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Horizontal Circular Motion-Tension, Speed- Quick Help Needed

  1. Feb 28, 2005 #1
    Can anyone give me some clues as how to do this problem. I have been on it for ever and am stumped. Any help greatly appreciated. Thanks

    Two wires are tied to the 2.0 kg sphere shown in the figure . The sphere revolves in a horizontal circle at constant speed.

    For what speed is the tension the same in both wires?(in m/s)
    What is the tension?
     

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  2. jcsd
  3. Feb 28, 2005 #2

    Andrew Mason

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    I posted an outline for the solution to this same problem a few weeks ago. See: https://www.physicsforums.com/showthread.php?t=64510

    Here are the details then:

    Let the top wire = L1 with tension T; the other is L2, T:

    [tex]Tcos(30) + Tcos(60) = mg[/tex]

    (1)[tex]T = 2*9.8/1.366 = 14.35 N[/tex]

    (2)[tex]Tsin(30) + Tsin(60) = m\omega^2r[/tex]

    (3)[tex]L_1cos(30) - L_2cos(60) = 1[/tex] (ie. 1 m.)

    [tex]L_1sin(30) = L_2sin(60) = r[/tex]

    So:
    [tex]L_2 = L_1sin(30)/sin(60)[/tex]

    Substitute for L_2 in (3):

    [tex]L_1(cos(30) - sin(30)cos(60)/sin(60)) = 1[/tex]

    [tex]L_1 = 1/(.866 - .5 * .5 /.866) = 1.54 m[/tex]

    [tex]r = .77m[/tex]

    I'll leave the rest to you to work out

    AM
     
  4. Feb 28, 2005 #3
    OK for the triangle give to you at first, find all teh angles and hence find all the sides

    Now for the ball, what forces are acting on it? (one starts with C and the other with G)

    G points straight down (as always), and C is 90 degrees to g , and points left

    Draw these two vectors.
    Now you have two triangles... (although it sounds more complicated it isnt)
    Now write the force equations (considering the components of force points up and down, and left and right separately). You should have no trouble hereafter
     
  5. Feb 28, 2005 #4
    Thanks to both of you. I got the answer correct
     
  6. Oct 15, 2011 #5
    Would it be for the same concept for the diagram below?
     

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