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+--- Thread: Science quick question. Need help! [More stuff to know!] (/showthread.php?tid=130556)
Science quick question. Need help! [More stuff to know!] - nOmnomnOm - 06-13-2015
So centripetal force...
governed by: F=(mv^2) / r
I know how this works when you have lets say a bucket of water that you are spinning with a string.
But how does the work for lets say a wheel?
Same thing?
Is it the radius of the wheel and then the mass of the whole wheel?
Thanks in advance for the help!
RE: Science quick question. Need help! - Alexander Walker - 06-13-2015
Maybe.
RE: Science quick question. Need help! - axisdeus - 06-13-2015
You'll have to apply the same principle and consider a point on the wheel for which you want a solution. In the case of the bucket of water on a string, that bucket of water is considered a material point with no friction nor air resistance for simplicity. A wheel, for a wheel, I might be horribly mistaking here, but for each spoke of the wheel you can consider a certain amount of weight, proportional to the number of spokes. So, all in all ,you need to consider a material point at the radius of the wheel, but it's weight must be proportional.
I hope it helped, and I hope I haven;t made a fool of myself.
[EDIT]: For a fairy wheel is rather simple, use the same principle but disconnected, each pod being your proverbial Bucket on a string. For a full body wheel, you'll have to use a more advanced mathematic engine, use integrals and all that.
RE: Science quick question. Need help! - nOmnomnOm - 06-13-2015
Like I'm actually serious though.
For example those rides on amusement parks that it is a big wheel with people in it. How does one calculate that?
Edit: ninja'd and reading above
RE: Science quick question. Need help! - Error - 06-13-2015
No, not the mass of the whole wheel, the mass of the object that the force F is acting upon. Using F=(mv^2)/r doesn't make all that much sense unless you have a point mass that is being held at a fixed distance from the rotational axis (i.e. the radius), such as the bucket you used as an example. If you wanted to know the force acting upon a part of the wheel a certain distance away from the rotational axis (i.e. the axle), I suppose you could find the mass of a very thin cylindrical slice of the wheel (i.e. the circumference) and put it into the formula to find a reasonable estimate.
PS: Or, to word it slightly differently: That formula only works when all three of those are constant. The mass of the entire wheel is constant, but the radius and rotational speed changes non-linearly across the radius of the wheel (the rotational speed increases as the radius increases) so you cannot apply it directly to the entire wheel like that.
Could you clarify what you're trying to find? The net centripetal force acting upon the wheel? That'd (ideally) be zero, as there's an equal opposite force acting upon all points in the wheel due to symmetry.
EDIT: If my post has any "horses" (read: forces) in it, blame FoxReplace; sorry. I hope the above makes a bit of sense, at least.
RE: Science quick question. Need help! - nOmnomnOm - 06-13-2015
(06-13-2015, 04:19 PM)axisdeus Wrote: You'll have to apply the same principle and consider a point on the wheel for which you want a solution. In the case of the bucket of water on a string, that bucket of water is considered a material point with no friction nor air resistance for simplicity. A wheel, for a wheel, I might be horribly mistaking here, but for each spoke of the wheel you can consider a certain amount of weight, proportional to the number of spokes. So, all in all ,you need to consider a material point at the radius of the wheel, but it's weight must be proportional.
I hope it helped, and I hope I haven;t made a fool of myself.
You got me confused.... but let me clarify a little what I am trying to do here.
I am making a model of a base i wana design that has parts of it that turn. The parts that turn are circular in shape where the people would live inside the outer circle. This is done in space and im reducing friction to minimal using magnets and all that jazz.
So to make an artificial gravity i need 9.81 for force.
The velocity is what I am trying to imagine.... how fast is too fast for a person???
The mass will be decided once i figure out the radius.... but i think i have 2 unknowns so far....
But here is the issue though... do i have to use the mass of the WHOLE wheel or can I use mass per meter squared? That would be way more simple to do for me to figure things out.
RE: Science quick question. Need help! - nOmnomnOm - 06-13-2015
(06-13-2015, 04:23 PM)Error Wrote: No, not the mass of the whole wheel, the mass of the object that the force F is acting upon. Using F=(mv^2)/r doesn't make all that much sense unless you have a point mass that is being held at a fixed distance from the rotational axis (i.e. the radius), such as the bucket you used as an example. If you wanted to know the force acting upon a part of the wheel a certain distance away from the rotational axis (i.e. the axle), I suppose you could find the mass of a very thin cylindrical slice of the wheel (i.e. the circumference) and put it into the formula to find a reasonable estimate.
PS: Or, to word it slightly differently: That formula only works when all three of those are constant. The mass of the entire wheel is constant, but the radius and rotational speed changes non-linearly across the radius of the wheel (the rotational speed increases as the radius increases) so you cannot apply it directly to the entire wheel like that.
Could you clarify what you're trying to find? The net centripetal force acting upon the wheel? That'd (ideally) be zero, as there's an equal opposite force acting upon all points in the wheel due to symmetry.
So wait can I use a different formula then? or do I have to pick a point and test it?
RE: Science quick question. Need help! - axisdeus - 06-13-2015
For that you'll have to consider the mass of an average individual at the radius of the station. Easy peasy. Only the individual will feel the effect of the centripetal force. The structure, as Error said, will be counterbalanced by itself.
Thus you have the F, the mass of an average person and the radius from construction...your velocity is easy solution now. Yes, the generic force will vary, but to the slightest, i beleive, as the most of your numerator comes form the velocity.
RE: Science quick question. Need help! - nOmnomnOm - 06-13-2015
(06-13-2015, 04:27 PM)axisdeus Wrote: For that you'll have to consider the mass of an average individual at the radius of the station. Easy peasy. Only the individual will feel the effect of the centripetal force. The structure, as Error said, will be counterbalanced by itself.
Oh i see so I account for the People only as a mass and not the entire structure?
Omg that makes life so much more easy then!
So do I also account for anything not fixed as well such as furniture? And do I just use 1 person or everyone on the station? One right? If that is the case then that is a heck of a lot more easier to do.
RE: Science quick question. Need help! - Error - 06-13-2015
(06-13-2015, 04:24 PM)nOmnomnOm Wrote: You got me confused.... but let me clarify a little what I am trying to do here.Ah, right, got it; that's a bit more complicated, but you could indeed just check on a point of the wheel to find the centripetal force there, as you suggested earlier. You don't even need to multiply by a mass in the first place if you just want to find the acceleration at some radius and rotational speed; a = (v^2)/r.
I am making a model of a base i wana design that has parts of it that turn. The parts that turn are circular in shape where the people would live inside the outer circle. This is done in space and im reducing friction to minimal using magnets and all that jazz.
So to make an artificial gravity i need 9.81 for horse.
The velocity is what I am trying to imagine.... how fast is too fast for a person???
The mass will be decided once i figure out the radius.... but i think i have 2 unknowns so far....
But here is the issue though... do i have to use the mass of the WHOLE wheel or can I use mass per meter squared? That would be way more simple to do for me to figure things out.