Torque about a point = rate of change of angular momentum about that point.

Say we're in an inertial frame and see a body accelerating and rotating and another point(either a part of this body or an external point) accelerating.

Can we apply T = dL/dt for that body about that accelerating point if we are in an inertial frame?

My guess is we can since angular momentum depends on the reference point. Is there any requirement for this point to be non-accelerating?

  • $\begingroup$ Oh, that was a typo. I'll correct it. $\endgroup$
    – xasthor
    Mar 22, 2018 at 18:02

3 Answers 3


The law you state is only valid for the center of mass, or for a fixed point in space.

Euler' law of rotation states:

Net torque of an object about the center of mass equals the rate of change of angular momentum measured at the center of mass.

$$ \boldsymbol{T}_C = \frac{{\rm d}}{{\rm d}t} (\boldsymbol{L}_C) = \frac{{\rm d}}{{\rm d}t} (\mathrm{I}_C \boldsymbol{\omega}) = \mathrm{I}_C \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times \boldsymbol{L}_C \tag{1} $$

where point C denotes the center of mass. This a direct equivalent to the fact that the net force on a body describes the motion of the center of mass only. The remaining motion (about the center of mass) is described by Euler's law.

The center of mass can be accelerating (and usually is) since usually both torque and force are considered at the same time

$$ \boldsymbol{F} = \frac{{\rm d}}{{\rm d}t}( \boldsymbol{p} ) = \frac{{\rm d}}{{\rm d}t}(m\, \boldsymbol{v}_C ) = m\, \dot{ \boldsymbol{v}}_C \tag{2} $$

So now what happens at a different location A? Consider the location vector $\boldsymbol{c}$ of the center of mass, relative to A

Angular momentum at A is

$$ \boldsymbol{L}_A = \boldsymbol{L}_C + \boldsymbol{c} \times \boldsymbol{p} \tag{3} $$

Net torque at A is

$$ \boldsymbol{T}_A = \boldsymbol{T}_C + \boldsymbol{c} \times \boldsymbol{F} \tag{4} $$

The total derivative of angular momentum at A is

$$ \require{cancel} \begin{aligned} \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) & = \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_C + \boldsymbol{c} \times \boldsymbol{p} ) = \boldsymbol{T}_C + \frac{{\rm d}\boldsymbol{c}}{{\rm d}t} \times \boldsymbol{p} + \boldsymbol{c} \times \underbrace{ \frac{{\rm d}\boldsymbol{p}}{{\rm d}t}}_{\boldsymbol{F}} \\ & = \boldsymbol{T}_A + \underbrace{( \boldsymbol{v}_C - \boldsymbol{v}_A ) \times \boldsymbol{p} }_{(\boldsymbol{v}_C-\boldsymbol{v}_A) \times (m\,\boldsymbol{v}_C) = -\boldsymbol{v}_A \times m\,\boldsymbol{v}_C = \boldsymbol{p} \times \boldsymbol{v}_A} \end{aligned} \tag{5}$$

I produce the following law (if no one else claims it, call it the ja72 law).

The rate of change of angular momentum at a non-fixed arbitrary point A equals the net torque at A, plus cross product of linear momentum with the speed of A.

$$ \boxed{ \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) = \boldsymbol{T}_A + \boldsymbol{p} \times \boldsymbol{v}_A } \tag{6} $$

The conditions where the derivate of angular momentum is exactly the net torque at a point _A_on a rigid body are as follows:

