It has been suggested that rotating cylinders about 10 mi long and 5 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?
I would say:
a_centripetal=v^2/r or rW^2
The Attempt at a Solution
I don't really get what it's asking. Free-fall acceleration is -9.8 m/s^2 right? So that means the centripetal acceleration has to equal -9.8?
I converted the radius of 5mi to meters and got 8047m.
If what I state above is true, then 9.8=8047*W^2 and the angular speed would equal .0349 rad/s?
Despite what I posted above, I think that you need to convert 9.8 (a tangential acceleration) to centripetal acceleration. So would I have to plug 9.8 for in a in this equation to get the angular accel? a=r(alpha) Angular accel is not the same as centripetal accel. right? So would this be wrong?