Physics: Circular Motion Question?

Question

The rotor has a Radius of 6m. If the Ride spins @ 20m/s, what is the minimum coefficient of friction that must exist between a rider and the wall?
[Fc is caused by Fnorm in this problem, therefore
Fc=Fnorm]

Answer 1

The force of gravity W is puling the rider down and the force of friction f is apposing the gravity. Now we have
F=ma where
a= V^2/R

The force of friction f is then
f=uF=mg

u= mg / ma= g/a
u=g/[V^2/R]
u=gR/V^2
Finally coefficient of friction is
u= 9.81x 6 / (20)^2=0.15

Answer 2

The weight of the person is mass x acceleration due to gravity or mg. That is set equal to the centripetal force of the person against the wall times the coef. of friction of the Rotor wall e.g. ...

mg = cmrw^2

where c is the coef of friction, w = angular speed, r is the radius.

w = v/r (where v is the tangential velocity)

mass drops out and your left with

g = cv^2/r

solving for c you get:

c = gr / v^2

c = (9.8 met/sec^2) (6 met) / (20 met/sec)^2

c = 0.147