt only works using v-u. How then does one obtain [itex]\sum\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}[/itex]? My thought would be to proceed exactly as was done here, yet it seems that they reached the same (incorrect) result as I did in my first post. Are they wrong in saying that

    [tex]\mathbf{p}_{\mathrm{initial}} = M\mathbf{v}[/tex]

    [tex]\mathbf{p}_{\mathrm{final}} = (M - dm)(\mathbf{v} + d\mathbf{v}) + dm(\mathbf{v} + d\mathbf{v} + \mathbf{u})[/tex]

    or is their problem elsewhere? Thanks again DH!
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