It has been suggested that rotating cylinders about 12 mi long and 5.4 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?
R = 2.7 miles = 1.426*10^4 ft
R w^2 = g = 32.2 ft^2/s
Solve for the angular speed, w, in radians/s
w = sqrt (g/R) = sqrt (32.2/1.4264*10^4)
= sqrt 2.259*10^-3 = 4.74*10^-2 rad/s
To find the angular speed of the cylinder, we need to equate the centripetal acceleration at its surface to the gravitational acceleration on Earth.
The centripetal acceleration can be calculated using the formula:
ac = ω^2 * r
Where 'ω' is the angular speed and 'r' is the radius of the cylinder.
The gravitational acceleration on Earth is approximately 9.8 m/s^2.
Given that the length of the cylinder is 12 mi, we need to convert it to meters. 1 mile is approximately equal to 1609.34 meters.
Length of the cylinder = 12 mi * 1609.34 m/mi = 19312.08 meters
The radius of the cylinder is half the diameter. So, the diameter is given as 5.4 mi, which needs to be converted to meters.
Diameter of the cylinder = 5.4 mi * 1609.34 m/mi = 8696.35 meters
Radius of the cylinder = Diameter / 2 = 8696.35 meters / 2 = 4348.18 meters
Now, we can equate the centripetal acceleration to the gravitational acceleration and solve for the angular speed (ω).
ω^2 * r = g
(ω^2) * (4348.18 meters) = 9.8 m/s^2
ω^2 = 9.8 m/s^2 / 4348.18 meters
ω^2 = 0.002251
Taking the square root of both sides:
ω ≈ √(0.002251) ≈ 0.0474 radians/second
Therefore, the angular speed of the cylinder should be approximately 0.0474 radians/second to have a centripetal acceleration at its surface equal to the free-fall acceleration on Earth.