Circular motion force problem

1. Nov 29, 2004

trixid

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.

2. Nov 29, 2004

Zlex

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.

3. Nov 29, 2004

Physics_wiz

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.

4. Nov 29, 2004

Physics_wiz

Hehe Zlex, we posted in the same exact minute

5. Nov 29, 2004

trixid

Ah I've got it now. Thanks a bunch.

Does 1.71 rev/s sound about right?

Last edited: Nov 29, 2004
6. Nov 29, 2004

Physics_wiz

That's not what I got, but I might have done it wrong. How did you get to that answer?

7. Nov 29, 2004

trixid

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?

8. Nov 29, 2004

trixid

Ah! Something is horriblely wrong with what I did. The speed has nothing to do with it.

9. Nov 29, 2004

trixid

Actually, I think I'm just confusing myself. I think my answer was right...maybe?

10. Nov 29, 2004

Physics_wiz

Yep it's the circumference that's giving you the pain

11. Nov 29, 2004

HawKMX2004

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?

12. Nov 29, 2004

Physics_wiz

The problem is that the circumference is NOT 47.1 m.

13. Nov 29, 2004

HawKMX2004

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 )

14. Nov 29, 2004

Physics_wiz

He got the velocity right but the circumference he got wrong.

15. Nov 29, 2004

trixid

What's the formula for circumference?

16. Nov 29, 2004

Physics_wiz

You really need to read your book if you don't know what the formula for circumference is. Anyways, it's 2*pi*r

17. Nov 29, 2004

trixid

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.

Last edited: Nov 29, 2004
18. Nov 29, 2004

Physics_wiz

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

19. Nov 29, 2004

trixid

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.

20. Nov 29, 2004

Physics_wiz

Nope.

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