Find period from centripetal motion that provides its own gravity

[JSmithDawg]

Homework Statement

The cylindrical space station, d = 280m in diameter, rotates in order to provide artificial gravity of g for the occupants. How much time does the station take to complete one rotation?

Homework Equations

velocity= (2*pi*r)/period
acceleration = v^2/r

The Attempt at a Solution

It's a multiple choice problem, so the possible answers are 34s, 38s, 24s, and 4s.
I'm thinking that to provide it's own gravity, the acceleration should be about equal to the Earth's acceleration due to gravity (9.8 m/s^2). Thus, I've found the acceleration that'll occur for each time.
a(38s)=[(2pi140m)/38s]^2/140m
a(38s) = 3.83 m/s^2
a(34s)=[(2pi140m)/34s]^2/140m
a(34s) = 4.78 m/s^2
a(24s)=[(2pi140m)/34s]^2/140m
a(24s) = 9.60 m/s^2
a(4s)=[(2pi140m)/34s]^2/140m
a(4s) = 345 m/s^2
Now, when the period is four seconds, the acceleration is far greater than the Earth's acceleration due to gravity, so it must generate it's own gravitational field, right? So would the answer be four seconds?

Also, if I'm given this problem in the future on a free response, how do I do it without plugging in the given periods to the formula?

 

[SteamKing]

Homework Statement

The cylindrical space station, d = 280m in diameter, rotates in order to provide artificial gravity of g for the occupants. How much time does the station take to complete one rotation?

Homework Equations

velocity= (2*pi*r)/period
acceleration = v^2/r

The Attempt at a Solution

It's a multiple choice problem, so the possible answers are 34s, 38s, 24s, and 4s.
I'm thinking that to provide it's own gravity, the acceleration should be about equal to the Earth's acceleration due to gravity (9.8 m/s^2). Thus, I've found the acceleration that'll occur for each time.
a(38s)=[(2pi140m)/38s]^2/140m
a(38s) = 3.83 m/s^2
a(34s)=[(2pi140m)/34s]^2/140m
a(34s) = 4.78 m/s^2
a(24s)=[(2pi140m)/34s]^2/140m
a(24s) = 9.60 m/s^2
a(4s)=[(2pi140m)/34s]^2/140m
a(4s) = 345 m/s^2
Now, when the period is four seconds, the acceleration is far greater than the Earth's acceleration due to gravity, so it must generate it's own gravitational field, right? So would the answer be four seconds?

What does this mean? AFAIK, the only thing which generates a gravitational field is a large mass.

Think about which acceleration is closest to the acceleration due to gravity on the Earth's surface.

What would your weight be if you were on a planet where the acceleration due to gravity was 345 m/s².

What would your weight be? Could you even move? Would your body even be able to survive?

 

[ehild]

Homework Statement

The cylindrical space station, d = 280m in diameter, rotates in order to provide artificial gravity of g for the occupants. How much time does the station take to complete one rotation?

Homework Equations

velocity= (2*pi*r)/period
acceleration = v^2/r

The Attempt at a Solution

It's a multiple choice problem, so the possible answers are 34s, 38s, 24s, and 4s.
I'm thinking that to provide it's own gravity, the acceleration should be about equal to the Earth's acceleration due to gravity (9.8 m/s^2).

Yes, the centripetal acceleration of the occupants is about g. So you have the equation g=v²/r . You need the time period T. There is the other equation v=(2πr)/T. Can you write an expression of T in terms of g and r? T=2πr/v, v=√(gr), substitute the second equation for v into the first equation.

Now, when the period is four seconds, the acceleration is far greater than the Earth's acceleration due to gravity, so it must generate it's own gravitational field, right? So would the answer be four seconds?

Also, if I'm given this problem in the future on a free response, how do I do it without plugging in the given periods to the formula?

At T=4 s, the acceleration is much more than g. The astronauts feel a very high centrifugal force, about 35 g, that presses them to the outward wall of the space station. It is not easy to survive! https://answers.yahoo.com/question/index?qid=20070101085015AAok4JJ