Rotation period of a space station

T ) ² ) / rNow use the quadratic equation:T=4*a*r^2In summary, for a centrifuge with a diameter of 13.5 meters, the rotation period needed to produce a 2 g centripetal acceleration is 8.13 seconds.f
  • #1

Homework Statement


To simulate the extreme accelerations during launch, astronauts train in a large centrifuge. If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of
If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of 2 g? of 5g?

Homework Equations


T=2*pi*r/v

The Attempt at a Solution


i tried solving for v using a=v^2/r and then plugging it in but i got the wrong answer
 
  • #2

Homework Statement


To simulate the extreme accelerations during launch, astronauts train in a large centrifuge. If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of
If the centrifuge diameter is 13.5m , what should be its rotation period to produce a centripetal acceleration of 2 g? of 5g?

Homework Equations


T=2*pi*r/v

The Attempt at a Solution


i tried solving for v using a=v^2/r and then plugging it in but i got the wrong answer
Please post your working, or we can't tell if or where you went wrong.
(Are you sure it asks for a centripetal acceleration of that magnitude, not a net g-force corresponding to it?)
 
  • #3
i did 9.8=v^2/6.75, v^2=(9.8)(6.75)=66.15 v=8.13
then i plugged v into the equation i gave above
 
  • #4
i did 9.8=v^2/6.75, v^2=(9.8)(6.75)=66.15 v=8.13
then i plugged v into the equation i gave above
It says 2g and 5g, not 1g.
 
  • #5
You need to introduce T into the root equation.
You have a = v ² / r
But v = ( 2 * π * r ) / T
 
  • #6
then what equation should i use to solve for v since i don't know T?
 
  • #7
then what equation should i use to solve for v since i don't know T?
I think dean barry is suggesting you eliminate v between the two equations so that you can go straight to finding T without having to calculate v. That's good advice generally, since it serves to reduce accumulation of rounding errors, but I don't think it matters here.
Do you have a response to my post #4?
 
  • #8
i did accidently forget to do 2g instead of just 9.8.
 
  • #9
i did accidently forget to do 2g instead of just 9.8.
So does that resolve your issue, or do you still have the wrong answer?
 
  • #11
so after solving for omega i plug it into T=2*pi*r/v? i get 407 when i do this and this seeems a little high
 
  • #12
so after solving for omega i plug it into T=2*pi*r/v? i get 407 when i do this and this seeems a little high
Yes, that's way too high. Please post all your steps.
 
  • #13
Take: a = v ² / r
You know:
v = ( 2 * π * r ) / T
( which introduces T into the game )
You get: a = ( ( 2 * π * r ) / T ) ² ) / r
Transpose for T
 

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