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Real Question This Time about a Merry go Round

  1. Jun 6, 2007 #1
    A disk-shaped merry-go-round of radius 2.74 m and mass 168 kg rotates freely with an angular speed of 0.737 rev/s. A 51.7 kg person running tangential to the rim of the merry-go-round at 2.86 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round?
     
    Last edited: Jun 6, 2007
  2. jcsd
  3. Jun 6, 2007 #2
    Actually, ignore all of this...the kinetic energy is not relevant...i need to focus on angular momentum

    therefore the equation should be:
    rmv(child) + Iw = (i + mr^2)(w)

    is this correct?
     
  4. Jun 6, 2007 #3
    So, when I tried to solve this i got:

    (2.74)(51.7)(2.86) + (1/2(168(2.74^2)(0.737) = [(.5(168)(2.74^2) + (51.7 x 2.74^2)] w

    then i solved for w and got 0.8539 which is not correct

    what am i missing?
     
  5. Jun 7, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    In order for Iw to equal angular momentum, w must be given in standard units of radians/sec (not rev/sec).
     
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