The velocity of a particle moving in a plane in polar coordinates is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]{\bf{v}} = v_r {\bf{\hat r}} + r\omega \hat \theta[/tex]

where [tex]v_r = \frac{{dr}}{{dt}}[/tex] and [tex]\omega = \frac{{d\theta }}{{dt}}[/tex].

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

[tex]{\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[/tex]

(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 [tex]\omega ^2 r[/tex] get into the derivative? I assume the [tex]\bf{\hat r}[/tex] is a unit vector, so the derivative of [tex]v_r[/tex] should just be [tex]{\frac{{dv_r }}{{dt}}[/tex] right? Anyway, if anyone could just explain that detail to me, I'll be on my way.

Thanks!

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# Vector differentiation of velocity (polar coord.)

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