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Conservation of Angular Momentum of merry-go-round

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A merry-go-round of radius 2m has a moment of inertia 250kg*m^2, and is rotating at 10rpm on a frictionless axle. Facing the axle and initially at rest, a 25kg child hops on the edge of the merry-go-round and manages to hold on. What will be the new angular velocity of the merry-go-round after the child jumps on?

    2. Relevant equations

    Li (system) = Lf (system)

    3. The attempt at a solution

    Let g = merry-go-round and c = child.

    Li = Lf, so (IW)g-initial = (IW)g+c-final.

    Solving for Ic:

    Ic = mr^2 = (25kg)(2m)^2 = 100kg*m^s

    Solving for Ig+c:

    Ig+c = Ig + Ic = 350kg*m^s

    Solving for final angular velocity of system:

    W(g+c)-final = (IW)g-initial / Ig+c = [(250kg*m^2)(10rpm)] / 350kg*m^s = 7.14rpm.


    My question is, should I be treating the child as a point-mass, allowing me to use I = mr^2 -- or should I approximate its shape in determining I? (Treating it as a cylinder - I = .5mr^2 - I get 8.33rpm.)

    Any insight into this will be much appreciated!
     
  2. jcsd
  3. Nov 30, 2013 #2

    tiny-tim

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    hi nx01! :smile:
    yes :smile:
    no, it would be I = .5mr2 + mR2

    where r and R are the radius of the cylinder and of the merry-go-round, respectively :wink:
     
  4. Nov 30, 2013 #3
    Thank you!
     
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