Simple question on proving polar coordinates

  • #1
A2Airwaves
9
0

Homework Statement


Prove: $$\frac{d\hat{r}}{dt} = \dot{\phi} \hat{\phi }$$ and $$\frac{d\hat{\phi}}{dt} = -\dot{\phi} \hat{r }$$


Homework Equations




The Attempt at a Solution


I solved this for an Analytical Mechanics assignment a month ago, and completely forgot how it goes..
$$\hat{r} ⊥ \hat{\phi}$$
An change from r1 to r2 will create a ##Δ\phi## that is in the ##\hat{\phi}## direction...
and because ##\hat{r} ⊥ \hat{\phi}##, we can say the same happens for a change from ##\phi1## to ##\phi2## except in the ##-\hat{r}## direction. Assuming the change is infinitesimal, we can write ##Δr## or ##Δ\phi## as d/dt.

But then I'm confused because, why are we assuming a change from r1 to r2 is a rotation by ##Δ\phi,## and not a change of the length r..? Am I getting something completely wrong here?
 

Answers and Replies

  • #2
geoffrey159
535
72
Go back to definition:
##\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j ##
##\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j ##

And use time derivative
 
  • #3
geoffrey159
535
72
Go back to definition:
##\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j ##
##\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j ##

And use time derivative
 
  • #4
A2Airwaves
9
0
Go back to definition:
##\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j ##
##\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j ##

And use time derivative
Thank you so much!
My professor emphasized the geometric interpretation of the answers that I completely forgot about those definitions.
Worked like magic, problem solved.

By the way, is there a reason why you're writing ##\vec{i}## and not ##\hat{i}##?
I'm used to ##\hat{i}## as a notation for unit vectors, but do you mean the same thing or are you referring to something else?
 
  • #5
geoffrey159
535
72
Yes you're right, I meant ##\hat i## and ## \hat j ##.
For clear explanations and nice drawings, look up Kleppner and Kolenkow first chapter.
 

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