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Please excuse the stupidity in the question if there is any. Why can’t we use the average acceleration formula for centripetal acceleration? Why do we need to derive another formula for it. In its derivation (in school) we even equated it to the average acceleration formula. Derivation written in school notebook

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    $\begingroup$ Well, there are certainly direct parallels between linear motion and circular motion. But, torque isn't force, and mass isn't moment of inertia, so showing they are directly parallel takes a small bit of doing. And showing they are directly parallel by the above should be comforting - the universe behaves in the "same" way. $\endgroup$ – Jon Custer Mar 22 at 19:20
  • $\begingroup$ @JonCuster that was a very concise explanation and it helped me a little bit. Could you care to elaborate a little more, maybe even post an answer! $\endgroup$ – snow_razer Mar 22 at 19:25
  • $\begingroup$ Note the "when Δt is very small" line in your comments. It is not equated to the "average acceleration formula", but to the derivative of the speed. $\endgroup$ – fishinear Mar 30 at 12:41
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Because here you assume that the object doesn't start and end at the same point. Say I were to start and end at the same point on the circular path. Then $\Delta v$ = 0, as the magnitude and direction are the same, and therefore the average acceleration would also be 0. But that cannot be the case as the velocity is continuously changing (in its direction in this case). Don't forget that velocity is a vector quantity, meaning that it has magnitude/length, and direction. The change in velocity is towards the center, as the object is being "pulled" to the center which is why it continues in a circular motion. This is why you must take into account that this is in fact not a linear displacement but a motion of rotation.

Hope this helps!

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