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In this answer to the question "Intuitive explanation for why centripetal acceleration is $(v_T)2/r$": https://physics.stackexchange.com/a/190532/262601 (excerpt from the answer is attached below), I don't understand why "we multiply the length of the velocity vector by v/r, just as before, to get the acceleration". I understand why we multiply the position vector, r, by v/r to obtain the velocity vector, v: r(v/r) = v. However, I don't understand why we are multiplying v by v/r to get the acceleration.


A point is moving around a circle. It has a blue position vector and a red velocity vector, like this: enter image description here

The position vector stays the same length and rotates around and around in a circle. Because the position vector is changing, it has a derivative. That derivative is the velocity.

Because we're always going the same speed, the velocity vector also stays the same length. Because the velocity is always 90 degrees rotated from the position, the velocity is also going around in a circle. In other words, the velocity vector is doing exactly the same thing as the position vector is doing; rotating and staying constant length. The only difference between the position and velocity is that we rotated by 90 degrees and multiplied the length by v/r.

(Note: it does not matter where we draw a vector; no matter where we draw it, the vector is the same. We draw the velocity vector at the end of the position vector, so it looks like the velocity vector is moving around in space. That's not the point. We can re-draw the velocity vector so it also starts at the origin and then it will not move around at all. What matters is just the magnitude and the direction. The velocity vector, even if we make it always start at the origin, spins around in a circle at the same speed the position vector does because they are always 90 degrees apart. So the velocity vector really looks just like a new position vector, just with a different magnitude and a direction 90 degrees ahead.)

Acceleration is the derivative of velocity, and we know how to take the derivative. Since the velocity is doing exactly the same thing the position is, we can take the derivative of the velocity in exactly the same way we did with the position.

We rotate the velocity by 90 degrees and get a vector pointed back in towards the center of the circle. That's the direction of the acceleration. Next we multiply the length of the velocity vector by v/r, just as before, to get the acceleration, which is v∗v/r=v^{2}/r.

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Since $\vec v$ is always orthogonal to $\vec r$ and has a constant magnitude, it means vector $\vec v$ rotates with the same angular velocity. The end of vector $\vec v$ will cover “distance” $\Delta = 2\pi v$ during the period of $T=2\pi r/v$. Thus, it moves with the “speed” $\Delta/T = v/(r/v) = v^2/r$.

Another way to look at it, is to see that $v_1$, $v_2$ and $a\,dt$ form an isosceles triangle which is similar to $r_1$, $r_2$, $v\,dt$. Thus, $a\,dt/v = v\,dt/r$.

enter image description here

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