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I just learned the derivation of the acceleration vector in circular motion. I know that acceleration vector has two components which are centripetal acceleration($\omega^2ra_r$) and tangential acceleration($r\alpha a_t$). But since these components are perpendicular, should't the net acceleration ($\vec a$) always be zero.

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$$\vec a= -\omega^2ra_r+r\alpha a_t$$

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    $\begingroup$ I'm not following. What's the logic by which you go from "these components are perpendicular" to "the net acceleration should be zero"? $\endgroup$ – David Z Oct 16 '15 at 14:20
  • $\begingroup$ if we have two vectors $\vec A$ and $\vec B$, the vector sum is $ABcos \theta$, cos90 is 0, so the vector sum should be zero. Same should apply to the components of acceleration. $\endgroup$ – Abhishek Mhatre Oct 16 '15 at 14:24
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    $\begingroup$ The vector sum of $\vec A$ and $\vec B$ is not $AB\cos(\theta)$. That'd be the scalar product. $\endgroup$ – ACuriousMind Oct 16 '15 at 14:26
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    $\begingroup$ Oh, Abhishek, I mean you should put that into the question. And @ACuriousMind that should be an answer. $\endgroup$ – David Z Oct 16 '15 at 14:32
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    $\begingroup$ @Aniket That's not true. Besides, I don't see what bearing it would have on the question if it were true. $\endgroup$ – garyp Oct 16 '15 at 15:49
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If two vectors $\vec{a_1}$ and $\vec{a_2}$ are perpendicular $\implies$ $\vec{a_1}.\vec{a_2} = 0$

To counter your intuition, say $\vec{a_1}$ represent your displacement in x-direction (say east) and $\vec{a_2}$ is displacement in y direction (say north). Do you think if you move 3 steps east and two steps north will bring you back at the same position!

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Because the two components are perpendicular, the net change in speed is zero - but since the object changes direction, the velocity (which is speed with direction) is not constant.

Think about the velocity at the top of the circle (where it moves in the -X direction) and the bottom of the circle (where it moves in the +X direction). It should be obvious that an object that changed direction like that must have undergone acceleration.

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  • $\begingroup$ I don't understand your first sentence. A tangential acceleration will produce a change in speed. $\endgroup$ – Bill N Oct 16 '15 at 19:36

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