I have two questions:-

  1. If we consider a scenario where a body is rotating and transalatingenter image description here simultaneously. Now say we are in the ground frame and see a point particle transplanting perpendicular to the radius vector joining the point particle to any axis of the body rotating. How do we calculate the angular momentum of the particle about the moving point(rigid body as seen in diagram)?

My approach:- We can first assume the body is not translating and find the angular momentum of the point using mvr and then add the angular momentum due to the effect of rotation of body. My question is do we have to add the angular momentum of the body transalating about a point on the ground? If yes why and what difference would it make if we didn’t do that?.

2nd question:- If we take two points whose velocities are only restrained to the x-y plane. Say the are moving parallel to each other and we want to find angular momentum of 2nd particle wrt to the first. Do we have to use the same logic used in the first question?

  • $\begingroup$ Angular momentum needs to be summed about an inertial point. Otherwise, the results are meaningless. $\endgroup$ Commented Dec 10, 2022 at 19:08
  • $\begingroup$ So actually we are finding angular momentum about a point on a stationary ground analogous to relative motion? $\endgroup$ Commented Dec 11, 2022 at 1:53
  • $\begingroup$ @JohnAlexiou Could you clarify? $\endgroup$ Commented Dec 11, 2022 at 3:12
  • $\begingroup$ Think of measuring the (linear) momentum of a free particle from an accelerating rocket. Momentum is going to change over time, even though there are no external forces acting on the particle. Newton's second law no longer applies now. The same is for angular momentum. It makes little sense to measure it from a rotating or accelerating point of view. $\endgroup$ Commented Dec 11, 2022 at 15:20
  • $\begingroup$ Also pure translation (all points move with the same velocity) does not contribute to angular momentum. Only rotations about the point of summation does. And any rigid body motion can be decomposed into some velocity and rotation about an arbitrary point. $\endgroup$ Commented Dec 11, 2022 at 15:51

1 Answer 1


For a continuous mechanical system, the angular momentum w.r.t. a pole $H$ reads

$\boldsymbol{\Gamma}_H = \displaystyle \int_V \rho ( \mathbf{r} - \mathbf{r}_H) \times \mathbf{v}$,

being $\mathbf{r}$ the points of the mechanical system, $\mathbf{v}$ their velocity, and $\mathbf{r}_H$ the position of the pole.

For a rigid body, we can write the velocity of a point as $\mathbf{v} = \mathbf{v}_P + \boldsymbol{\omega} \times (\mathbf{r} - \mathbf{r}_P)$, being $\boldsymbol{\omega}$ the angular velocity of the rigid body, and $P$ a reference material point of the rigid body, whose velocity is $\mathbf{v}_P$. Usually one chooses this reference point with a criterion of convenience, namely a point where you can easily measure the velocity, or the center of mass.

Using the kinematics of rigid motion, we can recast the angular momentum of a rigid system as

$\boldsymbol{\Gamma}^{rigid}_H = \displaystyle \int_V \rho ( \mathbf{r} - \mathbf{r}_H) \times [ \mathbf{v}_P + \boldsymbol{\omega} \times (\mathbf{r} - \mathbf{r}_P) ] = \\ \qquad \ =\displaystyle \int_V \rho(\mathbf{r} - \mathbf{r}_H) \times \mathbf{v}_P + \int_V \rho(\mathbf{r} - \mathbf{r}_P) \times [\boldsymbol{\omega} \times (\mathbf{r} - \mathbf{r}_P) ] + \int_V \rho(\mathbf{r}_P - \mathbf{r}_H) \times [\boldsymbol{\omega} \times (\mathbf{r} - \mathbf{r}_P) ] = \\ \qquad \ = m (\mathbf{r}_G - \mathbf{r}_H) \times \mathbf{v}_P + \mathbb{I}_P \cdot \boldsymbol{\omega} + m(\mathbf{r}_P - \mathbf{r}_H) \times [\boldsymbol{\omega} \times (\mathbf{r}_G - \mathbf{r}_P)]$.

where $\mathbb{I}_P$ is the tensor of inertia of the rigid system w.r.t. the point $P$. Using the center of mass $G$ as the reference point $P \equiv G$ we get

$\boldsymbol{\Gamma}^{rigid}_H = m (\mathbf{r}_G - \mathbf{r}_H) \times \mathbf{v}_G + \mathbb{I}_G \cdot \boldsymbol{\omega} = \\ \qquad \ = (\mathbf{r}_G - \mathbf{r}_H) \times \mathbf{Q} + \mathbb{I}_G \cdot \boldsymbol{\omega}$

being $\mathbf{Q} = m \mathbf{v}_G$ the momentum of the system.

  • $\begingroup$ Is Angular Momentum Relative? $\endgroup$ Commented Dec 12, 2022 at 2:00
  • $\begingroup$ Please clarify the question $\endgroup$ Commented Dec 12, 2022 at 2:02
  • $\begingroup$ Should I clarify the answer? $\endgroup$
    – basics
    Commented Dec 12, 2022 at 8:18
  • $\begingroup$ No pls clarify if angular momentum can be relative $\endgroup$ Commented Dec 12, 2022 at 18:18
  • $\begingroup$ Please respond……………… $\endgroup$ Commented Dec 14, 2022 at 15:36

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