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Velocity in Polar Coordinates

  1. May 28, 2008 #1
    Hi, I've just gone through a derivation and would like some confirmation that my reasoning is correct:

    Say the position of a particle is expressed in polar coordinates as


    If we want to describe it's velocity v we need to differentiate both components(angular and radial) with respect to time, as well as adding a directional component and giving each magnitude(speed(?)).

    So, differentiating individual components we say rate of change of the radial component is


    and hence velocity of radial component = [tex]\dot{r}[/tex][tex]\hat{r}[/tex]

    where we define [tex]\hat{r}[/tex] to be the unit vector pointing outwards in the positive direction along the radial component.

    secondly we need to define the rate of change of the angular component

    we can say this is [tex]\frac{d\theta}{dt}[/tex]

    and to give it direction we define [tex]\hat{\theta}[/tex] to be the unit vector pointing perpendicular to the radial line in the counterclockwise direction.

    now here is the more delicate bit: we need to add a factor of r to this in order to give the angular motion a magnitude, otherwise particles close to the origin would be moving at the same velocity as those far from it. This naturally only works if we are using radians.

    hence the final velocity is described as


    So what we've done is shifted from polar to vectorial system with the vector components of the velocity at the position of the particle at any time, adding to give the speed and direction.

    I may post this in other forums since it falls under more than one category, thanks in advance.
  2. jcsd
  3. May 28, 2008 #2
    I'm probably not wrong when I think that you have first seen this result somewhere, and then figured out a way to justify it? :wink: I believe be your reasoning is ok, and not useless because it makes it easier to remember the result, but I still suggest also calculating it more explicitly. It could be that some day you need to do something similar with a more complicated case, where you cannot simply see the result from a picture.

    If the Cartesian basis is [tex](\hat{e}_1, \hat{e}_2)[/tex], then for each fixed [tex]\theta[/tex] the basis in polar coordinates is

    \hat{r} = \cos\theta\;\hat{e}_1 + \sin\theta\;\hat{e}_2

    \hat{\theta} = -\sin\theta\; \hat{e}_1 + \cos\theta\;\hat{e}_2

    You can solve the Cartesian basis out of this, and get

    \hat{e}_1 = \cos\theta\;\hat{r} - \sin\theta\;\hat{\theta}

    \hat{e}_2 = \sin\theta\;\hat{r} + \cos\theta\;\hat{\theta}

    Then suppose you have some path in the plane, parametrized by the time [tex]t[/tex]. In Cartesian coordinates the path is [tex](x(t), y(t))[/tex], and in the polar coordinates it is [tex](r(t),\theta(t))[/tex], and these coordinates are related by

    x(t) = r(t)\cos\theta(t)

    y(t) = r(t)\sin\theta(t)

    Take the time derivative of both sides, and you get

    \dot{x}(t) = \dot{r}(t)\cos\theta(t) - r(t)\dot{\theta}(t)\sin\theta(t)

    \dot{y}(t) = \dot{r}(t)\sin\theta(t) + r(t)\dot{\theta}(t)\cos\theta(t)

    Now substitute these into

    v(t) = \dot{x}(t)\hat{e}_1 + \dot{y}(t)\hat{e}_2

    and you should be getting the same result as you already got.
  4. May 28, 2008 #3
    Yes this way of doing it seems more elegant, I can follow it easily but lack the confidence with maths to switch to cartesian to solve the problem and then back to polar. As you assumed the problem I had had the result stated (as in "prove that.."). The problem with a question like that with the result given beforehand is I guess that it tends to limit the originality of your solution, since the result gives you an early indication of where you need to go.

    Would you say that the approach you used is the more useful one for solving polar type problems?

    Also, are results such as the polar->Cartesian transformations so useful that you simply know them off by heart?

    Thanks for the reply.
  5. May 28, 2008 #4
    There's a dot missing above ... probably just a typo:


    If you are unsure of what you are done, I'd pick a simpler example to verify. Example, it should (and does with the typo fix) hold for a plane:

    z = a e^{i\theta}

    z' = a' e^{i\theta} + i\theta' a e^{i\theta} = a' \hat{z} + \hat{\theta} \theta' z
  6. May 28, 2008 #5
    Hi, yes the lack of a dot was a mistake!

    Thanks for the help.

    I've received another response in the other forum (calculus) that showed yet another really good way of doing this, it's nice when all the maths comes together.
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