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Centrifugal astronaut

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    we have a spaceship with a spinning wheel artificial gravity thingy on-board with radius 25m, period ~ 12s, a=6.87

    it is spinning anticlockwise if we take a perspective looking down on it (in the bottomleft is point A,bottom right point B,and at the top point C)

    an astronaut is running (like a hamster in a wheel) on the inside of this thing. find the astronauts (mass = 75kg) apparent weight if he runs in the direction ABC AND ACB.


    2. Relevant equations

    centrifugal force F = -mw2r

    apparent force in rotating frame F=F(external) - F(centrifugal)

    V=r*w
    a = v*w
    w=2pi/T

    3. The attempt at a solution


    external force = 75kg*6.87(angular acceleration of wheel) = 515.25N

    In direction ABC


    w = (angular velocity of wheel + extra angular velocity of antronaut running)
    = (0.52+0.2) = 0.72

    so F(Centrif) = 75*(0.72)2*25 = 972

    therfore astronauts weight = 515.25 - 972

    this cant be right can it?
     
  2. jcsd
  3. Dec 7, 2008 #2

    Doc Al

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    Staff: Mentor

    I'm a bit confused as to what you mean by "external force" here. Why not just view things from an inertial frame? The centripetal acceleration is ω²R, where ω is the total angular speed. The normal force producing that acceleration is mω²R, which is the apparent weight.
     
  4. Dec 7, 2008 #3
    to be fair im not sure what i mean by it also!

    so for an astronaut standing in this wheel his apparent weight is just mw2r - that makes enough sense - but i i really dont know how this changes when he is running along the wheel - all i have been told is that it is easier to run one way round the ring than the other way around
     
  5. Dec 7, 2008 #4

    Doc Al

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    Staff: Mentor

    Sure, since mω²R depends on which way he runs. (Where ω is the total angular speed.) In one direction, ω = ω0 + ω1; in the other, ω = ω0 - ω1. ω0 is the speed of the ring; ω1 is the speed of the man with respect to the ring.
     
  6. Dec 7, 2008 #5
    thanks for that, think ive got it nailed
     
  7. Dec 7, 2008 #6
    my calculation shows that the astronauts weight is larger when he is running with the angular rotation of the cylinder, and lighter when he is running against it. is this correct? because it doesnt make logical sense to me.


    the next part deals with coriolis force.

    the astronaut climbs a ladder from the edge of the cylinder to the centre of rotation at 1m/s. calculate direction and magnitude of experienced corolis force

    if we look down on the cylinder, and take the w of the cylinder to go anticlockwise - the direction of the force experienced by the astronaut moving from the edge to the middle of the cylinder is clockwise right?

    i am stuck on how to calculate the magnitude of this force. i understand that the force is independant of position and dependant upon the velocity of the object, but if the equation i use is F=-2mwv - surely w is dependant upon the astronauts radius from the origin? but the radius of the astronaut to the origin is constantly changing because of the astronauts 1m/s speed? it has me confused!
     
  8. Dec 7, 2008 #7

    Doc Al

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    Staff: Mentor

    Yes.
    Why? Do you not agree that the faster his total ω, the greater the centripetal force required to accelerate him? And thus the greater the force he feels pushing up on his feet.

    No. The coriolis force is given by [tex]-2m\vec{\omega}\times\vec{v}[/tex]. The direction of ω is out of the page if the rotation is anticlockwise. Using the right hand rule and the minus sign will tell you that the direction of the coriolis force is also anticlockwise.

    The angular speed of the ring is the same everywhere--ω does not depend on radius. (Maybe you're mixing it up with tangential speed, which does depend on radius: v = ωr.)
     
  9. Dec 7, 2008 #8
    how can w be out of the page when it is rotating anti-clockwise? w is angular velocity?
     
  10. Dec 7, 2008 #9
    but yeah youve made sense of the first part and yeah i was getting mixed up with tangential speed,thanks for that
     
  11. Dec 7, 2008 #10

    Doc Al

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    Staff: Mentor

    I was talking about the angular velocity vector, which is used to describe the direction of rotation using a right hand rule. Read this: Rotation Vectors
     
  12. Dec 7, 2008 #11
    right i see, wikid - for |F(cor)| i get 2*75kg*(0.52rad/s *1m/s) = 78N

    thanks for your help
     
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