  1. Body undergoes pure rotation with zero linear momentum, $\boldsymbol{p}=0$
  2. Point A is fixed in space, or instaneneously fixed, $\boldsymbol{v}_A=0$
  3. Point A is on the center of mass, making its motion parallel to momentum, $\boldsymbol{v}_A \parallel \boldsymbol{p}$
  4. Point A lies on a line parallel to the rotation axis, but through the center of mass, $\boldsymbol{v}_A \parallel \boldsymbol{v}_C$
  • $\begingroup$ In the original derivation of Torque = dL/dt, nowhere did we use that assumption. phys.libretexts.org/TextMaps/Classical_Mechanics_TextMaps/… $\endgroup$
    – xasthor
    Mar 23, 2018 at 2:43
  • $\begingroup$ @xasthor - Isn't the correct form of the time derivative $$\dot{\bf L} = \sum_i ( \dot{\bf r}_{i}\times {\bf p}_{i}+{\bf r}_{i}\times \dot{\bf p}_{i})$$ The cross product is linear operator and the regular product rule applies for derivatives. $\endgroup$ Mar 23, 2018 at 14:23
  • $\begingroup$ @xasthor - also what is ${\bf F}_{i}$ and ${\bf F}_{ij}$ that their sum equals velocity? $\endgroup$ Mar 23, 2018 at 14:27
  • $\begingroup$ @xasthor - Euler's laws of rotation motion apply at the center of mass only. You cannot pick an arbitrary location for the net torque since a body rotates only if the line of action of a force is not at the center of mass. Also, angular momentum not at the center of mass includes the moment of momentum component which is a function of linear motion. See The derivation of Equations of Motion not at the center of mass to see why $\frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) \neq \boldsymbol{T}_A$ at any point A. $\endgroup$ Mar 23, 2018 at 14:54
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    $\begingroup$ In my comment above I made a simplification that is confusing things. The definition of $L_A$ is based on $r_i$ the location of the particle relative to observer, and $v_i$ the velocity of the particle (from the observers point of view) as follows $$ \boldsymbol{L}_A = \sum_i (\boldsymbol{r}_i-\boldsymbol{r}_A) \times m_i \boldsymbol{v}_i $$ and now the velocity is clearly just $\boldsymbol{v}_i = \frac{\rm d}{{\rm d}t} \boldsymbol{r}_i$ $\endgroup$ Mar 3, 2021 at 13:29

This is not an answer. I post to clarify the mistakes in the answer from John.

Now, lets forcus on your above answer. I show your clearly the errors.

Eq. (3) in this post:

Eq.(3) in John's post Eq.(3) in John's post.

Where your defined $\mathbf{c}$ as : Consider the location vector c of the center of mass, relative to A. The point A has a velocity $\mathbf{v}_A$ refering to the origin O. And $\mathbf{p} = m \mathbf{v}_C$ is the velocity of the center of mass measured in frame O.

With these definitions, your Eq.(3) is NOT correct. Because A is moving with velocity $\mathbf{v}_A$, therefore, in frame A, the total linear momentum of the rigid body is $m (\mathbf{v}_C - \mathbf{v}_A)$. This error led you into wrong conclusion in Eq.(5) and Eq. (6).

Since $\mathbf{v}_C$ and $\mathbf{p}$ is measured in frame O and $\mathbf{c}$ is measured in frame A. The Eq. (3) should written as:

$$ \mathbf{L}_A = \mathbf{L}_C + \mathbf{c} \times m\left(\mathbf{v}_C - \mathbf{v}_A \right) $$

With all quanties measured in frame A. This will correct your results in Eq.(5) and Eq.(6), reders $\frac{d\mathbf{L}_A}{dt} = \mathbf{\tau}_A$.

Following John, I change few notations:

  • C : the center of mass frame
  • $\vec{R}$ : the position of center of mass w.r.t. frame A ($\vec{c}$ of John.)
  • frame A : another inertial frame, which has a relative constant velocity w.r.t C-frame.
  • $\vec{v}_A$ : velocity of the point measured in frame A.
  • $\vec{v}_C$ : velocity of the point measured in frame C.
  • $\vec{r}_A$ : position of the point measure in frame A.
  • $\vec{r}_C$ : position of the point measure in frame C.

The relation of positions:

$$ \tag{1} \vec{r}_A = \vec{R} + \vec{r}_C. $$

Derive Eq.(1) leads to the relation in velocity: $$ \tag{2} \vec{v}_A = \vec{V} + \vec{v}_C. $$ where $\vec{V}$ is the velocity of COM in frame A. It must be a constant for both frames to be inertial.

