# Expression for velocity

1. May 31, 2006

### littleHilbert

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

Last edited: May 31, 2006
2. May 31, 2006

### Andrew Mason

Where do you get this expression from? The units of $L^2/m^2r^2$ are [itex]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

3. May 31, 2006

### littleHilbert

Oh, Jesus sorry...I meant (anglular) impulse of force of course! :)...I've corrected it.

4. May 31, 2006

### Gokul43201

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

Last edited: May 31, 2006
5. May 31, 2006

### littleHilbert

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