Lagrange equation of motion solutions

[bobred]

Hi
I have the Lagrangian

\(\displaystyle L \left(r,\dot{r},\dot{\phi} \right)=\frac{m}{2} \left[ \left(1+\alpha^{2} \right)\dot{r}^{2}+r^{2}\dot{\phi}^{2} \right]-mg \alpha r\)

I am asked to find the equations of motion, which gives

\(\displaystyle \ddot{r}+\alpha^{2}\ddot{r}-r\dot{\phi}^{2}+g\alpha =0\)
and
\(\displaystyle \frac{d}{dt}\left( r^{2}\dot{\phi} \right)=0\)

Also that the equation of motion for \(\displaystyle \phi\) implies that \(\displaystyle r^2 \dot{\phi}=K\) which by integrating the above equation does. Using this, \(\displaystyle \phi\) is eliminated from the equation of motion for r.
The part that is troubling me is I am asked to show that there is a solution of the equations of motion where

and

take constant values

and

respectively. Would this be setting \(\displaystyle \ddot{r}\) equal to a constant \(\displaystyle r_0\) and doing the same for \(\displaystyle \dot{\phi}\)?
James

 

[bobred]

I was thinking about the equations in isolation to each other and not as part of a system. So uniform motion is a constant solution for both

and

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[Ishuda]

Hi
I have the Lagrangian

\(\displaystyle L \left(r,\dot{r},\dot{\phi} \right)=\frac{m}{2} \left[ \left(1+\alpha^{2} \right)\dot{r}^{2}+r^{2}\dot{\phi}^{2} \right]-mg \alpha r\)

I am asked to find the equations of motion, which gives

\(\displaystyle \ddot{r}+\alpha^{2}\ddot{r}-r\dot{\phi}^{2}+g\alpha =0\)
and
\(\displaystyle \frac{d}{dt}\left( r^{2}\dot{\phi} \right)=0\)

Also that the equation of motion for \(\displaystyle \phi\) implies that \(\displaystyle r^2 \dot{\phi}=K\) which by integrating the above equation does. Using this, \(\displaystyle \phi\) is eliminated from the equation of motion for r.
The part that is troubling me is I am asked to show that there is a solution of the equations of motion where

and

take constant values

and

respectively. Would this be setting \(\displaystyle \ddot{r}\) equal to a constant \(\displaystyle r_0\) and doing the same for \(\displaystyle \dot{\phi}\)?
James

As I understand it, they are just saying
\(\displaystyle K\, =\, r_0^2\, *\, \Omega;\)
\(\displaystyle r\, =\, r_0\, \text{(a constant)};\)
\(\displaystyle \overset{\bullet}{\phi}\, =\, \Omega\, \text{(another constant)}\)
So, yes.