Say you're looking at the piece of paper on your desk; that is the $xy$ plane. You place a dot in the center of the paper; that's your origin.
Your angular momentum is $\vec{L}=\vec{r}{\times}m\vec{v}$. For this example, $\vec{\omega}$ points in the same direction as your angular momentum, because $\vec{L}=mr^2\vec{\omega}$.
The way I remember the directions for the right-hand rule are as follows:
- Thumb points up (this is $\vec{L}$, the cross product)
- Index finger points forward (this is $\vec{r}$, the left vector being crossed)
- Middle finger points to the left (this is $\vec{v}$, the right vector being crossed)
Let's say the particle is on the paper, directly above the origin, and moving counterclockwise. Then, $\vec{v}$ points to the left edge of the page. The direction of $\vec{r}$ goes from the origin to the point, so it points to the top of the page.
My thumb points up, so this is the direction of $\vec{L}$, and hence the direction of $\vec{\omega}$.
Crane your hand around so that your middle finger points right and your index finger continues to point to the top of the page, and your thumb points down, which is the direction of $\vec{\omega}$ for clockwise motion.
1) Does this mean that for any particle on the rotating body the angular velocity is the same?
On a rigid rotating body, yes, the angular velocity is the same for every point in that body.
2) Does this mean that when angular momentum is described, we are technically still describing a relationship between linear velocity and mass (mv), only now the linear velocity depends on angular velocity and radial distance from axis?
In effect, yes. What you are setting up is an equation of momentum for every infinitesimal mass element of your body. You see the analogy between linear and angular momentum:
$$p = mv$$ and $$L = I\omega$$ where $I$ depends on the distribution of mass, not just on the total mass itself.
3) this would mean that linear velocity would be less for particles close to the axis of rotation, but angular velocity would be the same?
That's exactly what's happening. To visualize this, simply imagine spinning a weight fixed to a string over your head. If you spin one weight with a certain angular speed $\omega$ and then release the string, it will fly off at a certain speed. Do the same now with a shorter string but the same angular speed. The weight will fly off at a slower velocity.
4) Then why would something like a pulsar rotate faster as its matter get closer to the center?
Since angular momentum is conserved, decreasing the moment of inertia increases angular speed: $L = I \omega$. As an analogy, consider a pirouette of an ice skater. If the ice skater has her arms outstretched (big moment of inertia) and rotates at a certain angular speed $\omega$, after she pulls in her arms, she will spin at a faster rate. This is because the angular momentum she had before is the same as afterwards.
All in all, angular speed/momentum follows a neat analogy with linear speed/momentum.
Best Answer
First of all, ask to yourself
The property of positive and negative only depends on your coordinate system, i.e how you are looking to the system.Similarly, the matter of some quantity is decreasing or increasing also depends on your coordinate axis.Nonetheless, they describe the $\textbf{same}$ thing.
Addition to that, think "decreasing" as increasing in the negative direction.