Finding revolution time at surface

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ross moldvoer
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Homework Statement


For a 0.90km radius cylinder, find the time for one revolution if "gravity" at the surface is to be 9.8 m/s2.

Homework Equations


rω^2=a

The Attempt at a Solution


i tried solving for omega but i couldn't find a solution that i only had 1 variable in it.
 
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ross moldvoer said:

Homework Statement


For a 0.90km radius cylinder, find the time for one revolution if "gravity" at the surface is to be 9.8 m/s2.

Homework Equations


rω^2=a

The Attempt at a Solution


i tried solving for omega but i couldn't find a solution that i only had 1 variable in it.
Which other variable do you think you have no value for?
 
haruspex said:
Which other variable do you think you have no value for?
i don't know the velocity for solving for omega, is this what you meant?
 
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)?
 
ross moldvoer said:
i don't know the velocity for solving for omega, is this what you meant?
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?
 
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.
 
ross moldvoer said:
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.
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.
 
wow can't believe i missed that. so then 450ω^2=9.8, ω=.133
i converted .9 km to 900 m
 
Last edited:
haruspex said:
Ok, but why did you halve it to 450? Either way, I don't get .133.
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)
 
haruspex said:
It says radius.
my bad

then i get .104
 
so after this do i just plug it into the equation? the answer i get seems high