Question

A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire) as shown in the figure(Figure 1) . The circle formed by the tube has a diameter of about 1.1 km. What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the Earth (say 0.90 g) is to be felt?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A projected space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire) as shown in figure below. The circle formed by the tube has a diameter of about D = 1.26 km. What must be the rotation speed (in revolutions per day) if an effect equal to gravity at the surface of the Earth (1 g) is to be felt?

Answer

As the tube rotates, the normal reaction from the outer wall creates the sense of earth-like gravity. This normal reaction sustains the centrifugal accleration required to stay in space station. Thus, this normal reaction should be equal to the weight at earth ( mg).

N=mg

And , also N=mv²R, where R is radius of spca ship. v is the angular rotation speed

mv²r=mg

Therefore, v=sqrt(g/r) = sqrt (9.8*2/1260) = 0.1247 rad/s = 0.1247/2pi rev/s = 0.01985 rev/s.

1 rev in 50.38 sec

=24*60*60/50.38 rev per day

=1715 rev/day approx