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B Having problem with centripetal and centrifugal force

  1. Jul 15, 2016 #1
    Hi,I was reading about the centripetal and centrifugal forces then I felt a bit of contradiction in the explanations. It is said that the centripetal and centrifugal forces are the action-reaction pair of Newton's 3rd law and so they never act upon the same body.Centripetal force is acted upon the body that is rotating in a circular path and the centrifugal force is acted upon the centre of the circular path.But we can see the effect of the centrifugal force not the centripetal force on the rotating body apparently. Because, it's the centrifugal force that tries to draw the body outwards the circle. Please make it clear.
  2. jcsd
  3. Jul 15, 2016 #2

    Doc Al

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    Staff: Mentor

    Be very careful here!

    In typical physics usage, "centrifugal force" is a fictitious (or inertial) force that is only used when viewing things from an accelerated frame of reference. (Read about it here: https://en.wikipedia.org/wiki/Centrifugal_force)

    In very atypical usages (in my opinion) some folks use a concept called "reactive centrifugal force", which is the definition that applies to what you read. (Read about that here: https://en.wikipedia.org/wiki/Reactive_centrifugal_force) To call that "centrifugal force" will confuse many physics folks. Don't do it!

    So, imagine a ball tied to a string, whirled in a circle. Since there is a centripetal acceleration, there is a centripetal force: The force of the string pulling on the ball. No centrifugal forces involved since we are viewing things from an inertial frame of reference.

    It's certainly true that the ball pulls back on the string, following Newton's third. No special name for that force is needed.

    Viewed from a frame that rotates with the ball, the ball is not accelerating. (It's just sitting there.) To "fix" things so that Newton's 2nd law can be used, we add a fictitious outward centrifugal force that acts on the ball and "cancels" the inward centripetal force (a "real" force that exists in all frames). But unless you are studying non-inertial reference frames, I would skip it.
  4. Jul 15, 2016 #3
    Thank you.Now, I've become quite clearer in this topic.Hope that I'll be able to solve many problems in this regard more efficiently.
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