A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of 1.1 km. What must be the rotation speed (revolutions per day) in an effect nearly equal to gravity at the surface of the Earth (say,.90 g) is to be felt

Question

Josiah

Physics

5 March, 17:13

A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of 1.1 km. What must be the rotation speed (revolutions per day) in an effect nearly equal to gravity at the surface of the Earth (say,.90 g) is to be felt

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Madeleine Carney · Answer 1

1736rev/day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = 1.1km/2 = 0.55 km = 0.55 * 10^3 m

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

Now use the centripetal force equation

F = mg

But to balance the gravitational force

F = mv²/r ... plug in this for F

Then we have ...

mv²/r = m0.9g

Solve for v, we then have

v = √0.9gmr/m

v = √0.9gr

v = √ 0.9 * 9.81 * 0.55 * 10^3

v = 69.68465m/s

We now use the equation for frequency of a period

f = v/2 π r

f = 69.68465m/s / 2 * π0.55 * 10^3

f = 0.0201 rev/seconds

Therefore revolution for a day will be

0.0201 * 24hrs * 3600

f = 1736rev/day