Max % Change in Station Angular Speed Due to People Movement

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SUMMARY

The discussion centers on calculating the maximum percentage change in angular speed of a rotating space station due to the radial movement of 500 people, each with an average mass of 70.0 kg. The space station has a radius of 82.5 m and a moment of inertia of 3.00 x 109 kg m2. Participants emphasize the importance of using the conservation of angular momentum rather than energy conservation, as energy is not conserved during the movement of individuals within the station. The correct approach involves calculating the moment of inertia at both the outer edge and the axis to determine the change in angular speed.

PREREQUISITES
  • Understanding of angular momentum conservation principles
  • Familiarity with moment of inertia calculations
  • Basic knowledge of rotational dynamics
  • Ability to perform calculations involving mass and radius
NEXT STEPS
  • Calculate the moment of inertia for the space station with people at the outer edge (r=82.5 m)
  • Calculate the moment of inertia for the space station with people at the axis (r=0 m)
  • Determine the initial and final angular speeds using the conservation of angular momentum
  • Calculate the maximum percentage change in angular speed
USEFUL FOR

Physics students, aerospace engineers, and anyone interested in the dynamics of rotating systems and artificial gravity in space environments.

franknew
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Homework Statement


A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is 82.5 m. The moment of inertia of the station without people is 3.00 x 10^9 kg m^2. Suppose 500 people, with an average mass of 70.0 kg each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station’s angular speed due to the radial movement of the people?


Homework Equations



I think it is:
mgh + 1/2mv^2 + 1/2Iw^2 = 1/2mv(f)^2 + 1/2Iw(f)^2

The Attempt at a Solution



I have an example that the teacher did that I think is sort-of the same thing using the above formula. I'm just not sure how to use it with my question.

Any help is much appreciated.
Thanks,
Frank
 
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You're using a conservation of energy equation. That has a few problems:

(1) Energy is not necessarily conserved; some energy is lost when the astronauts move along the spaceship.
(2) mgh is gravitational potential energy, but this artificial gravity is non-uniform; it doesn't have a constant g.

Instead, use the conservation of angular momentum, which always applies in the absence of outside torque. Of course, the station's moment of inertia changes when people move from the outside to the axis, and you have to take that into consideration.
 


ideasrule said:
You're using a conservation of energy equation. That has a few problems:

(1) Energy is not necessarily conserved; some energy is lost when the astronauts move along the spaceship.
(2) mgh is gravitational potential energy, but this artificial gravity is non-uniform; it doesn't have a constant g.

Instead, use the conservation of angular momentum, which always applies in the absence of outside torque. Of course, the station's moment of inertia changes when people move from the outside to the axis, and you have to take that into consideration.

So instead of using Ekin=1/2mv^2, I will use Ekin=1/2Iw^2?

I think I need to do the calculation when r=82.5m for when the people are at outer edge of the cynlindrical satellite and then do another when r=0m for when the people are at the axis. I'm just not sure how I can do this with the equation I have.
 

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