I'm trying to solve an exercise in physics and there is a formula about linear acceleration that includes the term $\vecω\vec{r}_{AB}$, where this operation is an inner product.

Firstly, the problem is 2-dimensional but we know that the angular velocity is always perpendicular to our plane. Since $\vec{r}_{AB}$ is a vector in our plane, shouldn't their inner product be $0$? But according to the book it's not.

Also I find it a bit weird that the angular velocity is always vertical vector. I mean it's a 2D problem but it's like the angular velocity is always in the 3rd axis.

Here is the formula $$\vec{a}_{B}=\vec{a}_{A}+ (d\vec{ω}/dt)\times\vec{r}_{AB} - ω^2\cdot\vec{r}_{AB}.$$

Can you help me understand why?

I add a picture, the problem in question is the first figure. Ignore the rest, i couldn't make the picture smaller. AB, BC and CD are rods.

Picture of the problem, the first figure

  • $\begingroup$ What is "a type"? That term doesn't make sense in this context. $\endgroup$ – Bill N Jan 10 '18 at 19:47
  • $\begingroup$ A formula. It's about general plate motion? I'm sorry I'm from Greece and I'm not sure how to translate things in English. If it is any help, the formula is $\endgroup$ – Thomas Jan 10 '18 at 19:50
  • $\begingroup$ If it helps the formula is $\vec{a}_{B}=\vec{r}_{A}+ (d\vec{ω}/dt)x\vec{r}_{AB} - ω^2\vec{r}_{AB}$ where A and B are 2 points of the plane and by x i mean the outer product $\endgroup$ – Thomas Jan 10 '18 at 19:58
  • $\begingroup$ Edit your question to show the whole formula. The correct MathJax for a dot product would include a \cdot statement. The cross product would have a \times statement (inside a Math environment). Regarding angular velocity, yes, it is something called an axial vector or a bi-vector. $\endgroup$ – Bill N Jan 10 '18 at 19:59
  • $\begingroup$ I don't see an inner product as mentioned in the question. And I believe you left a double time derivative off of the $\vec{r}_A$ term. $\endgroup$ – Bill N Jan 10 '18 at 20:01

The $\omega^2 \vec{r}$ term is not an inner product. It is a scaler times a vector. The $\omega^2$ term itself is $\vec{\omega}\cdot \vec{\omega}$, but that is done before multiplying by $\vec{r}$. That term (with the negative sign) is commonly called the centripetal term. It is radially directed.

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    $\begingroup$ Ohhhhh my god. You are so right. Thank you very much for helping me! And sorry for bad post in the beginning ;) $\endgroup$ – Thomas Jan 10 '18 at 20:39

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