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?
Best Answer
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 $$~\dot R=\tilde\omega\,R\\\dot R^T=-R^T\,\tilde\omega$$
$\Rightarrow$
$$\left(\tilde\omega\,R\,J_P\,R^T-R\,J_P\,R^T\,\tilde\omega\right)\omega= \omega\times J_I\,\omega$$