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Merry Go Round

  1. Apr 5, 2006 #1
    A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 230 kgm2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round?

    Little bit of a problem with this one. How does the moment of inertia change when the 23 kg child steps on...Do you need to use angular momentum conservation?
     
  2. jcsd
  3. Apr 5, 2006 #2

    Doc Al

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    Staff: Mentor

    Treat the child as a particle. (What's the moment of inertia of particle at a distance from an axis?) Just add the child's moment of inertia to that of the merry-go-round.

    Yes.
     
  4. Apr 5, 2006 #3
    So the moment of inertia for a particle is Mi*ri = (23)(1.2)=27.6

    So I of the system is (230+27.6)=257.6. So now we take conservation of angular momentum:

    L initial = L Final
    L Initial = I*angular speed = 230*1.05 (10 rev/min = 20pi/min = 1.05rad/s?)
    L Final = 257.6*angular speed, so we set this equal to each other

    230*1.05 = 257.6w
    w=.9375?
     
  5. Apr 5, 2006 #4

    Doc Al

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    No. Does this expression even have the right units for moment of inertia?
     
  6. Apr 5, 2006 #5
    whoops, r should be squared right...so (23)(1.2^2) = 33.12

    So 230(1.05)=263.12w
    w=,874?
     
  7. Apr 5, 2006 #6
    Right sorry I always forget..it says I am off by a power of 10, why is this
     
  8. Apr 5, 2006 #7

    Doc Al

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    Right. But be sure to include proper units when stating a physical quantity.

    Check your arithmetic. Also, why not express the new angular speed in the same units as the original?
     
  9. Apr 5, 2006 #8
    oh wow whoops so w=.917 so converting back we take (.17 * 60)/2*pi giving 8.75 rev/min. Thanks for the help
     
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