Question

A space station consists of a giant rotating hollow cylinder of mass 10^6 kg, including people on the station, and a radius of 100.00 m. It is rotating in space at 3.30 rev/min to produce artificial gravity. If 100 people of an average mass of 65.00 kg spacewalk to an awaiting spaceship, what is the new rotation rate when all the people are off the station?

a) 3.30 rev/min
b) 2.70 rev/min
c) 2.10 rev/min
d) 1.50 rev/min

Answer 1

The new rotation rate when all the people are off the space station is 3.30 rev/min.

Step-by-step explanation:

To find the new rotation rate when all the people are off the space station, we can use the principle of conservation of angular momentum. The total angular momentum of the space station and people before the spacewalk is equal to the total angular momentum after the spacewalk. The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Before the spacewalk, the moment of inertia is the moment of inertia of the space station alone plus the moment of inertia of the people. The moment of inertia of a hollow cylinder is given by I = mr^2, where m is the mass and r is the radius. So, the moment of inertia of the space station before the spacewalk is (10^6 kg)(100.00 m)^2.

After the spacewalk, the moment of inertia is just the moment of inertia of the space station alone, since there are no people. So, the moment of inertia of the space station after the spacewalk is (10^6 kg)(100.00 m)^2.

By setting the initial angular momentum equal to the final angular momentum, we can solve for the final angular velocity. Plugging in the appropriate values, we have (10^6 kg)(100.00 m)^2(3.30 rev/min) = (10^6 kg)(100.00 m)^2(ω). Solving for ω, we find that the new rotation rate when all the people are off the station is 3.30 rev/min.