## Physics Friday 5

Spin Gravity and Coriolis Forces

Suppose we have a space craft or habitat using rotation to simulate gravity (as seen in 2001: A Space Odyssey, Mission to Mars, and Babylon 5). It is the fictitious centrifugal force that provides the apparent gravity, but for a uniformly rotating reference frame, there is another fictitious force: the Coriolis force.

Let us consider motion in a plane perpendicular to the rotation axis (as the component of motion parallel to the axis neither affects nor is affected by the fictitious forces). We call the angular velocity ω, and the distance from the axis r. Then the centrifugal force points outward from the axis, and produces a centrifugal accelleration in the rotating frame of .

Thus, we can figure out the angular velocity needed to produce a partiular strength of apparent gravity at a particular radius. Below is a table that for various radii gives the angular velocity ω and the rotation period  necessary to produce an apparent gravity of one gee (9.80 m/s2) for various radii:

 r (m) ω (s-1) T (s) 1 3.130 2.007 5 1.400 4.488 10 0.990 6.347 50 0.443 14.192 100 0.313 20.071 200 0.221 28.385 500 0.140 44.880 1000 0.099 63.470

Now, let us consider the Coriolis force. In three dimensions, the Coriolis force on a mass m moving with velocity  in the rotating reference frame is . Thus, if v is the maginude of the component of  perpendicular to the axis of rotation (given by ), then the magnitude of the Coriolis force is , and the direction is given by the right hand rule. It will be perpendicular to , and thus in the plane perpendicular to the axis. It will also be perpendicular to the velocity, and thus to the velocity component in the plane. Note that if we view the plane in the direction such that the rotation is counterclockwise, then the Coriolis force is in a direction ninety degrees clockwise from the velocity: motion on an outward radial path experiences a Coriolis force opposite the rotation direction, and motion inward experiences a force with the rotation.

Note, however, it is generally easier to work in the non-rotating reference frame, where trajectories are straight lines, than in the rotating frame. For example, suppose someone in our space station drops an object from a height h above the floor, which is at radius R. Let us define inertial frame polar coordinates
 and rotatng frame coordinates , with the two coinciding at t=0 and the thetas increasing in the direction of motion. Then . Our object has initial position of (R-h, 0) in the inertial frame, and initial velocity equal to the tangential velocity, or . Uniform linear motion in polar coordinates is given by  and 
In our case, this becomes  and .
Transforming, this becomes in the rotating frame:



We want the time to drop to the floor: t such that , which is .
Plugging this into , we find:

which is negative and independent of ω, as long as ω≠0! (We can also show that this angle is dependent only on the ratio of h to R).
The distance of “deflection” from the point straight down is 
(note that for , this can be approximated by ).

For example, dropping an object from a height of 125 cm above a floor at radius 100 m gives a deflection d=13.3 cm. At a floor of radius 500 m, this same height has deflection 5.9 cm. This is definitely enough to notice, particularly it a person stands up or bends over quickly; their head would experience some Coriolis deflection as a result.

Lastly, if someone were to “fall” with some small initial velocity from the axis of rotation, their motion in the inertial frame is , and so examining the transformation, the radial velocity in the rotating frame is also the small constant v0. However, note that when you reach radius R, your “horizontal” velocity in the rotating frame is the tangential velocity Rω.

If we have ω to maintain one gee at radius R, the tangential velocity at R is , and is given in this table:

 R (m) ω (s-1) V=Rω (M/s) 1 3.130 3.130 5 1.400 7.000 10 0.990 9.899 50 0.443 22.136 100 0.313 31.305 200 0.221 44.272 500 0.140 70.000 1000 0.099 98.995

For those not familiar with metric velocities, 1 m/s is about 2.24 mph, so at a radius of 50 m (approximately 164 ft), we have tangential velocities of almost 50 mph! So any injury from the fall would be from “sideways” motion, not “downward motion,” after a possibly rather slow fall.

Thus, we find the larger the rotating space, the less Coriolis force is felt by those within, but with higher tangential velocities.

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