Question #89bdd

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Question #89bdd

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Answer 1 · 2017-01-21T11:04:42

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Here it is given that the space station is spinning with an angular speed ( $ω$ $omega$ ) of 2 rpm

So $ω = 2 r p m = \frac{2 \times 2 π}{60} radian/s = \frac{π}{15} rad/s$ $omega=2rp m=(2xx2pi)/60"radian/s"=pi/15"rad/s"$

So centripetal acceleration produced $a = ω^{2} R$ $a=omega^2R$ , where R is the the radius of the space station. As per given condition this acceleration $a$ $a$ will be equal to the artificial acceleration due to gravity $g = 10 m/ s^{2}$ $g =10"m/"s^2$ experienced by an astronaut.

So $g = ω^{2} R$ $g=omega^2R$

$R = \frac{g}{ω^{2}} = \frac{10}{{(\frac{π}{15})}^{2}} m = \frac{2250}{π^{2}} m \approx 228 m$ $R=g/omega^2=10/(pi/15)^2m=2250/pi^2m~~228m$

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Question #89bdd

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