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

In summary: 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
  • #1
Redwaves
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7
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##.
 
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  • #2
Redwaves said:
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##.
 
  • #3
haruspex said:
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.
 
  • #4
Redwaves said:
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.
 
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Related to Angular momentum of a particles in the form of ##L = mr^2\omega##

1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of a particle or system of particles. It is defined as the product of the particle's moment of inertia and its angular velocity.

2. How is angular momentum related to the mass and distance of a particle?

Angular momentum is directly proportional to both the mass and the square of the distance of a particle from the axis of rotation. This means that increasing either the mass or the distance will result in an increase in angular momentum.

3. What is the formula for calculating angular momentum?

The formula for angular momentum is L = mr^2ω, where L is angular momentum, m is the mass of the particle, r is the distance from the axis of rotation, and ω is the angular velocity.

4. Can angular momentum be conserved?

Yes, angular momentum can be conserved in a closed system where no external torques act on the system. This means that the total angular momentum of the system remains constant, even as individual particles within the system may change their angular momentum.

5. How is angular momentum different from linear momentum?

Angular momentum and linear momentum are both measures of an object's motion, but they differ in the type of motion they describe. Linear momentum is a measure of an object's straight-line motion, while angular momentum is a measure of an object's rotational motion. Additionally, linear momentum is a vector quantity, while angular momentum is a vector quantity.

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