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Swinging bob

  1. Oct 20, 2009 #1
    Hi!

    I was just wondering regarding this.

    When we swing a bob tied to a thread, completely perpendicular to the vertical wall (that is, parallel to the floor), what force balances the weight of the bob?

    I don't think it can be mv^2/r because that centripetal force is perpendicular to the weight of the bob (thus mv^2 cos90 = 0N).

    What force balances the weight of the bob?

    Thanks!
     
  2. jcsd
  3. Oct 20, 2009 #2

    russ_watters

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    Staff: Mentor

    No force balances it - it is accelerating straight downward due to gravity.
     
  4. Oct 20, 2009 #3
    Thanks for the reply!

    But then how does it keep moving in a "straight" circle?

    Why doesn't it become a cone shaping circle?

    Thanks!
     
  5. Oct 21, 2009 #4
    Help?

    I am really feeling very anxious...

    Could it be air molecules?
     
  6. Oct 21, 2009 #5
    Are you describing the bob moving in a horizontal circle?If so the string can never get perfectly horizontal so it is the vertical component of the tension in the string that balances the weight of the bob, the horizontal component providing the centripetal force.Suppose you whirled the bob faster and faster ,if so the bob rises and the string gets closer to the horizontal but never quite reaches because this would require an infinite speed.
     
    Last edited: Oct 21, 2009
  7. Oct 22, 2009 #6
    Thanks for the replies!

    Dadface,

    Thus, ((mv^2)/r)sinx = mg.

    Even if it is agreed that perfectly horizontal isn't possible (till the speed is infinity), when we swing the bob, it does seem to be close to the eye. Thus, x is small (4-5 degrees at the most?). Thus, isn't x too small for sinx to balance mg?
     
  8. Oct 22, 2009 #7
    Hello newtonrulez,try the following:
    1.Draw a diagram of the arrangement when the string makes an angle theta to the horizontal.
    2.Mark in the two forces on the bob,the tension,T,which acts along the line of the string and the weight,mg,which acts vertically downwards.
    3.Resolve vertically from which Tsin theta=mg(it is T sin theta that supports the weight)
    4.Horizontally there is an unbalanced force which provides the centripetal force.We can write: Tcos theta=mvsquared/r
    5.Divide equation 3. by equation 4.This gives tan theta=gr/v squared.
    For the string to become perfectly horizontal tan theta must become equal to zero and from the equation v must become infinite or r become zero, both of these being impossible.If the bob is whirled very fast it may seem as if the string is horizontal but it never quite gets there.It is true that sin theta gets smaller as the speed is increased but T gets bigger.
     
  9. Oct 22, 2009 #8
    Thanks!!

    Wow, it all seems to fall into place now.
     
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