Help! Physics-Satellites in Circular Motion?

Question

In designing rotating space stations to provide for artificial-gravity environments, one of the constraints that must be considered is motion sickness. Studies have shown that the negative effects of motion sickness begin to appear when the rotational motion is faster than approximately 2.25 revolutions per minute. On the other hand, the magnitude of the centripetal acceleration at the astronauts' feet should equal the magnitude of the acceleration due to gravity on earth. Thus, to eliminate the difficulties with motion sickness, designers must choose the distance between the astronaut's feet and the axis about which the space station rotates to be greater than a certain minimum value. What is this minimum value?
I'm not even sure how to get started on this! Please help!

Answer 1

There are two formulas for figuring the acceleration of something going in a circle. One formula uses linear speed (v); the other uses angular speed (ω). Since they are talking about "revolutions per minute", they're talking about angular speed, so use that formula here.

acceleration = ω²r

Now, they want the acceleration to match earth's gravity, "g". So, they want this:

g = ω²r

Now, they also said they want the angular speed to be less than or equal to 2.25 revolutions per minute:

ω <= 2.25 revs/minute

From the equation "g = ω²r", solve for ω:
ω = sqrt(g/r)

and substitute into the inequality:
sqrt(g/r) <= 2.25 revs/minute

Now solve the inequality for r:
r >= (2.25rev/min)² / g

The final step is: whenever you are dealing with angles (angular speeds, etc.) in physics, you must use units of RADIANS for everything to work out right. A "revolution" is the same as 2π radians, so:

r >= (2.25(2π rad)/min)² / g

Also, it's good to use "seconds" instead of "minutes", since g will probably have "seconds" in it too:

r >= (2.25(2π rad)/(60sec))² / g

Now just use your calculator.

Answer 2

This is a balance of forces physics problem.

ID the forces: centripetal (P) and centrifugal (F) on the person of mass m. As the person on the wheel is fixed in relation to R, the radius of rotation, the forces P and F are equal, but opposite; so that f = ma = (P - F) = 0. f is the net force acting along the radius R on the person of mass m. Clearly, the person is not accelerating along R as a = 0 also.

Centrifugal force F = mv^2/R = mw^2R; where v = wR and w = 2.25 revolutions per min = 2pi*2.25 radians per min and 2pi*2.25/60 radians per second as there are 60 seconds/min. We need to multiply revolutions times 2pi to change them to radians, which is consistent with the R units. [Always use radians when R is considered because radians are in part defined by R.]

Now the issue is, how many G's can the person take? As f = ma = mw^2R; we have a = w^2 R. If we assume we'd like to be at a comfortable G = 1 = a/g; so that a = 9.81 m/sec^2, R needs to be R = a/w^2 = 9.81/(2.25*2pi/60)^2. You can do the math. [R is quite small it turns out.] As a = w^2 R, we can see that, for a given acceleration, w can be reduced considerably as R gets larger; so a larger R will keep w below the 2.25 revolutions per minute critical value.