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Original Question
It has been suggested that rotating cylinders about 12 mi long and 5.4 mi in diameter be placed in space and used as colonies....Asked by Alexander
It has been suggested that rotating cylinders about 10 mi long and 5.6 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration?
Answers
Answered by
Gil
take 5.6/2=2.8
convert 2.8miles into meters=4506
9.8/4506= .002174878
now squaroot that and you wil getthe answer which is
(.0466)
convert 2.8miles into meters=4506
9.8/4506= .002174878
now squaroot that and you wil getthe answer which is
(.0466)
Answered by
Your Teacher
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Answered by
Tutor Allan
r=5.6/2 = 2.8mi
convert from mi to m use
imi= 1609.34m
2.8mi? =1609.34*2.8=4506.152m
g=v^2/r , g=9.8, making v subject we get
v^2= 9.8*4506.152
v=210.14m/s
but w=v/r
w= 210.14/4506.152=0.047rad/s
for help eseyhelp @ gmail com (no spaces)
convert from mi to m use
imi= 1609.34m
2.8mi? =1609.34*2.8=4506.152m
g=v^2/r , g=9.8, making v subject we get
v^2= 9.8*4506.152
v=210.14m/s
but w=v/r
w= 210.14/4506.152=0.047rad/s
for help eseyhelp @ gmail com (no spaces)
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