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Uniform Circular Motion; bead on a string.

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    An 100g bead is free to slide along an 80cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40cm apart. When the pole is rotated, AB becomes horizontal.
    a) find the tension in the string
    b) find the speed of the bead at B.

    in the diagram, A is the point the string is attached at the bottom of the pole, B is where the bead is (horizontally in line with A), and C is where the string is attached at the top of the pole (40cm up from A).

    2. Relevant equations
    F=ma, a=v^2/r


    3. The attempt at a solution
    For a free body diagram I have drawn the forces acting on the bead; the weight of the bead (mg) acting vertically downwards, the y component of Ftens acting upwards on the bead, and the x component of Ftens acting horizontally towards the pole.
    I have figured that since the bead rotates in a horizontal plane that mg = Ftensy
    therefore: (0.1kg)(9.8N) = 0.98N = Ftensy

    I have also figured by the ole 3-4-5 rule that the horizontal part of the string (aka the radius) is 30cm and that the hypotenuse of the triangle is 50cm. Via trig, the angle at point C is 36.9o and the angle at point B (aka the bead) is 53.1o.

    I am stuck now, as I think the next step should be to figure out the radial acceleration via the net force, but I am unsure as to how to figure out the value for Ftensx...which, if Ftensy = mg (but in opposition), would be Fnet...?

    Note: I did find a couple of posts on this in the archive but was still unclear after reading through them, so any help is much appreciated!!
     
  2. jcsd
  3. Nov 29, 2009 #2

    cepheid

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    Hi ourheroine, welcome to PF!

    I have a question. If I am visualizing your set up correctly, then I understand that the segment of the string labelled AB is horizontal and therefore its tension contributes to the centripetal force. But isn't there also a horizontal component to the tension in the other segment of the string (BC) i.e. the hypotenuse of the triangle? After all, it is pulling both upward and (radially) inward.
     
  4. Nov 29, 2009 #3

    rl.bhat

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    Hi ourheroine, welcome to PF.
    If T is the tension in the string segment BC, then T*sinθ balances mg and T*cosθ adds to the centripetal force. Write down two equations and solve for T and v.
     
  5. Nov 29, 2009 #4

    cepheid

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    What about the tension in segment AB? It does nothing?
     
  6. Nov 29, 2009 #5
    Thanks for the welcome! :)

    Checking to see how far off I am...I'm getting a tension of 2.0N in the (total) string?
     
  7. Nov 29, 2009 #6

    cepheid

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    I get less than that. Can you post your calculation steps?
     
  8. Nov 29, 2009 #7
    Sure thing!

    Ftens*sin(theta) = mg
    Ftens = (0.1kg)(9.8N) / sin53.1o
    Ftens = 1.2255N
    So this would be the tension exerted by the portion of the rope BC

    Ftens*cos(theta) = Ftensx
    1.2255N*cos53.1o = Ftensx
    0.7358 = Ftensx

    I'm assuming where I went wrong in all this is because, under theory of the string's portion AB exerting force on the bead, I then added Ftensx to Ftens...:S Which really doesn't make a whole lot of sense but shouldn't that portion of the string do some pulling too??
     
  9. Nov 29, 2009 #8

    cepheid

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    Yeah, that was my result as well. I was confused because I thought that what you meant was that you calculated *this* to be 2.0 N.
     
  10. Nov 29, 2009 #9

    cepheid

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    Yeah, if we assume that the tension in AB is the same as the tension in BC (and I'm not sure if that is correct), then it seems to me that AB pulls inward as well. Do you know for sure that this leads to the wrong answer?

    EDIT: because if it is right, then you have all of the info you need for part b. You know F (centripetal) and you know r, therefore you can solve for v.
     
  11. Nov 29, 2009 #10
    Sorry! I should have been more clear with my calculations straight off :)

    I don't actually have the answer for this problem, so I cant say whether or not I am approaching this the right way...from my imagination of this system I think maybe that since AB is taut (horizontal) it is pulling inwards...and my thinking is that it pulls inwards the same amount as FtensBCx pulls inwards because both are horizontal tension exerted over the same distance on the same object by the same type of string...?
     
  12. Nov 29, 2009 #11

    cepheid

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    No, I think that what you said in post #7 is correct. I.e. the whole string is at tension T. Therefore, AB pulls inward with force T, and BC pulls inward with force Tcos(theta). These two combined comprise the centripetal force. The more I think about it, the more that seems to make sense.
     
  13. Nov 29, 2009 #12
    Sorry, just to clarify;

    Whole string tension = TensBC + Tens AB
    tensile force of AB = Tcos(theta)
    tensile force of BC = T
    (if T= 1.2255N and Tcos(53.1o) is 0.7358N, then the tension in the entire string is 2.0N)

    Centripedal force = TensBCx + Tens AB
    tensile force of AB = Tcos(theta)
    tensile force of BCx = Tcos(theta)
    (so the centripedal force = 2(0.7358N) = 1.5N)
     
    Last edited: Nov 29, 2009
  14. Nov 29, 2009 #13

    rl.bhat

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    Bead is free to move in the string. It moves in a curved path. Component of weight moves it down along tangent to the curve. Component of centripetal force moves it up along the tangent to the curve. The component of weight and the component of centripetal force contribute to the tension in BC.
    In equilibrium position mg*sinθ = (m*v^2/R)*cosθ
    And tension in BC = mg*cosθ + (m*v^2/R)*sinθ
    If the bead slows down it falls. If the bead moves fast it rises up.
     
  15. Nov 29, 2009 #14
    I am unclear as to why we are taking the components of weight and centripetal force here, have you chosen the +y axis along BC?

    If I work backwards from isolating velocity I get a net force of 1.6321N, which I don't understand in terms of my beginning free body diagram using AB along the x-axis, with mg and FtensBCy along the y axis. :\
     
  16. Nov 29, 2009 #15

    rl.bhat

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    When the bead is at rest, it hangs vertically downwards. In that position tension in each segment of the string is equal and it is equal to mg/2. As the bead starts moving in the circular orbit, the centripetal force causes to move up. The tensions in the strings are caused by the weight and the centripetal force. So to find the tension in the string you have to take the components of these two forces.
    In the horizontal position of AB, there is no tension in AB due to weight. In this position tension is due to centripetal force.
     
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