Rotating Space Station: Friction Threshold and Spin-Up
A cylindrical space station has radius R = 50 m and moment of inertia I = 4.0×10^7 kg·m^2 about its rotation axis. A 2.0 kg equipment box rests against the inner wall, where the coefficient of static friction is μ_s = 0.30 and the coefficient of kinetic friction is 0.20. The station is initially at rest. Engineers propose spinning it up to a rotation rate that makes loose objects stay on the wall without clamps. During commissioning, a technician reports that at a lower rate the box slides along the wall, even though a 70 kg astronaut standing on the wall says she feels almost normal weight. The motor can deliver a constant torque of 1.2×10^5 N·m, and bearing losses are negligible. Treat the box as a point mass and the station's mass distribution as fixed. The team needs to know the minimum angular speed for the box, the energy required to reach it from rest, and whether the astronaut's sensation is consistent with the same rotation rate. They also ask whether the motor's torque is enough to reach that speed in 60 s.
1. Using the static friction limit, derive the minimum angular speed at which the box does not slide, and compute the rotational kinetic energy needed to spin the station from rest to that speed.
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At angular speed ω the wall normal force on the box supplies the centripetal force, so N = m_b ω²R. The maximum static friction is μ_s N. To hold the box against gravity, require μ_s m_b ω²R ≥ m_b g, hence ω_min = sqrt(g/(μ_s R)). Numerically, ω_min = sqrt(9.8/(0.30×50)) = sqrt(0.653) = 0.808 rad/s. The box mass cancels. Using kinetic friction would be wrong here because the box is not sliding at the threshold.
The station rotational energy is K = 1/2 I ω² = 0.5(4.0×10^7)(0.653) = 1.3×10^7 J, about 13 MJ. This ignores the small kinetic energy of the box and motor losses, as specified.
2. At the minimum speed from part (a), what effective gravitational acceleration does the astronaut feel? Reconcile this with her report of almost normal weight and explain the technician's observation.
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Apparent gravity at the wall is a = ω²R = (0.653)(50) = 32.7 m/s², or 32.7/9.8 = 3.3 g. That is much larger than normal weight, not nearly 1 g.
For a 1 g sensation, ω would be sqrt(g/R) = sqrt(9.8/50) = 0.44 rad/s, only about half the box requirement. At that lower ω the normal force is N = m_b(9.8), and maximum static friction is 0.30 m_b(9.8), less than m_b g; so the box slides. Thus both reports can be true at different rotation rates: the astronaut's comfortable rate is below the no-slide threshold for the box.
3. The motor supplies 1.2×10^5 N·m. Determine whether it can bring the station to 0.808 rad/s in 60 s, showing the angular acceleration and time required.
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The angular acceleration is α = τ/I = 1.2×10^5 / 4.0×10^7 = 3.0×10^-3 rad/s². In 60 s the speed gain would be αt = (3.0×10^-3)(60) = 0.18 rad/s, far below 0.808 rad/s.
The time to reach ω_min from rest at constant torque is t = ω_min/α = 0.808/0.0030 = 2.7×10^2 s, about 270 s. Therefore the motor cannot meet the 60 s requirement; a larger torque, lower inertia, or longer startup time would be needed.