A circular space station (shaped like a big wheel) rotates 100 times in one hour. The apparent weight of an astronaut standing on the inside surface of the outer wall is equal to her weight on earth. How far from the axis of rotation is the inside surface of the station's outer wall?
A) 322 m
B) 102 m
C) 56 m
D) 1012 m

Question

Quizplus · Accepted Answer

The apparent weight of the astronaut is produced by the centripetal force required to keep her in circular motion. This force depends on the radius of the circle (distance from the axis of rotation), the mass of the astronaut, and the square of the angular velocity. The formula for the centripetal force is $F = \frac{mv^2}{r}$, where $m$ is mass, $v$ is velocity, and $r$ is radius. However, since we're dealing with rotational motion, it's more appropriate to use the formula involving angular velocity, $F = mr\omega^2$, where $\omega$ is the angular velocity in radians per second.Given that the station rotates 100 times in one hour, we first convert this to radians per second:$$100 \, \text{revolutions per hour} = \frac{100}{60} \, \text{revolutions per minute} = \frac{100}{3600} \, \text{revolutions per second}$$Since each revolution is $2\pi$ radians,$$\omega = \frac{100 \times 2\pi}{3600} \, \text{radians per second}$$The apparent weight on Earth is due to gravity, $F = mg$, where $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$). Setting this equal to the centripetal force gives us:$$mg = mr\omega^2$$$$g = r\omega^2$$Solving for $r$, we get:$$r = \frac{g}{\omega^2}$$Substituting the values for $g$ and $\omega$,$$r = \frac{9.8}{\left(\frac{100 \times 2\pi}{3600}\right)^2}$$Calculating this gives a radius of approximately 322 meters, which means the astronaut must be 322 meters from the axis of rotation for her apparent weight to equal her actual weight on Earth.

A Circular Space Station (Shaped Like a Big Wheel) Rotates