How Is the Coefficient of Friction Calculated for an Amusement Park Ride?

[chicagobears]

Homework Statement

In an amusement park ride, people are spun at 5 radians per sec in a 3 m cylinder. What is the coefficient of friction to prevent people from falling down?

Homework Equations

Centripetal Force= (mv^2)/r
Force of friction=(mu)(mg)
Velocity=(r)(ω)

The Attempt at a Solution

V=(3 m)(5 rad/sec)=15 m/s
Centripetal Force=(M(15^2))/3=75M
Centripetal force=force of friction...
75M=9.8(mu)M
75/9.8=(mu)=7.653?

 

[chicagobears]

No need to answer. I get this now...

 

[Delphi51]

Is it a vertical cylinder of radius 3 m? (A 3m cylinder suggests the diameter is 3.)
With the people above the floor of the cylinder?
If so, the centripetal force is the normal force, not the force of friction.

 

[chicagobears]

yea that was my mistake...