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A space station, in the form of a spoked wheel 120 m in diameter, rotates to provide an artificial gravity of 3.00 m/s² for persons who walk around on the inner wall of the outer rim. Find the rate of the wheel’s rotation in revolutions per minute (rpm) that will produce this effect.

User Grapes
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Answer:

2.12 rpm

Step-by-step explanation:


a_c= Centripetal\ acceleration=3.00\ m/s^2\\r=radius=\frac {120}{2}=60\ m\\a_c=\frac {v^2}{r}\\\Rightarrow v^2=a_c* r\\\Rightarrow v^2=3* 60=180\\\Rightarrow v=√(180)=13.41\ m/s\\


v=\omega r\\\Rightarrow \omega=\frac {v}{r}\\\Rightarrow \omega=\frac {13.41}{60}\\\Rightarrow \omega=0.2236\ rad/s\\1\ rad/s=9.55\ rpm\\\Rightarrow 0.2236\ rad/s=2.12\ rpm\\\therefore Wheel\ rotations=2.12\ rpm

User Bobby Cannon
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