What Is the New Friction Force if Angular Speed Is Doubled?

[vaio-911]

Homework Statement

A barrel of fun consists of a large vertical
cylinder that spins about the vertical axis.
When it spins fast enough, any person inside
will be held up against the wall.

http://i.snag.gy/Wh1B9.jpg

Question:

If you double the angular speed (i.e., $\omega _2=2\cdot \omega _1$) what is the new friction force $f_2$?

Answer choices: $2\cdot F_1,1/3,1/4,1/2,3,1.5,1/3.5,1,2.5,4$

Homework Equations

$F_f=\mu_k\cdot F_N$

$v=r\cdot \omega$

$a_c=\dfrac{v^2}{r}$

The Attempt at a Solution

The frictional force is defined as $F_f=\mu_k\cdot F_N$.

The normal force of the object is the centripetal force that is pointing towards the center of the cylinder (perpendicular to surface).

Thus frictional force is

$\begin{align*}F_f&=\mu_k\cdot (m\cdot a_c)\\&=\mu_k\cdot \left(m\cdot \dfrac{v^2}{r}\right)\\ &=\mu_k\cdot\left(m\cdot\dfrac{(r\cdot\omega)^2}{r}\right) \end{align*}$

This means that $\boxed{F_f\propto \omega^2}$

so $F_2=4\cdot F_1$, but when I submitted it, it said that I was wrong.

What am I doing wrong?

 

[NascentOxygen]

Hi vaio-911. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

The frictional force is defined as $F_f=\mu_k\cdot F_N$.

I don't think it is, actually. I think that is the maximum frictional force before relative motion can occur. Isn't that so?

The actual frictional force to secure each person against gravity is their weight, and that doesn't change just because you spin them faster!

Maybe this is a trick questionhttp://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif Is your professor a barrel of fun?