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

Abhishek Mhatre
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    I'm not following. What's the logic by which you go from "these components are perpendicular" to "the net acceleration should be zero"? – David Z Oct 16 '15 at 14:20
  • 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. – Abhishek Mhatre Oct 16 '15 at 14:24
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    The vector sum of $\vec A$ and $\vec B$ is not $AB\cos(\theta)$. That'd be the scalar product. – ACuriousMind Oct 16 '15 at 14:26
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    Oh, Abhishek, I mean you should put that into the question. And @ACuriousMind that should be an answer. – David Z Oct 16 '15 at 14:32
  • so is the value of |$\vec a| $ = $\sqrt{\omega^4r^2+r^2\alpha^2 +- \omega^2r.r\alpha.cos90 }$ – Abhishek Mhatre Oct 16 '15 at 14:32
  • If two vectors are orthogonal i.e. perpendicular to each other, it means they are independent of each other. – SchrodingersCat Oct 16 '15 at 15:07
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    @Aniket That's not true. Besides, I don't see what bearing it would have on the question if it were true. – garyp Oct 16 '15 at 15:49
  • @garyp Why do you say so? Why two orthogonal vectors are not independent? If the radial force on a particle is increased or decreased, how would it affect the tangential force acting on it? Don't limit the scope of your answer to circular motion only. – SchrodingersCat Oct 16 '15 at 17:12
  • @Aniket $\vec{A} = a\hat{x}$ and $\vec{B}=a\hat{y}$, for real $a$, are orthogonal but not independent. – garyp Oct 16 '15 at 17:48
  • @garyp I wanted to mean that if two vectors are orthogonal, then one's component won't influence the other (component $\times \cos 90$, you know that). BTW is there any other meaning for independence which I am missing? – SchrodingersCat Oct 16 '15 at 18:01
  • @Aniket If two vectors are orthogonal, they are linearly independent of each other in the sense of a vector space. You won't be able to express one of them as a linear vector sum involving the other (with the understanding that multiplying by zero is excluded). That word linearly is important. – Bill N Oct 16 '15 at 19:29
  • @BillN Thanks. Thats what I meant. Linearly independent. – SchrodingersCat Oct 16 '15 at 19:30

2 Answers2

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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!

Manish Kumar Singh
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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.

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