How to Derive the Simplified Form of Modified Euler Equations?

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SUMMARY

The discussion focuses on deriving the simplified form of the Modified Euler Equations, specifically transforming the equation M_x=(I_0-I)\dot\Psi^2\sin\theta\cos\theta+I_0\dot\Phi\dot\Psi\sin\theta into the expression for \dot\Psi. The final result is \dot\Psi=\displaystyle\frac{I_0\dot\Phi}{2(I-I_0)\cos\theta} \left[1\pm \left( {1-\displaystyle\frac{4M_x(I-I_0)\cos\theta}{I_0^2\dot\Phi^2\sin\theta}}\right)^{1/2}\right]. The key insight is the application of the quadratic formula to isolate \dot\Psi, confirming the simplification process outlined in the referenced textbook.

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  • Knowledge of trigonometric identities involving sine and cosine
  • Basic principles of dynamics and rotational motion
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Students and professionals in physics, particularly those focusing on dynamics and rotational motion, as well as anyone looking to deepen their understanding of the Modified Euler Equations and their applications.

Telemachus
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Homework Statement


Hi there. I'm not sure if this question corresponds to this subforum, but I think you must be more familiarized with it. The thing is I don't know how to get from:

[tex]M_x=(I_0-I)\dot\Psi^2\sin\theta\cos\theta+I_0\dot\Phi\dot\Psi\sin\theta[/tex]

to:
[tex]\dot\Psi=\displaystyle\frac{I_0\dot\Phi}{2(I-I_0)\cos\theta} \left[1\pm \left( {1-\displaystyle\frac{4M_x(I-I_0)\cos\theta}{I_0^2\dot\Phi^2\sin\theta}}\right)^{1/2}\right][/tex]
I don't know how to get Phi from the first, but this is the simplification given on my book, but I don't know which intermediate steps to give.

Bye, and thanks.
 
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Looks like a pretty straightforward application of the quadratic formula to find Psi-dot.
 
Right. Thanks.
 

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