Expression for velocity

  • #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:
[itex]{\vec{l}}^2=l^2=(\vec{r}\times{m\dot\vec{r}})^2=m^2|\vec{r}|^2|\dot\vec{r}|^2-(\vec{r}\bullet{m\dot\vec{r}})^2=
m^2r^2{\dot\vec{r}}^2-(\vec{r}\bullet{m{\dot\vec{r}})^2[/itex]
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?

Thanks
 
Last edited:

Answers and Replies

  • #2
Andrew Mason
Science Advisor
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littleHilbert said:
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 momentum
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:

[tex]\dot\vec{r}^2=\dot{r}^2+\left(\frac{l}{mr}\right)^2[/tex]

AM
 
  • #3
Oh, Jesus sorry...I meant (anglular) impulse of force of course! :)...I've corrected it.
 
  • #4
Gokul43201
Staff Emeritus
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littleHilbert said:
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.

[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]
 
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  • #5
Ahhh! Of course - polar coordinates! I've got it now!
Thanks guys! :-)
 

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