The definition of the angular momentums:

$$ \vec{L}_A = m \vec{r}_A \times \vec{v}_A\\ \vec{L}_C = m \vec{r}_C \times \vec{v}_C\\ $$

Their relation can be found from Eq.(1) and Eq.(2): $$ \vec{L}_A = m \left(\vec{R} + \vec{r}_C \right) \times \left( \vec{V} + \vec{v}_C \right)\\ = m \left(\vec{R} + \vec{r}_C \right) \times \vec{V} + m \vec{R} \times \vec{v}_C + m \vec{r}_C \times \vec{v}_C\\ = \left\{ \vec{L}_C + m \vec{R} \times \vec{v}_C \right\} + m \left(\vec{R} + \vec{r}_C \right) \times \vec{V}\\ = \{ \text{John's term} \} + \{ \text{missed term.} \} $$

The term inside the curry braket is show in John's equation. The last term is missed from his angular momentum relation. It is indeed $m \vec{r}_A \times \vec{V}$, the extra agular momentum of the particle in the frame A due to relative motion between frame A and C.

The incorrect relation in angular momentum leads to his wrong conclusion. His conclusion is a very serious accusation against the equivalent principle of all inertial frames - a very imprortant base concept of Newtonian mechanics.

$$ \frac{d\vec{L}_A}{dt} = m \frac{d\vec{r}_A}{dt} \times \vec{v}_A + m \vec{r}_A \times \frac{d\vec{v}_A}{dt}\\ = 0 + m \vec{r}_A \times \frac{d\vec{v}_A}{dt}\\ = m \vec{r}_A \times \frac{d}{dt} ( \vec{V} + \vec{v}_C )\\ = \vec{r}_A \times \frac{d ( m \vec{v}_C ) }{dt}\\ = \vec{r}_A \times \vec{F} = \vec{\tau}_A. $$

As long as the relative motion is a constant velocity $\frac{d\vec{V}}{dt} = 0$, the rate change of angular momentum is equal to torque.

Describe a many-bodys motion in a certain inertial frame:

$$ \vec{L}_{total} = \sum_i m_i \vec{r}_i \times \vec{v}_i\\ $$

And the rate change: $$ \frac{d\vec{L}_{total}}{dt} = \sum_i m_i \frac{d\vec{r}_i}{dt} \times \vec{v}_i + \sum_i \vec{r}_i \times \frac{d \vec{p}_i}{dt}\\ = \sum_i m_i 0 + \sum_i \vec{r}_i \times \vec{F}_i\\ = \sum_i \vec{\tau}_i = \vec{\tau}_{total} $$

If you try to argue that away from frame of center of mass:

$$ \frac{d\vec{r}_i}{dt} \times \vec{v}_i \ne 0. $$

You have to do much better that a hand-wave saying.

I will illustrate wuth a simple example, a dumbell of two point mass points ($2kg$) separate by massless wire (2m). Its center mass is moving with velocity $\vec{V} = 4 m/s \vec{x} + 3 m/s \vec{y}$. In the center mass frame C, they rotate around the center with a frequency $\nu = 1/s$.

Moving and rotation of a dumbbell.

Describe the motion of these two masses in the center of mass (C frame) $\vec{r}_{Ci} = (x_i', y_i')$ for $i = 1, 2$:

$$ \begin{matrix} x_1' = \cos(2\pi t); & y_1'= \sin(2\pi t) \\ x_2' = -\cos(2\pi t); & y_2'= -\sin(2\pi t) \\ \text{Velocity:} & \\ v_{1x}' = -2\pi \sin(2\pi t); & v_{1y}'= 2\pi\cos(2\pi t) \\ v_{2x}' = 2\pi\sin(2\pi t); & v_{2y}'= -2\pi\cos(2\pi t) \\ \end{matrix} $$

The angularmomentum in C $L_c = (0, 0, L_z') = (0, 0, L_{1z}'+L_{2z}')$: $$ L_{1z}' =2 x_1' v_{1y}' -2 y_1' v_{1x}' = \cos(2\pi t) 4\pi\cos(2\pi t) - \sin(2\pi t) \{-4\pi \sin(2\pi t)\} = 4\pi.\\ L_{2z}' = 4\pi, \text{ similarly}\\ $$ The total angular momentum in fram C is $8\pi$, a constant in time.

