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.

Question

Kendra

Physics

12 July, 02:14

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.

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Harold · 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