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In this link about finding equations of motion of cart pole problem, There is an equation about acceleration of center of mass of the pole. Screenshots of them below.

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I don't understand why they have more than two parts about angular acceleration - $\varepsilon \times r_p$ and $\omega \times (\omega \times r_p)$?

If I'm being right, first one is torque, and second one is acceleration of a point in circular movement.

I guess in some part I'm being incorrect, but I don't understand why they put two angular acceleration of it? It's copy of them, aren't they?

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When you transfer velocity from one point to another in a rigid body you end up with an equation like

$$ \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r}_P $$

Acceleration is just the time derivative of the above with

$$ \begin{aligned} \tfrac{\rm d}{{\rm d}t} \boldsymbol{v}_P &= \boldsymbol{a}_P \\ \tfrac{\rm d}{{\rm d}t}\boldsymbol{v}_C & = \boldsymbol{a}_C \\ \tfrac{\rm d}{{\rm d}t} \boldsymbol{\omega} &= \boldsymbol{\epsilon} \\ \tfrac{\rm d}{{\rm d}t} \boldsymbol{r}_P &= \boldsymbol{\omega} \times \boldsymbol{r}_P \end{aligned} $$

The last part is because $\boldsymbol{r}_P$ is a fixed vector riding along with the rigid body.

The transformation of acceleration is thus

$$ \boldsymbol{a}_P = \boldsymbol{a}_C + (\tfrac{\rm d}{{\rm d}t}\boldsymbol{\omega} )\times \boldsymbol{r}_P + \boldsymbol{\omega} \times (\tfrac{\rm d}{{\rm d}t} \boldsymbol{r}_P) $$

$$\boldsymbol{a}_P = \boldsymbol{a}_C + \boldsymbol{\epsilon} \times \boldsymbol{r}_P + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{r}_P) $$

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That last term is the centripetal acceleration associated with the rotation of the rod.

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