# Vector differentiation of velocity (polar coord.)

1. Dec 15, 2009

### quozzy

The velocity of a particle moving in a plane in polar coordinates is

$${\bf{v}} = v_r {\bf{\hat r}} + r\omega \hat \theta$$

where $$v_r = \frac{{dr}}{{dt}}$$ and $$\omega = \frac{{d\theta }}{{dt}}$$.

By differentiating w.r.t. time, show that the acceleration of the particle is

$${\bf{a}} = \left( {\frac{{dv_r }}{{dt}} - \omega ^2 r} \right){\bf{\hat r}} + \left( {2\omega v_r + r\frac{{d\omega }}{{dt}}} \right)\hat \theta$$

(The no-subscript v should be bold, as should the a and the r's with hats.)

Okay, I'm confident I can work this one out, except for one thing: how does that $$\omega ^2 r$$ get into the derivative? I assume the $$\bf{\hat r}$$ is a unit vector, so the derivative of $$v_r$$ should just be $${\frac{{dv_r }}{{dt}}$$ right? Anyway, if anyone could just explain that detail to me, I'll be on my way.

Thanks!

2. Dec 15, 2009

### Mindscrape

I like writing in the dot notation because it helps me to remember the dependencies of each variable.

$$\dot{\mathbf{r}}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\mathbf{\theta}}$$

So r depends on time and theta depends on time, what are all the variables that now have r in them and theta in them? (Hint: unit vectors might count!)

Last edited: Dec 15, 2009