# Circular Motion and artificial gravity

1. Nov 11, 2007

### mkwok

1. The problem statement, all variables and given/known data
A space station, in the form of a wheel 135 m in diameter, rotates to provide an "artificial gravity" of 3.80 m/s$$^{2}$$ 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.

2. Relevant equations
a$$_{c}$$=$$\frac{v^{2}}{R}$$
VT=2piR

3. The attempt at a solution

3.8=$$\frac{v^{2}}{135/2}$$
v=$$\sqrt{3.8/67.5}$$=16.0156m/s

then since VT=2piR
T=2piR/v = 2pi(67.5)/16.0156 = 26.4813 rev/sec
converting that to rev/min is (26.4813 rev/sec)*(60sec/min) = 1588.88rev/min

however, I think I am doing something wrong, my online homework website tells me this is incorrect

Last edited: Nov 11, 2007
2. Nov 11, 2007

### hage567

This gives you seconds/revolution, not revolutions/second. Check your units. You're calculating the period here, not the frequency.