Uniform Circular Motion Space Station question

[winnayy]

Homework Statement

A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), as shown in the figure (http://session.masteringphysics.com/problemAsset/1057181/4/GIANCOLI.ch05.p048.jpg). The circle formed by the tube has a radius of about 1.9 km. What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the Earth (1.0g) is to be felt?

g = 9.8 m/s²
r = 1900m/2 = 950m

Homework Equations

g = (4\pir²/T²)

The Attempt at a Solution

Solve for T:
T = \sqrt{}(4\pir²/g)
T = 2\pi\sqrt{}(r/g)

Plug in known values:
T = 2\pi\sqrt{}(950m/9.8m/s²) = 61.86266614 sec/rev

Seconds in a day:
(24hr/day)(60min/hr)(60sec/min) = 86400sec

(86400s/day)/(61.86266614sec/rev) = 1396.642936 rev/day

Rounded to two significant figures, this is 1400 rev/day, but MasteringPhysics keeps telling me I'm wrong. I've tried another way finding the velocity and diving that into the circumference to get T, and I get the same answer... so I'm confused as to where my mistake lies.

Thanks in advance for the help!

 

[tiny-tim]

welcome to pf!

hi winnayy! welcome to pf!

(have a pi: π )

no, g isn't 4πr²/T²

(and is r 1900 or 950? )

 

[winnayy]

Thank you, Tim =).

g = 9.8 m/s²
r = 950m
(Sorry for the confusion.)

Using the formula for centripetal acceleration:
a = v²/r
9.8 = v²/950
v = 96.4883416 m/s

Rotation speed:
c = 2\pir = 5969.926942m
P = c/v = 5969.926942m / 96.4883416m/s = 61.86266592 s

Number of revolutions per day:
(24hr/day)(60min/hr)(60sec/min) = 86400s/61.86266592s = 1396.642041 rev/day

I still get the same number using a different equation? :(

 

[tiny-tim]

hi winnayy!

your solution looks ok to me

 

[winnayy]

ah~ It looks okay to me, too, so I can't figure out why Mastering Physics won't accept it x___x. Thank you, though, Tim!

edit: /facepalm. You were right to question what the radius was! I was reading incorrectly this whole time and blew the number they gave us to be the diameter, sigh. Thank you for your help, Tim =).