For a circular motion centripetal acceleration can be expressed as $$a_{c}=\frac{v^2}{R} \hat{u_N}\tag{1}$$ Where $\hat{u_n}$ is the normal unit vector.

Nevertheless in the expression for acceleration in rotating reference frames I found the term

$$\vec{\omega}\times(\vec{\omega} \times \vec{r})\tag{2}$$

Is there any conceptual difference between $(1)$ and $(2)$? Or the acceleration is the same besides the fact that in $(1)$ it is expressed in curvilinear coordinates, while the expression $(2)$ is frame-indipendent?

  • $\begingroup$ @Soren .That is more of a mathematical question than a physics question( for more details see vector triple product). $\endgroup$
    – oops
    Apr 20, 2016 at 13:19
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    $\begingroup$ Expression (1) comes from expression (2). Let the opposite of the normal to $\vec{\omega}$ component of $\vec{r}$ be the vector $-\vec{r}_\bot=R\hat{u_N}$, where $R$ its length and $\hat{u_N}$ a unit vector. Note that $v=\omega R$. Make use of $\mathbf{a}\boldsymbol{\times}\left(\mathbf{b} \boldsymbol{\times}\mathbf{c}\right)=\left(\mathbf{a}\circ \mathbf{c}\right)\mathbf{b}-\left(\mathbf{a}\circ \mathbf{b}\right)\mathbf{c}$ and so if $\mathbf{n}$ is a unit vector then $\mathbf{r}_\bot =(\mathbf{n}\times\mathbf{r})\times \mathbf{n}$ is the component of $\mathbf{r}$ normal to $\mathbf{n}$. $\endgroup$
    – Frobenius
    Apr 20, 2016 at 14:23
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    $\begingroup$ @Frobenius that should probably be an answer $\endgroup$
    – David Z
    Apr 20, 2016 at 14:49

1 Answer 1


The first figure shows the parallel and normal components of a vector $\mathbf{r}$ relatively to a direction $\mathbf{n}$. Based of this, the second figure shows the centripetal acceleration. In case of plane circular motion $\omega R = v$.enter image description here

enter image description here


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