# Finding acceleration of center of mass in cart pole problem

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.

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?

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

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

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

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