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:
[itex] r(t) = r_0 + v_0t [/itex]
[itex]\phi(t) = \omega_0t [/itex].
Determine the angular momentum of m (in spherical coordinates).

The attempt at a solution
Angular momentum is defined as [itex] \vec{l} = m\cdot\vec{x(t)}\times\vec{v(t)} [/itex].
Here, I would say, [itex] \vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi} [/itex].
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, [itex] \vec{x(t)} = r(t)\hat{r} [/itex]. 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, [itex] \vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi} [/itex].

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 [itex]\hat{r}[\itex]. Here is a picture of it dimensionally:
http://i.gyazo.com/6685ade5d3ab158b496077f7481551ea.png

 

[haruspex]

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

But it is dimensionally wrong. x and r have dimension of length, φ does not.
##\vec r## means the same as ##\vec x##. They are both the vector representing the point.

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?
Maybe you are getting confused with ##\dot{\vec x}=\dot{\vec r}=\dot r\hat r+r\dot \phi\hat\phi##.

 

[stevendaryl]

Here, I would say, [itex] \vec{x(t)} = r(t)\hat{r} + \phi(t)\hat{\phi} [/itex].

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

[itex]\frac{d \vec{r}}{dt} = \frac{d}{dt} r \hat{r} = \frac{dr}{dt} \hat{r} + r \frac{d \hat{r}}{dt}[/itex]

[itex]\phi[/itex] comes into play because [itex]\frac{d\hat{r}}{dt} = \frac{d\phi}{dt} \hat{\phi}[/itex]

So

[itex]\frac{d \vec{r}}{dt} = \frac{dr}{dt} \hat{r} + r \frac{d \phi}{dt} \hat{\phi}[/itex]

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

[itex]\frac{d \hat{\phi}}{dt} = - \frac{d\phi}{dt} \hat{r}[/itex]
[itex]\frac{d \hat{r}}{dt} = \frac{d\phi}{dt} \hat{\phi}[/itex]