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System of ODEs in a rotating coord. system

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data

    pb14Q4Q.png

    imgur link: http://i.imgur.com/pb14Q4Q.png

    2. Relevant equations


    3. The attempt at a solution

    The thing I don't understand is where the first two terms of each 2nd order ODE came about.

    I understand that they are there because the coordinate system is rotating, but when I set a rotating coord. system and try to get [itex]x{''}_1 = x_1 + 2x{'}_2[/itex] and [itex]x{''}_2 = x_2 - 2x{'}_1[/itex] I get [itex]x{''}_1 = \alpha x{'}_2[/itex] and [itex]x{''}_2 = - \alpha x{'}_1[/itex] where [itex]\alpha[/itex] is the constant angular velocity.

    My reasoning is, let r = 1 (the distance between origin and any (x,y)), let A be the initial angle prior to some rotation and [itex]\alpha t[/itex] be the rotation rate by time.

    [tex]x = \cos{A-\alpha t}[/tex]

    [tex]x{'} = \alpha \sin{A-\alpha t}[/tex]

    [tex]x{''} = -\alpha^2 \cos{A-\alpha t}[/tex]

    Do the same thing for y and wind up with [itex]x{''}_1 = \alpha x{'}_2[/itex] and [itex]x{''}_2 = - \alpha x{'}_1[/itex]

    That looks like it's on it's way to being [itex]x{''}_1 = x_1 + 2x{'}_2[/itex] and [itex]x{''}_2 = x_2 - 2x{'}_1[/itex]...but I'm missing something.
     
  2. jcsd
  3. Feb 11, 2016 #2

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