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Homework Help: Projectile subject to air resistance

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a projectile subject to air resistance. The drag force is [itex]F=-\hat{v} C v^{a}[/itex] where C and a are constants and v=[itex]\dot{r}[/itex]. Restrict the problem to 2-D and take y in the vertical direction. Take the magnitude of initial velocity to be [itex]v_{0}[/itex] and the initial position to be the origin. Introduce dimensionless coordinates X,Y,Z by relations

    [itex]x=\frac{v_{0}^{2}}{g}X[/itex]
    [itex]y=\frac{v_{0}^{2}}{g}Y[/itex]
    [itex]t=\frac{v_{0}}{g}T[/itex]

    Given the above definitions show that

    [itex]\frac{d^2Y}{dT^2}=-1-k V^{a-1} \frac{dY}{dT}[/itex]
    [itex]\frac{d^2X}{dT^2}=-k V^{a-1} \frac{dX}{dT}[/itex]

    where


    [itex]k=\frac{C v_{0}^a}{m g}[/itex]

    [itex]V=\sqrt{(\frac{dY}{dT})^2+(\frac{dX}{dT})^2}[/itex]

    3. The attempt at a solution


    [itex]x=\frac{v_{0}^{2}}{g}X[/itex]
    [itex]dx=\frac{v_{0}^{2}}{g}dX[/itex]

    [itex]y=\frac{v_{0}^{2}}{g}Y[/itex]
    [itex]dy=\frac{v_{0}^{2}}{g}dY[/itex]

    [itex]t=\frac{v_{0}}{g}T[/itex]
    [itex]dt=\frac{v_{0}}{g}dT[/itex]

    finding d?/dT equations
    [itex]\frac{dx}{dt}=\frac{\frac{v_{0}^{2}}{g}dX}{\frac{v_{0}}{g}dT}
    =v_0\frac{dX}{dT}
    [/itex]

    [itex]\frac{dy}{dt}=\frac{\frac{v_{0}^{2}}{g}dY}{\frac{v_{0}}{g}dT}
    =v_0\frac{dY}{dT}
    [/itex]

    finding [itex]\frac{d^2?}{dT^2}[/itex]
    [itex]\frac{d^2x}{dt^2}=\frac{dx}{dt}\frac{dx}{dt}=v_0^2\frac{d^2X}{dT^2}[/itex]

    [itex]\frac{d^2y}{dt^2}=\frac{dy}{dt}\frac{dy}{dt}=v_0^2\frac{d^2Y}{dT^2}[/itex]


    From force equation:
    [itex]F_{d,x}=m \ddot{x}=-\hat{v} C v^{a}=\frac{-\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} C (\sqrt{\dot{x}^2+\dot{y}^2})^{a}[/itex]
    [itex]=-\dot{x} C \sqrt{\dot{x}^2+\dot{y}^2}^{a-1}[/itex]


    [itex]m v_0^2\frac{d^2X}{dT^2}=-(v_0\frac{dX}{dT}) C \sqrt{(v_0\frac{dX}{dT})^2+(v_0\frac{dY}{dT})^2}^{a-1}[/itex]

    [itex]m v_0^2\frac{d^2X}{dT^2}=-v_0^a \frac{dX}{dT} C \sqrt{(\frac{dX}{dT})^2+(\frac{dY}{dT})^2}^{a-1}[/itex]

    [itex]\frac{d^2X}{dT^2}=-\frac{v_0^a C}{m v_0^2} V^{a-1} \frac{dX}{dT} [/itex]

    [itex]\frac{d^2X}{dT^2}=-k \frac{g}{v_0^2} V^{a-1} \frac{dX}{dT} [/itex]
    Which is not what I'm supposed to get. I must be messing up a rule or two somewhere.

    Thank you.
     
    Last edited: Feb 19, 2012
  2. jcsd
  3. Feb 19, 2012 #2

    ehild

    User Avatar
    Homework Helper

    Do the second derivatives again. They are not the squares of the first derivatives, as you wrote.

    ehild
     
  4. Feb 19, 2012 #3
    So then would the derivative be

    [itex]\frac{d^2x}{dt^2t}=\frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt}(v_0 \frac{dX}{dT})=\frac{d}{dT}(\frac{g}{v_0})(v_0 \frac{dX}{dT})[/itex]

    [itex]\frac{d^2x}{dt^2}=g\frac{d^2X}{dT^2}[/itex]

    similarly

    [itex]\frac{d^2y}{dt^2}=g\frac{d^2Y}{dT^2}[/itex]

    Thanks for your help
     
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