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I know that the angular momentum $\mathbf{L}_{cm}$ with respect to the centre of mass of a rigid body can be expressed as $I\boldsymbol{\omega}$ where $I$ is the inertia matrix and $\boldsymbol{\omega}$ is a column vector representing angular velocity. I am intuitively convinced that the resulting torque $\sum\boldsymbol{\tau}_{cm,\text{ext}}=\frac{d\mathbf{L}_{cm}}{dt}$ -where the equality is a consequence of the cardinal equations of dynamics- uniquely determines the angular acceleration of a rigid body and I have tried to prove it to myself in the following way, which I am going to explain with some details of calculation in order to be clearer.

[EDIT: There was my proof here, but ja72 has suggested to me to post it as an answer, below, and I accept the suggestion]

I such a proof correct? I $\infty$-ly thank you!

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    $\begingroup$ Where is the torque coming from? Is it supposed to be an external torque? $\endgroup$ – CuriousOne May 26 '15 at 7:35
  • $\begingroup$ @CuriousOne I know that the torque due to internal forces is 0, therefore the resulting torque $\sum\boldsymbol{\tau}_{cm}$ is identical to the resulting torque due to the external forces $\sum\boldsymbol{\tau}_{cm,\text{ext}}$. I have obtained $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}$ from what I know (from Gettys-Keller-Skove's Physics) by the name second cardinal equation of dynamics: $\sum\boldsymbol{\tau}_{P,\text{ext}}=\frac{d\mathbf{L}_P}{dt}+\mathbf{v}_P \times\mathbf{P}$ ... $\endgroup$ – Self-teaching worker May 26 '15 at 10:20
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    $\begingroup$ See physics.stackexchange.com/a/113896/392 which is where I think you are going with this. $\endgroup$ – John Alexiou May 26 '15 at 12:54
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    $\begingroup$ As a note, pure angular motion does not have a location. So the statement "the angular acceleration with respect to the centre of mass" is a little confusing. All parts of a rigid body move with the same angular motion. Only linear motion needs to have a location specified in order to be fully qualified. $\endgroup$ – John Alexiou May 26 '15 at 13:06
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    $\begingroup$ For being self taught you have come a long way. I've been hesitant to vote for close this question as duplicate because a) your efforts should be rewarded and b) the duplicate question does not have an accepted answer. Best course would be for you to answer your own question with your own equations (and referencing my answer too maybe) and award it to yourself. $\endgroup$ – John Alexiou May 26 '15 at 20:12
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As an answer by ja72, whom I heartily thank, confirms, my following calculations prove that $\frac{d\boldsymbol{\omega}}{dt}=I^{-1}\big(\frac{d\mathbf{L}_{cm}}{dt}-\boldsymbol{\omega}\times(I\boldsymbol{\omega})\big)$.

By using a moving Cartesian coordinate system, dextrogyre as usual when calculating angular quantities, with basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ solidally moving with the rigid body, whose vectors I use as the columns of a matrix $E(t)=\left( \begin{array}{ccc}\mathbf{i}(t)&\mathbf{j}(t)&\mathbf{k}(t) \end{array} \right)$ for the sake of brevity, we have that$$I=EI_mE^{-1},\quad\boldsymbol{\omega}=E\boldsymbol{\omega}_{m},\quad I\boldsymbol{\omega}=EI_m\boldsymbol{\omega}_{m}$$where I have used the index $m$ to mark the inertia matrix and, respectively, the triplet of the coordinates of angular velocity with respect to basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$. By differentiating I get$$\frac{d\mathbf{L}_{cm}}{dt}=\frac{dE}{dt}I_m\boldsymbol{\omega}_{m}+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$and, by taking Poisson formula, which give us $\frac{d}{dt}\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)=\left( \begin{array}{ccc}\boldsymbol{\omega}\times\mathbf{i}&\boldsymbol{\omega}\times\mathbf{j}&\boldsymbol{\omega}\times\mathbf{k} \end{array} \right)$, into account and calling $I_m^{(i)}$ the $i$-th row of $I_m$,$$\frac{d\mathbf{L}_{cm}}{dt}=\boldsymbol{\omega}\times( I_m^{(1)}\cdot\boldsymbol{\omega}_m\mathbf{i})+\boldsymbol{\omega}\times( I_m^{(2)}\cdot\boldsymbol{\omega}_m\mathbf{j})+\boldsymbol{\omega}\times( I_m^{(3)}\cdot\boldsymbol{\omega}_m\mathbf{k})+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$$$=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}$$and since $I_m$ does not depend upon time$$\frac{d\mathbf{L}_{cm}}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+EI_m\frac{d\boldsymbol{\omega}_{m}}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I E\frac{d(E^{-1}\boldsymbol{\omega})}{dt}$$$$=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I\bigg(\frac{d(\boldsymbol{\omega}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{k})}{dt}\mathbf{k}\bigg)$$but Poisson formulae give $\frac{d\mathbf{r}}{dt}=\frac{d(\mathbf{r}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\mathbf{r}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\mathbf{r}\cdot\mathbf{k})}{dt}\mathbf{k}+\boldsymbol{\omega}\times\mathbf{r}$ for any differentiable vector function and therefore $\frac{d\boldsymbol{\omega}}{dt}=\frac{d(\boldsymbol{\omega}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{k})}{dt}\mathbf{k}+\boldsymbol{\omega}\times\boldsymbol{\omega}$ where $\boldsymbol{\omega}\times\boldsymbol{\omega}=\mathbf{0}$, which give$$\frac{d\boldsymbol{\omega}}{dt}=I^{-1}\bigg(\frac{d\mathbf{L}_{cm}}{dt}-\boldsymbol{\omega}\times(I\boldsymbol{\omega})\bigg)$$where all the quantities are calculated with respect to the external inertial frame, and where $I=EI_mE^\text{T}$ with $E^{\text{T}}$ as the transpose matrix of $E$.

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