Indeterminate velocity components of a particle at the center of a polar coordinate system

Question

Given coordinates of a particle and components of its velocity vector in Cartesian coordinates, I need to convert this data into polar coordinates. I had no problem to deduce the formulas which give $v_r$ and $v_\theta$ in terms of $v_x$ and $v_y$ (see, e.g., the answer to this question). The result is: $$v_r = \dot r = \frac{x\dot x+y\dot y}{\sqrt{x^2+y^2}}, \qquad v_\theta = r\dot\theta = \frac{x\dot y - \dot x y}{\sqrt{x^2+y^2}}.$$ But what if a particle happens to be exactly at the center of the coordinate system? In Cartesian coordinates a particle might perfectly have $x=0$ and $y=0$, and , say $v_x=1$, $v_y=1$, so that the velocity vector has the magnitude of $\sqrt{2}$ and is directed to the center of the first quadrant. Now, when I switch to the polar system, I know that the velocity magnitude should be the same ($\sqrt{2}$), but in this case the two components are indeterminate (both are $0/0$)! How can I circumvent this "coordinate artifact" and to get sensible values of the velocity components in polar coordinates?

Answer 1

If you are considering the velocity of a point particle at the centre of a polar coordinate system then you know that all of the velocity is directed radially outwards from this centre.

Therefore $v_r = v_{total}$ and $v_{\theta} = 0$.

Mathematically we have: $$\dot x = \dot r \cos \theta - r\dot\theta \sin\theta = \dot r\cos\theta$$ $$\dot y = \dot r \sin \theta + r\dot\theta \cos\theta = \dot r\sin\theta$$ therefore, $$ \dot r = \sqrt{\dot x^2+\dot y^2}$$ which is generally not true but is in this case because $v_{\theta} = 0$ by definition.