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Artificial Gravity

  1. Dec 3, 2005 #1
    Hi,

    Question about fictitious force, The set-up is a torus rotating in space of radius r.
    First part involves working out the speed of rotation v, for which an artificial gravity of 1g will be created in the torus.
    Which is just the centripetal acceleration (centrifugal force?)
    For r = 100m at g = 9.81ms-2, v = 31.3 ms-1 (to 3.s.f), w = 0.313 revs s-1. Acting radially outwards.

    Next part I'm not sure, we were asked if someone was moving along the corridor (still gravity = g, r = 100m ) at a speed of V = 1ms-1 (in either direction). What will be the magnitude and direction of the acceleration the walker experiences.
    I think it's the corolis force, a = 2(V X w) (cross-product), but I'm unsure of the angle between the axis of rotation and the direction the person is walking in.
    Any help would be appreciated.
     
  2. jcsd
  3. Dec 3, 2005 #2

    lightgrav

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    Homework Helper

    first, the velocity vector is along the torus outside rim, perpendicular to the r-vector.
    Angular velocity omega is along the axis of the torus, perp to r-vector and v-vector.
    Centripetal acceleration vector points toward the center of the torus.

    If somebody moves along the rim of the torus, their angular velocity relative to the torus (v=rxw) must be parallel to the angular velocity of the torus itself. so, this is not a coriolus force situation.

    But, the moving people's REAL tangential velocity will be 30.3 [m/s] or 32.3 [m/s] at a radius of 100 [m] ... what centripetal accelerations do they need to not sink thru the floor?
     
  4. Dec 3, 2005 #3

    Astronuc

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    Staff: Mentor

    Why not simply consider that the person moving with speed V will experience a force related to v-V or v+V if running opposite or with the toroidal rotation. The force on the runner depends on his velocity.

    [itex]\omega[/itex]r = v

    Angular acceleration = v2/r or [itex]\omega[/itex] v

    useful reference - http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#rotcon
     
  5. Dec 3, 2005 #4
    Tx, that helps alot.
    Also, the question asked about playing table tennis. Assuming the ball is hit @ 30 [m/s] along the corridor. Therefore the centripetal acceleration is a = 37.6 [m/s^2] . This would make the game impossible too play but is there solution that would make the game playable even if not perfect?
     
  6. Dec 3, 2005 #5

    lightgrav

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    Homework Helper

    If you start thinking in 3-d, you'll realize that the coriolis force is the same for both players, if they're both right-handed.

    Your playing parallel and antiparallel to v_vector is unfair because one player has a = 4g , while the other has a = g/600 .
     
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