# Merry Go Round conservation of angular momentum

## Homework Statement:

Why is angular momentum conserved for the following situation

## Relevant Equations:

Iwi=Iwf
So, I was reading my textbook in the section regarding net torque, and they gave an example of a seesaw with one person at each end, and they said that there is a net external torque due to the force of gravity on each person. I completely understand that; however, when I was reading another section of my textbook, they were talking about the conservation of angular momentum, which only happens when the net torque is zero. So they gave an example of a circular disk that rotates freely in a horizontal plane about a frictionless, verticle axle. On top of that, there is a person who walks slowly from the rim of the disk towards the center. The textbook modeled this problem as conservation of angular momentum problem, but why? Isn't the weight of the person causing an external torque just how the weight of the two people in the seesaw caused a net external torque?

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Filip Larsen
Gold Member

Isn't the weight of the person causing an external torque just how the weight of the two people in the seesaw caused a net external torque?
You are correct, but when you think torque you second thought should be "around what?". In this case, try think about which axis the weight is creating a torque and how that axis relates to the axis of rotation for the merry-go-round. The problem text implies a certain configuration or condition of the two mentioned axis such that the net torque on the rotation axis is zero even as the person walks around. Can you see what that condition is?

Well, I know that if torque were to be zero in this case, then the force of gravity has to be perpendicular to the axis of rotation... but I can't see it. For example, I drew the person standing straight up on the x axis (the x axis being the floor of the Merry Go Round) and I drew the y axis as the axis of rotation. Mg is parallel to the y axis, which then leads to torque due to the definition of torque.

Filip Larsen
Gold Member
Well, I know that if torque were to be zero in this case, then the force of gravity has to be perpendicular to the axis of rotation
Almost. The weight of the person would create a torque at the axle point (the origo in your diagram) around a direction that indeed is perpendicular the rotation axis (e.g. pointing out of the paper in your diagram, lets call that the z-axis). Perhaps that is what you meant.

There are two parts to understand why the torque from that weight can be ignored in the problem.

First, just like for linear momentum and force, angular momentum and torque takes 3 degrees of freedom to describe in full for an actual mechanical systems. We say that these quantities are vectors (i.e. consist of 3 numbers), as opposed to, say energy, which is a scalar (a single number). By using symmetries, actual systems can quite often be modeled in a much simpler way where only one degree of freedom is analyzed and the other two degrees of freedom (under some conditions) is assumed to be independent and therefore can be ignored.

Secondly, there is another (unstated) assumption in the problem text, namely that the rotation axis is fixed vertical. Physically this means there must be a bearing of some kind that only allow free (torque-free) rotation around the vertical axis and transports the torque around all directions perpendicular to that axis (i.e. all directions that lie in the xz-plane in your diagram) into the ground (a merry-go-round need a fairly solid ground support for this reason compared with a seesaw). With the assumption that the disc is attached to such an axle, we almost by definition have that the net torque around the axle direction is zero, and hence we have conservation of angular momentum.

Note that if some of all these assumptions used to make a simple model are not valid for a particular merry-go-round then there may very well not be (complete) conservation of angular momentum. Two likely issues that would creep up in an actual merry-go-round would be that the axle bearing has some friction meaning there would be a small frictional torque "opposing" any rotation making it slow and stop without anyone around to spin it up, and that the axis of rotation is slightly off-vertical meaning that any unbalanced weight on the disc (like a single person sitting near the edge) would have to be lifted slightly up and down at the disc rotates making an "oscillatory" variation in net torque around the rotation axis. In this later case (off-vertical axis) you can see that the weight of a person would indeed matter if the model and analysis of the mechanics has to capture this.