Recently, when asking for the derivative of angular momentum, I gave my own solution$$ \frac {d \vec L} {dt} = \frac {d \vec r_{(t1)}} {dt} \times \frac {d \vec r_{(t2)}} {dt} + \vec r_{(t1)} \times \frac {d^2 \vec r_{(t1)}} {dt} \tag 1$$How to calculate the derivative of the angular momentum vector $ d\vec L = d(\hat I \vec \omega)?$
Even though for a few examples I have counted, the derivative gave the correct result the question remained: is it always true?
I already have an answer, but first I have to ask a question, these are the rules here
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$\begingroup$Here we are dealing with some rigid body $B$ with density $\rho$.
I will use $r,z$ as vectors, $I,A,X$ as matrices in 3-dimensional space.
Such a rigid body is usually normalized in that its center of mass is zero and its principal axes to the inertia tensor $I$ are the coordinate axes, that is, $I$ is diagonal, where$$ I=\int_B\rho(z) ((z^Tz)-zz^T)\,d{\rm vol}(z). $$
This body moves with some displacement and rotation $r(t)+A(t)z$, $z\in B$. The path in the orthogonal matrices has a derivative $\dot A=AX$ where $X$ is an anti-symmetric matrix whose operation on vectors can be given as $Xz= ω\times z$. Then $\ddot A=\dot A X+A\dot X=AX^2+A\dot X$, so that\begin{align} \ddot Az=A(ω×(ω×z))&=A((ω^Tz)ω-|ω|^2z)\\ z^TA^T\ddot Az&=ω^Tzz^Tω-|ω|^2|z|^2\\ \int_B\rho(z)z^TA^T\ddot Az\,d{\rm vol}(z)&=-ω^TIω \end{align}
The total angular momentum in an external or lab frame of the body is then\begin{align} L_e&=\int_B\rho(z)(r+Az)\times (\dot r+\dot A z)\,d{\rm vol}(z)\\ &=mr\times \dot r + \int_B\rho(z)(Az)\times (AX z)\,d{\rm vol}(z)\\ &=mr\times \dot r + A\int_B\rho(z) (z× (ω× z))\,d{\rm vol}(z)\\ &=mr\times \dot r + A\int_B\rho(z) ((z^Tz)ω- zz^Tω))\,d{\rm vol}(z)\\ &=mr\times \dot r + AIω.\\ \end{align}One could now name $L_b=Iω$ the internal moment of inertia, relative to the body frame.
Then the derivative of the external moment, which is what is driving the mechanics in reacting to external forces, is\begin{align} \text{Then }~\dot L_e&=mr\times \ddot r + \dot AIω + AI\dot ω\\ &=mr\times \ddot r + AXIω + AI\dot ω\\ &=mr\times \ddot r + A[(ω×(Iω)) + I\dot ω] \end{align}Again one can split off the second term and relate it to the time evolution of the internal moment. I would not call it the time derivative,...
$\endgroup$ 4 $\begingroup$$$\forall m \in \Bbb R \land \forall \vec r_{(t1)}= (r_{1x},r_{1y},r_{1z}) \land \forall \vec r_{(t2)}= (r_{2x},r_{2y},r_{2z}) \land \forall \vec r_{(t3)}= (r_{3x},r_{3y},r_{3z})$$$$(\frac {d \vec L} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} )$$
velocity and acceleration vectors
$$\frac {r_{2x} - r_{1x}}{dt} = \vec v_{x(t1)} ; \frac {r_{2y} - r_{1y}}{dt} = \vec v_{y(t1)}; \frac {r_{2z} - r_{1z}}{dt} = \vec v_{z(t1)} \tag 1 $$
$$\frac {r_{3x} - r_{2x}}{dt} = \vec v_{x(t2)} ; \frac {r_{3y} - r_{2y}}{dt} = \vec v_{y(t2)}; \frac {r_{3z} - r_{2z}}{dt} = \vec v_{z(t2)} \tag 2 $$
$$\frac {v_{2x} - v_{1x}}{dt} = \vec a_{x(t1)} ; \frac {v_{2y} - v_{1y}}{dt} = \vec a_{y(t1)}; \frac {v_{2z} - v_{1z}}{dt} = \vec a_{z(t1)} \tag 3 $$
angular momentum vectors
$$\vec L_{(t1)}= \vec r_{(t1)} \times m \vec v_{(t1)} $$$$(r_{1y}v_{1z} - r_{1z}v_{1y}, r_{1z}v_{1x}-r_{1x}v_{1z},r_{1x}v_{1y} -r_{1y}v_{1x}) \tag 4 $$$$\vec L_{(t2)}= \vec r_{(t2)} \times m \vec v_{(t2)} $$$$(r_{2y}v_{2z} - r_{2z}v_{2y}, r_{2z}v_{2x}-r_{2x}v_{2z},r_{2x}v_{2y} -r_{2y}v_{2x}) \tag 5 $$change of angular momentum vector with time$$\vec L_{(t2)}-\vec L_{(t1)} \tag 6$$
$$\frac {d l_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {6.a}$$$$\frac {d l_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {6.b}$$$$\frac {d l_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {6.c}$$
we calculate the components of the derivative
$$\vec v_{(t1)}\times \vec v_{(t2)}=(v_{1y}v_{2z} - v_{1z}v_{2y}, v_{1z}v_{2x}-v_{1x}v_{2z},v_{1x}v_{2y} -v_{1y}v_{2x}) \tag 7$$
$$v_{1y}v_{2z} - v_{1z}v_{2y}=r_{2y}v_{2z}-r_{1y}v_{2z} -r_{2z}v_{2y}+r_{1z}v_{2y}\tag {7.a}$$$$v_{1z}v_{2x} - v_{1x}v_{2z}=r_{2z}v_{2x}-r_{1z}v_{2x} -r_{2x}v_{2z}+r_{1x}v_{2z}\tag {7.b}$$$$v_{1x}v_{2y} - v_{1y}v_{2x}=r_{2x}v_{2y}-r_{1x}v_{2y} -r_{2y}v_{2x}+r_{1y}v_{2x}\tag {7.c}$$
$$\vec r_{(t1)}\times \vec a_{(t1)}=(r_{1y}a_{1z} - r_{1z}a_{1y}, r_{1z}a_{1x}-r_{1x}a_{1z},r_{1x}a_{1y} -r_{1y}a_{1x}) \tag 8$$
$$r_{1y}a_{1z} - r_{1z}a_{1y}=r_{1y}v_{2z}-r_{1y}v_{1z} -r_{1z}v_{2y}+r_{1z}v_{1y}\tag {8.a}$$$$r_{1z}a_{1x} - r_{1x}a_{1z}=r_{1z}v_{2x}-r_{1z}v_{1x} -r_{1x}v_{2z}+r_{1x}v_{1z}\tag {8.b}$$$$r_{1x}a_{1y} - r_{1y}a_{1x}=r_{1x}v_{2y}-r_{1x}v_{1y} -r_{1y}v_{2x}+r_{1y}v_{1x}\tag {8.c}$$We count the derivative$$\frac {d \vec L`} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} \tag 9$$
$$\frac {d l`_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {9.a}$$$$\frac {d l`_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {9.b}$$$$\frac {d l`_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {9.c}$$
$$(6)=(9)$$
C.K.D.
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