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Angular momentum and a skateboarder's halfpipe

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A skate boarder with height 1.5m and weight 60 kilograms
    drops down a half-pipe with a radius of 4 meters using her
    frictionless skateboard. At first, she carries her center of
    gravity really low on the skateboard, say 1/4 her height.
    At the bottom of the half-pipe, she pushes her center of
    gravity up to 1/2 her height. She then continues up the other
    side of the pipe. How much air does she catch (above the other side
    of the half-pipe)?


    2. Relevant equations
    1/2mv2 + mgh = 1/2mv2 + mgh + Wapp
    L = I Ω (omega)
    Ω = v/r
    I = m r
    L = mr (v /r ) = mr (v /r )
    r1v1 = r2v2

    3. The attempt at a solution
    what do i start with to derive to the following equation?!

    d = (√(2g (r + h/4)) ( r-h/4)/(r-h/2))^2/ 2g
    d = (r + h/4) (( r-h/4)/(r-h/2))^2
    d = 4m + (1.5m/4)((4m-(1.5m/4))/(4m-(1.5m/2)))^2/(2 (9.81m/s^2))
     
  2. jcsd
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