A space station is to provide artificial gravity to support long-term stay of as

Question

A space station is to provide artificial gravity to support long-term stay of astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feet will be on the inside of the outer wall of the space station and their heads will be pointing toward the hub). After the space station is assembled in orbit, its rotation will be started by the firing of a rocket motor fixed to the outer rim, which fires tangentially to the rim. The radius of the space station is R = 73.7 m, and the mass is M = 2.63 · 105 kg. If the thrust of the rocket motor is F = 169 N, how long should the motor fire?

Explanation / Answer

If we treat the space station as a thin hoop, its moment of inertia I = mr2 = (2.63 * 105 kg)(73.7 m)2 = 1.43 * 109 kg m2.

The torque from the rocket = Fr = (169 N)(73.7 m) = 1.25 * 104 N m.

To match gravity, we need centripetal acceleration 2r = g, therefore

= (g/r)½ = ((9.81 m/s2)/(73.7 m))½ = 0.365 s-1

The angular accerelation from the torque = /I = (1.25 * 104 N m)/(1.43 * 109 kg m2) = 8.72 * 10-6 s-2

but we know from kinematics that = t. Solving for t, t = /

= (0.365 s-1)/(8.72 * 10-6 s-2) = 4.18 * 104 s

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