Deriving Velocity Expression for Angular Impulse in Polar Coordinates

[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:
${\vec{l}}^2=l^2=(\vec{r}\times{m\dot\vec{r}})^2=m^2|\vec{r}|^2|\dot\vec{r}|^2-(\vec{r}\bullet{m\dot\vec{r}})^2=<br /> m^2r^2{\dot\vec{r}}^2-(\vec{r}\bullet{m{\dot\vec{r}})^2$
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

 

[Andrew Mason]

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 [itex]m^2/sec^2[/tex] so this cannot be correct.<br /> <br /> It should be:<br /> <br /> $$\dot\vec{r}^2=\dot{r}^2+\left(\frac{l}{mr}\right)^2$$<br /> <br /> AM[/itex]

 

[littleHilbert]

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

 

[Gokul43201]

We can't simply write: ${\dot\vec{r}}^2={\dot{r}}^2$, since then l=0. But why? Which rule forbids that equality.

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

 

[littleHilbert]

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