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Angular velocity from diving board

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.34 complete rotations in 1.23 s. If his rotational inertia decreases by a factor of 2.56 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?


    2. Relevant equations
    I1W1=I2W2


    3. The attempt at a solution
    I dont know how to solve the problem using the equation, Im not sure how to plug in the information.
     
  2. jcsd
  3. Nov 2, 2011 #2

    cepheid

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    You are given that I1/I2 = 2.56 (where I2 is after the tuck, in the more compact position).

    You can use the angular distance Δθ and time interval Δt that are given to calculate the angular speed ω2 of the diver during the tuck.
     
  4. Nov 3, 2011 #3
    so I do (2.56)W1=W2


    W1 is V/R
    How do I do that part?
     
  5. Nov 3, 2011 #4

    cepheid

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    Again, just like I said before, you can solve for ω2 because you are given angular distance and time, and therefore you can find the angular speed. Once you know ω2, you can use the conservation of angular momentum equation that you wrote above to find ω1, which is what the problem is asking for.
     
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