Find the velocity vector in the form of polar coordinates

Question

Lets say for an object doing a projectile motion, in order to find the velocity vector in the function of t;

R is constant.

$$\vec r(t) = R\hat r + \theta (t) \hat \theta \\ \vec V(t) = \frac{d\vec r(t)}{dt} \\ \vec V(t) = 0\hat r + \theta '(t) \hat \theta \\ \vec V(t) = \theta '(t) \hat \theta$$,which is not true.

So, where is the mistake in this calculation ?

Answer 1

Note that $R$ may not be constant. Besides, there is no $\hat{\theta}$ component of $\vec{r}$ in polar coordinates.

The position of the particle at time $t$ is specified by $r(t)$ and $\theta(t)$ through

$$\vec{r}(t)=r(t)\hat{r}\left(r(t),\theta(t)\right).$$

Note that $\hat{r}$ is a function of $t$ through $$\hat{r}\left(r(t),\theta(t)\right)$$

$$\vec{v}(t)=\frac{dr}{dt}\hat{r}(t)+r(t)\frac{d\hat{r}}{dt}$$ $$=\dot{r}(t)\hat{r}(t)+r(t)\left(\frac{\partial \hat{r}}{\partial r}\dot{r}+\frac{\partial \hat{r}}{\partial \theta}\dot{\theta}\right)$$ Since $$\frac{\partial \hat{r}}{\partial r}=\vec{0}, \frac{\partial \hat{r}}{\partial \theta}=\hat{\theta}$$ we have $$\vec{v}(t)=\dot{r}(t)\hat{r}(t)+r(t)\dot{\theta}(t)\hat{\theta}(t)$$