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Inertia confusion

  1. Oct 15, 2003 #1
    Thanks to gnome for answering my last physic question. However, I am still confused about inertia. Here is my next question maybe someone can alleviate my confusion:


    Q: A merry-go-round (a piece of playground equipment)consists of a disk 10 ft in diameter that weighs 250 lbs. The disk turns on a low-friction bearing.

    A kid who weighs 100 lbs sits on the edge of the merry-go-round. The merry-go-round is turning at 1 rev every 3 seconds. If the kid moves to the center of the merry-go-round, how fast will the merry-go-round be turning?

    A: I know that when the kid moves to the center the merry-go round is going to move at a faster speed. I am not sure how to figure it mathematically. I am confused how to calculate the inertia when the kid is on the outer part of the merry-go-round and when the kid moves toward the center.

    5ft = 1.52ft; 250lb = 113.5kg; 100lb = 45.39kg; w = .667rad/s
    I think the inertia of a disk is I =½ m r^2, I do not know how to account for the kid when is on the outside and when he is on the inside.

    Then I was going to set the KE equations when the kid is on the outside equal when the kid is in the middle to find the speed.
    KE(kid on outside) = KE(kid on inside)
    ½ I(kid on outside) (.667rad/s)^2 = ½ I(kid on inside) w^2
     
  2. jcsd
  3. Oct 15, 2003 #2
    Seems OK sofar. Now what is I of the total system: disk + kid on the outside and the total I of the disk + kid in the centre. If the kid is sitting in the centre then obviously there is much 'r' left. :smile:
     
  4. Oct 15, 2003 #3
    Andre,

    I am sorry I still don’t understand. Did you mean that when the kid is in the center that you do not calculate radius in the inertia equation (I=½mr^2), or do you not square the radius? Thank you for your quick response!!!
     
  5. Oct 15, 2003 #4
    No, and I take back the "seems to be alright". let's focus on the kid first. What would be his I, when sitting on the edge of the disk? Think of a mass concentrated in one point. And what would be his turning energy over there when the merry go round goes round?

    And incidentally perhaps recheck your formulas. What is the definition of I? the turning momentum and what is the formula for energy of a turning mass?
     
  6. Oct 15, 2003 #5
    Inertia with kid on side = 45.39kg (1.52m)^2 =104.8kgm^2
    Inertia with kid in the middle = 158.9 (1.52m)^2 = 367kgm^2

    Is that right?

    So, you think it would be better to use the momentum equation L = I*omega or to use the KE=½*I*omega^2

    I am sorry about my confusion.
     
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