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Dot product of acceleration and velocity

  1. Jul 24, 2012 #1
    Just wondering...

    If the dot product of the acceleration and velocity vectors is zero, then does v2/r = 0 have to be true?

    If this is true, is it possible to prove it? If the statement is false, is it possible to prove that as well?
     
  2. jcsd
  3. Jul 24, 2012 #2
    I guess what I'm saying is that if the dot product of the acceleration and velocity vectors is zero (like in circular motion), does that imply v2/r?
     
  4. Jul 24, 2012 #3

    mfb

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    v^2/r=0 implies v^2=0 and that implies v=0. However, [itex]a \cdot v = 0[/itex] just means that a and v are orthogonal to each other, v can be different from 0. It also implies that v^2 is constant: [itex]\frac{dv^2}{dt}=0[/itex].

    Where do you see any r here? A circular motion is just a special case of [itex]a \cdot v = 0[/itex].
     
  5. Jul 24, 2012 #4
    I'm not completely sure what you mean here, but I'm going to take it to mean this: If the acceleration and velocity are perpendicular, does the acceleration have to equal v^2/r as it does in circular motion?

    I think the answer is yes, sort of, but you have to first ask what r means if the motion is not a circle. You could say it is the radius of curvature at the point in question, which is like saying it is the radius of the circular path that we imagine the particle is instantaneously travelling along. But how would you find this radius of curvature? Well if the particle were to continue along a circular path with the speed it has at the instant in question, then its acceleration would obey the usual v^2/r, so rearranging tells you that the radius of curvature is the speed squared divided by the magnitude of the acceleration. So yes the acceleration can be written as v^2/r but that statement is really just a definition of r, the radius of curvature, it is true by definition, it doesn't need a proof.
     
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