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?
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:
(19.5.3) and (19.5.4). If you try to go through the same proof with a point S riding on the body and you get my law. The difference is that the location of each point j is $r_j = r_S + r_{j,S}$ so in (19.5.4) $\frac{{\rm d}\vec{r}j}{{\rm d}t} = \vec{v}_S + \vec{v}{j,S}$ or $$ \frac{{\rm d}\vec{r}_j}{{\rm d}t} \times \vec{p}_j = \vec{v}_S \times \vec{p}_j \neq 0$$ I have added in my answer the conditions which would make the above zero.
– John Alexiou
Mar 23 '18 at 19:50
In my scenario. I am standing on the ground holding a long stick pointing at different locations in space. I am looking at the center of mass C or some other arbitrary point A observing what is happening at that location.
– John Alexiou Mar 02 '21 at 13:24This 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.
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:
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 $\nu = 1/s$.
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:
$$ \vec{R} \times m\vec{v}_1' = \hat{z} \{ 16 t\pi\cos(2\pi t) + 12 t\pi \sin(2\pi t)\}\\ L_{1z}' + [\vec{R} \times m\vec{v}_1']_z = 4\pi + 16 t\pi\cos(2\pi t) + 12 t\pi \sin(2\pi t) \ne L_{1z} $$
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.
