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

AI Thread Summary
The discussion focuses on how a man's movement affects the angular speed of a merry-go-round, emphasizing the conservation of angular momentum. As the man moves from a radius of 1.4 m to 0 m, the total moment of inertia changes, which in turn affects the angular velocity. The initial angular speed remains at 0.12 rev/s, but as the man approaches the center, the system's dynamics shift, raising questions about the implications for kinetic energy. Participants debate whether the man can be treated as a particle and how this impacts calculations of angular momentum. The conversation highlights the complexities of rotational motion and the interplay between linear and angular velocities.
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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

theres 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.
 
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