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There is the popular question “What is the gravity at the center of the Earth?”. And the answer is zero, because the forces cancel out. And then the gravity increases linearly as you move to the surface. Would the same be true for a rotating cylinder?

I mean, if, for example, I have a cylinder with a radius 270 of meters, rotating at 2 rpm, you can calculate, that the gravity will be about 1.21 g on the surface of the cylinder, right? And then, if I am inside and I go deeper (towards the center) in the cylinder while it is still spinning, would the gravitational force decrease to zero (for example, when I'm 80 meters in, so in about 190 meters distance from the center, would be in a place with 0.8 g)?

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  • $\begingroup$ Are you asking about the gravitational force ( which we'd need the mass or density to work out ) or the centrifugal effect due to rotation ( or both ) ? And could you please show your own efforts, as that's the rule for homework-related questions. $\endgroup$ – StephenG Jan 7 '17 at 12:19
  • $\begingroup$ Cross posted: space.stackexchange.com/questions/19707/… $\endgroup$ – David Hammen Jan 7 '17 at 13:34
  • $\begingroup$ You are talking about the artificial gravity you create by rotating the cylinder, correct? Why don't you just take your equation that you used to get 1.21 g, and substitute other radial locations? $\endgroup$ – Chet Miller Jan 7 '17 at 13:41
  • $\begingroup$ Qmechanics edited my tags. That's not a homework related question. I am asking about the centrifugal effect due to rotation $\endgroup$ – Gale Staneva Jan 7 '17 at 14:16
  • $\begingroup$ And I thought about taking the same equation, but it's not what happens with, for example, the Earth, right? (If I understand correctly, you are asking to just plug in different radiuses in the (AngularVelocity^2)*R formula, right?) $\endgroup$ – Gale Staneva Jan 7 '17 at 14:19
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As I guess you know, it's not really gravity, it's just a force being exerted because you are in an non inertial system.

We have $$F=mω^2R$$ at the surface. and we have $$F=mω^2r$$ at distance $r$ from the center of the cylinder. (Of course I assumed that inside your cylinder is full of mass with same density everywhere which of course changes the exact equations but the answer of your question stays intact.)

So the distance to the center is $r-R$ so: $$\partial F / \partial (R-r) = (mω^2).(-1)$$ as you see, $$ \partial F / \partial (R-r) <0$$ So, the $F$ decreases while the distance decrease

Of course I didn't calculate the gravitational force exerted due to the mass of the cylinder. Still the answer doesn't change but the equations change. It's simple to apply gravity. But you should notice that the force on you at distance $R$ is only because the mass which is at $R>r$.

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Yes, at the center of mass of the cylinder. At that point, you can assume all the mass to be concentrated and thus, no gravitational net force there.

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  • $\begingroup$ Why has this been downvoted? It answers the question "would the gravitational force decrease to zero?" $\endgroup$ – Rippr Jan 7 '17 at 17:52

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