Question

A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons who walk around on the inner wall of the outer rim. find the rate of the wheel's rotation in revolutions per minute that will produce this effect.

Answer 1

Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)= 11.44 m/s.
After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= 32.68 rev/s
Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)= 1960.74 rev/min