System of ODEs in a rotating coord. system

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kostoglotov
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Homework Statement



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

Homework Equations

The Attempt at a Solution


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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.
 
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