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Slight confusion about centripetal acceleration

  1. Jul 1, 2008 #1
    I just read in my book that the farther you are from the axis of rotation the greater the centripetal acceleration. But when dealing with circular motion the centripetal acceleration decreases as you increase the radius. Am I missing something? Or did I just confuse 2 different things?

    Thanks for the clarification.
     
  2. jcsd
  3. Jul 1, 2008 #2

    Hootenanny

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    Could you quote the passage in question from the book and perhaps give us some idea of the context in which the comment was made?
     
  4. Jul 1, 2008 #3
    Well the textbook just gave the centripetal acceleration =(omega^2)r. Then it says thus the centripetal acceleration is greater the farther you are from the axis of rotation. It then talks about a carousel. Is that enough?
     
  5. Jul 1, 2008 #4

    Kurdt

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    If you're thinking of centripetal acceleration in terms of [itex]\frac{v^2}{r}[/itex] then you have to remember that v is also a function of r.
     
  6. Jul 1, 2008 #5

    Hootenanny

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    Yes that is enough, thank you. The text is correct, but could perhaps be a little clearer by specifying that the angular velocity is fixed. For example, the quote could be re-written thus:

    Given that the centripetal acceleration may be written as [itex]a_c = \omega^2 r[/itex], then for a fixed [itex]\omega[/itex], increasing the radius results in an increased centripetal acceleration.

    In the case of the carousel, each point on the floor of the carousel has the same angular velocity (since the angular velocity is uniquely defined for a rigid body). Hence, the further you are from the centre of the carousel, the faster your linear velocity and the greater your centripetal acceleration. This can perhaps be seen more explicitly if you note that [itex]\omega = v r \Rightarrow v= \omega/r[/itex].

    Does that make sense?

    Edit: Kurdt beat me to it.
     
  7. Jul 1, 2008 #6
    Ah I got it. I thought of angular velocity and linear velocity as interchangeable for some reason. Thanks for the help guys. And thanks for the help yesterday Hootenanny.
     
  8. Jul 1, 2008 #7

    George Jones

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    The other situation you might be thinking about is circular orbital motion. In the this case, centripetal acceleration is proportional to [itex]1/r^2[/itex], and thus decreases as [itex]r[/itex] increases.

    As your book, Kurdt, and Hootenanny have stated, in the carousel case, centripetal acceleration is proportional to [itex]r[/itex], and thus increases as [itex]r[/itex] increases.

    [EDIT]I see I'm wrong.[/EDIT]
     
  9. Jul 1, 2008 #8

    Hootenanny

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    A pleasure :smile:
     
  10. Jul 1, 2008 #9
    Isn't v=r times omega. Not omega over r?
     
  11. Jul 2, 2008 #10

    Hootenanny

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    You are indeed correct, an embarrassing typo. I need to stop posting at 1am... :zzz:
     
    Last edited: Jul 2, 2008
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