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A circular motion problem

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    Centripetal Force= (mv^2)/r
    Force of friction=(mu)(mg)

    3. 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...
  2. jcsd
  3. Nov 13, 2011 #2
    No need to answer. I get this now....
  4. Nov 13, 2011 #3


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    Homework Helper

    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.
  5. Nov 13, 2011 #4
    yea that was my mistake...
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