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Say that a bead is on a wire that has shape $x^2+z^2/\kappa^2 =1$, where the wire can rotate about the $z$-axis. Gravity acts upon the bead. I can take $\theta,\phi$ as two generalized coordinates. $\theta$ is the angle from positive $z$, anticlockwise. $\phi$ is rotation in the $x-y$ plane.

I am confused about two things: 1) Do two generalized coordinates actually determine this system? If the wire was a circle, then the length from the origin to the wire would be constant, but it is an ellipse, so the length is not invariant. Is it because $\theta$ determines the length? So we only need $(l,\theta,\phi)$ but $l(\theta)$ is the length, dependent on $\theta$?

2) How is the kinetic energy found? I'd think we would take $\frac12 mv^2$, but I am not sure how to do that, in terms of the generalized coordinates. Does one still write $\vec{r}=(x,y,z)$ and derive this, and take the norm, or do we work this out as $q=(q_1,q_2)=(\theta,\phi)$ and $T(q_1,q_2)=\frac12m\|\dot{q}\|^2$?

Do I take $(x,y,z)=(\cos(-\theta+\frac\pi2),\cos(\phi),\sin(-\theta+\frac\pi2))=(\sin(\theta),\cos(\phi),\cos(\theta))$ and $$\vec{r}=(\cos(\theta)\dot{\theta},-\sin(\phi)\dot{\phi},-\sin(\theta)\dot{\theta})$$ $$\|\dot{\vec{r}}\|^2 = \dot{\theta}^2+\sin^2(\phi)\dot{\phi}^2$$ $$T=\frac12m(\dot{\theta}^2+\sin^2(\phi)\dot{\phi}^2)$$ Is that what I want?

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  • $\begingroup$ Hi! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. You might want to consider posting elsewhere, e.g., at the PhysicsForums. $\endgroup$ – stafusa Oct 15 '17 at 12:24
  • $\begingroup$ @stafusa It's not actually homework, but I thought the tag included homework type problems. My actual question is more or less question 1 and question 2 could perhaps be rephrased as: do we find kinetic and potential energy entirely in the original coordinates, and then translate. Which isn't a worked example type, but I wanted to demonstrate my attempt $\endgroup$ – Terry E Oct 15 '17 at 12:27
  • $\begingroup$ I agree that the question might be conceptual enough in nature to be on topic (and, for what it's worth, I retracted my 'close' vote). The way it's written, tough, can make it look too much like it's about a particular problem (which is discouraged). Usually more conceptual and general questions attract better answers, and I'd suggest always keeping that in mind when writing questions. $\endgroup$ – stafusa Oct 15 '17 at 12:59
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Three coordinates to locate your bead minus one constraint makes 2 degrees of freedom, hence two generalized coordinates. Moreover, you need to realize that the $y$ coordinate is implicit for otherwise the statement about the rotation makes no sense as the rotation of $x$ about $z$ will give you a combination of $x$ and $y$.

It is always simplest to compute the kinetic term in Cartesian, and then covert. Here, the natural coordinate system is cylindrical since gravity acts along the down axis. Then use \begin{align} x(t)=\rho(t) \cos(\phi(t))\, ,\quad y(t)=\rho(t)\sin(\phi(t))\, ,\quad z(t)=\kappa \sqrt{1-\rho^2(t)} \end{align} From this, you simply compute $$ \dot{x}= \cos(\phi(t)) \dot{\rho}-\rho(t)\sin(\phi(t))\dot{\phi} $$ etc. and form $T=\frac{1}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2)$. The end result, as far as I can compute, is something like $$ T=\frac{1}{2m}\dot{\rho}^2\left(1+\frac{\kappa^2\rho^2(t)}{1-\rho^2(t)}\right)+\frac{1}{2m} \rho^2(t)\dot{\phi}^2 $$

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  • $\begingroup$ Brilliant. This is very helpful, and answers my questions. $\endgroup$ – Terry E Oct 15 '17 at 14:29
  • $\begingroup$ I don't understand the math you have here apparently, although the text (in your answer) does indeed answer my questions. Shouldn't the $x,y,z$ be $x(t) = \rho(t)\sin(\theta(t)),z(t)=\kappa(1-x^2)$? (and $y(t)$ something). This is with the $\theta$ giving the angle from the $z$-axis clockwise. Where the $\phi$ is for the revolution around the $z$-axis? $\endgroup$ – Terry E Oct 15 '17 at 15:22
  • $\begingroup$ The confusion is the way your constrait is written, with $x$ instead of $x$ and $y$. See the image of this question physics.stackexchange.com/questions/362848/… for related intuition. The actual location of the bead is in terms of $x,y,z$ so you really have $z^2=\kappa (1-(x^2+y^2))$. In other words, the shape is given by $z$ as a function of $x$ but a point on the rotating wire must be a function of $x$ and $y$ since the rotation moves the bead in the $xy$ plane. $\endgroup$ – ZeroTheHero Oct 15 '17 at 15:29
  • $\begingroup$ @TerryE In other words you need to account for the fact that the shape of the wire $z^2=\kappa (1-x^2)$ is rotating. If you express $z$ in terms of $x$ and $y$, you automatically account for the constraint so you don’t need the $\theta$ angle as the $z$-coordinate is determined from $\rho$ and $\phi$ through the constraint. $\endgroup$ – ZeroTheHero Oct 15 '17 at 15:32
  • $\begingroup$ Thanks for your help I got it. Just a side question though, since I don't have the reputation to leave comments on other questions. In that place you linked me. What are the generalized coordinates? One is $\rho(t)$, what is the other? I would like to say it is more or less $\Omega t$, but I am unsure. (In the cylindrical coordinates you suggested, I mean) $\endgroup$ – Terry E Oct 16 '17 at 22:42
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1.Yes, this system is described by the two parameters completely. And it is call the spherical coordinate. The distance between the origin and the reference point is decided only by $\theta$ . Work it out by following steps:

  • Calculate the equation of the curve surface
  • Substitute x, y, z by new parameters and then you will find once they are decided, one can easily work out the length

I suggest you draw a picture of it to make a peculiar analysis.

2.Make a substitution and find its derivative with respect to time.

You may google it for the substitution of the two coordinate systems (Cartesian and spherical). But the more technical way is: Draw a vector from the origin in a Cartesian coordinate. Then find where is $\theta$, $\phi$, length, and its relation with x, y, z. We may make it since high school!

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  • $\begingroup$ You almost got it: Greek symbols, and math notation in general, is made writing between dollar symbols, since that's "laTex equation". \$\theta\$ $\endgroup$ – FGSUZ Oct 15 '17 at 10:56

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