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A merry go around problem

  1. Dec 14, 2009 #1
    What do I need to do with this problem?

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

    A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgm2 and is rotating at 15.0 rev/min about a frictionless vertical axle. Facing the axle, a 35.0 kg child hopes onto the merry-go-round. What is the new angular speed of the merry-go-round?
     
  2. jcsd
  3. Dec 14, 2009 #2

    DaveC426913

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    Gold Member

    Again, show your work.
     
  4. Dec 15, 2009 #3
    How does this look?

    Ok so moment of intertia =35kg(2^2)=140
    15rev/m*pi/30=1.57 rad/s

    Linitial=Lfinal
    Linitial=I*angular speed
    250(1.57)=263.12w
    392.5=263.12w
    1.49 rad/s=w

    Converting back to rev/min:
    (1.49*60)/2pi=14.23 rev/min
     
  5. Dec 15, 2009 #4
  6. Dec 15, 2009 #5
    I think this now is the correct answer.
    Final moment of inertial (if)= 250kgm^2+(35kg)(2m)^2=390kgm^2

    IiWi=IfWf
    (250)(15)=390wf
    3750=390wf
    = 9.62 rev/min
     
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