Finding revolution time at surface

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1. Feb 3, 2015

ross moldvoer

1. The problem statement, all variables and given/known data
For a 0.90km radius cylinder, find the time for one revolution if "gravity" at the surface is to be 9.8 m/s2.

2. Relevant equations
rω^2=a

3. The attempt at a solution
i tried solving for omega but i couldnt find a solution that i only had 1 variable in it.

2. Feb 3, 2015

haruspex

Which other variable do you think you have no value for?

3. Feb 3, 2015

ross moldvoer

i dont know the velocity for solving for omega, is this what you meant?

4. Feb 3, 2015

Staff: Mentor

ω is a velocity: an angular velocity. How is ω related to the frequency and the period of one "cycle" (one revolution in this case)?

5. Feb 3, 2015

haruspex

I don't know where you are stuck. The first step is to calculate ω - can you do that from the formula you quoted? Next, as gneill says, you can find the time of one revolution from ω. Is that part your problem?

6. Feb 4, 2015

ross moldvoer

i know that ω = v/r but i do not know v. I guess i am confused about what to do in order to solve for ω.

7. Feb 4, 2015

haruspex

In the OP you quoted the right equation to be using. That has three variables and you are given the values of two of them. Forget v for now.

8. Feb 4, 2015

ross moldvoer

wow can't believe i missed that. so then 450ω^2=9.8, ω=.133
i converted .9 km to 900 m

Last edited: Feb 4, 2015
9. Feb 4, 2015

haruspex

Ok, but why did you halve it to 450? Either way, I don't get .133.

10. Feb 4, 2015

ross moldvoer

the cylinder's diameter is 900m and i need the radius. 900/2=450. then i did 9.8/450=ω^2=.022, ω=.147
(i had a math error in the previous post, this is what i actually got)

11. Feb 4, 2015

haruspex

12. Feb 4, 2015

ross moldvoer

then i get .104

13. Feb 4, 2015

haruspex

Sounds right.

14. Feb 4, 2015

ross moldvoer

so after this do i just plug it into the equation? the answer i get seems high

15. Feb 4, 2015