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Is it fine to express the centripetal acceleration as a cross product?
a=v X w (where a is centripetal acceleration, v is magnitude of velocity, w is angular velocity)

And is it v X w or w X v?

What I think:
Since centripetal acceleration requires tangential (perpendicular) velocity, I start thinking about cross products, and was able to express the acceleration vector as 2 other vectors.
Fiddling around with my right hand, I think that a=v X w and not a=w X v.
Where the convention is
-angular velocity towards me implies positive anticlockwise movement
-centripetal acceleration upwards is taken as positive
-velocity has to move in a way to cause anticlockwise movement

Thing is, I've been searching this up on the Internet but couldn't find any resources for confirmation.

Is it true that centripetal acceleration can be represented as the cross product of velocity and angular velocity? v X w

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2 Answers 2

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$$\vec{a}=\vec{\omega} \times \vec{v}$$ This delivers the goods for a point moving at speed $v$ in a circular path with angular velocity $\vec\omega$ about the centre of the circle. Because the circle lies in one plane, the direction of $\vec\omega,$ as well as its magnitude, is constant.

We can indeed derive the centripetal acceleration formula rather neatly starting with $$\vec{v}=\vec{\omega} \times \vec{r}$$ So$$\frac{d\vec{v}}{dt}=\frac{d\vec{\omega}}{dt} \times r + \vec\omega \times \frac{d\vec r}{dt}$$ The first term on the right disappears because $\vec\omega$ is a constant for a body moving in a circle at constant speed, so you're left with$$\vec a=\vec\omega \times \vec v$$

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    $\begingroup$ Rohit's answer, back-substituting $\vec{v}=\vec{\omega} \times \vec{r}$ gives a much neater equation without the cross product or the pseudo-vector $\omega.$ $\endgroup$
    – Philip Wood
    Commented Jan 18, 2019 at 18:56
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    $\begingroup$ Suppose that body is moving anticlockwise on a table top, seen from above. Thus $\vec\omega$ is upwards (using right hand screw rule). Agreed? Consider the instant when the body is East of the circle centre, that is $\vec r$ is Eastwards} So $\vec{\omega} \times \vec r$ is Northwards, confirming that the body really is going anticlockwise. The rest follows by algebra. If you want to check just the final result, consider the same instant for the same set-up and evaluate $\vec{\omega} \times \vec v$. It gives an Eastward vector, that is towards the circle centre. $\endgroup$
    – Philip Wood
    Commented Jan 18, 2019 at 19:38
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    $\begingroup$ My answer and comment assumed a right hand rule for the definition of $\vec\omega$ and for the definition of the vector product. If, instead, we used left hand rules $for\ both,$ we'd get the same result! $\endgroup$
    – Philip Wood
    Commented Jan 18, 2019 at 21:40
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    $\begingroup$ (a) "v is towards my right on the middle finger. Centripetal acceleration is upwards on thumb. For v X w, v is towards the left " Why are you changing the direction for $v$? (b) You can erase and replace a wrong comment by clicking on the red blob that appears when you sweep the mouse past the end of your comment (c) If I am to help, please use the set-up that I described in the first 3 lines of my comment starting "Suppose...". Then I'll understand what you mean more clearly. $\endgroup$
    – Philip Wood
    Commented Jan 19, 2019 at 9:10
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    $\begingroup$ No cause for apology. I'm glad you now understand. $\endgroup$
    – Philip Wood
    Commented Jan 19, 2019 at 9:46
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If you are using the right hand rule(convention) for evaluating the cross product, that is $\hat{x}\times\hat{y}=\hat{z}$ for the cartesian co-ordinate unit vectors, then the centripetal acceleration seems to be $\bf{\omega}\times\bf{v}$. Furthermore, it can be simplified to be $\bf{\omega}\times\bf{v}=\bf{\omega}\times[\bf{\omega}\times\bf{r}]=-|\omega|^2\bf{r}$. Refer to the following reference for more details. Hope this answers your question.

Reference: Derivation using vectors https://en.wikipedia.org/wiki/Centripetal_force

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  • $\begingroup$ Why is the centripetal acceleration negative? I know that it is obtained from the vector triple product expansion, but is there another way to explain why it is negative? $\endgroup$
    – helpme
    Commented Jan 18, 2019 at 19:17
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    $\begingroup$ The acceleration is along $-\bf{r}$ direction. This is because, for an object to in the circular motion, it needs to be "pulled" inwards towards the origin(pivot) by some force(for example the tension of a string, gravitational force, etc), which causes an inward acceleration. $\endgroup$
    – Rohit
    Commented Jan 19, 2019 at 6:03
  • $\begingroup$ Is the inward acceleration taken as negative by convention? $\endgroup$
    – helpme
    Commented Jan 19, 2019 at 8:12
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    $\begingroup$ yes, you are right. That's the convention, with direction of $\hat{r}$ being radially outward (hence positive), pointing away from the origin. $\endgroup$
    – Rohit
    Commented Jan 20, 2019 at 6:19

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