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Taylor Mechanics Section 3.2

  1. Oct 6, 2013 #1
    Hello,

    I am reading section 3.2, concerning the analyzation of a moving rocket with a changing mass. (I couldn't find a preview of the book in google books, so hopefully someone out there has this textbook.) Here is an except from the book, but be warned that I am adding notes in brackets:

    "At time t, the momentum [of the rocket] is P(t) = mv [m and v are the mass and velocity of the rocket at time t]. A short time later at t + dt, the rockets mass is (m + dm), where dm is negative, and its momentum is (m+dm)(v+dv). The fuel ejected in the time dt has mass (-dm) and velocity v - vex [v is the velocity of the rocket as viewed by some stationary person on earth, and vex is rate at which the fuel flows out, relative to the rocket]. Thus, the total momentum (rocket plus the fuel just ejected) at t + dt is
    P(t+dt) = (m + dm)(v+dv) - dm(v - vex)."

    As one might notice, as I did, they accounted for the momentum of the fuel at time t+dt, which is - dm(v - vex) (there's a negative because the momentum is in the opposite direction); but at time t, they did not, for the expression is P(t) = mv. Where is the momentum term for the fuel at time t?
     
  2. jcsd
  3. Oct 6, 2013 #2

    AlephZero

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    Before it burns, the fuel has the same velocity as the rocket, so its momemtum is part of the P(t) = mv.

    You could write
    (m+dm)v - (dm)v
    if you really want to separate out "the momentum of the fuel of mass -dm that you are going to burn next" and "the momemtum of everything else". But of course (m+dm)v - (dm)v = mv.
     
  4. Oct 6, 2013 #3
    Isn't the velocity of the fuel at any instant v - vex?
     
  5. Oct 6, 2013 #4

    AlephZero

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    The fuel has velocity v before it burns, and v - vex after it burns.

    At time t, the bit of fuel with mass dm hasn't burned yet. At time t+dt, it has burned.
     
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