# Mass rotating on ellipse track

1. Apr 21, 2014

### skrat

1. The problem statement, all variables and given/known data
A body with mass $m$ can move without any friction on ellipse that $(x/a)^2+(y/b)^2=1$ describe. In $y$ direction homogeneous gravity field $g$ is present. For generalized coordinate we take angle $\alpha$ defined with $x=acos\alpha$, $y=bsin\alpha$. Find equilibrium position and frequency of oscillation around that position.

2. Relevant equations

3. The attempt at a solution

$L=T-V$

$T=\frac{1}{2}m(\dot{r}^2+r^2\dot{\alpha }^2)$ and $V=mgsin\alpha$.

$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\alpha }^2)-mgsin\alpha$

$\frac{\partial L}{\partial \alpha }-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\alpha }}=0=-mgbcos\alpha -2mr\dot{r}\dot{\alpha }-mr^2\ddot{\alpha }=0$

So finally

$\ddot{\alpha }+\frac{2\dot{r}}{r}\dot{\alpha }+\frac{gb}{r^2}cos\alpha =0$.

For equilibrium position:

$cos\alpha =0$ so $\alpha =-\frac{\pi}{2}$

Now for frequency, I am guessing I can write:

$\ddot{\alpha }+\frac{2\dot{r}}{r}\dot{\alpha }+\frac{gb}{r^2}\alpha =0$

But this will be like horrible to calculate, since $r=\sqrt{a^2cos^2\alpha +b^2sin^2\alpha }$...

Hmmm, is that ok? :/

Last edited: Apr 21, 2014
2. Apr 21, 2014

### Staff: Mentor

Your generalized coordinate is not the polar angle, don't mix them.

At some point you'll have to approximate the kinetic and potential energy for small oscillations, otherwise you won't find a meaningful solution.

3. Apr 21, 2014

### skrat

4. Apr 22, 2014

### Staff: Mentor

You can use the more general form described above that (and simplify it), but then you need a different variable and you don't use the one the problem statement suggested.

5. Apr 24, 2014

### skrat

hmmm...

$\vec{r}=(acos\alpha ,bsin\alpha )$ and $r^2=a^2cos^2\alpha +b^2sin^2\alpha$

$\dot{\vec{r}}=\dot{\alpha }(-asin\alpha ,bcos\alpha )$ and $\dot{r}^2=\dot{\alpha }^2(a^2sin^2\alpha +b^2cos^2\alpha )$.

Than $L=T-V=\frac{1}{2}m(\dot{r}^2+r^2\dot{\alpha }^2)-mgbsin\alpha$

$L=\frac{1}{2}m(\dot{\alpha }^2(a^2sin^2\alpha +b^2cos^2\alpha )+\dot{\alpha }^2(a^2cos^2\alpha +b^2sin^2\alpha ))-mgbsin\alpha$

which leaves me with $L=\frac{1}{2}m\dot{\alpha }^2(a^2+b^2)-mgbsin\alpha$

Now $\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\alpha }}-\frac{\partial L}{\partial \alpha }=m\ddot{\alpha }(a^2+b^2)+mgcos\alpha=0$

and finally $\ddot{\alpha }+\frac{gb}{a^2+b^2}cos\alpha=0$.

Which leaves me with equilibrium position at $\pi/2$ and therefore frequency $\omega ^2=\frac{gb}{a^2+b^2}$

However, this has to be wrong (don't ask me why), but if $a=b$, than the track would be a simple circle and I would expect frequency something like $\omega ^2=\frac{g}{R}$... Hmmm?

6. Apr 24, 2014

### skrat

One more question, I would expect that $T$ is independent of cartesian or polar coordinates, but obviously this is not the case:

$T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)=\frac{1}{2}m\dot{\alpha }^2(a^2sin^2\alpha + b^2cos^2\alpha )$ and

$T=\frac{1}{2}m(\dot{r}^2+r^2\dot{\alpha }^2)=\frac{1}{2}m\dot{\alpha }^2(a^2+b^2)$.

How is that so?

7. Apr 24, 2014

### dauto

Your second equation is wrong because, as explained, α is NOT the polar coordinate.

Last edited: Apr 24, 2014
8. Apr 24, 2014

### skrat

Aaaaaaaaaaaaaaaaaaaaaaaaa,

Now I can see what the problem is! Aha! HA! Ok! I can't believe it took me so long.

Well than let's forget about polar coordinates and do everything in cartesian.

This brings me to $\omega ^2=\frac{gb}{a^2}$