Rotation of cylindrical habitat in space

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SUMMARY

The discussion centers on the calculation of the rotation speed required for a cylindrical habitat, as proposed by G.K. O'Neill in 1974, to simulate Earth's gravitational field. The habitat has a diameter of 6.07 km and a length of 30 km. The key formula used is centripetal acceleration, expressed as rw² = g, where g is 9.8 m/s². The calculated frequency of rotation is 0.009043859 revolutions per second, confirming that the length of the cylinder does not affect the centripetal acceleration, which depends solely on the radius and angular velocity.

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  • Understanding of centripetal acceleration and its formula
  • Familiarity with angular velocity and frequency calculations
  • Basic knowledge of rotational dynamics
  • Ability to convert units (e.g., km to meters)
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tubworld
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I have this question on this article by G.K. O'Neill,1974. who proposed having a cylindrical habitat in space. Then the article posed a question asking me to ponder on the speed of the rotation of the cylinder such that it wld imitate Earth's gravitational field at the walls of the cylinder given that it is 6.07 km in diameter and 30km long. It rotates abt its long axis. My working is as follows but am unsure if I am right cos I don't seem to use the information on the length of the cylinder.

radius = 6.07/2km
Since rw^2 = 9.8, and w = 2PI * f, where w=angular velocity and f=frequency,

we solve for f to get the ans.

From what I have the ans for f is 0.009043859 rev/s . am i right? If not any hints? Cos i don't seem to get the ans.
 
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You seem to be OK. The centripetal acceleration does not get influenced by the length of the cylinder. It is only a function of the radius and the angular velocity.
 

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