1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rocket Motion in a drag force

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider a rocket is subject to linear resistive force, [tex] f = -bv [/tex]. [tex] \dot m[/tex] is constant. Use the equation: [tex] m \dot{v} = -\dot{m }v + f [/tex] to determine the velocity of the rocket :

    since the rate of mass lost is constant
    let [tex] \dot{m} =k [/tex]
    vex = nuzzle velocity

    [tex]v = \frac{k}{b} vex (1 - (\frac{m}{m_0})^{\frac{b}{k}})[/tex]
    2. Relevant equations
    Already given above


    3. The attempt at a solution
    let [tex] m= \frac{dm}{dt} \cdot dt = k \cdot dt[/tex]

    [tex] k dt \frac{dv}{dt} = -kvex - bv[/tex]
    I don't know if cancelling [tex] dt [/tex]'s are allowed but when I solve this equation, it doesn't resemble the expected answer at all.
     
    Last edited: Jan 16, 2012
  2. jcsd
  3. Jan 16, 2012 #2
    You say 'm' is constant... but you have a non-zero m derivative.
    Also, [tex]m = \int \frac{dm}{dt} dt[/tex]
     
  4. Jan 16, 2012 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Make it clear if k is the rate of loss of mass, which means dm/dt=-k or k=dm/dt.
    m means the mass of the racket at time t. You need m(t), the expression of mass in terms of time. The original mass is m0. The loss of mass is constant, k. What is m(t), the mass t time after start? It is certainly not kdt as you wrote.


    ehild
     
    Last edited: Jan 16, 2012
  5. Jan 17, 2012 #4
    I believe I figured it out. Sorry for the confusion. dm/dt =k.

    for future reference, if one is looking at this problem, they can easily solve it by raping calculus, i.e.

    m dv/dt = m dm/dm dv/dt = m dm/dt dv/dm = m k dv/dm

    Now the only dependencies are on m and v, thus by separation of variables the problem becomes solvable.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rocket Motion in a drag force
  1. Drag Force (Replies: 4)

  2. Rocket Motion (Replies: 1)

Loading...