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.
To solve this problem, we need to consider the conservation of angular momentum. The angular momentum of the system before and after 100 people move to the center of the station should be equal.
Angular momentum is given by the equation: L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Before the people move to the center, the moment of inertia and the angular speed are related by the equation: I₁ω₁ = I₂ω₂, where the subscripts 1 and 2 refer to the initial and final states, respectively.
Given the values of the moment of inertia and the radius of the space station, we can calculate the initial angular speed (ω₁) as follows:
I₁ = 5.09 × 10^8 kg · m²
R = 105 m
Using the equation for the moment of inertia of a rotating solid wheel: I = (1/2)MR², where M is the mass and R is the radius, we can calculate the mass of the space station:
M = (2I₁) / R²
Substituting the values:
M = (2 × 5.09 × 10^8 kg · m²) / (105 m)²
M = 970476.19 kg
Now, we can calculate the initial angular speed (ω₁) using the equation:
I₁ω₁ = I₂ω₂
(5.09 × 10^8 kg · m²) × ω₁ = (970476.19 kg) × ω₂
Simplifying the equation, we get:
ω₁ = (970476.19 kg × ω₂) / (5.09 × 10^8 kg · m²)
After 100 people move to the center of the station, the moment of inertia decreases because the mass is now concentrated closer to the rotation axis. The new moment of inertia (I₂) can be calculated as follows:
I₂ = (1/2)M'(R/2)²
Where M' is the new mass of the space station. Since 100 people with an average mass of 65.0 kg move to the center of the station, the new mass is:
M' = M - (100 × 65.0 kg)
Substituting the values:
M' = 970476.19 kg - (100 × 65.0 kg)
M' = 970476.19 kg - 6500 kg
M' = 963976.19 kg
Now, we can calculate the final angular speed (ω₂) using the equation:
I₁ω₁ = I₂ω₂
(5.09 × 10^8 kg · m²) × ω₁ = (963976.19 kg) × ω₂
Simplifying the equation, we get:
ω₂ = (5.09 × 10^8 kg · m² × ω₁) / (963976.19 kg)
Finally, we can calculate the apparent acceleration experienced by the managers remaining at the rim using the equation:
a = R × ω₂²
Substituting the values:
a = (105 m) × (5.09 × 10^8 kg · m² × ω₁)² / (963976.19 kg)²
Calculating the numerical value of the expression will give us the acceleration experienced by the managers remaining at the rim.