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Brain Buster Need Help

  1. Nov 11, 2004 #1
    One of the amusements at many carnivals is a rotating cylinder. The customers step inside and stand with their backs to the wall. The cylinder spins very rapidly, and at some angular velocity the floor is pulled away. The thrill-seekers now hang like flies on the wall. If the radius of the cylinder is 5.6 m, and the coefficient of static friction between the people and the wall is 0.5, what is the minimum angular velocity, w, at which the floor can be withdrawn?
  2. jcsd
  3. Nov 11, 2004 #2
    i used the equation W=√g/µr but it was incorrect
  4. Nov 11, 2004 #3


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    This problem is pretty straightforward if you draw an FBD from the rotating reference frame. The normal force exerted by the cylinder is going to provide the centripetal acceleration.

    [tex]F_{friction}=\mu m a_c[/tex]
    [tex]\mu m a_c = m g[/tex]
    [tex]\mu r \omega^2 = g[/tex]
    [tex]\omega=\sqrt{\frac{g}{\mu r}}[/tex]

    Looks like you were right.
    Now plugging in numbers:
    [tex]\omega=\sqrt{\frac{9.81}{0.5 \times 5.6}} \approx 2 s^{-1} [/tex]
    (0.5 has only one sig fig).
  5. Nov 12, 2004 #4
    thanks for the help i used that equation but for some reason the computer didnt like it the course tells me to use the significant figure rule then doesnt use it later so the accepted answer was 1.872s-1 but thanks for the help again i really appreciate it.
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