Finding revolution time at surface

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Homework Help Overview

The discussion revolves around calculating the time for one revolution of a cylinder with a radius of 0.90 km, given that the gravitational acceleration at its surface is 9.8 m/s². The problem involves concepts from rotational dynamics and angular motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between angular velocity and other variables, such as linear velocity and radius. There are attempts to isolate variables and clarify the use of given equations. Questions arise regarding missing values and the proper interpretation of the problem setup.

Discussion Status

Participants are actively engaging with the problem, with some providing calculations and others questioning the steps taken. There is a recognition of errors in previous calculations, and a few participants are exploring the implications of their findings without reaching a consensus on the final answer.

Contextual Notes

There are mentions of confusion regarding the use of radius versus diameter, and some participants express uncertainty about the values needed to proceed with calculations. The discussion reflects a collaborative effort to clarify the problem's requirements and assumptions.

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:
ross moldvoer said:
wow can't believe i missed that. so then 450ω^2=9.8, ω=.133
i converted .9 km to 900 m
Ok, but why did you halve it to 450? Either way, I don't get .133.
 
  • #10
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)
 
  • #11
ross moldvoer said:
the cylinder's diameter is 900m
It says radius.
 
  • #12
haruspex said:
It says radius.
my bad

then i get .104
 
  • #13
ross moldvoer said:
my bad

then i get .104
Sounds right.
 
  • #14
so after this do i just plug it into the equation? the answer i get seems high
 
  • #15
ross moldvoer said:
so after this do i just plug it into the equation? the answer i get seems high
You know what I'm going to say, surely? Please post your working and answer.
 

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