# Shouldn't the net acceleration in circular motion always be zero?

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.

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

• 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
• 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
• 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
• @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

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!