- #1
Gamdschiee
- 28
- 2
Hello,
I have a question about polar coordinates.
It is
[tex]\vec r = \begin{pmatrix}r cos\phi \\ rsin\phi \\ z\end{pmatrix}=r\cdot \vec e_r + z\cdot \vec e_z[/tex]
and than is
[tex]\ddot{\vec r} = (\ddot{r}-r\dot{\phi}^2)\vec e_r + (r\ddot{\phi} +2\dot{r}\dot{\phi})\vec e_{\phi} + \ddot{z}\vec e_z[/tex]
The following I got on the exercise
[tex]m\cdot \begin{pmatrix}\ddot{x} \\ \ddot{y} \\ \ddot{z}\end{pmatrix} =m\cdot \begin{pmatrix}0 \\ 0 \\ -g\end{pmatrix} + 2\lambda \begin{pmatrix}-x\cdot cos^2 \alpha \\ -y\cdot cos^2 \alpha \\ z\cdot sin^2 \alpha\end{pmatrix}[/tex]
I have three components now. One component is an equation(I, II and III).
I also got following hint:
[tex]I:\cos\phi + III: \tan\alpha[/tex]
[tex]z=\frac{r}{\tan \alpha}[/tex]
When I use this hints, I got this:
[tex]\frac{\ddot x}{\cos \phi}+\frac{\ddot{z}}{\tan \alpha} = -\frac{g}{\tan \alpha}[/tex]
Until this point, it should be all clear to me, I guess.
Apprently you will get following, if you go further:
[tex]\frac{(\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi}{\cos \phi}+\frac{\ddot{r}}{\tan^2 \alpha}+ \frac{g}{\tan \alpha}=0[/tex]
My Questions:
1. Why is [tex]\ddot{x} = (\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi[/tex], apperently?2. And why is [tex]2\dot{r}\dot{\phi}+r\ddot{\phi}=0[/tex]?
I hope someone can help me with this case.
Kind regards,
Gamdschiee
I have a question about polar coordinates.
It is
[tex]\vec r = \begin{pmatrix}r cos\phi \\ rsin\phi \\ z\end{pmatrix}=r\cdot \vec e_r + z\cdot \vec e_z[/tex]
and than is
[tex]\ddot{\vec r} = (\ddot{r}-r\dot{\phi}^2)\vec e_r + (r\ddot{\phi} +2\dot{r}\dot{\phi})\vec e_{\phi} + \ddot{z}\vec e_z[/tex]
The following I got on the exercise
[tex]m\cdot \begin{pmatrix}\ddot{x} \\ \ddot{y} \\ \ddot{z}\end{pmatrix} =m\cdot \begin{pmatrix}0 \\ 0 \\ -g\end{pmatrix} + 2\lambda \begin{pmatrix}-x\cdot cos^2 \alpha \\ -y\cdot cos^2 \alpha \\ z\cdot sin^2 \alpha\end{pmatrix}[/tex]
I have three components now. One component is an equation(I, II and III).
I also got following hint:
[tex]I:\cos\phi + III: \tan\alpha[/tex]
[tex]z=\frac{r}{\tan \alpha}[/tex]
When I use this hints, I got this:
[tex]\frac{\ddot x}{\cos \phi}+\frac{\ddot{z}}{\tan \alpha} = -\frac{g}{\tan \alpha}[/tex]
Until this point, it should be all clear to me, I guess.
Apprently you will get following, if you go further:
[tex]\frac{(\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi}{\cos \phi}+\frac{\ddot{r}}{\tan^2 \alpha}+ \frac{g}{\tan \alpha}=0[/tex]
My Questions:
1. Why is [tex]\ddot{x} = (\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi[/tex], apperently?2. And why is [tex]2\dot{r}\dot{\phi}+r\ddot{\phi}=0[/tex]?
I hope someone can help me with this case.
Kind regards,
Gamdschiee