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Angular Speed and Energy

  1. Nov 14, 2008 #1
    A 3.0-m-diameter merry-go-round with rotational inertia 120 kg*m2 is spinning freely at 0.60 rev/s. Four 25-kg children sit suddenly on the edge of the merry-go-round. (a) Find the new angular speed in rev/s. (b) Determine the total energy lost to friction between the children and the merry-go-round in J.


    Okay so I know that Li[tex]\omega[/tex]i=Lf[tex]\omega[/tex]f. That means that (120 kg*m^2)(0.60 rev/s)=(mass of merry-go-round + 4(25kg))(1.52)[tex]\omega[/tex]f.

    To find the mass of the merry-go-round I took the Inertia and divided it by R2 to get m = (120 kg*m2)/(1.52) = 53.3.

    When I plugged it into the equation I found the new angular speed to be .21 rev/s.

    My problem is with part (b)...

    So I know that J is (kg*m2)/s2, but I am unsure as to how to solve for it...A little help, please?
  2. jcsd
  3. Nov 14, 2008 #2
    Find the angular velocity before the children get on, then find rotational energy use (1/2)I(omega^2).

    Then find the angular velocity after the children get on. Take initial rotational energy - final rotational energy. The energy loss went to friction.
  4. Nov 14, 2008 #3


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    While you got the correct answer for (a), you did make two errors in the process ... which fortunately cancelled out.

    First, let's clear up those equations:

    Li = Lf
    Imerryωi = (Imerry+Ichildrenf

    It's not necessary to find the mass of the merry-go-round, you can just use the Imerry value here to find ωf.

    Actually, this is 0.5*m, since I = 0.5 m R^2. But that's okay, since 0.5*m is what should have been used in your earlier expression,
    (mass of merry-go-round + 4(25kg))

    So the answer is 0.21 rev/s as you got.
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