Help with problem..."rotor ride"

[delta_mu]

OK I was assigned this homework sheet on Uniform Circular Motion and Law of Universal Gravitation...and there is only one problem I can not get. If you could, please help me. Thanks in advance.

Here it is:
The "Rotor" ride is the one which presses you against the walls of the spinning rotor as the floor drops away. The coefficient of static friction between the wall and the riders is 0.0580. If the floor dropped away for 2.00 minutes, and during this time the rotor has spun 100 times at a constant rate, what must be the radius of the rotor?

I recieved .006 meters for the radius

[Chen]

You know that the friction needs to cancel the gravity force so the person doesn't fall:
f = mg = N\mu
But what is N? You can find it by finding the radial acceleration of the device:
\Sigma F_r = ma_r = m\omega ^2r
But what is \omega ? You know that the device rotates 100 times in 120 seconds, which makes for a frequency of 0.833Hz. And you know that:
\omega = 2\pi f
And therefore:
\Sigma F_r = ma_r = 4\pi ^2f^2mr
This is equal to the normal force, so:
mg = N\mu = 4\pi ^2f^2mr\mu
And finally:
r = \frac{g}{4\pi ^2f^2\mu}
I get a radius of about 6.17 meters.

[delta_mu]

thanks a lot...it helped out a lot

[delta_mu]

Wait...I am so sorry, I realized that I have no idea as to what equations you are using. The equations I was given are the Force of Friction(Ff=mu times the normal force) and the Centripital Force(Fc=mass times the velocity squared divided by the radius. Thanks again to whoever answers.

[Chen]

Fc=mass times the velocity squared divided by the radius is:
F_c = ma_r = m\frac{v^2}{r}
But since v = \omega r this becomes:
F_c = ma_r = m\omega ^2r
Which is what I used. :smile: