# It's a question involving angular velocity....?

It has been suggested that if humans are to travel to the nearest star systems, that rotating cylinders be used to simulate gravity and provide a living space during this 20 - 30 year long interstellar travel. If the cylinders are about 12 mi long and 6.4 mi in diameter. What angular speed(omega) must such a cylinder have so that the centripetal acceleration at its inner surface equals the free-fall acceleration on Earth?

### 4 Answers

- ?Lv 72 months ago
Of course the length of the cylinder is immaterial. You need the centripetal acceleration to equal gravity:

ω²*r = g

ω = √(g/r) = √(32.2ft/s² / (½*6.4*5280 ft)) = 0.044 rad/s

- JimLv 72 months ago
First, that's 20,000 to 30,000 years to get to our closest neighbor at the fastest speed we can go. Anyway... feet and miles are outlawed in space since the Mars Orbitor crashed.

- az_lenderLv 72 months ago
r*omega^2 = 9.8 m/s^2 =>

omega^2 = (9.8 m/s^2)/(10300 m) =>

omega = sqrt(9.515 x 10^(-4)) rad/s = 0.031 rad/s.