8
$\begingroup$

I am trying to convert an equation from Cartesian to polar coordinates. I know for a given $x$ and $y$ Cartesian, we can get polar $r=\sqrt{x^2+y^2}$ and $\theta = \arctan(y/x)$.

However, for a given $v_x$ and $v_y$, I want to get $v_r$ and $v_\theta$. Can we write $v_r = \sqrt{v_x^2+v_y^2}$ and $v_\theta = \arctan{(v_y/v_x)}$?

I know its' probably not that simple, but a simple derivation and explanation will be really helpful.

$\endgroup$

2 Answers 2

7
$\begingroup$

If $r(t) = \sqrt{x(t)^2+y(t)^2}$ , then $$\dot r = \frac{x\dot x+y\dot y}{\sqrt{x^2+y^2}}$$

If $\theta(t) = \arctan\left[y(t)/x(t)\right]$ , then $$\dot\theta = \frac{x\dot y - \dot x y}{x^2+y^2}$$

Alternatively:

If $x(t) = r(t)\cos\left[\theta(t)\right]$ , then $$\dot x = \dot r \cos \theta - r\dot\theta\sin\theta$$

If $y(t) = r(t)\sin\left[\theta(t)\right]$ , then $$\dot y = \dot r\sin\theta + r\dot\theta \cos\theta$$

$\endgroup$
3
  • $\begingroup$ So, Can we write $v_r = \sqrt{v_x^2+v_y^2}$ and $v_\theta = \arctan{(v_y/v_x)}$? $\endgroup$
    – Nosrati
    Commented Sep 25, 2017 at 19:25
  • $\begingroup$ @Nosrati I know I am very late to the party but I just wanted to point out that expression for $v_\theta$ does not even have units of velocity.. $\endgroup$
    – Luismi98
    Commented Mar 5, 2021 at 22:55
  • $\begingroup$ Also, I think this may be useful for latecomers: note the conversion from $(v_x,v_y)$ to $(v_r,v_\theta)$ is nothing but a rotation of the axis clockwise by $\theta$. The modulus of the velocity is $\sqrt{v_x^2+v_y^2}$ in $v_x v_y$ coordinates, but the $v_r v_\theta$ coordinates are nothing but a rotation of the latter, and hence the modulus of the velocity must also be $\sqrt{v_r^2+v_\theta^2}$. Therefore, $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}=\sqrt{v_r^2+v_\theta^2} \neq v_r$. $\endgroup$
    – Luismi98
    Commented Mar 5, 2021 at 23:02
5
$\begingroup$

In polar coordinates the position and the velocity of a point are expressed using the orthogonal unit vectors $\mathbf e_r$ and $\mathbf e_\theta$, that, are linked to the orthogonal unit cartesian vectors $\mathbf i$ and $\mathbf j$ by the relations:

$$ \mathbf e_r=\mathbf{i}\cos \theta +\mathbf{j}\sin \theta $$

$$ \mathbf e_\theta=-\mathbf{i}\sin \theta +\mathbf{j}\cos \theta $$

The position of a point is given by $\mathbf r=r\mathbf e_r$, where $r=\sqrt{x^2+y^2}$. from this we can find the velocitiy as:

$$ \mathbf v= \frac{d}{dt}(r\mathbf e_r)=\dot r\mathbf e_r+r\dot{\mathbf e}_r $$

with abit of calculus we see that this gives: $$ \mathbf v=\dot r\mathbf e_r+r\dot \theta \mathbf e_\theta $$

so the two components of this vector, in polar coordinates, are: $$ v_r=\dot r \qquad v_\theta=r\dot \theta $$

and you can see that

$$ v_r=\dot r=\frac{d}{dt}\sqrt{x^2+y^2}=\frac{x\dot x+y\dot y}{\sqrt{x^2+y^2}}\ne \sqrt{\dot x^2+\dot y^2}=\sqrt{v_x^2+v_y^2} $$

You can find by yourself the expression of $v_\theta$ in terms of the cartesian components of the velocity (that is not simple as in your question).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .