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Merry-Go-Round physics homework

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data

    A merry-go-round in a playground has a radius of about 2 m and a mass of about 400 kg. Starting from rest, a parent begins to rotate it with constant angular acceleration. After 5 s, it reaches its final angular velocity of about pi/2 rad/s, and continues to rotate at that angular velocity

    a.
    Treat the merry-go-round as a uniform disk (the moment of inertia of a uniform disk is mr2/2); the final rotational kinetic energy of the merry-go-round is about ___ J.

    b.
    5 children, each with a mass of about 20 kg, step on the merry-go-round and stay near its edge. As they step on it, its angular velocity drops to about ____ rad/s. Ignore friction.



    2. Relevant equations

    I_disk=0.5mr^2
    KE_rotational=0.5*I*w^2


    3. The attempt at a solution

    For part a I did...

    KE=(0.5)(0.5)(400)(2^2)(pi/2)^2=987 J

    For part b I did...

    KE_i=KE_f

    987 = 0.5(100)(2^2)(w^2) + 0.5(0.5)(400)(2^2)(w^2)

    and get w=1.3 but the answer is supposed to be 1.0. Is mr^2 the correct way to do the kid's intertia?
     
    Last edited: Mar 5, 2007
  2. jcsd
  3. Mar 5, 2007 #2
    Why do you think that kinetic energy will remain the same?

    What physical quantity you are sure that will remain constant? Maybe angular momentum, who knows!?
     
  4. Mar 5, 2007 #3
    Yes, think about what gyro is stating.
    Also, your KE eqn is incorrect, there should be an omega squared. You seemed to take this into account in your soln. though.*edit* it is fixed now
     
    Last edited: Mar 5, 2007
  5. Mar 5, 2007 #4
    Used conservation of angular momemtum and got the correct answer. Though, why is KE not conserved?

    Thanks,
    Brandon
     
  6. Mar 5, 2007 #5
    When the kids jump to the merry-go-round they exert forces on it. Forces cause change in kinetic energy, do not they?
     
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