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trixid
Nov29-04, 06:57 PM
A device for training astronauts and jet fighter pilots is a designed to rotate the trainee in a horizontal circle of radius 10.0 m. If the force felt by the trainee is 7.75 timers her own weight, how fast is she rotating? Express your answer in both m/s and rev/s.

I got the main problem but I can't get the rev/s. :redface:

Zlex
Nov29-04, 07:15 PM
Well, think about it.

You know how many meters it will tavel in one second.

You want to change that into how many revolutions will occur in one second

Consider how many meters are in a revolution.

Physics_wiz
Nov29-04, 07:15 PM
Find the circumference of the circle first. After that, change the meters/seconds speed that you have in circumferences/second because you know how many meters are in one circumference.

Physics_wiz
Nov29-04, 07:16 PM
Hehe Zlex, we posted in the same exact minute :biggrin:

trixid
Nov29-04, 07:19 PM
Ah I've got it now. Thanks a bunch.

Does 1.71 rev/s sound about right?

Physics_wiz
Nov29-04, 07:46 PM
That's not what I got, but I might have done it wrong. How did you get to that answer?

trixid
Nov29-04, 07:53 PM
The speed came out to be 27.6 m/s. The circumference was 47.1 m. I divided the circumference by the speed. Maybe I used the wrong numbers?

trixid
Nov29-04, 07:55 PM
Ah! Something is horriblely wrong with what I did. The speed has nothing to do with it.

trixid
Nov29-04, 07:58 PM
Actually, I think I'm just confusing myself. I think my answer was right...maybe?

Physics_wiz
Nov29-04, 07:59 PM
Yep :biggrin: it's the circumference that's giving you the pain :approve:

HawKMX2004
Nov29-04, 08:02 PM
Using those numbers, i assume your trying to figure out revolutions per second with a speed of 27.6 and a circumfrence of 47.1 ?? You take the speed and divide by the circumfrence. Do it like this, forget about the circle, and Say you go 27.6 m/s constantly. how much of the 47.1 would you go in 1 second. (Hint: Speed / Distance = ? ) I'll tell you what I got after you try for yourself, but yes speed has something to do with it. Also think of this equation

Xf = Xi + ViT + 1/2aT^2
a = acceleration = 0 (constant velocity)
T = Time = 1 sec
Xf = Meter part of the speed.

Xf = 0 + 26m/s(1s)
Xf = 26 meters

So think of it this way, what percent of 47.1 is 27.6?

Physics_wiz
Nov29-04, 08:08 PM
The problem is that the circumference is NOT 47.1 m.

HawKMX2004
Nov29-04, 08:13 PM
Oh, ok, i just worked it out and got 62.8 meters for the circumference. Is that what you got? ( I hadnt tried to find circumference or velocity, just going with his numbers )

Physics_wiz
Nov29-04, 08:14 PM
He got the velocity right but the circumference he got wrong.

trixid
Nov29-04, 08:18 PM
What's the formula for circumference?

Physics_wiz
Nov29-04, 08:21 PM
You really need to read your book if you don't know what the formula for circumference is. Anyways, it's 2*pi*r

trixid
Nov29-04, 08:24 PM
OOPS! I was mixing up problems. I used 7.5 as r, from a different problem entirely.
I got 2.28 rev/s. I'm hoping that's right.

Physics_wiz
Nov29-04, 08:42 PM
Nope, at least that's not what I got.

You seem to be going in the right track so I'll just show you how I did it:

cintripital force = 7.75*weight of astronaut

(mv^2)/r = 7.75*mg

V^2/r = 7.75g

v^2 = 7.75gr

v = sqrt(7.75gr) = sqrt(7.75*9.98*10) = sqrt(773.45) = 27.81 m/s

circumference = 2*pi*r = 62.83 m

27.81m/s * 1circumference/62.83m = .4426 rev/s

trixid
Nov29-04, 08:50 PM
But if it goes around 30 m in 1 second then it should go around a 62.8 m "track" at least twice, shouldn't it?
No scratch that, it would go less than once, like your answer shows, because it doesn't go a whole revolution in one second. I think i'm catching on...slowly. This is sad, I'm confused on the part that's pretty much 8th grade math. :redface:

Physics_wiz
Nov29-04, 08:52 PM
Nope.

If it can only travel 30 m per second then it can only travel HALF the 60 m track in one second.

trixid
Nov29-04, 08:59 PM
That's what I said, right? (Just checking to make sure I've got it.)

Physics_wiz
Nov29-04, 09:00 PM
You got it.

trixid
Nov29-04, 09:02 PM
Yay! Thanks a bunch.