Asked by Mike Joseph
A projected space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire) as shown in the figure . The circle formed by the tube has a diameter of about 1.1-km
g = R w^2 = 9,8 m/s^2
R = 550 m
How do I convert the answer with this information? I cannot get the proper conversion to solve the problem. Please help me
g = R w^2 = 9,8 m/s^2
R = 550 m
How do I convert the answer with this information? I cannot get the proper conversion to solve the problem. Please help me
Answers
Answered by
Elena
If acceleration =g, then the centripetal acceleration = 9.8 m/s²:
g = (4•π2•r)/T2
T2 = (4•π2•r)/g
T = sqrt(4π2r/g) = 2•π•sqrt(r/g) = 2•π•sqrt(550 m/9.8 m/s²) = 47.0704 sec/rev
Now this is sec per revolutions, and they asked for rev/day, so there are (24hr/day)(3600sec/hr) = 86400 sec/day, so
rev/day = (86400 sec/day)/(47.0704 sec/rev) = 1835.547245 = 1800 rev/day
g = (4•π2•r)/T2
T2 = (4•π2•r)/g
T = sqrt(4π2r/g) = 2•π•sqrt(r/g) = 2•π•sqrt(550 m/9.8 m/s²) = 47.0704 sec/rev
Now this is sec per revolutions, and they asked for rev/day, so there are (24hr/day)(3600sec/hr) = 86400 sec/day, so
rev/day = (86400 sec/day)/(47.0704 sec/rev) = 1835.547245 = 1800 rev/day
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