I know moment results from a force applied. Momentum is quantity of motion of all particles. Does this momentum acts like force so it makes moment? I understand that angular momentum is the quantity of rotation of a body, but I don't actually understand this.

  • $\begingroup$ I've never heard it being called so. I've always used the term "angular momentum" and not "moment of momentum". $\endgroup$ Jan 20, 2018 at 18:15
  • 5
    $\begingroup$ Just like moment of force is torque, moment of linear momentum is angular momentum. $\endgroup$
    – Mitchell
    Jan 20, 2018 at 19:00
  • $\begingroup$ Ahmed, you are apparently familiar with the term "moment" meaning a pure torque (or possibly, the moment of a force). I know this usage exists, but it is an abuse of terminology. The moment of (linear) momentum, $\vec r \times \vec p$, where $\vec p = m\vec v$ is typically called angular momentum. The moment of a force, $\vec r \times \vec F$, is typically called torque. The first moment of mass of a collection of particles, $\sum_i m_i \vec x_i$, is the total mass times the center of mass position. And so on. The concept of moments is very generic, and very useful. $\endgroup$ Jan 20, 2018 at 20:30
  • $\begingroup$ More on the rationale for the use of the prefix 'moment': Why is a dipole moment called a dipole moment? $\endgroup$ Jan 20, 2018 at 21:33
  • $\begingroup$ thanks everybody for answering me . great respect for you all $\endgroup$ Jan 20, 2018 at 23:29

2 Answers 2


Ahmed, you are apparently familiar with the term "moment" as shorthand for the moment of a force. I know this usage exists, but it is an abuse of terminology. A much better term is "torque". This abuse of terminology (using "moment" to mean a torque) is a source of confusion, as evidenced by your question.

The generic concept of a "moment" in this context is (from dictionary.com) "the product of a physical quantity and its directed distance from an axis". Other uses beyond the moment of a force include the first moment of mass (mass times center of mass), the second moment of mass (typically called the moment of inertia), and the moment of (linear) momentum (typically called angular momentum). The concept of moments is very generic, and very useful. Statisticians have a similar concept of moments.

The term "moment of momentum" is now used very rarely. From looking at google ngrams and google scholar, "moment of momentum" briefly held sway over "angular momentum" for a short period around 1900.

Even a point particle can have angular momentum with respect to some other point; it is $\vec L = \vec r \times \vec p = \vec r \times (m\,\vec v)$ where $m$ is the mass of the point particle, $r$ is the vector from the point in question to the point particle and $v$ is the time derivative of $r$. That angular momentum is indeed the moment of linear momentum jumps right out in this context.

It's not so clear in the context of a rotating solid body, where the angular momentum with respect to the center of mass is $L = \mathrm{I}\,\vec \omega$. This still is the moment of momentum: $\mathrm I \vec \omega$ is indeed $\int_V \vec x \times (\rho(\vec x) \dot{\vec x})\, d\vec x$ for a rigid body rotating in three dimensional space. There's a lot of stuff hiding in the moment of inertia tensor and in the assumption that the body is rigid that makes this the case.

  • 1
    $\begingroup$ Google ngram $\endgroup$
    – valerio
    Jan 20, 2018 at 23:19
  • 1
    $\begingroup$ @valerio92 That's an extremely curious ngram. $\endgroup$ Jan 21, 2018 at 1:54
  • $\begingroup$ Interestingly, my textbook "Orbital Mechanics for Engineering Students" distinguished the momentum of net force from angular momentum, with moment of net force being $r \times F_{net}$ and angular momentum as the time integral of moment of net force. $\endgroup$
    – noodles
    Apr 13, 2022 at 23:51

$$\text{Angular moment}(L) = \text{moment of inertia}(I) \cdot\text{Angular velocity}(o)$$

$$O \cdot \text{radius}=\text{tangential velocity}(V)$$

$$L= MR^2\cdot \frac VR=MRV$$

$MV$ IS linear momentum $MV$ times $r$ mean moment of liner momentum


Your Answer

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

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