A space station in the form of a large wheel,

Question

173 m in diameter, rotates to provide an “ar-
tificial gravity” of 6
.
6 m
/
s
2
for people located
on the outer rim.
Find the rotational frequency of the wheel

Answer 1

v^2/r = 6 m/s^2
so
v^2 = 86.5 *6
v = sqrt (86.5*6)

T = pi d /v = pi(173)/sqrt(86.5*6)
f = frequency = 1/T
omega = 2 pi f = 2 pi/T
= 2 sqrt(86.5*6) / 173

Answer 2

so it that 6.6m/s^2?

a=w^2 r
w=2PI f= sqrt (a/r
f=.159*sqrt(6.6/(173/2))

Answer 3

I gave you the frequency and the rotational speed because I do not know what "rotational frequency" means.