Finding the Lagrangian for a 3D Spring Pendulum

[fluidistic]

Ok before putting an end to this thread, I'd like some help to get the initial conditions.
The initial position of the mass is (0,1,0) in Cartesian coordinates. So it makes $$(1,\pi /2,\pi /2)$$ in spherical.
The initial velocity of the mass is (0,0,-1/2) in Cartesian coordinates. I'm not sure how to convert this into spherical ones.
An attempt of mine is to write $$\vec v_0=-\frac{1}{2}\hat z=\frac{\sin (\theta) \hat \theta}{2}-\frac{\cos (\theta)\hat r}{2}$$. I don't know how to proceed from here.

Edit: Hmm, is it just $$(1/2,0, \pi)$$?
Edit 2: Well I think so. This would mean $$\dot r (0)=1/2$$, $$\dot \phi (0)=0$$ and $$\dot \theta (0)=\pi$$. I hope someone can confirm this.

 

[fluidistic]

Any idea? I'm not sure I have the right initial conditions. I hope someone can confirm if they're good.

 

[jdwood983]

It does look fine to me, the (0.5,0,pi):

$$r=\sqrt{x^2+y^2+z^2}=\sqrt{1/4}=1/2$$

$$\theta=\cos^{-1}\left[\frac{z}{r}\right]=\cos^{-1}\left[1\right]=0$$

$$\phi=\tan^{-1}\left[\frac{x}{y}\right]\rightarrow\pi$$EDIT: Technically, $\phi$ is undefined because x=0 divided by y=0 is undefined, but given your position it should be okay (note that my $\phi$ is your $\theta$, I was taught coordinates this way and always use it)

 

[fluidistic]

Ok thank you jdwood983. Problem solved. I'll now tackle the numerical part!