Circular Motion and artificial gravity

[mkwok]

Homework Statement

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.
answer must be in rev/min

Homework Equations

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

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

 

[hage567]

T=2piR/v = 2pi(67.5)/16.0156 = 26.4813 rev/sec

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