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?

  • $\begingroup$ 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, 2017 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, 2017 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, 2017 at 12:59

2 Answers 2


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 $$ T=\frac{1}{2m}\dot{\rho}^2\left(1+\frac{\kappa^2\rho^2(t)}{1-\rho^2(t)}\right)+\frac{1}{2m} \rho^2(t)\dot{\phi}^2 $$

  • $\begingroup$ Brilliant. This is very helpful, and answers my questions. $\endgroup$
    – Terry E
    Oct 15, 2017 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, 2017 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. $\endgroup$ Oct 15, 2017 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. $\endgroup$ Oct 15, 2017 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, 2017 at 22:42

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!

  • $\begingroup$ 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, 2017 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.