Question

Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 100 m diameter rotating about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?

Answer 1

Given Information:

Diameter = d = 100 m

Required Information:

Period = T = ?

Answer:

Period = T = 14.2 seconds

Step-by-step explanation:

We know that a station revolving at an angular velocity ω, have an acceleration given by

α = ω²r

Where ω is angular velocity and r is the radius of cylindrical space station.

Normal gravity means α = g = 9.8 m/s²

ω² = α/r

ω = √(α/r)

The radius is given by

r = d/2

r = 100/2

r = 50 m

ω = √(9.8/50)

ω = 0.4427 rad/sec

We also know that

ω = 2πf

f = ω/2π

f = 0.4427/2π

f = 0.0704 rev/sec

Finally time period is given by

T = 1/f

T = 1/0.0704

T = 14.2 sec

Therefore, the rotation period is 14.2 seconds.