homework and exercises - Write Equation of Motion in Polar Coordinates

a point of mass $m$ moves on the $xy-$Plane and has the potential energy $U(x,y)=\frac{A}{2}(x^2+y^2) + B$ with $A,B > 0$ being two constants.

Let $\vec{r}$ denote the position of the mass on the $xy-$Plane.

We knoe that $m\ddot{\vec{r}}=-\nabla U(x,y)$

We first write $U(r,\varphi)=\frac{A}{2}r^2 + B$

so we get $\nabla U(r,\varphi)=\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

We now calculate $\ddot{\vec{r}}$ in polar coordinates.

We have

$\vec{r}=r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}$

$\dot{\vec{r}}=\dot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi}$

$\ddot{\vec{r}}=\ddot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + 2\dot{r} \begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi} - r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}\dot{\varphi}^2 + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\ddot{\varphi}$

we put everything together:

$\ddot{\vec{r}}=\ddot{r}\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix} + 2\dot{r} \begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\dot{\varphi} - r\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}\dot{\varphi}^2 + r\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}\ddot{\varphi}=\frac{-1}{m}\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

(*)Since $\begin{pmatrix}\cos(\varphi) \\ \sin(\varphi)\end{pmatrix}=\hat{e}_r$ and $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=\hat{e}_\varphi$ and also $\begin{pmatrix} Ar \\ 0 \end{pmatrix}=Ar\hat{e}_r + 0\cdot \hat{e}_\varphi$ we get:

(**)$\begin{pmatrix}\ddot{\vec{r}}-r\dot{\varphi}^2 \\ 2\dot{r}\dot{\varphi} + r\ddot{\varphi}\end{pmatrix}=\frac{-1}{m}\begin{pmatrix} Ar \\ 0 \end{pmatrix}$

My lecture notes now write it down like this:

(***)$\begin{pmatrix}\ddot{\vec{r}} \\ r\ddot{\varphi}\end{pmatrix}=\begin{pmatrix} \frac{-Ar}{m} + r\dot{\varphi}^2\\ -2\dot{r}\dot{\varphi} \end{pmatrix}$

Question 1: Is $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=\hat{e}_\varphi$ true or is it rather $\begin{pmatrix}-\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}=-\hat{e}_\varphi$?

Question 2: Is there a reason why we wrote ( ** ) as ( * ** )? Does the left side of (***) tell us something?

Question 3: Is it possible that my potential depends on $r$ and $\varphi$? And if so, could I write down the general formula like that:

$\begin{pmatrix}m\ddot{\vec{r}} \\ mr\ddot{\varphi}\end{pmatrix}=\begin{pmatrix} K_r(r,\varphi) + r\dot{\varphi}^2\\ K_\varphi(r,\varphi)-2m\dot{r}\dot{\varphi} \end{pmatrix}$

With $K(r,\varphi)$ being a conservative force field (so $K=-\nabla U$ applies)

Question 4: Why don't I have to transform the constants?

Answer

Question 1:

It's up to you to chose a basis to your problem. They can differ only by a global fase (such a minus sign, since $e^{i\pi}=-1$).

Question 2:

As you might see, both expressions are the same. It's up to you chose the one you prefer. However, when you get something equals to a constant is always a good thing, since means that a quantity is conserved (in your case, $\frac{d}{dt}2\dot{r}\dot{φ}+r\ddot{φ}$)

Question 3:

you can get (at least, theoretically) a potential $V=V(r,\varphi)$ and there will be a general formula (take into account that $\nabla U=\frac{\partial U}{\partial r}+\frac{1}{r}\frac{\partial U}{\partial \varphi}$)