Determine the angular momentum in polar coordinates

[welssen]

Hi there,

I've been trying to solve the following problem, which I found looks pretty basic, but actually got me really confused about the definition of angular momentum.

Problem
The trajectory of a point mass m is described by the following equations, in spherical coordinates:
$r(t) = r_0 + v_0t$
$\phi(t) = \omega_0t$.
Determine the angular momentum of m (in spherical coordinates).

The attempt at a solution
Angular momentum is defined as $\vec{l} = m\cdot\vec{x(t)}\times\vec{v(t)}$.
Here, I would say, $\vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi}$.
I would take the cross product of this vector with its derivative (the speed) multiplied by mass to get the angular momentum.

But apparently this is wrong. In the given solutions, $\vec{x(t)} = r(t)\hat{r}$. The angular term is absent.

Could someone explain why the angular term is set to 0 in the cross product of angular momentum?
Thank you.

 

[haruspex]

Here, I would say, $\vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi}$.

Does that make sense dimensionally?

 

[welssen]

Does that make sense dimensionally?

Yes it makes sense: in polar coordinates [itex]\hat{\phi} [\itex] is a unit vector perpendicular to $\hat{r}[\itex]. Here is a picture of it dimensionally: <br /> <a href="http://i.gyazo.com/6685ade5d3ab158b496077f7481551ea.png" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://i.gyazo.com/6685ade5d3ab158b496077f7481551ea.png</a>$[/itex]

 

[haruspex]

Yes it makes sense: in polar coordinates [itex]\hat{\phi} [\itex] is a unit vector perpendicular to $\hat{r}[\itex].$[/itex]

[itex][itex] But it is dimensionally wrong. x and r have dimension of length, φ does not.<br /> $\vec r$ means the same as $\vec x$. They are both the vector representing the point.<br /> <br /> Or think about it this way: if you start at the origin and go distance r in the direction of the vector r, won't you be at the point?<br /> Maybe you are getting confused with $\dot{\vec x}=\dot{\vec r}=\dot r\hat r+r\dot \phi\hat\phi$.[/itex][/itex]

 

[stevendaryl]

Here, I would say, $\vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi}$.

No, that's not correct. If you have a point $P$ in 2-D space, the associated vector $\vec{r}$ is the vector from the origin to the point $P$. The coordinate $r$ is just the magnitude of $\vec{r}$, and $\hat{r}$ is the unit vector $\frac{1}{r} \vec{r}$. So $\hat{\phi}$ doesn't appear at all in the position vector. It does appear in the velocity vector, though:

$\frac{d \vec{r}}{dt} = \frac{d}{dt} r \hat{r} = \frac{dr}{dt} \hat{r} + r \frac{d \hat{r}}{dt}$

$\phi$ comes into play because $\frac{d\hat{r}}{dt} = \frac{d\phi}{dt} \hat{\phi}$

So

$\frac{d \vec{r}}{dt} = \frac{dr}{dt} \hat{r} + r \frac{d \phi}{dt} \hat{\phi}$

I'm not sure what's the best way to demonstrate these facts, but you can show that:

$\frac{d \hat{\phi}}{dt} = - \frac{d\phi}{dt} \hat{r}$
$\frac{d \hat{r}}{dt} = \frac{d\phi}{dt} \hat{\phi}$