# Expression for velocity

Hey guys!

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

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

I thought we could do it like this:
${\vec{l}}^2=l^2=(\vec{r}\times{m\dot\vec{r}})^2=m^2|\vec{r}|^2|\dot\vec{r}|^2-(\vec{r}\bullet{m\dot\vec{r}})^2= m^2r^2{\dot\vec{r}}^2-(\vec{r}\bullet{m{\dot\vec{r}})^2$
We can't simply write:${\dot\vec{r}}^2={\dot{r}}^2$, 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

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Andrew Mason
Homework Helper
littleHilbert said:
Hey guys!

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

$\dot\vec{r}=\dot{r}+\frac{l^2}{m^2r^2}$, where l is the angular momentum
Where do you get this expression from? The units of $L^2/m^2r^2$ are $m^2/sec^2[/tex] so this cannot be correct. It should be: $$\dot\vec{r}^2=\dot{r}^2+\left(\frac{l}{mr}\right)^2$$ AM Oh, Jesus sorry...I meant (anglular) impulse of force of course! :)...I've corrected it. Gokul43201 Staff Emeritus Science Advisor Gold Member littleHilbert said: We can't simply write: [itex]{\dot\vec{r}}^2={\dot{r}}^2$, since then l=0. But why? Which rule forbids that equality.

$$\vec{r} = r\hat{r}$$
$$\implies \dot \vec{r} = \frac{d}{dt} (r\hat{r}) = r\frac{d \hat{r}}{dt} + \frac{dr }{dt} \hat{r} = r \dot \theta \hat{\theta} + \dot r \hat{r}$$

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Ahhh! Of course - polar coordinates! I've got it now!
Thanks guys! :-)