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2 coupled EDOs

  • Thread starter quasar987
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  • #1
quasar987
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I must solve the following two coupled EDOs in the context of a Lagrangian mechanics problem (a rigid pendulum of lenght l attached to a mass sliding w/o friction on the x axis). The problem statement does not mention that we can make small angle approximation. It says "find the equations of motion and solve them for the following initial conditions:...". Is this feasable?

[tex](m_1+m_2)\ddot{x}+m_2l\ddot{\theta}\cos(\theta)-m_2l\dot{\theta}^2\sin(\theta)=0[/tex]

[tex]l\ddot{\theta}+\ddot{x}\cos(\theta)+g\sin(\theta)=0[/tex]

They can be uncoupled but there remains a second order non-linear ODE to solve.

Is this doable analytically?

And an annexed question (perhaps this one is more of a physical nature): why can we say that [itex]\dot{\theta}\approx 0[/itex] in the small angle approximation? The angle can be small and nevertheless vary furiously fast. What indicates that if theta is small, the so is its derivative?
 
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  • #2
quasar987
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Same question but with

[tex]\ddot{y}+ay+b\cos(y)=0[/tex]

Solvable? (a,b are constants)
 
  • #3
dextercioby
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The one in the second post is doable:

[tex] t+C_{2}=\int \frac{dy}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} [/tex]

It remains to compute the integral.

Mathematica returns

[tex] \int \frac{dy}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} =-\frac{2}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} F\left[\frac{1}{4}\left(\pi -2y\right), \frac{4b}{-2C_{1}+ay^2 +2b}\right]\sqrt{\frac{2C_{1}-ay^{2}-2b\sin y}{2C_{1}-ay^{2}-2b}} [/tex]

Daniel.
 
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  • #4
cva
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So how do you inverse the function t?
 
  • #5
dextercioby
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By an appropriate use of Jacobi's elliptic functions. In this case i think sine amplitudinis.

Daniel.
 

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