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Centripetal Force of a tied ball

  1. Nov 1, 2006 #1
    A 0.50 kg ball that is tied to the end of a 1.0 m light cord is revolved in a horizontal plane with the cord making a 30° angle, with the vertical (See Fig. P7.52.)

    I managed to get my answer down to solving for theta however I dont know how to do the math:

    I solved for T = mg / cos(x)
    The Fc is = Tsin(x) = (mg)[tan(x)]
    (mg)[tan(x)] = (mv^2) / r
    R (radius) = 1sin(x)

    The following equation is where I am stumped...
    sin^2(x) / cos(x) = 8 / 4.9
     

    Attached Files:

  2. jcsd
  3. Nov 1, 2006 #2

    rsk

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    Are you trying to find the speed?

    Calculate Fc from Tsin(x) = (mg)[tan(x)]

    Then calculate r, and you can find v.

    I don't understand why you're stumped at that point, since you know what x is.
     
  4. Nov 1, 2006 #3
    sorry for not clarifying my question. I am trying to solve for x. The question is: If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical?
     
  5. Nov 1, 2006 #4

    rsk

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    OK - sorry for delay - site wouldn't let me post

    sin^2(x) / cos(x) = 8 / 4.9

    Gives you 4.9 sin^2 x = 8 cos^2 x

    Remember sin^2 x = (1 - cos^2 x) - substitute this and you should be able to solve for cos x (might be a quadratic you have to solve)
     
  6. Nov 1, 2006 #5
    thank you so much.
     
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