The inertia tensor is clearly constant the in a frame moving with the rigid body. But what is the simplest way to see why its columns can be considered rotating vectors in space with the angular velocity of the rotating rigid body?


1 Answer 1


This has been answered below, but consider the rotation matrix $\mathrm{R}$ whose columns represent the local $\hat{x}$, $\hat{y}$ and $\hat{z}$ axis in the world coordinates.

$$ \mathrm{R} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \end{vmatrix}\tag{1}$$

This defines the changing inertia tensor from the local inertia tensor $\mathcal{I}_{\rm body}$

$$ \mathcal{I} = \mathrm{R}\, \mathcal{I}_{\rm body} \mathrm{R}^\top \tag{2}$$

Now that rate of change of the rotation matrix is by definition described by the derivative on a rotating frame

$$ \begin{aligned} \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{x}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{x}} \\ \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{y}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{y}} \\ \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{z}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{z}} \\ \end{aligned}$$

or the shorthand rotation $$\dot{\mathrm{R}} = \boldsymbol{\omega} \times \mathrm{R} \tag{3}$$

Now the rate of change of the inertia tensor is the time derivative of (2) and the product rule

$$\require{cancel} \begin{aligned}\dot{\mathcal{I}} & =\mathrm{\dot{R}}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}+\mathrm{R}\,\cancel{\dot{\mathcal{I}}_{{\rm body}}}\mathrm{R}^{\top}+\mathrm{R}\,\mathcal{I}_{{\rm body}}\dot{\mathrm{R}}^{\top}\\ & =\left(\boldsymbol{\omega}\times\mathrm{R}\right)\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}+\mathrm{R}\,\mathcal{I}_{{\rm body}}\left(\boldsymbol{\omega}\times\mathrm{R}\right)^{\top}\\ & =\boldsymbol{\omega}\times\left(\mathrm{R}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}\right)-\mathrm{R}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}\left(\boldsymbol{\omega}\times\right)\\ & =\boldsymbol{\omega}\times\mathcal{I}-\mathcal{I}\,\boldsymbol{\omega}\times \end{aligned} \tag{4}$$

with the slight of hand notation $\omega \times$ meaning worry about the cross product later when the tensor is acted upon a vector.

So to derive the equations of motion from the change in angular momentum $\boldsymbol{L} = \mathcal{I} \boldsymbol{\omega}$ you

$$ \tfrac{{\rm d}}{{\rm d}t}\left(\mathcal{I}\boldsymbol{\omega}\right)=\mathcal{I}\dot{\boldsymbol{\omega}}+\left(\boldsymbol{\omega}\times\mathcal{I}-\mathcal{I}\boldsymbol{\omega}\times\right)\boldsymbol{\omega}=\mathcal{I}\dot{\boldsymbol{\omega}}+\boldsymbol{\omega}\times\mathcal{I}\boldsymbol{\omega}-\mathcal{I}\left(\cancel{\boldsymbol{\omega}\times\boldsymbol{\omega}}\right) \tag{5} $$

  • $\begingroup$ Thanks! I should have seen $\vec{M}_C = \mathcal{I}_C \vec{\alpha} + \vec{\omega} \times \mathcal{I}_C \vec{\omega}$ in some books before! Why don't the authors write it and let me live in ignorance? Can you tell me some title? $\endgroup$
    – user291161
    May 4, 2022 at 21:40
  • $\begingroup$ en.wikipedia.org/wiki/Euler%27s_equations_(rigid_body_dynamics) $\endgroup$
    – JAlex
    May 4, 2022 at 22:34
  • $\begingroup$ Page 19 - dma.ing.uniroma1.it/users/lss_da/MATERIALE/Textbook.pdf $\endgroup$
    – JAlex
    May 4, 2022 at 22:38
  • $\begingroup$ Thank you. So the operator $(\boldsymbol{\omega}\times\mathrm{R})^T$ and the operator $-\mathrm{R}^T \boldsymbol{\omega}\times$ are the same thing? I find this step difficult. $\endgroup$
    – user291161
    May 6, 2022 at 21:15
  • $\begingroup$ The transpose of a product reverses the order of operations $(A B)^\top = B^\top A^\top$. So if you consider $\omega \times$ matrix 3×3 skew symmetric cross product matrix and the fact that the transpose of skew matrix is minus the matrix and you can derive the above. $\endgroup$
    – JAlex
    May 7, 2022 at 1:34

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