Which other variable do you think you have no value for?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.
i don't know the velocity for solving for omega, is this what you meant?haruspex said:Which other variable do you think you have no value for?
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?ross moldvoer said:i don't know the velocity for solving for omega, is this what you meant?
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.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.
Ok, but why did you halve it to 450? Either way, I don't get .133.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
the cylinder's diameter is 900m and i need the radius. 900/2=450. then i did 9.8/450=ω^2=.022, ω=.147haruspex said:Ok, but why did you halve it to 450? Either way, I don't get .133.
It says radius.ross moldvoer said:the cylinder's diameter is 900m
my badharuspex said:It says radius.
Sounds right.ross moldvoer said:my bad
then i get .104
You know what I'm going to say, surely? Please post your working and answer.ross moldvoer said:so after this do i just plug it into the equation? the answer i get seems high
Revolution time at the surface refers to the amount of time it takes for a planet or celestial body to complete one full orbit around its parent star or central body while staying at the same distance from the central body's surface.
Calculating revolution time at the surface allows us to understand the orbital dynamics of a planet or celestial body. This information is crucial in predicting and understanding astronomical events, such as eclipses, and in studying the evolution of planetary systems.
Revolution time at the surface is calculated using Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. This means that revolution time at the surface can be calculated by knowing the distance between the planet and its parent star, and the mass of the parent star.
Yes, revolution time at the surface can change over time due to various factors such as gravitational interactions with other objects, tidal forces, and changes in the planet's orbit. However, these changes are usually very small and can only be observed over long periods of time.
Revolution time at the surface refers to the time it takes for a planet to complete one orbit around its parent star, while rotation time refers to the time it takes for a planet to complete one full rotation on its axis. These two times can be different, as some planets have longer or shorter days than their years due to their rotation speeds and orbital distances.