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Differential equations: applying force to projectile

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The following series of differential equations represents a projectile's path when solved (g=9.81):
    nANP9I9.png
    Modify this series of differential equations to account for an additional force F with vector components a and b acting on the projectile.

    Here is a sample plot of this system:
    TeUvVr2.png

    2. Relevant equations
    See above.

    3. The attempt at a solution

    Knowing that gravity is a force with a positive i-component and negative j-component, I attempted to apply the same logic to F. This resulted in:

    vdot = -g*sin(theta) + a*cos(theta) + b*sin(theta)
    and
    thetadot = -g/v*cos(theta) + a/v*sin(theta) + b/v*cos(theta)

    However, that solution didn't appear to be correct, as when I plotted this out with F with a positive i-component and a negative j-component, I got this plot:
    ovgeY6g.png
    From intuition, the positive i-component should have caused the projectile path to move to the right more (hard to describe in words, but I hope you get what I mean), but instead, with my system (somehow), the projectile somehow happens to move to the left and go to the negative x-axis. Because my attempt to apply how gravity was represented in this system to how an arbitrary force would be represented, well, failed, I am currently stuck.

    As this is my first post, I'm not sure whether it belongs in the introductory or advanced section (as it involves differential equations, but is a simple system). Sorry in advance if it's wrong.
     
  2. jcsd
  3. Apr 9, 2015 #2

    haruspex

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    Check the signs in your thetadot equation.
     
  4. Apr 9, 2015 #3
    Flipping the sign of the second term appears to make it be more reasonable, and I understand why that makes sense from a directional perspective. However, I'm unsure if the actual magnitude is correct, if you get what I'm saying...

    Here's what it looks like with a flipped second term:
    BQrw8pV.png
    It's more reasonable, but I don't understand it from a trigonometry perspective. In addition, if that's the case, shouldn't I also flip the sign in the vdot equation? If not, why?
     
    Last edited: Apr 9, 2015
  5. Apr 9, 2015 #4

    haruspex

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    Patterns like
    +sin +cos
    -cos +sin
    arise frequently, since as a matrix it represents a rotation.
    While moving up and to the right, v is in the first quadrant but ##\dot \theta## is in the second quadrant, so a force horizontally to the right is positive for ##\dot v## but negative for ##\ddot \theta##.
     
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