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Expression for velocity

  1. May 31, 2006 #1
    Hey guys!

    I've got a question. How do we get this expression for the velocity:

    [itex]\dot\vec{r}=\dot{r}+\frac{l^2}{m^2r^2}[/itex], where l is the angular impulse of force

    I thought we could do it like this:
    We can't simply write:[itex]{\dot\vec{r}}^2={\dot{r}}^2[/itex], since then l=0. But why? Which rule forbids that equality. Similarly we can't treat the scalar product above as we would wish to. So how should one proceed in this case?

    Last edited: May 31, 2006
  2. jcsd
  3. May 31, 2006 #2

    Andrew Mason

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    Where do you get this expression from? The units of [itex]L^2/m^2r^2[/itex] are [itex]m^2/sec^2[/tex] so this cannot be correct.

    It should be:


  4. May 31, 2006 #3
    Oh, Jesus sorry...I meant (anglular) impulse of force of course! :)...I've corrected it.
  5. May 31, 2006 #4


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    [tex]\vec{r} = r\hat{r} [/tex]
    [tex]\implies \dot \vec{r} = \frac{d}{dt} (r\hat{r}) = r\frac{d \hat{r}}{dt} + \frac{dr }{dt} \hat{r} = r \dot \theta \hat{\theta} + \dot r \hat{r} [/tex]
    Last edited: May 31, 2006
  6. May 31, 2006 #5
    Ahhh! Of course - polar coordinates! I've got it now!
    Thanks guys! :-)
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