First derivative of the moment of inertia in time as a physical parameter My question is addressed to the reputable community of physicists in connection with the ignorance of some of the subtleties of mechanics. Perhaps it will be interesting to other users.
Moment of inertia of a three-dimensional rigid body relative to a certain center of rotation $O$ can be found by the formula (Huygens-Steiner theorem) [1]:
$$J = m_{l} \cdot i \left( t \right) \cdot i \left( t \right)^T +E \left( t \right) \cdot J_2 \cdot E \left( t \right)^T$$
where $i(t)$ - three dimensional vector, that include coordinates of center of mass;
$E(t)$ - matrix of rotation;
$m_l$ and $J_2$ - body mass and basic tensor of inertia;
If we find the derivative of the moment of inertia with respect to time, we get the formula:
$$\frac{\mathrm dJ}{\mathrm dt} = m_{l}\,{\frac {\rm d}{{\rm d}t}}i \left( t \right) \cdot i \left( t \right)^T +m_{l}\,i \left( t \right) \cdot {\frac {\rm d}{{\rm d}t}}i \left( t \right)^T +{\frac {\rm d}{{\rm d}t}}E \left( t \right) \cdot J _2 \cdot E \left( t \right)^T +E \left( t \right) \cdot J_2 \cdot {\frac {\rm d}{{\rm d}t}}E \left( t \right)^T$$

My question is: what parameter did we get in the end? What is the physical meaning of the derivative of the moment of inertia with respect to time: consumption of rotational mass?
 A: 
Huygens-Steiner theorem ( parallel axes transformation) is:
$$J_P=J_C-m\,\tilde{r}\,\tilde{r}\tag 1$$
where

*

*$J_C$ is the inertia tensor  in coordinate system that locate at the center of mass


*m is the total mass


*$\vec{r}$ is the vector from the CM to point P, the components of the vector r are given in the CM coordinate system.


*$J_P$ is the inertia tensor  in coordinate system that locate at point P and is parallel to the coordinate system of the CM
with
$$\tilde{r}\tilde{r}=\vec{r}\,\vec{r}^T-\vec{r}^T\,\vec{r}\,I_3$$
in equation (1)
$$J_P=J_C-m\,\left(\vec{r}\,\vec{r}^T-\vec{r}^T\,\vec{r}\,I_3\right)\tag 2$$
to obtain  the angular momentum $\vec{L}=J_I\,\vec{\omega}$ in Inertial system, you have to transformed the inertia tensor that  given in body fixed system to inertial system
$$J_I=R\,J_P\,R^T\tag 3$$
where $R$ is the transformation matrix from body fixed system to inertial system.
the equation of motion are:
$$\frac{d}{dt}\vec{L}=\frac{d}{dt}\left(J_I\,\vec{\omega}\right)=J_I\vec{\dot{\omega}}+\frac{d}{dt}\,\left(J_I\right)\,\vec{\omega}
=J_I\vec{\dot{\omega}}+\vec{\omega}\times (J_I\,\vec{\omega})=\vec{\tau}\tag 4$$
here is where you need the derivative of the inertia tensor

Appendix
$$\vec{r}= \left[ \begin {array}{c} x\\ y\\ 
z\end {array} \right] \quad,
\tilde{r}=\left[ \begin {array}{ccc} 0&-z&y\\ z&0&-x
\\-y&x&0\end {array} \right] 
$$
edit
$$\frac{d}{dt}\left(R\,J_P\,R^T\right)\,\omega=
\left(\dot R\,J_P\,R^T+R\,J_P\,\dot R^T\right)\omega$$
with $$~\dot R=\tilde\omega\,R\\\dot R^T=-R^T\,\tilde\omega$$
$\Rightarrow$
$$\left(\tilde\omega\,R\,J_P\,R^T-R\,J_P\,R^T\,\tilde\omega\right)\omega=
\omega\times J_I\,\omega$$
A: Since you're asking about physical intuition, I'd start with a simpler, more intuitive formula:
$$L = I\omega$$
in a suitable coordinate system (i.e., origin at the center of mass, coordinate axes chosen such that the rotation is an a plane).  A change in the moment of inertia would contribute to the change in angular momentum:
$$ \frac{dL}{dt} = \frac{dI}{dt}{\omega} + 
I\frac{d{\bf {\bf \omega}}}{dt}$$
You've specified a rigid body, so the only way for $I$ to change is via a change in mass.  It depends on the mechanism by which the mass was changing, but if angular momentum were conserved, the effect would be that the rotational speed $\bf \omega$ would change.  In other words, the derivative of the moment of inertia would tell you how much the angular velocity would need to change in order to keep angular momentum constant.
$$\frac{dI}{dt} = -\frac{I}{\omega} \frac{d\omega}{dt}$$
