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

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Homework Help Overview

The discussion revolves around the concept of inertia in the context of a merry-go-round, specifically how the position of a child affects the rotational speed of the system. The problem involves a disk-shaped merry-go-round with a specified diameter and weight, along with a child who moves from the edge to the center.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to calculate the moment of inertia for both the child on the edge and at the center of the merry-go-round. There are discussions about the formulas for inertia and kinetic energy, as well as the implications of the child's position on the system's rotational dynamics.

Discussion Status

Some participants are attempting to clarify the calculations of inertia and the relevant equations for rotational motion. There is an ongoing exploration of the definitions and formulas involved, with no clear consensus reached yet.

Contextual Notes

Participants express confusion regarding the application of the inertia formula when the child moves, and there are references to potential errors in the initial calculations. The discussion is framed within the constraints of a homework assignment, which may limit the information available for resolving the problem.

tucky
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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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