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Non constant acceleration due to one constant force, and one non constant force

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    This is actually related to another https://www.physicsforums.com/showthread.php?t=531596"
    basically, I have a Constance force of gravity, and a non constant force of a rocket going the opposite way.

    the net force would be something like F(booster) -Force(gravity). Assumed (-) direction, and Force(booster) is smaller. I know when acceleration isn't constant we have to integrate the equation.



    2. Relevant equations
    a = dx/dt


    3. The attempt at a solution

    a=( F(booster)- F(gravity) )/m

    a=( F(booster)- mg) )/m

    but I'm not sure how i should go about integrating this mess.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 20, 2011 #2
    Hi there...
    This is a tad more complex than you put it.
    A rocket, in your case has to have a constant rate of ejection of its gases, lets call it, u;
    Now, recall the proper definition of a force; You're right to state that F=ma, when the mass is constant, but in our case, our rocket propels upwards, and in doing that it loses its fuel, therefore, F has to be taken as: [itex]\large \vec{F}=\frac{d\vec{p}}{dt}=-mg, p = m(t)\cdot\vec{v} [/itex]
    Loss of mass means: m = m_0-u*t;
    Differentiating:
    [itex]\large \vec{F}= \frac{d}{dt}m(t)\vec{v}(t) = \frac{dm}{dt}v(t)+m(t)\cdot\frac{d\vec{v}}{dt} = -uv+ma=mg [/itex]
    Assuming motion only on the vertical axis, which we'll call y, we get a very pleasant differential equation:
    -uy'+(m_0+ut)*y'' = (m0-ut)g;
    That you can solve, either by exponential substitution, or, simply through this:
    http://www.wolframalpha.com/input/?i=-u*y'[t]+++(m0+-+u*t)*y''[t]+==+(m0-u*t)*g"
    But you probably know better(feel free to ask about the technique of resolution to be employed here, if you care to do this manually).
    Don't forget to specify your initial conditions,
    Daniel
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 20, 2011 #3
    Also, note that this solution does not account for what happens after all the fuel is exhausted. So its only valid till t=m_0/u; afterwards, standard projectile motion with an initial velocity laws apply...
    Daniel
     
  5. Sep 20, 2011 #4
    Thanks this helps a lot more than the lack of info in my text!!!

    And yes, my book doesn't have any of that in it. I'm "Supposed to figure that out"
     
  6. Sep 20, 2011 #5
    Glad to be of service,
    Do you know how to slove such differential equations? It might be worth your while to learn, if you're in the mood?
    Daniel
     
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