elegysix
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I'm trying to find the equation of motion of a bead, which is constrained to roll along the bottom half of this frictionless ellipse, by using langrange's method - L(\phi,\dot{\phi}).
Here's the setup:
given the bottom half of an ellipse:
\mathbf{r}(\phi) = acos(\phi)\mathbf{\hat{i}}-bsin(\phi)\mathbf{\hat{j}}
where 0<\phi<\pi, and \mathbf{\hat{i}} , \mathbf{\hat{j}} are the unit vectors for x and y, and \phi is measured from the positive x-axis counterclockwise.
my question is should I use
1) KE=\frac{m}{2}(\dot{r}^{2}+r^{2}\dot{\phi}^{2})
or
2) KE=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})
where \dot{x} \:\&\: \dot{y} come from x=acos(\phi) , and y=-bsin(\phi)
which one?
I'm confused because I have polar type coordinates, r(\phi) in terms of Cartesian components.
Assuming that the second equation for KE is not appropriate to use here,
If I use the first the first equation for KE, then is it 'Legal' to find and use \mathbf{\dot{r}} like this? or is this wrong?
\mathbf{\dot{r}}=-a\dot{\phi}sin(\phi)\mathbf{\hat{i}}-b\dot{\phi}cos(\phi)\mathbf{\hat{j}}
--------------------------------------------------------------------------------------------------------------
As my only other option I'm aware of, I know that the definitely correct, brute force approach looks like this:
\mathbf{r}(\phi)_{polar}=|\mathbf{r}(\phi)_{cartesian}|\mathbf{\hat{r}}=\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{r}}
which means we're in polar coordinates, so
\mathbf{\dot{r}}(\phi)=\dot{r}\mathbf{\hat{r}}+r\dot{\phi}\mathbf{\hat{\phi}}
\mathbf{\dot{r}}(\phi)=\frac{d(\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)})}{dt}\mathbf{\hat{r}}+\dot{\phi}\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{\phi}}
and L=KE-PE... and then I have to solve \frac{dL}{d\phi}-\frac{d(\frac{dL}{d\dot{\phi}})}{dt}=0
which, that route is becoming a nightmare very quickly, and I don't want any trouble with the algebra police... any advice?
Here's the setup:
given the bottom half of an ellipse:
\mathbf{r}(\phi) = acos(\phi)\mathbf{\hat{i}}-bsin(\phi)\mathbf{\hat{j}}
where 0<\phi<\pi, and \mathbf{\hat{i}} , \mathbf{\hat{j}} are the unit vectors for x and y, and \phi is measured from the positive x-axis counterclockwise.
my question is should I use
1) KE=\frac{m}{2}(\dot{r}^{2}+r^{2}\dot{\phi}^{2})
or
2) KE=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})
where \dot{x} \:\&\: \dot{y} come from x=acos(\phi) , and y=-bsin(\phi)
which one?
I'm confused because I have polar type coordinates, r(\phi) in terms of Cartesian components.
Assuming that the second equation for KE is not appropriate to use here,
If I use the first the first equation for KE, then is it 'Legal' to find and use \mathbf{\dot{r}} like this? or is this wrong?
\mathbf{\dot{r}}=-a\dot{\phi}sin(\phi)\mathbf{\hat{i}}-b\dot{\phi}cos(\phi)\mathbf{\hat{j}}
--------------------------------------------------------------------------------------------------------------
As my only other option I'm aware of, I know that the definitely correct, brute force approach looks like this:
\mathbf{r}(\phi)_{polar}=|\mathbf{r}(\phi)_{cartesian}|\mathbf{\hat{r}}=\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{r}}
which means we're in polar coordinates, so
\mathbf{\dot{r}}(\phi)=\dot{r}\mathbf{\hat{r}}+r\dot{\phi}\mathbf{\hat{\phi}}
\mathbf{\dot{r}}(\phi)=\frac{d(\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)})}{dt}\mathbf{\hat{r}}+\dot{\phi}\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{\phi}}
and L=KE-PE... and then I have to solve \frac{dL}{d\phi}-\frac{d(\frac{dL}{d\dot{\phi}})}{dt}=0
which, that route is becoming a nightmare very quickly, and I don't want any trouble with the algebra police... any advice?