How Does a Man's Movement Affect the Angular Speed of a Merry-Go-Round?

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Homework Help Overview

The problem involves a merry-go-round rotating at a specific angular speed with a man moving from a point away from the center to the center. The context includes concepts of angular speed, moment of inertia, and conservation of angular momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the man's movement on angular speed and moment of inertia, questioning whether the problem contains excess information. There is also consideration of the relationship between linear velocity and angular velocity.

Discussion Status

Participants are exploring the conservation of angular momentum and its effects on the system as the man moves. Some have raised questions about the treatment of the man as a particle and the implications for moment of inertia and angular speed.

Contextual Notes

There is uncertainty regarding the treatment of the man's moment of inertia as he moves to the center, and whether this affects the overall angular momentum of the system. The follow-up question regarding kinetic energy is noted but not the focus of the current discussion.

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Homework Statement




A merry-go-round rotates at the rate of
0.12 rev/s with an 86 kg man standing at
a point 1.4 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 45 kg
cylinder of radius of 1.4 m.
Answer in units of rad/s

there's a follow up question:
What is the change in kinetic energy due to
this movement?
Answer in units of J

but don't really worry about that right now.

Homework Equations


v = rw ?


The Attempt at a Solution


Is it just me being dumb or is there a lot of excess information/ kind of a trick question? The angular speed of the merry go round would still be 0.12 rev/sec (need to convert to rad). just the linear velocity changes with the radius... I think? maybe some of that other information has to do with the change in energy question?

I just thought about the follow up question for a minute, maybe I know more than I think, but I'm not confident/have been getting these wrong.
>would you use V= rw to get the linear velocity and then plug that into 1/2mv^2 , and when the radius is 0 the energy goes to 0 since v=rw?
 
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Angular momentum is conserved.

As the man walks to the centre the total moment of inertia about the centre is changing. Hence, by the principle of conservation of angular momentum, the angular velocity will also change.
 
so if L= Iw and I = 1/2 MR^2
then you get L = (0.5) (131) (1.4) (0.7539822).
but wouldn't w need to go to infinite if R is 0?
 
Note that the system consists of the merry-g-round wheel TOGETHER with the man!

Hence altough the moment of inertia of the wheel remains the same value , yet the moment of inertia of the man is changing.

Now I am at a loss what to do!

Because if the man were to be considered as a particle (!) his new moment of inertia about the centre of the wheel would be 0. But can one consider the man as a particle?
 
grzz said:
Note that the system consists of the merry-g-round wheel TOGETHER with the man! Hence altough the moment of inertia of the wheel remains the same value , yet the moment of inertia of the man is changing.

Because if the man were to be considered as a particle (!) his new moment of inertia about the centre of the wheel would be 0.
The total angular momentum is the sum of the angular momentum of the wheel and the man. If the man walks to the center of the wheel and can be considered a particle (or a point mass), then all of the angular momentum is in the wheel. The total angular momentum continues to remain constant.
 
Last edited:

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