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Conservation of angular momemtum

  1. Nov 1, 2003 #1
    Seems easy enough - but could I get a check on this.


    A 1.8 m radius merry go round has a mass of 120 kg and is rotating with an angular velocity of .5 rev/s.

    What is its angular velocity after a 22 kg childs gets on its edge which was initially at rest???

    First, I determined that this was an Inelastic collision with conservation of momentum


    Iw1 + Iw2 = Iw (final)

    I for a disc was determined to be (1/2)MR^2

    I converted the .5rev/s to 3.14rad/s

    So the work is as follows:

    (1/2)(120 kg * 1.8m^2)(3.14 rad/s) + 0 = ((22kg + 120 kg) * 1.8m^2)W(final)

    so w(final) = 1.33 rad/s

    Thanks again
  2. jcsd
  3. Nov 1, 2003 #2
    I think the equation should read
    (1/2)*120 kg*(1.8m)^2*(3.14 rad/s) + 0 = ((1/2)*120 kg*(1.8m)^2 + 22 kg*(1.8m)^2)W(final)

    Remember, your final I is a sum of the merry-go-round's (whose mass is distributed over a disk), and the child's (whose mass isn't.)
  4. Nov 2, 2003 #3
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