It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station in constructed as a 1170.0 m diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period (in s) will provide "normal" gravitational acceleration of 9.80 m/s2 at deck level?
please help.
3 answers
hlep
Acceleration(g) = (Velocity^2)/(Radius) Velocity^2 = (Radius(g))
Velocity= sqrt(Radius (g))
Diameter= 2(radius)
radius = Diameter/2
Velocity = sqrt(1170.0 m)/2(9.80m/s^2)
Velocity = 75.7165 m/s ~ [75.7m/s] 3sig
Velocity= sqrt(Radius (g))
Diameter= 2(radius)
radius = Diameter/2
Velocity = sqrt(1170.0 m)/2(9.80m/s^2)
Velocity = 75.7165 m/s ~ [75.7m/s] 3sig
First, find the velocity: since a=(v^2)/r, and a=g, substitute and simplify to get v=root(g*r). remember to divide the diameter by 2 to get the radius. normal g=9.8m/s^2. The period can be found by dividing the circumpherence by that velocity. C=2pi(r).
1 answer