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I Problem with polar coordinates

  1. Nov 16, 2016 #1
    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
     
  2. jcsd
  3. Nov 17, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You get the first expression if you differentiate x = r cos(ϕ) twice.
    Where does the second equation come from? It is not true in general.
     
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