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Understanding a book example about centripetal force

  1. May 7, 2017 #1
    1. The problem statement, all variables and given/known data
    f871aca6837c4aa88848c3bfbc284ad8.png
    2ac4c1bcf7f24af7b7d5a36450d2ad50.png
    If the centripetal force is at the left side of the equation; that means if we move it over to the other side, it'll have a negative sign, which means it is opposite in sense in relation to [itex]N[/itex] and [itex]mg[/itex]. But how is that possible, considering that the centripetal force always points to the center of the circumference?

    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2017 #2

    kuruman

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    The centripetal force is the force that provides the centripetal acceleration. In this case, it is the vector sum of ##\vec{n}_{bot ~or ~top} +m\vec{g}##. Note that it points in the same direction as the centripetal acceleration in both diagrams which is all that is required of it.
     
  4. May 7, 2017 #3

    haruspex

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    Centripetal force is not an applied force, like N and mg are. It is the radial component of the resultant force.
    If you shift it to the other side and combine it with real applied forces as though it is another such then in effect you are switching to a non-inertial view, replacing the centripetal force with a centrifugal one. That acts away from the centre.
     
  5. May 7, 2017 #4

    FactChecker

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    Since gravitational force is down and it has a positive sign, down is positive. The acceleration on the left side of the equation is downward when at the top of the loop, so it needs to be positive. The rest follows.
     
  6. May 7, 2017 #5

    haruspex

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    The first quoted equation has ΣFr=mv2/r, so I would say the book treats it as centripetal at that point. But see post #3.
     
  7. May 7, 2017 #6

    FactChecker

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    Yes. Sorry. I was correcting my post as you responded to it..
     
  8. May 9, 2017 #7
    OK. I understand it now. Thanks for the inputs.
     
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