Polar velocity given cartesian components

I would like to calculate the polar velocity components given the position $(x,y)$ and velocity $(u_x,u_y)$ in cartesian coordinates. First of all, $r=\sqrt{x^2+y^2}$ and $\theta={\rm atan}{(y/x)}$. By now, I know the angle and radius in the global cylindrical coordinate system. I assume that $u=u_re_r+u_\theta e_\theta$. Is it correct to write $u_r=\cos(\theta)u_x + \sin(\theta)u_y$ and $u_\theta=-\sin(\theta)u_x + \cos(\theta)u_y$ ?The problem is that I would like to calculate the components in polar coordinates (not angular velocity though) given that a particle is moving from $(x_1,y_1)$ to $(x_2,y_2)$. Thank you in advance for the help given

• Everything looks correct to me. – Ingix Jun 13 '18 at 14:19

$$\left( \begin{array}{c} x'(t) \\ y'(t) \end{array} \right) = \left( \begin{array}{cc} \cos (\theta (t)) & -r(t) \sin (\theta (t)) \\ \sin (\theta (t)) & r(t)\cos (\theta (t)) \end{array} \right)\cdot \left( \begin{array}{c} r'(t)\\ \theta '(t) \end{array} \right)$$
• $\theta'(t)$ is the angular velocity (rad/s) or the linear velocity (m/s) ? – dimpep Jun 13 '18 at 14:40
• The units for $\theta'(t)$ are [rads/s] – Cesareo Jun 13 '18 at 14:42
• Thus $r(t)\theta'(t)$ is the linear one I assume – dimpep Jun 13 '18 at 17:39
• You can do that assuming a referential $\hat{n},\hat{\tau}$. In that referential we have $\dot r\hat{n}+r\dot\theta\hat{\tau}$ – Cesareo Jun 13 '18 at 18:04