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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?

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2 Answers 2

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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 $$~\dot R=\tilde\omega\,R\\\dot R^T=-R^T\,\tilde\omega$$

$\Rightarrow$

$$\left(\tilde\omega\,R\,J_P\,R^T-R\,J_P\,R^T\,\tilde\omega\right)\omega= \omega\times J_I\,\omega$$

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  • $\begingroup$ Thank you for your answer, but I was interested in another question: the physical meaning of the time derivative of the moment of inertia. $\endgroup$
    – dtn
    Feb 10, 2020 at 14:42
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    $\begingroup$ @AndrewSol the inertia tensor eq. (2) is constant thus the time derivative is zero. the inertia eq. (3) $J_I=R^T(t)\,J_P\,R(t)$ so the time derivative is $\dot{J}_I=\dot{R}^T(t)\,J_P\,R(t)+R^T(t)\,J_P\,\dot{R}(t)$ $\endgroup$
    – Eli
    Feb 10, 2020 at 15:01
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    $\begingroup$ @FourierFlux I put more information in my answer for you $\endgroup$
    – Eli
    Jul 9, 2021 at 12:45
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    $\begingroup$ Thanks, so the key is that the crossproduct of $w$ with itself is 0 knocking out the second term. $\endgroup$ Jul 9, 2021 at 14:51
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    $\begingroup$ the torque is $\overrightarrow{\tau }=\overrightarrow{r}\times m\vec{\ddot{R}}_{coM}$ and you have also the NEWTON equation (translation) $\endgroup$
    – Eli
    Jul 11, 2021 at 7:01
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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.

$$\frac{dI}{dt} = -\frac{I}{\omega} \frac{d\omega}{dt}$$

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    $\begingroup$ $\omega$ is a vector, you can’t divide $\endgroup$
    – Eli
    Feb 10, 2020 at 16:29
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    $\begingroup$ @Eli take ω as the vector magnitude. You can do this because, by construction, the rotation is in a plane. Because it's a rigid body, the rotation unit vector will not change. Note that I'm assuming the change in mass is such that the CM remains the same. If it didn't, the algebra would be a bit more complicated, but the physical intuition would be the same. $\endgroup$
    – Richter65
    Feb 10, 2020 at 16:40
  • $\begingroup$ This answer suits me, it became clearer. $\endgroup$
    – dtn
    Feb 10, 2020 at 17:15
  • $\begingroup$ But, pay attention to the fact that if we have a system of several interacting bodies, and we calculate the generalized forces using the Lagrange equations, then the variable moment of inertia complicates the algebra. It is necessary to provide some tools for simplifying expressions. $\endgroup$
    – dtn
    Feb 10, 2020 at 17:17
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    $\begingroup$ Yes, that's effectively the same as relaxing the rigid body constraint and choosing an arbitrary origin. But even in the more general case, the moment of inertia still acts as a linear operator that transforms ${\vec \omega}$ into $\vec{L}$. You just have to treat is as a second-rank tensor with dynamically varying components, hence the complicated algebra. $\endgroup$
    – Richter65
    Feb 10, 2020 at 18:09

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