# I Problem with polar coordinates

1. Nov 16, 2016

### Gamdschiee

Hello,

I have a question about polar coordinates.

It is
$$\vec r = \begin{pmatrix}r cos\phi \\ rsin\phi \\ z\end{pmatrix}=r\cdot \vec e_r + z\cdot \vec e_z$$
and than is
$$\ddot{\vec r} = (\ddot{r}-r\dot{\phi}^2)\vec e_r + (r\ddot{\phi} +2\dot{r}\dot{\phi})\vec e_{\phi} + \ddot{z}\vec e_z$$

The following I got on the exercise
$$m\cdot \begin{pmatrix}\ddot{x} \\ \ddot{y} \\ \ddot{z}\end{pmatrix} =m\cdot \begin{pmatrix}0 \\ 0 \\ -g\end{pmatrix} + 2\lambda \begin{pmatrix}-x\cdot cos^2 \alpha \\ -y\cdot cos^2 \alpha \\ z\cdot sin^2 \alpha\end{pmatrix}$$

I have three components now. One component is an equation(I, II and III).

I also got following hint:
$$I:\cos\phi + III: \tan\alpha$$
$$z=\frac{r}{\tan \alpha}$$

When I use this hints, I got this:
$$\frac{\ddot x}{\cos \phi}+\frac{\ddot{z}}{\tan \alpha} = -\frac{g}{\tan \alpha}$$

Until this point, it should be all clear to me, I guess.

Apprently you will get following, if you go further:
$$\frac{(\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi}{\cos \phi}+\frac{\ddot{r}}{\tan^2 \alpha}+ \frac{g}{\tan \alpha}=0$$

My Questions:
1. Why is $$\ddot{x} = (\ddot{r}-r\dot{\phi}^2)cos \phi-(2\dot{r}\dot{\phi}+r\ddot{\phi})\sin \phi$$, apperently?

2. And why is $$2\dot{r}\dot{\phi}+r\ddot{\phi}=0$$?

I hope someone can help me with this case.

Kind regards,
Gamdschiee

2. Nov 17, 2016

### 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.