System of ODEs in a rotating coord. system

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SUMMARY

The discussion centers on the derivation of second-order ordinary differential equations (ODEs) in a rotating coordinate system. The user attempts to express the equations as x{''}_1 = x_1 + 2x{'}_2 and x{''}_2 = x_2 - 2x{'}_1, but instead arrives at x{''}_1 = α x{'}_2 and x{''}_2 = -α x{'}_1, where α represents the constant angular velocity. The confusion arises from the need to incorporate centrifugal and Coriolis forces, which are essential in rotating frames of reference. The user is encouraged to review these concepts to clarify the derivation process.

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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 x{''}_1 = x_1 + 2x{'}_2 and x{''}_2 = x_2 - 2x{'}_1 I get x{''}_1 = \alpha x{'}_2 and x{''}_2 = - \alpha x{'}_1 where \alpha 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 \alpha t be the rotation rate by time.

x = \cos{A-\alpha t}

x{'} = \alpha \sin{A-\alpha t}

x{''} = -\alpha^2 \cos{A-\alpha t}

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

That looks like it's on it's way to being x{''}_1 = x_1 + 2x{'}_2 and x{''}_2 = x_2 - 2x{'}_1...but I'm missing something.
 
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