Basic question pertaining to Polar Coordinates & how to employ them

[warhammer]

I have a question that might be considered vague or even downright idiotic but just wanted to know that once we find out the velocity & acceleration of a body in angular motion in plane polar coordinates, and are asked to integrate the expressions in order to find position at some specified time 't' , would we proceed to integrate the velocity expression (say) like we do usually with respect to time when the said expression involves r cap as well as theta cap; since the polar unit vectors are not "fixed" unlike the unit vectors of i,j,k in Cartesian system?

 

[pasmith]

The position vector is $\mathbf{r} = r\mathbf{e}_r(\theta)$. You need both $r(t)$ and $\theta(t)$ to fully specify the position. The velocity vector is $$\dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta$$ so if $\dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta$ then $$\begin{align*} r &= \int v_r\,dt \\ \theta &= \int \frac{v_\theta}{r}\,dt \end{align*}$$

(You can always resolve components because $\mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1$ and $\mathbf{e}_r \cdot \mathbf{e}_\theta = 0$ are constant.)

 

[warhammer]

The position vector is $\mathbf{r} = r\mathbf{e}_r(\theta)$. You need both $r(t)$ and $\theta(t)$ to fully specify the position. The velocity vector is $$\dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta$$ so if $\dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta$ then $$\begin{align*} r &= \int v_r\,dt \\ \theta &= \int \frac{v_\theta}{r}\,dt \end{align*}$$

(You can always resolve components because $\mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1$ and $\mathbf{e}_r \cdot \mathbf{e}_\theta = 0$ are constant.)

Thank you so much for your help sir