Mass rotating on ellipse track

[skrat]

Homework Statement

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.

Homework Equations

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? :/

 

[mfb]

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.

 

[skrat]

How is $\alpha$ not the polar angle? http://en.wikipedia.org/wiki/Ellipse#General_parametric_form

Than where is $\alpha$?

 

[mfb]

Note that the parameter t (called the eccentric anomaly in astronomy) is not the angle of (X(t),Y(t)) with the X-axis.

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.

 

[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?

 

[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?

 

[dauto]

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

 

[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 $\omega ^2=\frac{gb}{a^2}$