Posted by **Mike Joseph** on Sunday, June 17, 2012 at 7:59pm.

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

- Physics Gravity -
**Elena**, Monday, June 18, 2012 at 3:22am
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

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