Question
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.
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Final Answer

2.30 rad/s2.30 \textrm{ rad/s}

Solution video

OpenStax College Physics, Chapter 10, Problem 39 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We have a merry-go-round that's rotating at an angular velocity of 3.1416 radians per second. That's the result of converting half of a revolution per second into radians per second by multiplying by two pi radians in every revolution. Revolutions cancel there leaving us with radians per second. A child is going to go on to the merry-go-round, right on the edge and so the after picture looks like this, the merry-go-round with the child on the edge and the merry-go-round has a certain mass and so does the child. They each contribute a term to the total moment of inertia after the child is on the merry-go-round. So initially the moment of inertia of the system is just that of the merry-go-round, and we have a formula for the moment of inertia of a disk which is mass times its radius squared divided by two. Then after the child is on the merry-go-round, we have a moment of inertia i prime which is that of the disk plus the moment of inertia of the child which we model as a point mass rotating about an axis of rotation a distance r away. So, we have angular momentum is conserved and so that means the moment of inertia multiplied by the angular velocity initially, equals the new moment of inertia multiplied by the new angular velocity after the child goes on the merry-go-round. We can rearrange this to solve for omega prime which is what this question is asking for, what is the angular velocity after the child is on the merry-go-round? We end up with omega prime is i omega over i prime and now we make substitutions for i and for i prime and then plug in numbers. So we have 120 kilogram mass of the merry-go-round, times 1.8 meter radius squared, divided by two, multiplied by 3.1416 initial angular velocity, that's in radians per second, and divide that by 120 kilograms times 1.8 meters squared over two plus the moment of inertia contribution of the child which is 22 kilograms times 1.8 meters squared. This gives 2.3 radians per second will be the angular velocity after the child is on the merry-go-round. We expected this number to be less than the initial angular velocity since we have increased the moment of inertia by this much.

Comments

Shouldnt the final momentum include the momentum of the linear velocity of the child. He starts at rest so not only is the moment of inertia changed but momentum is also taken from having to speed up the child. Am I correct in this observation?

Hello dr#, thank you for the question. Since the child is moving in a circle their momentum is modelled as "angular momentum", not linear momentum. The velocity of the child is continually changing direction, so it isn't useful to think of the velocity as "linear", although you're correct if you're thinking that their instantaneous velocity is tangential to the circle. Angular momentum is conserved independently from linear momentum. To say "momentum is taken from having to speed up the child" sounds like the focus is on the momentum of the merry-go-round - however it's better to consider the momentum of the system, which consists of both the child and merry-go-round. The angular momentum of the system is conserved, and this is the important concept of this type of question. While the angular momentum of the merry-go-round is indeed reduced, the reduction is compensated for exactly by the increase in angular momentum of the child. Having the total angular momentum of the system stay the same is what it means to say angular momentum of the system is conserved.
Hope this helps,
Shaun