Circular motion force problem

  • Thread starter trixid
  • Start date
  • #1
trixid
12
0
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:
 

Answers and Replies

  • #2
Zlex
40
1
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.
 
  • #3
Physics_wiz
227
0
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
Physics_wiz
227
0
Hehe Zlex, we posted in the same exact minute :biggrin:
 
  • #5
trixid
12
0
Ah I've got it now. Thanks a bunch.

Does 1.71 rev/s sound about right?
 
Last edited:
  • #6
Physics_wiz
227
0
That's not what I got, but I might have done it wrong. How did you get to that answer?
 
  • #7
trixid
12
0
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
trixid
12
0
Ah! Something is horriblely wrong with what I did. The speed has nothing to do with it.
 
  • #9
trixid
12
0
Actually, I think I'm just confusing myself. I think my answer was right...maybe?
 
  • #10
Physics_wiz
227
0
Yep :biggrin: it's the circumference that's giving you the pain :approve:
 
  • #11
HawKMX2004
36
0
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
Physics_wiz
227
0
The problem is that the circumference is NOT 47.1 m.
 
  • #13
HawKMX2004
36
0
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
Physics_wiz
227
0
He got the velocity right but the circumference he got wrong.
 
  • #15
trixid
12
0
What's the formula for circumference?
 
  • #16
Physics_wiz
227
0
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
trixid
12
0
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:
  • #18
Physics_wiz
227
0
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
trixid
12
0
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:
 
  • #20
Physics_wiz
227
0
Nope.

If it can only travel 30 m per second then it can only travel HALF the 60 m track in one second.
 
  • #21
trixid
12
0
That's what I said, right? (Just checking to make sure I've got it.)
 
  • #22
Physics_wiz
227
0
You got it.
 
  • #23
trixid
12
0
Yay! Thanks a bunch.
 

Suggested for: Circular motion force problem

  • Last Post
Replies
6
Views
211
  • Last Post
Replies
18
Views
331
  • Last Post
Replies
4
Views
242
Replies
7
Views
269
Replies
6
Views
285
Replies
4
Views
308
Replies
11
Views
117
Replies
11
Views
318
Top