Asked by laura
A space station, in the form of a wheel 110 m in diameter, rotates to provide an "artificial gravity" of 3.00 m/s2 for persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions per minute) that will produce this effect.
Thank you
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Answers
Answered by
Damon
3 = v^2/r
3 * 55 = v^2
v = 12.8 m/s
Time for revolution = 2 pi r/v
= 2 pi * 55/12.8 = 26.9 seconds = .448 minutes
1/.448 = 2.23 revs/minute
3 * 55 = v^2
v = 12.8 m/s
Time for revolution = 2 pi r/v
= 2 pi * 55/12.8 = 26.9 seconds = .448 minutes
1/.448 = 2.23 revs/minute
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