Finding the Lagrangian for a 3D Spring Pendulum

[fluidistic]

Hmm, I got

$$\ddot{\theta}=\sin\theta\cos\theta\dot{\phi}-\frac{g\sin\theta}{r}-\frac{2\dot{r}\dot{\theta}}{r}$$



$$\ddot{r}=r\dot{\theta}^2+r\sin^2\theta\dot{\phi}^2+g\cos\theta-\frac{k}{m}\left(r-1\right)$$

and

$$\ddot{\phi}=-\dot{\phi}\left[\frac{2\dot{r}}{r}+2\cot\theta\right]$$

I keep getting the same result I had. Did you start with the same motion equations as mine? My $$\ddot \theta$$ is worth $$- \ddot \theta$$ of yours. The same apply for $$\ddot r$$. And for $$\ddot \phi$$, I still get this "r" term in front of the $$\cot (\theta )$$ term.

 

[jdwood983]

Well your Lagrangian is

$$L=\frac{1}{2}m\left(\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2\right)+mgr\cos\theta-\frac{1}{2}k\left(r-1\right)^2$$

So then the Euler-Lagrange equations come from

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)-\frac{\partial L}{\partial q}=0$$

where q is your coordinate. Looking at $\phi$ (because it's the most simple)

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\phi}}\right)-\frac{\partial L}{\partial\phi}=\frac{d}{dt}\left(mr^2\sin^2\theta\dot{\phi}\right)=0$$

Clearly m doesn't change, this gives that

$$\frac{d}{dt}\left(mr^2\sin^2\theta\dot{\phi}\right)=2mr\dot{r}\sin^2\theta\dot{\phi}+2mr^2\sin\theta\cos\theta\dot{\theta}\dot{\phi}+mr^2\sin^2\theta\ddot{\phi}=0$$

So then solving for $\phi$-double-dot:

$$mr^2\sin^2\theta\ddot{\phi}=-2mr\dot{r}\sin^2\theta\dot{\phi}-2mr^2\sin\theta\cos\theta\dot{\theta}\dot{\phi}$$

Canceling m and dividing by r-squared & sine-squared

$$\ddot{\phi}=-2\frac{\dot{r}\dot{\phi}}{r}-2\cot\theta\dot{\theta}\dot{\phi}$$

So it seems (being more careful now) that I forgot a term ($d\theta/dt$), but that r does not belong in front of the cotangent term.

 

[jdwood983]

$$L=\frac{1}{2}m\left(\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2\right)+mgr\cos\theta-\frac{1}{2}k\left(r-1\right)^2$$

Then looking at the r-coordinate,

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{r}}\right)-\frac{\partial L}{\partial r}=0$$

we get

$$\frac{d}{dt}\left(m\dot{r}\right)-mr\dot{\theta}^2-mr\sin^2\theta\dot{\phi}^2-mg\cos\theta+k\left(r-1\right)=0$$

Which gives

$$\ddot{r}=r\dot{\theta}^2+r\sin^2\theta\dot{\phi}^2+g\cos\theta-\frac{k}{m}\left(r-1\right)$$

Similarly for $\theta$:

$$\frac{d}{dt}\left(mr^2\dot{\theta}\right)-mr^2\sin\theta\cos\theta\dot{\phi}^2+mgr\sin\theta=0$$

$$\ddot{\theta}=\sin\theta\cos\theta\dot{\phi}^2-\frac{g\sin\theta}{r}-\frac{2\dot{r}\dot{\theta}}{r}$$

 

[fluidistic]

Hey jdwood983, I agree with your 2 last posts. To convince myself I've redone all the arithmetic from the Lagrangian and I get exactly the same result as yours now. My previous errors were in applying E-L equation.
I think the problem is solved. The remaining part is numerical.
Thanks for all to both.

 

[jdwood983]

Not a problem, glad I (we) could help!

 

[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!