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Uniform circular motion

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A 100 g bead is free to slide along an 80 cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated
    about its axis, AB becomes horizontal.
    (a) Find the tension in the string.
    (b) Find the speed of the bead at B.


    2. Relevant equations



    3. The attempt at a solution

    I am just wondering how to find the length of AB and BC without knowing any of the angles of the triangle besides the right angle one. can anyone help me figure that out?
     
  2. jcsd
  3. Feb 24, 2009 #2

    LowlyPion

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    It says that AB is horizontal and apparently AC is vertical.

    So BC would be the hypotenuse apparently without seeing your drawing.

    You know what AC and you know what the others add to.

    Think Pythagoras.
     
  4. Feb 24, 2009 #3
    does this make sense:
    BC = 80-AB
    then
    80-AB = 40 + AB
    40 = 2AB
    AB = 20
     
  5. Feb 24, 2009 #4

    LowlyPion

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    Does that satisfy the Pythagorean relationship?
     
  6. Feb 24, 2009 #5
    if i cancel out the squares than yes but i dont know if i can do that or not.
     
  7. Feb 24, 2009 #6

    LowlyPion

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    It is a right triangle.

    402 + AB2 = BC2

    AB + BC = 80

    Just solve.
     
  8. Feb 24, 2009 #7
    ok so i got AB = 30 cm
    so i know that i need to break the tension into x and y components.
    Fy = mg = 0.1 kg(9.8) = 0.98 N
    Fx = m (v^2/r), but we dont have the velocity. i know that v = 2(pi)r/T, but i dont know the T. what can i do?
     
  9. Feb 24, 2009 #8

    LowlyPion

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    What must the vertical component of the Tension be?
     
  10. Feb 24, 2009 #9
    the vertical component is equal to the mass of the bead x gravity
    Ty = mg = 0.1 (9.8)
    = 0.98 N
     
  11. Feb 24, 2009 #10

    LowlyPion

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    And you know the cosθ because you know the ratio directly from 40 cm/50 cm.

    So what's holding you back from the answer for T?
     
  12. Feb 24, 2009 #11
    so i can do cos 36.9 = 0.98/Ft
    Ft = 1.23 N ?
    but dont i have to account for the centripetal acceleration at all for finding the tension in the string?
    also, is there a second string force acting out from the horizontal because it is one peice of string?
     
  13. Feb 25, 2009 #12
    can someone please tell me if this is correct or not?
    is there two tensions to the string since the string is attached at two points on the pole and there are two parts of the string divded by the bead.
     
  14. Feb 25, 2009 #13

    LowlyPion

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    The string is free to move in the bead, so the string isn't accelerating relative to the bead. The Tension in the string then from the top = the tension horizontally. And sure the centripetal acceleration is translated outward and is part of the tension. But they don't give you ω, so I dare say you would have to figure it already from the Tension from the m*g anyway.

    As to the numbers, if you are satisfied you have the right value for Cos, then ... I'm not doing your math for you.
     
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