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Ball on a String

  1. Dec 28, 2009 #1
    This isn't actually a homework problem, but I thought it would be relevant to post it here since it is a problem afterall, and I need some help at the solution. Thank you everyone!

    1. The problem statement, all variables and given/known data

    There is a ball mass M hanging on a massless string length L, from a nail in the wall. When the ball is projected with initial horizontal velocity V, it swings up, but then falls back down and hits the nail. Find V in terms of M L and g.

    2. Requested Help

    I would appreciate it if someone could see if my method is problematic, and if my answer is correct if you would want to work it out.

    3. The attempt at a solution

    For ball to leave a circle and transit to a parabolic motion, Tension in string must equal to zero at point of transition. Thereafter, the parabolic motion will coincide with the nail. I have 3 equations and 3 unknowns - velocity at moment of transition, time taken for parabolic flight, and angle from vertical which transition occurs.

    Using the above, I got an angle of 65.53 deg
    I got v^2 = gl cos(angle)
    And I eliminated t.

    Finally, after using conservation of energy, I arrive at projection velocity V = Sqrt(3.24Lg)
     
  2. jcsd
  3. Dec 28, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi melancholy2! Welcome to PF! :smile:

    (have a square-root: √ and a degree: º and try using the X2 tag just above the Reply box :wink:)
    Yes, that's correct. :smile: But it's very difficult to check your final result without seeing your full calculations. :redface:
     
  4. Dec 31, 2009 #3
    Hi everyone,

    I've relooked at my workings and managed to come up with a new answer which I think is correct. Angle is now 54.7 degrees. Please see the attachment solution. Thanks!
     

    Attached Files:

  5. Dec 31, 2009 #4

    tiny-tim

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    Hi melancholy2! :smile:

    (I haven't actually checked the last 6 lines :redface:, but apart from that …)

    Yes, that looks fine. :smile:

    (except you could have saved yourself a little trouble if you'd noticed that lcosθ + lsin2θ/cosθ = l/cosθ :wink:)
     
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