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Rotational Kinematics of Running

  1. Apr 24, 2006 #1
    I need some help.
    The question is:

    At the local swimming hole, a favourite trick is to run horizontally of a cliff that is 8.3m above the water. One diver runs off the edge of the cliff, tucks into a ball and rotates on the way down with an average angular speed of 1.6rev/s. Ignore air resistance and determine the number of revolutions she makes on the way down.

    Now i've done it like this but the anser is wrong.........

    ∆Ө = 1.6rev[(2 pie rad) ÷ (1 revolution)] = 10.05 radians
    Well, gravity pulls her down at 9.8m/s². So in the first second she will fall 9.8m.
    She only needs to fall 8.3m to the water.
    So 8.3m ÷ 9.8m = 0.85s
    It will take her only 0.85s to reach the water.
    And she makes 1.6rev/s.
    So 1.6 ÷ 100 = 0.016
    Then 0.016 x 85 = 1.36
    Thus she will make 1.36 revolutions on her way down.

    What am i missing?
  2. jcsd
  3. Apr 24, 2006 #2
    9.8ms^-2 means that the body will achieve the velocity of 9.81ms^-1 in a second, not 9.81m.
    I think the equation, s=ut+0.5at^2 should be used instead.
  4. Apr 24, 2006 #3


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    You are assuming her speed was a constant 9.8 m/s which is not true.
    As Harmony said, use s= ut+ 0.5at^2. Here u (initial speed) is 0 and a (acceleration) is -9.8. Since she is diving down 8.3 m, -8.3= -(0.5)(9.8)t^2.
  5. Apr 24, 2006 #4
    Thanx alot, i dont know how i could have made such a stupid mistake......

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