Calculating RPM Required to Simulate Earth Gravity on 100m Space Habitat

[galoshes]

So i was presented this question and I would not even like the answer but just the path to walk on. Can you help?

In order to simulate the Earth gravity on a space habitat of 100m diameter, what spin rate of a donut-shape space habitat, in terms of RPM, should be.

conversion of RPM is 2(pi)/60 radians/sec

where do I even begin?

 

[quarkman]

I think you should look into the radial acceleration of a rotating body and the "centrifugal force" associated with a body rotating in a circle. Any freshman physics text could help. An extended (large) donut would be difficult to use though...

Good luck!

 

[JohnDubYa]

There is no centrifugal force. Instead, you want the normal force exerted on the people inside the space colony to match the normal force that a person on Earth would experience.

A "weight" scale is really nothing more than a normal force o'meter. And when a person steps on a weight scale on flat land with no acceleration, then N - mg = 0 (Newton's second law) so that N = mg. (NEVER ASSUME N = MG! USE NEWTON'S SECOND LAW TO PROVE IT.)

So you want N = mg on the space colony as well. Find the spin rate that produces that amount.

If you need more help, let us know. But try to finish the problem off to the best of your ability first.

 

[quarkman]

galoshes I do not want to spoil your fun, so don't look if you haven't satisfied yourself yet.

There is no centrifugal force. Instead, you want the normal force exerted on the people inside the space colony to match the normal force that a person on Earth would experience.

...

So you want N = mg on the space colony as well. Find the spin rate that produces that amount.

Are we not saying the same thing? The g produced by the rotation of the donut would be the same as the radial (centripetal) acceleration caused by rotation in a circular path. Something else which confuses me: wouldn't the mass of the the donut produce some gravitational force, and thus acceleration? I thought since the dounout is not spherical, then you can't consider it's mass to effectively be located at the geometric center!

 

[JohnDubYa]

Yes, we are saying pretty much the same thing.

RE: "The g produced by the rotation of the donut would be the same as the radial (centripetal) acceleration caused by rotation in a circular path."

But this brings up the problem we are discussing in a separate thread: The person in the colony is radially accelerating, but the person on the Earth is not (to any significant extent). So if you tell them to produce a radial acceleration equal to g, then that is going to confuse the Hell out of them. That is why I prefer to talk in terms of normal force instead of acceleration. And this makes even more sense if you think about what actually happens when you weigh yourself. You are not measuring g, or even m*g. Instead you are measuring the normal force.

Even if you define weight as w = mg, a bathroom scale is not a weight scale. It measures the normal force, which happens to equal mg in some situations.

We can ignore the mass of the donut, since the gravitational attraction of the two bodies is negligible. After all, it takes an object having the mass of the Earth to produce a force of mg on the person.