I am currently reading about angular position, angular velocity, and angular acceleration. I came across this paragraph that was particularly confusing, and was wondering if someone could perhaps help me. Here is the paragraph:

"If a particle rotates in the $xy$ plane, the direction of $\vec{\omega}$ for the particle is out of the plane of the diagram when the rotation is counterclockwise and into the plane of the diagram when the rotation is clock-wise. To illustrate this convention, it is convenient to use the right-hand rule demonstrated in Figure 10.3. When the four fingers of the right hand are wrapped in the direction of rotation, the extended right thumb points in the direction of $\vec{\omega}$The direction of $\vec{\alpha}$ follows from its definition $\vec{\alpha}=d\vec{\omega}/dt$. It is in the same direction as $\vec{\omega}$ if the angular speed is increasing in time, and it is antiparallel to $\vec{\omega}$ if the angular speed is decreasing in time."

The above text in bold is what I am having most difficulty with. What do they mean? Also, how do I color words?

  • 1
    $\begingroup$ If you imagine your computer screen as a 2D plane, then if you look at the analog clock app your computer has it is rotating around clockwise, then the angular momentum vector points into your computer screen (like out the back). If something rotates the other way (counter-clockwise) there is an arrow pointing at you out of the screen which represents the angular momentum vector, and is orthogonal to the screen. I hope this helps. PS you should quick search the forum and see if you can find this question before asking, it's a pretty typical confusion :) $\endgroup$
    – kηives
    Nov 21, 2012 at 17:53
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    $\begingroup$ I'd recommend you leave that for now and concentrate on the right-hand rule, which is more intuitive and easier to apply. Once you have a feeling for that, go back to the plane and the normal. $\endgroup$ Nov 21, 2012 at 18:17
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Sep 2, 2017 at 4:49

2 Answers 2


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.


The least ambiguous way to describe it would be using cartesian vector notation. so we have $\hat x$, $\hat y$ and $\hat z$. If the particle is moving in a circle from the $\hat x$ axis to the $\hat y$ axis, it's moving counter clockwise, like moving from 3 to 12 on an analog clock. We say it's angular velocity is in the $+ \hat z$ direction, or up out of the clock.

If it's moving from the $\hat x$ axis to the $-\hat y$ axis, it's moving clockwise, like moving form 3 to 6 on an analog clock. It's convention to say it's angular velocity is in the $- \hat z$ direction, or down into the clock. These relations fall out of the right hand rule and the cross-product in vector arithmetic, since angular velocity is $\hat r \times \hat V=\hat \omega$.

  • $\begingroup$ I've replaced $x$ by $\times$. $\endgroup$ Nov 21, 2012 at 18:48

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