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Acceleration perpendicular to velocity in 2D

  1. Sep 8, 2010 #1
    If the magnitude of acceleration is constant, and acceleration is perpendicular to velocity, is speed constant? Also, is speed not constant when the magnitude of acceleration is not constant? How would I show this?

    I tried to do this:

    If position is [tex]p(t)=(x(t),y(t))[/tex], then velocity is [tex]p'(t)=(x'(t),y'(t))[/tex] and acceleration is [tex]p''(t)=(x''(t),y''(t))[/tex]. If the magnitude of acceleration is constant, [tex]|p''(t)|=k[/tex]. If acceleration and velocity are perpendicular, [tex]p'(t) \cdot p''(t) = x'(t)x''(t) + y'(t)y''(t) = 0 [/tex].

    But I'm stuck here.

    How do I show [tex]|p'(t)|=c[/tex] for some constant?
     
  2. jcsd
  3. Sep 8, 2010 #2

    rock.freak667

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    If the acceleration is perpendicular that means your object is moving in a circular path. So there is a centripetal force and hence acceleration acting.

    a=v2/r, so 'a' and 'r' are constants.
     
  4. Sep 8, 2010 #3
    OK, but can you show me how you would deduce the fact that the object is moving with circular motion, given the assumption that acceleration is perpendicular to velocity?
     
  5. Sep 8, 2010 #4

    rock.freak667

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    Force is in the direction of acceleration, meaning that normally the acceleration and velocity would lie in the same plane. As far as I know, the only time acceleration is perpendicular to velocity is during circular motion.
     
  6. Sep 8, 2010 #5

    G01

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    Assuming that there is no external forces other than the one causing the acceleration.

    Use the work-kinetic energy theorem:

    In that case:

    [tex]\Delta T =- \Delta W = -\int \vec{F}\cdot d\vec{r}[/tex]

    From here you should be able to show that T doesn't change if a is always perpendicular to the velocity of the particle (write dr=v(t)dt), and thus speed stays the same.
     
    Last edited: Sep 9, 2010
  7. Sep 9, 2010 #6
    If acceleration and velocity are perpendicular:
    a1v1+a2v2=0
    but that is nothing else than the time derivative of v2, that is speed is costant whenever a and v are perpendicular.

    But I like G01 explanation. It focus on the physics of the system.
    If force and velocity are perpendicular (forse perpendicular to the direction of motion), work made on the system is zero, which implies that the kinetic energy (and so speed) is costant.
     
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