Space Station - Conservation of Angular Momentum

AI Thread Summary
A space station with a radius of 109m and a moment of inertia of 5.07x10^8 kgm^2 is initially rotating to provide an apparent acceleration of 0.8g for 180 crew members. When 140 crew members move to the center, the angular speed changes, affecting the apparent acceleration for the remaining 40 at the rim. The calculations involve conservation of angular momentum, leading to a new apparent acceleration of approximately 1.15g for those at the rim. The final result was confirmed to be around 1.148g. The discussion highlights the application of physics principles in a rotating space environment.
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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 centripital 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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