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Whirling stone - Tension/Velocity

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    A stone of mass 500g is whirled in a vertical circle at a constant speed of 4m/s on a string of length 75cm.

    a) At what point in the circle is the tension in the string maximum and what is its value?
    b) At what point in the circle is the tension in the string minimum and what is its value?
    c) At what minimum speed shoud the stone be rotated if the string is to remain taut throughout its motion?


    2. Relevant equations
    F=Tcosθ=mv2/r
    Tsinθ=mg
    v=ωr


    3. The attempt at a solution

    I am really bad at things whirling around type questions. I must be missing some vital understanding :uhh:
    I have to find maximum and minimum tensions. Considering gravity acting downwards vertically, I would say that the maximum tension is when the stone is moving downwards and the opposite for minimum.
    I think I need to consider centripetal force but don't have the radius. This is where I always get stuck:confused:

    Please help
     
  2. jcsd
  3. Feb 7, 2010 #2

    kuruman

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    Say again, you don't have the radius?
     
  4. Feb 13, 2010 #3
    a) The max. Tension will be at the bottom of the circle and is max. there because of the effect of the object moving in the same direction as gravity, so gravity must be included in the calculation.
    T=(mv2/r) + mg


    b)The min. Tension will be at the top of the circle and is min. there because of the effect of the object moving in opposite direction to gravity, so gravity must be excluded in the calculation.
    T=(mv2/r) - mg

    c) v2min=rg

    But I still don't see the radius here. The 75cm is the hypotenuse, and I don't have an angle to calculate the radius nor another side neither:confused:
     
  5. Feb 13, 2010 #4

    kuruman

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    Try this.

    1. Get yourself a string.
    2. Tie a stone (or other object) at the end of the string.
    3. Cut the string so that its length from the stone (or object) to the other end is 75 cm.
    4. Whirl the string in a vertical circle.

    What is the radius of the circle?
     
  6. Feb 13, 2010 #5
    Kuruman. I think you taking the mickey out of me now. That is not a doable experiment.
    So you are saying that the length of the string is the same as the radius of the circle formed that a whirling stone/pencil/chalk piece attached to it would make in the air when it is whirled?
     
  7. Feb 13, 2010 #6
    Not exactly, but close. It depends on how you swing it,
     
  8. Feb 13, 2010 #7

    kuruman

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    That is what the problem wants you to say. It gives you the radius of the circle as 75 cm.
     
  9. Feb 14, 2010 #8
    Wow! Why does that seem so improbable to me?

    a) The max. Tension will be at the bottom of the circle and is max. there because of the effect of the object moving in the same direction as gravity, so gravity must be included in the calculation.
    T=(mv2/r) + mg
    (0.5x42)/0.75+(0.5x10)=15.67N


    b)The min. Tension will be at the top of the circle and is min. there because of the effect of the object moving in opposite direction to gravity, so gravity must be excluded in the calculation.
    T=(mv2/r) - mg
    (0.5x42/0.75-(0.5x10))=5.67N

    c) v2min=rg
    √0.75x10=8.7m/s (2s.f.)
     
  10. Feb 14, 2010 #9

    PhanthomJay

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    This appears correct, however, you seemed to have used the concept of centrifugal or pseudo forces, which can often get you into trouble. At the bottom of the circle, the tension force acts opposite to the gravity force, and the object is accelerating in the opposite direction of gravity. Thus per newton 2 , T -mg =mv^2/r. At the top of the circle, gravity and the tension force both act in the same direction, and the object is accelerating in the same direction as gravity, thus, per newton 2, T +mg =mv^2/r. This is the same result as your method of approach, but just be sure to understand what is happening. You can solve it with either approach, just be careful ....
     
  11. Feb 14, 2010 #10
    So, my calculations are correct but I should say that:
    At the bottom the tension force is acting opposite to weight so will be subtracted from the tension force.
    At the top the tension force is acting in the same direction with weight, and therefore should be added to the tension force.
    That seems a lot clearer.
    thank you PJ
     
  12. Feb 14, 2010 #11
    Just out of interest I would like to know how a stone on a string can be whirled in a vertical circle with a steady speed.Did the question compiler forget the PE to KE changes of the stone and the fact that it tends to speed up as it goes down and slow down as it goes up?
     
  13. Feb 14, 2010 #12
    He has explained during a lecture that it doesn't actually go at a steady speed. Although the question didn't say it, I think we were supposed to assume it just for the purpose of this question.
    I believe him to be a very good lecturer.
    But thank you for the point.
     
  14. Feb 14, 2010 #13
    Hello lemon,
    I knew you were supposed to assume it goes at a steady speed.It is in fact a fairly common question and I have seen it crop up in text books.I'm glad your lecturer explained that the speed is not really steady,the books I have seen seem to overlook this.
     
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