Radius to angular velocity relationship (constant angular accel)

[siderealtime]

Homework Statement

I'm trying to create a graph showing the relationship between radius and angular velocity for a toriod space station that maintains a constant angular acceleration of Earth's gravity (9.8 m/s^s) on it's rim.

Homework Equations

ac = r$$\omega$$^2

The Attempt at a Solution

I'm having trouble understanding how to show this relationship correctly. I'm proposing putting radius on the x-axis and the solving for $$\omega$$ on the y-axis with a constant ac. So my x just goes from zero to whatever and my y is sqrt(9.8/r) labled $$\omega$$.

This seems odd to me, if anyone could suggest a proper way to represent this relationship I would appreciate it.

 

[Delphi51]

I am probably misunderstanding . . . but isn't the angular velocity the same regardless of radius? If it is making one revolution a minute at the rim, it will be one revolution/minute at the center, too.

 

[siderealtime]

It's maintaining a constant centripetal acceleration of 9.8m/s^2 on the rim of the station.

So, the relationship between radius and angular velocity when this condition is true.

 

[Delphi51]

I realize that the angular velocity is increasing with time, but at any given time it is the same for all radii. Now if you mean the tangential velocity, that is v = r*ω so v does vary with r. At the center of rotation v = 0 and it linearly increases with radius. But ω is the same at all radii.

 

[siderealtime]

Solving for ω in the centripetal acceleration equation ac = rω², keeping ac constant 9.8m/s².

9.8 = rω²

solve for ω

ω = sqrt( 9.8 / r )

It's clear from the above that varying r will vary ω.

Am I incorrect?

 

[Delphi51]

Well, you've got me sweating now! I may well have misunderstood the question.

If we are talking about a particular space station with rim radius R and "at the rim" a = 9.1 then we have ω = sqrt( 9.8 / R).
I put R here to indicate it is the radius of the rim.
At other radii on the same space station, the acceleration will be less than 9.8, but ω the same for the whole space station.

If however, you are haven't built the space station yet and are trying to decide what its rim radius should be, you have
ω = sqrt( 9.8 / R)
and you see that for bigger values of R the ω will be smaller. Okay, this must be what you are doing. Tricky to graph that - looks like a combination of a quadratic and inverse. You'll have to make a table of values and plot the points, perhaps on a spreadsheet, to get it right. It looks pretty much like an inverse square function.

Sorry I got confused!

 

[siderealtime]

Thanks for your help. The relationship that you pointed out is the only one I can come up with. Representing it in a more complex way doesn't seem like a good idea so I'll stick with the way I have it.

I've noted that these calculations ignore the mass of the station that would increase with and increasing radius, so the conclusion that a larger radius would be easier to operate is not true. You have to then consider the torque required to rotate the station and the associated tangential thrust required would increase with increased moment of inertia. So there's a sweet spot somewhere but that isn't part of the question to find it.

Does this second part sound legitimate?

 

[Delphi51]

Do you have a curved graph for w vs r? If not, it doesn't seem right. The graph should clearly show that the w decreases with increasing r, but not linearly. Just a matter of popping that equation into a spreadsheet formula to get a table of values, then telling it to make an x-y scatter plot.

I'm not keen on torque as the measure of difficulty. A small torque acting for a long time could spin up anything. Maybe the kinetic energy required to spin it would be a better measure. It isn't clear to me offhand if it would take more energy to spin the bigger one up to a smaller angular velocity. Would be interesting to work it out. Have to guess at the mass increase, I guess. Or would it be about the same for the same amount of living space?

 

[siderealtime]

Thanks a ton Delphi.

Yes, my graph is exactly as you described.

I'll give the KE idea some thought.

 

[Delphi51]

Most welcome!