Asked by Anonymous
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
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
Answers
Answered by
Damon
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
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
Answered by
bobpursley
so it that 6.6m/s^2?
a=w^2 r
w=2PI f= sqrt (a/r
f=.159*sqrt(6.6/(173/2))
a=w^2 r
w=2PI f= sqrt (a/r
f=.159*sqrt(6.6/(173/2))
Answered by
Damon
I gave you the frequency and the rotational speed because I do not know what "rotational frequency" means.
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