Now, examine the angular momentum observed in Frame A, $\vec{r}_{iA} = (x_i, y_i)$, and the position of CM in frame A $\vec{R} = (4 t, 3 t)$: $$ \begin{matrix} x_1 = 4 t + \cos(2\pi t); & y_1 = 3 t + \sin(2\pi t) \\ x_2 = 4 t -\cos(2\pi t); & y_2 = 3 t - \sin(2\pi t) \\ \text{Velocity:} & \\ v_{1x} = 4 - 2\pi \sin(2\pi t); & v_{1y}=3 + 2\pi\cos(2\pi t) \\ v_{2x} = 4 + 2\pi\sin(2\pi t); & v_{2y}= 3- 2\pi\cos(2\pi t) \\ \end{matrix} $$

Now, check the angular momentum in frame A: $$ \begin{matrix} L_{1z} = 2 \{ 4 t + \cos(2\pi t) \} \{3 + 2\pi\cos(2\pi t)\} - 2 \{3 t + \sin(2\pi t)\} \{4 - 2\pi \sin(2\pi t)\} \\ =4\pi + 6 \cos(2\pi t) + 16 t \pi\cos(2\pi t) - 8\sin(2\pi t) + 12 t\pi \sin(2\pi t).\\ L_{2z} = 2\{ 4 t - \cos(2\pi t) \} \{3 - 2\pi\cos(2\pi t)\} -2\{3 t - \sin(2\pi t)\} \{4 + 2\pi \sin(2\pi t)\} \\ =4\pi - 6 \cos(2\pi t) - 16 t \pi\cos(2\pi t) + 8 \sin(2\pi t) - 12 t\pi \sin(2\pi t).\\ \end{matrix} $$

Finally, the resultant angular momentum in Frame A:

$$ L_{Az} = L_{1z} + L_{2z} = 8 \pi. $$

It is also a constant in time, even though frame A has relative motion with Frame C.

Check John's identity for mass 1:

$$ \vec{R} \times m\vec{v}_1' = \hat{z} \{ 16 t\pi\cos(2\pi t) + 12 t\pi \sin(2\pi t)\}\\ L_{1z}' + [\vec{R} \times m\vec{v}_1']_z = 4\pi + 16 t\pi\cos(2\pi t) + 12 t\pi \sin(2\pi t) \ne L_{1z} $$

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    $\begingroup$ Thank you for the detailed explanation. I am reviewing now, and I see there are some subtle differences in interpretation. For example $\vec{v}_C$ and $\vec{v}_A$ for is the velocity of a rigid body of whatever particle is at locations C and A respectively, both measured from the same inertial frame (the world frame). $\endgroup$ Feb 26, 2021 at 21:52
  • $\begingroup$ Where i show only for one particle. Description of many particles will add a sum over all particles. Each of them has similar form in momentum relation between two frames. $\endgroup$
    – ytlu
    Feb 27, 2021 at 0:53
  • $\begingroup$ I added a simple example continuing in my post, All qantity can be check in the simple model. $\endgroup$
    – ytlu
    Feb 27, 2021 at 2:24
  • $\begingroup$ Can you please review this post and see if you agree, This is the foundation on which I am building my rebuttal. The argument from the particle summation point of view. Also this similar post and goes into Newton;s laws also. $\endgroup$ Feb 28, 2021 at 16:27
  • $\begingroup$ In the first post, I cannot find the subjects relatied to your angular momentum relation which was wrong. $\endgroup$
    – ytlu
    Feb 28, 2021 at 23:27

If the body is in un-constrained motion, using the center of mass (CM) as the reference point, the sum of the torques from the real forces equals the change in the angular momentum, even if the CM is accelerating. If the body is constrained to rotate about a point other than the CM and that point is accelerating, then fictictious forces/torques must be considered using that point as the reference point. This is discussed at some length in the text Mechanics, by Symon.


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