Space Station - Conservation of Angular Momentum

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SUMMARY

The discussion centers on the conservation of angular momentum in a rotating space station with a radius of 109 meters and a moment of inertia of 5.07x10^8 kgm². Initially, 180 crew members experience an apparent acceleration of 0.8g while rotating. When 140 crew members move to the center, the angular speed changes, resulting in a new apparent acceleration of approximately 1.148g for the remaining 40 crew members at the rim. The calculations utilize the relationships between angular momentum, moment of inertia, and centrifugal force to derive the new acceleration.

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Question:

A space station shaped like a giant wheel has a radius 109m and a moment of inertia of 5.07x10^8 kgm^2. A crew of 180 is living on the rim, and the station is rotating so that the crew experiences an apparent acceleration of .8g. When 140 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume an average mass of 64.0 kg for all the inhabitants.

What I have done thus far:
I = 5.07x10^8
N = 180
N`= 40
m = 64.0
A = .8

L = L`
L = (I + NmR^2)w
L` = (I + N`mR^2)w`

w' = [(I + NmR^2)w] / (I + N`mR^2)

Problem: How do I solve for radial acceleration, when I do not even have the radial speed or the initial speed.

W = Initial + AT ---- Are they asking for radial acceleration or the force felt pushing the inhabitants down on the rim.
Any help is appreciated.
 
Last edited:
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Alright I solved it

Basically I did the following:

Acceleration is the centrifugal force with is equal in magnitude to the centripetal force.

Ar = V^2 / 2
V = Root (Ar*R)
w = Root (Ar*R) / R

Substuting that for w` and w I arrived at the equation

Ar` = [([(I + NmR^2) * Root (Ar*R) / R] / (I + N`mR^2)) R ] ^2 / R
Ar` = 1.15g Can anyone double check for me.
 
Check. I got 1.148g
 

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