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 \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>

 

[haruspex]

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

<br /> 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 $\dot{\vec x}=\dot{\vec r}=\dot r\hat r+r\dot \phi\hat\phi$.

 

[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}