Kinetic energy for generalized coordinates?

Question

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?

Answer 1

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

Answer 2

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!