This question is created to get additional information for my students from a high school.

We know that the angular moment is given by:

$$\mathbf{L}=\mathbf{r}\times \mathbf{p}=m(\mathbf{r}\times \mathbf{v}) \tag 1$$ where: \begin{array} {cc}\mathbf{p}=m\mathbf v \quad \text{it is the momentum with a velocity } \mathbf v \\ \mathbf{r} \quad \text{ it is the displacement vector} \\ \hline \end{array}

This vector $\mathbf{L}$ is perpendicular to the plane identified by the two vectors $\mathbf{r}$ and $\mathbf{v}$ and is outgoing from the sheet plane.

enter image description here

The moment of a force $\mathbf{M}$ or $\boldsymbol{\tau}$ applied to a rigid body ($\mathbf{F}$) gives the rotation of it with respect to a fixed point (for example $O$) or respect to a fixed axis,

$$\mathbf{M}=\mathbf{r}\times \mathbf{F} \tag 2$$ In addition

$$\frac{\sum \Delta \mathbf{L}}{\Delta t}=\mathbf{M}$$

The definition of momentum of a force I believe is easily intuitable. But could the definition of angular momentum be as follows?

The moment of a force generates the rotation while the angular momentum gives the intensity of the rotation?

I know, but not my students that

enter image description here

$$\mathbf{v}=\boldsymbol{\omega}\times \mathbf{r} \tag 2$$ and

$$\mathbf{L}=I\boldsymbol{\omega} \tag 3$$ where $I$ is the moment of inertia.

As can I easily derive the $(3)$ starting from the $(1)$ and to have a proof from the $(1)$ that $\mathbf v =\boldsymbol \omega \times \mathbf r$?

PS: I cannot use mixed products or double vector products, or in general the derivative, integral - my students are 16 years old.

Addendum: I have not yet explained work, conservation laws, and rotational dynamics.


2 Answers 2


I like your proposed definition as an intuitive explanation but not as a definition of angular momentum, primarily because:

  1. Angular momentum is more fundamental than torque, so defining it in terms of torque is presenting these in a confusing order.

  2. Of conserved quantities in mechanics (momentum, energy, etc) angular momentum is the easiest to observe and intuit (bicycle wheels, rolling objects, mobiles, etc), and defining it in terms of torque (a situation where it's not conserved) sets an awkward framework. And, conservation laws are very important in understanding physics more generally, so take advantage of it here.

defining angular momentum:

Rather than one ball, I would define angular momentum using two balls held apart by a mass-less rod. That is, this is the simplest thing that can have angular momentum without linear momentum, so it helps disambiguate the ideas (in a way that one ball doesn't).

Then you can introduce angular momentum with this as a demo (held by a string from the center). Maybe say something like, "we learned that momentum is conserved, and clearly something is being conserved here but it isn't the previous momentum we learned. What is it?"

(Then, later, you can also say that a single ball has angular moment. That's important to know as a fact and to solve some problems, but it not the usual way to think of angular momentum.)

torque and intensity of angular momentum:

For an intuitive understanding and a definition of torque, I think your idea that angular moment quantifies how "hard" it is to stop something spinning, where, by "hard" we mean (torque)$\times$ (time applied). So this nicely goes from intuitive to quantitative and is your definition of torque when rewritten as $\Sigma \Delta \mathbf{L}=\mathbf{M}{\Delta t}$. (Personally, I prefer "hard" to "intensity" since intensity has many technical meanings which aren't apt here, and hard shows you're after an intuitive understanding, even though it's quantitative in the end).

an advanced aside:

Given these two conserved quantities (linear and angular momentum), it seems enough examples to generalize this idea to note that space is invariant under both translation and rotation, and that's why we have a momentum associated with those two types of motions. (I find there to be something very beautiful about this idea, so for anyone who might also be caught by it, it seems potentially worth stating, but I don't know much about high school students.)

  • 2
    $\begingroup$ Very very nice and I have appreciated very much your efforts. +1. First I must to translate with the translator and then I have to print it out to understand better: I'm sorry :-(. It is very difficult to explain Physics when students do not have the proper knowledge of calculus. Thank you and I will thank you again. $\endgroup$
    – Sebastiano
    Commented Mar 5, 2021 at 12:37
  • 2
    $\begingroup$ @Sebastiano: Thanks! If in your subsequent reading you find something I said that was unclear or could have been phrased better, I'm happy to hear it. I'm confused by your Italian translation comment because it seems that your English is good, or is my writing particularly confusing? I occasionally answer questions here to work on my explanatory skills more than to learn physics (which I do in other ways), so your comments are welcome. (Also, I reverted your "\Sigma" to "\sum" change because I don't like the line space distortion of the "\sum", but if I change it back, it's OK with me.) $\endgroup$
    – tom10
    Commented Mar 5, 2021 at 18:08
  • $\begingroup$ Don't worry for my edit of LaTeX that I like very much (see my profile in TeX.SE). I am not good in English and I think in Italian first and then write if I don't use the translator. I realized that deepl translate does not translate well and I have to be careful if the translation is at least faithful to my thought in Italian. I will let you know if I have any problems. $\endgroup$
    – Sebastiano
    Commented Mar 5, 2021 at 21:26

Keep everything scalar and discuss components only.

  1. Translational momentum is $p = m\,v$
  2. Rotational momentum is $L = d\,p$, where $d$ is the moment arm (the perpendicular distance of the line of motion and origin).
  3. Translational speed is $v = \omega\, d$, also with the perpendicular distance
  4. Rotational momentum is thus $L = (m d^2) \omega$
  5. The thing inside the parenthesis is mass moment of inertia, and it is a geometrical quantity as it does not depend on the motion, but on the geometry of the problem.
  6. When a mass rotates or moves with a moment arm $d$, its mass moment of inertia is $I = m d^2$
  • $\begingroup$ Hi, and +1 also for you. I totally understand and thank you for your effort. The book is from an Italian publisher/writers and I don't like the textbook setting we adopt. Unfortunately I need the vector notation also because I have to adjust to the text, but not too much. Thank you and I will thank you again. $\endgroup$
    – Sebastiano
    Commented Mar 5, 2021 at 12:39
  • $\begingroup$ Then you have to introduce the cross product first. Simply put the cross product defines the moment arm in each situation. Later you can even talk the geometry of physics also due to the use of cross products. $\endgroup$ Commented Mar 5, 2021 at 13:07
  • $\begingroup$ Could you kindly post a response or edit yours by adapting it in my context? Otherwise I'm making a mixer in my head. I thank you if it would be possible. My students want often from me, simple explanations in math-mode. :-) $\endgroup$
    – Sebastiano
    Commented Mar 5, 2021 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.