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Upward force on a object spinning on on string

  1. Jul 24, 2013 #1
    Hi everyone,
    I was wondering what forces would counteract the force of gravity on an object which is spinning horizontally on a string. I realize its a component of the tension, but how does the force come about?
    Is this force also the reason that the object's plane of rotation rises when we start the spinning from when the object was at rest?

    Thanks,
    Ramana
     
  2. jcsd
  3. Jul 24, 2013 #2
    From what I think, yes, that is the reason the string would go up, i.e., would become slanted instead of horizontal.

    In suck a case, the vertical component of the tension would balance the weight and the horizontal component would act as the centrifugal force required for rotation.
     
  4. Jul 24, 2013 #3
    Thanks for the reply...
    What I was wondering was how a vertical component could exist for a horizontal force... Anything I'm missing here?
     
  5. Jul 24, 2013 #4
    I just realised something... Is it that as soon as the rotating plane tilts away from the horizontal, the vertical component "steps in" to ensure that the plane is set back?
     
  6. Jul 24, 2013 #5

    HallsofIvy

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    Staff Emeritus
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    An object "spinning on a vertical string" has NO vertical component. The force on the string is just the weight of the object. IF you mean an object moving in a horizontal circle, held by a string from a point above the center of the circle, then the object does NOT move outward on its own. There must have been a additional force initially to move the object horizontally. The vertical component is still the weight of the object.
     
  7. Jul 24, 2013 #6
    I believe there is some confusion... I am sorry if I wasn't clear.
    The situation is a uniform horizontal circular motion (parallel to a level ground). I was wondering why the plane of the circle stays at that level and does not drop, despite the obvious force of gravity 'pulling down' on the object. At this point, the vertical component of the tension on the string is zero, as there simply is no vertical component.
    I believe Siddharth answered the question by saying that at the point where the plane of the circle dips, the vertical component of the tension forces the plane back to its original position. This does seem logical to me... But I am not entirely sure..

    Thanks...
     
  8. Jul 24, 2013 #7

    jbriggs444

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    It is not that the plane of rotation rises up so that a rock-on-a-string is level with the hand that is swinging it. It is a matter of equilibrium.

    If the plane of rotation is far below the hand, the vertical component of tension will tend to cause it to rise.

    If the plane of rotation is level with the hand, gravity will tend to cause it to fall.

    Somewhere in between is an equuilibrium level where the force of gravity and the vertical component of tension are in balance. No matter how fast you spin the rock, this level will be somewhere below the hand.
     
  9. Jul 24, 2013 #8

    cjl

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    This is the part that is wrong. There is a vertical component. You will never get the string perfectly horizontal, at least not in a steady-state condition. The string will be slightly angled downwards, and the vertical component of the tension will balance the weight of the object.
     
  10. Jul 24, 2013 #9
    Ok... I understand...
    Thanks a lot guys... :D
     
  11. Jul 25, 2013 #10
    Exactly. It will never remain in the horizontal plane.
    Cheers.
     
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