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Angular Speed of a bob of mass

  1. Dec 8, 2014 #1
    1. Relevant information

    A bob of mass 0.5kg is attached to one end of a light inextensible string of length 0.500 m, whose other end is attached to a fixed pivot. The bob performs uniform circular motion in a horizontal plane, with the string making an angle of 30.0° with the vertical.

    2. Question

    What is the angular speed of this circular motion?
    Hint: you can begin by using Newton's second law in the vertical direction to find the tension in the string.)

    3. The attempt at a solution

    Tension in string
    F Cos 30 = 0.5 x 9.81
    F = (0.5 x 9.81) / Cos 30
    F = 5.66N

    Radius of circle must be 0.5x Sin 30 = 0.25m

    F = mass x radius x angular speed^2
    Angular speed = Sqrt ( 5.66 / (0.5 x 0.25))
    Angular speed = 6.72 rad s^-1

    Have I done this right?
     
  2. jcsd
  3. Dec 8, 2014 #2

    Doc Al

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    Staff: Mentor

    Almost. Realize that the centripetal acceleration is horizontal, so only the horizontal component of the tension provides the centripetal force.
     
  4. Dec 8, 2014 #3
    I am not sure what you mean by that. Have I used a wrong equation?
     
  5. Dec 8, 2014 #4

    Doc Al

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    Staff: Mentor

    When you used ##F = m \omega^2 r##, you used the entire tension as the force. But only the horizontal component of the tension creates the centripetal acceleration.
     
  6. Dec 8, 2014 #5
    F(x) = ((m x g ) x Xcomponent / length
    F (x) = (0.5 x 9.81 x 0.5 sin 30 ) / 0.5m = 2.22 rad s^-1
     
  7. Dec 8, 2014 #6

    haruspex

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    You seem to have taken a step backwards. Your original equation for tension(F) was correct:
    Doc Al is telling you this equation is wrong (assuming F is still the tension):
    Please try to post a correct version.
     
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