A circular motion problem

  • #1

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?
 
  • #2
No need to answer. I get this now...
 
  • #3
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.
 
  • #4
yea that was my mistake...
 

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