# Thread: I need a physics formula

1. ## I need a physics formula

I've looked for this for a really long time, like years, and I can't find it. I'm not sure what it would even be called, but basically what I need to be able to figure is for a rotating object, classic SF ring shaped space station, the relation between RPM, radius and g's.

How fast will a 500 meter radius object have to rotate to hit .25g?
How many g's will a 300 foot object rotating at 3 RPM generate?

That sort of thing. I know it's got to be stupid-simple, but it took me two tries to get through algebra 1. I need story problems or it just doesn't make sense. Physics carried a pre-requisite for being in calculus at my high school.

Thanks.

2. one of the formulas I remember is a=(v^2)/r where v is the velocity of the thing at the end of say a rope of length r.

3. There is a pretty thorough explanation (with diagrams and equations) here.

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Another way of coming at it is (Edit: referring back to Frogmarch's post; I overlapped with cope):

a = ω&#178;r

where a is the accleration experienced by your revolving object, ω is the angular velocity and r is the radius of gyration. If ω is in radians per second and r is in metres, you get a in metres per second squared (divide by 9.8 to get acceleration in g's).

If you know the rotation period (P) in seconds, divide by 2π and take the inverse to get ω. So

ω = 2π/P

and

P = 2π/ω

Grant Hutchison

5. Toq_, grant's formula is the same as Frogmarch's, since v = ωr

Originally Posted by Tog_
How fast will a 500 meter radius object have to rotate to hit .25g?
If g=10m/s2, then

a = ω2 r
a = (2π/P)2 r
P = 2π sqrt(r/a)
P = 2π sqrt(500/(.25 x 10)) or about 110 seconds

How many g's will a 300 foot object rotating at 3 RPM generate?

6. ...just under 1g.

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Originally Posted by hhEb09'1
Toq_, grant's formula is the same as Frogmarch's ...
Yes, thanks, I should have pointed that out to avoid confusion. Meant to. Didn't.

Grant Hutchison

8. Originally Posted by hhEb09'1
P = 2π sqrt(500/(.25 x 10)) or about 110 seconds
or less

9. Originally Posted by Ken G
...just under 1g.
Be careful of his trick questions. [Edit: Sorry, it was one of Tog's original questions. I assumed something else was afoot. ]

10. Originally Posted by hhEb09'1
Toq_, grant's formula is the same as Frogmarch's, since v = ωr

If g=10m/s2, then

a = ω2 r
a = (2π/P)2 r
P = 2π sqrt(r/a)
P = 2π sqrt(500/(.25 x 10)) or about 110 seconds
Thanks all for the replies. I think I'm doing something wrong here. Is the thing after the 2 in the bottom 2 pi?
If so, I get 88 seconds.
P= 2 pi sqrt(200)= 2 pi 14= 88

What did I do wrong?
For the second one I did get about 0.9 g's so if we rounded 100 yards to 100 meters that would be right.

11. Originally Posted by Tog_
For the second one I did get about 0.9 g's so if we rounded 100 yards to 100 meters that would be right.
Radius normally reprsents half the length, so be careful.

[Added: Yes, 2n is 2 pi.]

[Edit: Sorry, that was your question not hhEbo9'1's] Is the radius = 300 feet, or is it a 300 foot object that is rotating about its center with a radius of 150 ft?]

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Originally Posted by George
[Added: Yes, 2n is 2 pi.]
No, 2π is 2 pi. 2n is just 2 times a lower-case N.

Pesky san serif fonts ... It looks much better in Times New Roman: 2π.

Grant Hutchison

13. Originally Posted by grant hutchison
No, 2π is 2 pi. 2n is just 2 times a lower-case N.
Where n = pi only. Why is everything looking tricky all of a sudden to me? What is the radius of a 300 foot rotating object? [That should be a trick question, right? ] [Added: What if the c.g. is not the geometric center? What if the center is moving in an epicycle? ]

14. Be nice... My only physics instruction came from the show that came on after Sesame Street.

I guess 300 feet would be the full radius. I was just throwing out examples of what I wanted to be able to find out. And I meant to do meters, so it was one of those "whoops, we lost a Mars probe" sort of typos.

So it that 0.25 g 500 meter radius object making on revolution every 110 seconds, or every 88? If it is 110, where am I missing it?

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## syntax

And P is usually reserved for gas pressure, lower case p for momentum....it's T for period, the reciprocal of frequency, which is usually f, but is v(nu) in the case of light .

so ac = v2/r becomes when v =2 pi r/T...4 (pi)2r/T2

the whole thing over 9.8m/sec2...gives g's too.

16. Originally Posted by Tog_
So it that 0.25 g 500 meter radius object making on revolution every 110 seconds, or every 88? If it is 110, where am I missing it?
Sorry for the consternation, I found my mistake, which is why I added "or less"

Good job on checking!

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Originally Posted by trinitree88
And P is usually reserved for gas pressure, lower case p for momentum....it's T for period ...
Not so much in celestial mechanics, where T is reserved for some epoch or other ("Time") and P is the orbital period.

Grant Hutchison

18. Originally Posted by hhEb09'1
Sorry for the consternation, I found my mistake, which is why I added "or less"

Good job on checking!
No problem at all, I was just worried that I was doing it wrong.

As for the checking, I just tossed out some numbers as examples. I just wanted to be sure I knew how to do it for when it comes up.

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Originally Posted by grant hutchison
Not so much in celestial mechanics, where T is reserved for some epoch or other ("Time") and P is the orbital period.

Grant Hutchison
Grant. As I've not taken a course specifically designed in celestial mechanics, but work only with a variety of texts I'll defer. I just wish authors and publishers kept to a single set of symbols, it's easier on the kids if it's consistent.
see:http://selland.boisestate.edu/jbrenn...cs_symbols.htm
pete

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Originally Posted by trinitree88
Grant. As I've not taken a course specifically designed in celestial mechanics, but work only with a variety of texts I'll defer. I just wish authors and publishers kept to a single set of symbols, it's easier on the kids if it's consistent.
Yeah, someone actual wrote a poem about this, published in the last Annals of Improbable Research. Trouble is, there just aren't enough Roman and Greek letters available to keep consistency across all areas: best that can be done is to have consistency within certain fields of endeavour.
For rotary movement, I tend to flip into orbital mechanics. I pulled a few textbooks off the shelf for a sanity check when you made your post about the use of "P", and it does seem to be consistent that P = period in my small sample.

Grant Hutchison

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