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A mass point moves on the wall of a hollow circle cone under influence of the homogeneous gravitational field of earth.

Use spherical coordinates to solve this problem.

a)Set up the lagrange equations of second kind.

b)If the lagrange function L isn't dependent on a coordinate q but on the associated velocity, then q is a cyclic coordinate. Is any of the generalized coordinates in this case a cyclic coordinate?

c) Check if the momentum, the z-component of the angular momentum and the total energy are conserved.

First I tried drawing the situation as best as I could. This is what I got:

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I was sick the past few weeks so I basically missed everything regarding lagrangian formalism. Since we use a text book as a script to go along to I tried reading in it, but all the copies in the school library were borrowed. So I had to make use of wikipedia.

I'm not yet familiar with the terminology. The lagrange equations of second kind are the euler-lagrange equations, right?

To be honest this topic doesn't stick with me, especially the general coordinates and velocities.

Anyways, I tried setting up $L=T-V$ since this is the first step as far as I could tell from the wiki entry.

Generally it should be $L=\frac{1}{2}m(\dot{r}^2+\dot{z}^2+r^2\dot{\phi}^2)-mgz$, right?

$z=r\cot{\theta}$? Meaning, $L=\frac{1}{2}m((1+\cot^2(\theta))\dot{r}^2+r^2\dot{\phi}^2)-mgr\cot{\theta})$, correct? I hope someone could help me out here.

Edit:

@JohnHughes: I think you were right about r. Once I tried expressing the lagrangian in spherical coordinates as was asked it seemed reasonable to assume that r is actually the diagonal line closest to it.

So I got this:

a) $\vec{x}=r\begin{bmatrix}\cos{\phi}\sin{\theta}\\\sin{\phi}\sin{\theta}\\\cos{\theta}\end{bmatrix}$

Since I need the velocity for the lagrangian, I have:

$\dot{\vec{x}}=\dot{r}\begin{bmatrix}\cos{\phi}\sin{\theta}\\\sin{\phi}\sin{\theta}\\\cos{\theta} \end{bmatrix}+r\dot{\phi}\sin{\theta}\begin{bmatrix} -\sin{\phi}\\\cos{\phi}\\0\end{bmatrix}$.

Plugging that all I have: $T=\frac{m}{2}\dot{\vec{x}}^2=\frac{m}{2}(\dot{r}^2+r^2\dot{\phi}^2\sin^2{\theta})$ and $V=mgr\cos{\theta}$. $L=\frac{m}{2}(\dot{r}^2+r^2\dot{\phi}^2\sin^2{\theta})-mgr\cos{\theta}$.

That should conclude a), right?

About b): Just from the given definition in b) I can say that $\phi$ is a cyclic coordinate since the lagrangian only has $\dot{\phi}$, correct?

About c): I'm not really sure how to approach this one. My ideas so far were:

Momentum: I can imagine that the momentum isn't conserved since the mass point is accelerated due to the gravitational field, but I have no idea how to write that down mathematically correct.

Angular mometum: To be honest I'm pretty much lost here.

Total Energy: Again I would guess that the total energy is conserved since the expressions for the kinetic and potential energy don't have a dependency on t (time)?

  • I think you want $z = r \cos \theta$ rather than the cotangent. Cosine is adjacent (z) over hypotenuse (r). Otherwise it looks as if you're headed in the right direction. – John Hughes Jul 09 '15 at 17:29
  • I think r is supposed to be the radius of each individual circle. Or not? – Franie Cinna Jul 09 '15 at 17:34
  • That makes perfect sense, Franie -- I was reading it as marking the diagonal line that it's closest to, but in that case, it doesn't make sense in the Lagrangian that you've written. So now that I understand it (D'oh!) your lagrangian looks fine. If you write out the Lagrangian expression (the thing involving $\partial L / \partial q$ for each of the generalized coordinates $q$ (i.e., for $r$ and $\phi$) you should get a pair of coupled differential equations that your particle must satisfy. – John Hughes Jul 09 '15 at 17:54
  • @JohnHughes I played around with this problem for the last hour and I think you were right about r. I made an edit in my post, could you read through it and see if I'm correct? And I'm really not sure how to deal with c) yet. – Franie Cinna Jul 09 '15 at 19:36
  • Yeah, you're right. I was thinking of "cylindrical polars", where $r$ would be the horizontal line. Anyhow, it makes almost no difference in your formulas, since $\theta$ is a constant. I believe your answer to part "b" is correct, yes. As for part c -- I have no idea. Physics is not my strong suit, and Lagrangians were about the point where I stopped "getting it". :) – John Hughes Jul 09 '15 at 19:57
  • I feel the same way. I will try seeing if I can find similar questions with solutions on the internet. Maybe I will find something. By the way, thanks for all your help so far. I wouldn't have gotten so far without you. – Franie Cinna Jul 09 '15 at 20:07

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