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Centrifugal force and angular velocity

  1. Dec 13, 2013 #1
    According to the wikipedia page, when given an angular velocity, centrifugal force increases with radius. I always thought the larger the radius, the smaller the centrifugal force. I think I'm misunderstanding some term here (possibly "angular velocity"). Can someone please explain?
     
  2. jcsd
  3. Dec 13, 2013 #2

    jfizzix

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    Angular velocity [itex]\omega[/itex] is the rate at which angle changes with respect to time (revolutions per second, degrees per second, radians per second, etc). If this rate is a constant, then the larger your distance to the center [itex]r[/itex] is, the larger the centrifugal force [itex]F[/itex] you will feel.

    [itex]F = m \;r\;\omega^{2}[/itex]

    However, if you are saying your tangential velocity [itex]v = r\;\omega[/itex] is constant, then the farther out from the center you are, the smaller the centrifugal force you experience, but this also means your angular velocity is smaller too.

    [itex]F = m \frac{v^{2}}{r}[/itex]
     
  4. Dec 13, 2013 #3

    tiny-tim

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    Hi Lsos! :smile:
    Yes, if the angular velocity is fixed, the larger the radius, the more force you need to keep something in the circle.

    Loosely speaking, changing the velocity from v to -v in the same time is obviously a larger acceleration if v is larger! :wink:
    Angular velocity is angle per second.

    It's equal to revolutions per second times 2π. :smile:
     
  5. Dec 13, 2013 #4
    Hi,
    The centrifugal force is a fictitious force used in non-inertial reference frames. In general, it is given by $$ F_{\text{centrifugal}} = m\textbf{w} \times (\textbf{w}\times\textbf{r}). $$ Here, ## \textbf{w} ## is the angular velocity of the rotating reference frame and ## \textbf{r} ## is the position of the particle relative to the rotating frame's origin.

    Usually it will simplify to jfizzix' answer (the first equation).

    From a more qualitative perspective, imagine a large disc spinning. If you stand near the centre, you move in a small circle but not very fast. As you move out, you move faster and faster and you will find it more difficult to stay on the disc.

    :)

    PS: I used ##\textbf{w}## because for some reason the TeX code for omega isn't working... :\
     
  6. Dec 13, 2013 #5

    Andrew Mason

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    I think your use of the bold face text is the problem. Latex assumes what follows is text. You can use vectors like this$$ F_{centrifugal} = m\vec{\omega} \times (\vec{\omega} \times \vec{r}) $$

    AM
     
  7. Dec 14, 2013 #6
    Ok...yeah. I think I get it. Seems like a very simple concept, but I just need to wrap my aging mind around this. Thanks everyone!
     
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