Solving Coupled EDOs in Lagrangian Mechanics: Is it Feasible?

[quasar987]

I must solve the following two coupled EDOs in the context of a Lagrangian mechanics problem (a rigid pendulum of length 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?

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

l\ddot{\theta}+\ddot{x}\cos(\theta)+g\sin(\theta)=0

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 \dot{\theta}\approx 0 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?

 

[quasar987]

Same question but with

\ddot{y}+ay+b\cos(y)=0

Solvable? (a,b are constants)

 

[dextercioby]

The one in the second post is doable:

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

It remains to compute the integral.

Mathematica returns

\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}}

Daniel.

 

[cva]

So how do you inverse the function t?

 

[dextercioby]

By an appropriate use of Jacobi's elliptic functions. In this case i think sine amplitudinis.

Daniel.