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Centripetal Acceleration

  1. Mar 14, 2005 #1
    For centripetal acceleration, what is the relationship between the radius and the period?
     
  2. jcsd
  3. Mar 14, 2005 #2

    HallsofIvy

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    Do you mean the relationship for fixed acceleration?

    If an object is moving in a circle, at constant speed, we can write the vector equation as [itex]\vec r= R cos(\omega t)\vec i+ R sin(\omega t)\vec j[/itex] (ω is the angular speed- the object will do one full circle ([itex]2\pi[/itex] radians) in [itex]\frac{2\pi}{\omega}[/itex] seconds and so the period is: one full circle in [itex]\frac{\omega}{2\pi}[/itex] seconds: the period.

    The velocity vector is [itex]-R\omega sin(\omega t)\vec i+ R\omega cos(\omega t)\vec j[/itex] and the acceleration vector is [itex]-R\omega^2 cos(\omega t)\vec i- R\omega^2 sin(\omega t)[/itex] which has length [itex]\alpha = R\omega^2[/itex].
    That is [itex]\omega= \sqrt{\frac{\alpha}{R}}[/itex] and so the period is
    [tex]T= \frac{\omega}{2\pi}= \frac{1}{2\pi}\sqrt{\frac{\alpha}{R}}[/tex]

    You can also solve that for R:
    [tex]R= \frac{4\pi^2T^2}{\alpha}[/tex]
     
    Last edited: Mar 14, 2005
  4. Mar 14, 2005 #3
    oh man...too advanced for me....

    Can somone just tell me if as the radius increases the period is longer or something?
     
  5. Mar 14, 2005 #4

    HallsofIvy

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    Yes, the period increases as the square root of the radius: if the radius is 4 times as large, the period is twice as large.
     
  6. Mar 14, 2005 #5

    dextercioby

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    Yes,if the velocity (the linear/tangetial) is kept constant...

    Daniel.
     
  7. Mar 14, 2005 #6

    HallsofIvy

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    It's a little spooky to see responses to responses while you are editing!
     
  8. Mar 14, 2005 #7

    dextercioby

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    Would u mind if i told u that i didn't look at your post...?Before correcting it...


    Daniel.
     
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