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Effective gravity and an astronaut falling, in a spinning cylindrical space station

  1. May 1, 2012 #1
    A cylindrical space station of radius r with thin walls and mass M rotates at angular velocity ω such that the apparent gravity on the inner surface of the cylinder is equal to g.

    1) Radial spokes of negligible mass connect the cylinder to the centre of motion. An astronaut of mass m climbs a spoke to the centre. What is the fractional change in apparent gravity on the surface of the cylinder?

    2) If the astronaut climbs halfway up a spoke and lets go, how far form the base of the spoke will he hit the cylinder? Assume the astronaut is point like.

    Solution:

    1) For the first part I got that the ratio of apparent gravity before and after will be (1 - m/M)2, because angular momentum is invariant but the space station can change angular velocity, so Mrv1 = (M-m)rv2

    This is a little confusing though, because if the station is effectively losing mass m, so the new mass is M-m, then the angular velocity should have to go up so that angular momentum stays the same! Why doesn't that happen? Shouldn't the effective gravity also increase as a result of the increased angular velocity?

    2) The astronaut should fly off tangentially, and had to cover a distance (√3)r/2 to reach the edge of the cylinder, and he's traveling at ωr/2, so he should arrive √3/ω seconds later, right? Should the spoke stop turning he would arrive Pi/3 radians ahead of spoke.

    And in that time, the spoke turns through ωt radians, which gives √3 radians.

    So the angle subtending the arc between him and the spoke should be (Pi/3 - √3) radians, placing him (Pi/3 - √3)r away from the foot of the spoke.


    Can anyone confirm or deny that these methods and answers are accurate? They both feel wrong to me.

    Thank you for your help.
     
  2. jcsd
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