Angular momentum of a particles in the form of ##L = mr^2\omega##

[Redwaves]

Homework Statement

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the xy plane, with the sun at the origin and label the planet's position by polar coordinates ($r,\phi$). Show that the planet's angular momentum has magnitude $L = mr^2\omega$, where $\dot\phi$ = $\omega$

Relevant Equations

$\vec{L} = \vec{P} \times \vec{r}$

$\vec{L} = \vec{P} \times\vec{r}$

$L = mvr sin \phi$, where P = mv

Since $\vec{r}$ and $\vec{v}$ are always perpendicular, $\phi$ = 90.

Then, $L = mvr$

At this point, I don't see how to get $L = mvr = mr^2\omega$, using $\omega = \dot{\phi}$

I know that $\omega = \frac{v}{r}$. However, the way the question if written, I'm wondering if there's another by way using the angle $\phi$.

 

[haruspex]

Homework Statement:: Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the xy plane, with the sun at the origin and label the planet's position by polar coordinates ($r,\phi$). Show that the planet's angular momentum has magnitude $L = mr^2\omega$, where $\dot\phi$ = $\omega$
Relevant Equations:: $\vec{L} = \vec{P} \times \vec{r}$

$\vec{L} = \vec{P} \times\vec{r}$

$L = mvr sin \phi$, where P = mv

Since $\vec{r}$ and $\vec{v}$ are always perpendicular, $\phi$ = 90.

Then, $L = mvr$

At this point, I don't see how to get $L = mvr = mr^2\omega$, using $\omega = \dot{\phi}$

I know that $\omega = \frac{v}{r}$. However, the way the question if written, I'm wondering if there's another by way using the angle $\phi$.

You are confusing two uses of the variable name $\phi$.
In "where $\dot\phi$ = $\omega$", $\phi$ is the angle in the XY plane as a polar coordinate.
In $L = mvr sin \phi$, it is the angle between the vector $\vec{P}$ and the vector $\vec r$.

 

[Redwaves]

You are confusing two uses of the variable name $\phi$.
In "where $\dot\phi$ = $\omega$", $\phi$ is the angle in the XY plane as a polar coordinate.
In $L = mvr sin \phi$, it is the angle between the vector $\vec{P}$ and the vector $\vec r$.

Isn't $\dot{\phi} = \frac{d\phi}{dt}$, which is $\frac{\Delta\phi}{\Delta t}$ where t -> 0

Thus, we are using $\phi$ the angle.

 

[haruspex]

Isn't $\dot{\phi} = \frac{d\phi}{dt}$, which is $\frac{\Delta\phi}{\Delta t}$ where t -> 0

Yes, but the ${\phi}$ in $\omega=\dot\phi$ is the angle around the sun in the XY plane. I.e. if we write the coordinates in polar, the position of the planet at time t is $\vec {r(t)}=(r(t),\phi(t))$. Clearly this angle is not always 90°.

In the equation $L = mvr sin \phi$, ${\phi}$ has a completely different meaning. It is the angle between the vector $\vec v= \dot {\vec r }(t)$ and the vector $\vec r$. For circular motion, this angle is always 90°.

To avoid confusion, write $L = mvr \sin (\psi)$ instead.