Question

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 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.

Answer 1

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where 

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$