Vector differentiation of velocity (polar coord.)

[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!

 

[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!)