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
[Math] Polar velocity given cartesian components
polar coordinates
Related Solutions
$\vec v = v_r \hat r + v_\theta \hat \theta$, where $\hat r$ and $\hat \theta$ are the unit radial vector and unit rotational vector at $(r, \theta)$. Letting $\vec p = x\hat i + y \hat j$ be the position vector, and noting that $x = r\cos \theta, y = r\sin \theta$, we have $$\vec p = r\cos \theta \hat i + r\sin \theta\hat j$$. Now $\hat r$ is the direction that $\vec p$ changes when $r$ increases, and $\hat \theta$ is the direction that $\vec p$ changes then $\theta$ increases: $$\hat r = \frac{\frac{\partial\vec p}{\partial r}}{\left\|\frac{\partial\vec p}{\partial r}\right\|}\qquad\hat \theta = \frac{\frac{\partial\vec p}{\partial \theta}}{\left\|\frac{\partial\vec p}{\partial \theta}\right\|}$$
So $$\hat r = \cos \theta \hat i + \sin \theta\hat j\\\hat \theta = -\sin \theta \hat i + \cos \theta\hat j$$
Thus $$\begin{align}\vec v &= \pi r\hat \theta\\&=\pi(-r\sin \theta\hat i + r\cos \theta\hat j)\\&=-\pi y\hat i + \pi x \hat j\end{align}$$
Since you've not defined what $U$ and $V$ mean, I can't answer any further than that.
$\hat{\theta}$ tells you the angle you need to be at from the positive $x$-axis, and $\hat{r}$ tells you how far you need to walk out from the origin. The magnitude is most certainly given by the $\hat{r}$ component. Perhaps your issue is that you're not adding/multiplying vectors in polar form properly. You cannot simply take, for example, $v_1 = 1\hat{r} + \pi\hat{\theta}$, and conclude that $v_1 + v_1 = 2\hat{r} + 2\pi\hat{\theta}$. Notice that the angle has been changed, which shouldn't happen for two vectors that are colinear.
It is better to represent vectors as $re^{i\theta}$. From which in our example we have, $v_1 + v_1 = e^{i\pi} + e^{i\pi} = 2e^{i\pi}$, which has an $\hat{r}$ of 2, and $\hat{\theta}$ of still $\pi$.
Best Answer
Hint.
Remember that
$$ \left( \begin{array}{c} x'(t) \\ y'(t) \end{array} \right) = \left( \begin{array}{cc} \cos (\theta (t)) & -r(t) \sin (\theta (t)) \\ \sin (\theta (t)) & r(t)\cos (\theta (t)) \end{array} \right)\cdot \left( \begin{array}{c} r'(t)\\ \theta '(t) \end{array} \right) $$