A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely simila

Question

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments.If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

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Thu Nguyệt 5 years 2021-09-04T11:41:02+00:00 1 Answers 46 views 0

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    2021-09-04T11:42:14+00:00

    Answer:

    The required angular velocity (ω) will be 0.313~rads^{-1}.

    Explanation:

    Due to the rotation of the space station the astronauts experience a centripetal acceleration towards the centre of the space station. If ‘\large{a_{c}}‘, ‘ω’ and ‘R’ represent the centripetal acceleration, angular velocity of the space station and the radius of the space station respectively, then

    a_{c} = \omega^{2}.R

    As according to the problem the space station has to rotate in such an angular velocity that it produces the same “artificial gravity” as Earth’s surface, we can write

    a_{c} = g = 9.8 ms^{-2}

    Also given R = \dfrac{diameter~of~the~space~station}{2} = \dfrac{200 m}{2} = 100 m

    Therefore we can write,

    && a_{c} = g = \omega^{2}.R\\&or,& \omega = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{9.8 ms^{-1}}{100 m}} = 0.313~rads^{-1}

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