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Another Merry Go Round Question

  1. Jun 6, 2007 #1
    A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.615 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 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.

    (b) Calculate the initial and final kinetic energies for this system.
    Ki =
    Kf =

    So I am stuck on Kf, but for Ki i thought it should just be the sum of the kinetic energy of the man plus the kinetic energy of the disk

    I set up my equation like this:

    0.5mv^2(KE of person) + 0.5Iw^2(KE of disk) =
    0.5(59.4)(3.99^2) + (0.5)(0.5(155)(2.63^2))(0.615^2)

    I solved for this and did not get the correct answer which means I have probably set up the equation wrong...can anyone help me out with this?
  2. jcsd
  3. Jun 6, 2007 #2
    not sure if this is correct or not but i think you have to use angular momentum and then combine it with the energy formula. and solve for the unknown. so u have to use two equations on this one. whats the answer ur suppose to get?
  4. Jun 7, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Again, use correct units for angular speed: radians/sec.
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