A space station in the form of a large wheel,
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
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
so it that 6.6m/s^2?
a=w^2 r
w=2PI f= sqrt (a/r
f=.159*sqrt(6.6/(173/2))
I gave you the frequency and the rotational speed because I do not know what "rotational frequency" means.
To find the rotational frequency of the wheel, we can use the formula:
ω = √(g/r)
Where:
ω is the rotational frequency (in radians per second),
g is the acceleration due to gravity (in meters per second squared), and
r is the radius of the wheel (in meters).
In this case, we need to find the rotational frequency, so we can rearrange the formula to solve for ω:
ω = √(g/r)
Given:
g = 6.6 m/s^2 (acceleration due to gravity)
r = 173 m (radius of the wheel)
Substituting the values into the equation, we get:
ω = √(6.6/173)
Calculating this, we find:
ω ≈ 0.230 radians per second
Therefore, the rotational frequency of the wheel is approximately 0.230 radians per second.