Solve for wInertia Confusion: Calculating Speed of a Merry-Go-Round

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The discussion revolves around calculating the speed of a merry-go-round as a child shifts from the edge to the center. The initial parameters include a 10 ft diameter disk weighing 250 lbs and a child weighing 100 lbs, with the merry-go-round initially rotating at 1 revolution every 3 seconds. Participants clarify the concept of inertia, emphasizing the importance of calculating the moment of inertia for both the disk and the child in different positions. The equations for kinetic energy and angular momentum are suggested as methods to find the new speed when the child moves. The conversation highlights the need for accurate application of formulas related to rotational motion and inertia.
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Thanks to gnome for answering my last physics 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
 
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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:
 
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!
 
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?
 
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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