A space station shaped like a giant wheel has a radius of 151 m and a moment of inertia of 5.10 ✕ 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 apparent acceleration is experienced by the managers remaining at the rim? Assume that the average mass of each inhabitant is 65.0 kg.

1.57 g

Explanation:

Angular momentum is conserved, so:

I₁ ω₁ = I₂ ω₂

Solving for ω₂:

ω₂ = ω₁ (I₁ / I₂)

Calculating each moment of inertia

I₁ = 5.10*10⁸ kg m² + 150 * 65.0 kg * (151 m) ² = 7.32*10⁸ kg m²

I₂ = 5.10*10⁸ kg m² + 50 * 65.0 kg * (151 m) ² = 5.84*10⁸ kg m²

Therefore:

ω₂ = ω₁ (7.32*10⁸ / 5.84*10⁸)

ω₂ = 1.25ω₁

Relating angular velocity to centripetal acceleration:

a = v² / r

a = ω² r

r is constant, so acceleration is proportional to the square of angular velocity.

a₂ / a₁ = ω₂² / ω₁²

a₂ / a₁ = (ω₂ / ω₁) ²

Plugging in:

a₂ / 1g = (1.25) ²

a₂ = 1.57g

The new apparent acceleration experienced at the rim is 1.57 g.