A space station shaped like a giant wheel has a radius 105 m and a moment of inertia of 5.09 108 kg · m2. A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of 1g. When 100 people move to the center of the station for a union meeting, the angular speed changes. What acceleration is experienced by the managers remaining at the rim? Assume that the average mass of each inhabitant is 65.0 kg.
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To solve this problem, we can apply the principle of conservation of angular momentum. The initial angular momentum of the space station is equal to its final angular momentum.
The initial angular momentum, L1, is given by the moment of inertia (I) multiplied by the initial angular speed (ω1):
L1 = I * ω1
The final angular momentum, L2, can be calculated using the moment of inertia for the space station with 100 fewer people (since they moved to the center) and the final angular speed (ω2):
L2 = (I - (100 * m * r^2)) * ω2
Where:
- m is the mass of each inhabitant (65.0 kg)
- r is the radius of the space station (105 m)
Since angular momentum is conserved, we can equate L1 and L2:
I * ω1 = (I - (100 * m * r^2)) * ω2
Now, let's solve for the final angular speed (ω2):
ω2 = (I * ω1) / (I - (100 * m * r^2))
Substituting the given values:
- I = 5.09 * 10^8 kg·m^2
- ω1 = acceleration due to gravity (1g = 9.8 m/s^2)
- m = 65.0 kg
- r = 105 m
ω2 = (5.09 * 10^8 * 9.8) / (5.09 * 10^8 - (100 * 65.0 * 105^2))
ω2 ≈ 6.65 rad/s
Now, the acceleration at the rim can be calculated by multiplying the final angular speed (ω2) with the radius (r):
acceleration = ω2^2 * r
acceleration = (6.65)^2 * 105
acceleration ≈ 4,365 m/s^2
Therefore, the managers remaining at the rim experience an acceleration of approximately 4,365 m/s^2.