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Conical Pendulum

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a conical pendulum with a 78.0 kg bob on a 10.0 m wire making an angle of θ = 5.00° with the vertical. (Consider positive i to be towards the center of the circular path.)

    (a) Determine the horizontal and vertical components of the force exerted by the wire on the pendulum.
    ________N i + _________N j


    (b) What is the radial acceleration of the bob?
    ________m/s^2


    2. Relevant equations

    a = v^2 / R


    3. The attempt at a solution

    I found the vertical component of the force:
    [tex]\Sigma[/tex]Fy = Tcos(5) - 78(9.8)

    T = 78(9.8)/Cos(5)

    T = 761.491 N

    To find the horizontal component wouldn't you need to find the acceleration(part b) first?

    I used the following to find the velocity:

    [tex]\Sigma[/tex]F(radial) = -Tsin(5) = -m(v^2 / R)

    v^2 = R*Tsin(5) / m

    using trig and the given 10m wire length i found the radius to be .871557m

    v^2 = .871557(761.491)(sin(5)) / 78

    v = 7.60551 m/s

    Then i plugged that into the acceleration equation given in the relevant equations section but it was 66.3683 m/s^2 which does not sound reasonable for this problem.

    Any suggestions on what i did wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2007 #2

    Astronuc

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    Staff: Mentor

    Going back to v^2 = R*Tsin(5) / m

    to find angular acceleration, one simply takes v2/r = T sin(5)/m

    or 762 N * (0.08716) / 78 kg = __________
     
  4. Oct 14, 2007 #3
    Ok, I found the acceleration to be .851m/s^2. Now, to find the horizontal component of the force i would set that acceleration equal to T*Sin(5)?

    Tsin(5) = a
    T = .851 / sin(5)
    T = 9.764 N ?
     
  5. Oct 14, 2007 #4

    Astronuc

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    Staff: Mentor

    The tension was correctly calculated.

    With a = v2/r = T sin(5)/m

    one obtains 762 N * (0.08716) / 78 kg = 66.42 N/ 78 kg = 0.85 m/s2

    I think in the OP, one simply forgot to divide by 78.
     
  6. Oct 14, 2007 #5
    I entered 9.764 N for the horizontal component of the tension force and it said it was incorrect.
     
  7. Oct 14, 2007 #6

    Astronuc

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    Staff: Mentor

    is not correct. T is a force, a is an acceleration. There has to be a mass associated with a.

    Let T = 762 N and the horizontal force is T sin(5).
     
  8. Oct 14, 2007 #7

    Doc Al

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    Staff: Mentor

    You calculated the tension correctly in your first post. Just find the horizontal component of that, since you have the angle.

    You don't need to use the centripetal acceleration formula for this problem.
     
  9. Oct 14, 2007 #8
    so, the horizontal component of 761.491 N would be sin(5)*761.491 which equals 66.3683 N
     
  10. Oct 14, 2007 #9

    Astronuc

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    Staff: Mentor

    Yes.
     
  11. Oct 14, 2007 #10
    Alright, I get it now. Thanks for your help Astronuc and Doc Al!
     
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