Calculating Artificial Gravity on Rotating Space Station: Finding rA, rB, and g

[avenue]

The question says...
To create artificial gravity, the space station shown in the drawing (THE DRAWING SHOWS THREE CHAMBERS, WITH THE TWO OUTER CHAMBERS BEING A'S AND THE MIDDLE BEING B) is rotating at a rate of 0.720 rpm. The radii of the cylindrically shaped chambers have the ratio rA/rB = 4.00. Each chamber A simulates an acceleration due to gravity of 5.70 m/s2. Find values for (a) rA, (b) rB, and (c) the acceleration due to gravity that is simulated in chamber B.
Here's what I've done so far..
mg=mv^2/r and m's cancel out
therefore g=v^2/r
5.70m/s2 = v^2/r(a)
(.720rev/min)*(2pi*r(a)/1rev)*(1min/60 sec) = 2pi*r(a)/60s which is m/s
Since 5.70m/s2 = v^2/r, 5.70= [(2pi*r(a)/60)^2]/r(a)
Anyway after solving for r(a),i get 519.777
That's incorrect. I can't continue until I get r(a). CAN SOMEBODY OUT THERE PLEASE HELP ME. I'M DESPERATE.

 

[Integral]

How do you know it is incorrect?

If you are doing a web based system, you might try entering the number with the correct number of sig figs.

 

[avenue]

I've submitted the same problem with different numbers. I got it wrong. I only have one more try, with different numbers. I just worked it out the same with the new numbers and assumed it was wrong because I used the exact method.

You have a +/-2% margin of error on your answer - so i ususally just try to round as less as possible.

 

[avenue]

Can anybody out there help. I'm running out of time.

 

[Chi Meson]

(.720rev/min)*(2pi*r(a)/1rev)*(1min/60 sec) = 2pi*r(a)/60s which is m/s
Since 5.70m/s2 = v^2/r, 5.70= [(2pi*r(a)/60)^2]/r(a)

It seems you dropped the 0.720 rev/min.

 

[avenue]

hmmm. i did drop that. Is that my only mistake? Did I work it out right?

 

[avenue]

help me somebody. I'm desperate

 

[Chi Meson]

I believe that was your only mistake. I get (with 3 sigs) a VERY ROUND NUMBER as an answer for r(a).