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Centripetal acceleration and friction

  1. Oct 20, 2005 #1
    In a "rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." The room radius is 4.6 m, and the rotation frequency is .5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

    I am totally confused as to how to find that with only the frequency and radius provided. I was able to find the centripetal acceleration, but I am stuck now. I don't have the mass...so what about the forces?

    Thank you so much for all your help.
     
  2. jcsd
  3. Oct 20, 2005 #2
    ok, this might not be correct as I've only just covered this myself...

    Under the conditions you describe the ride should have a radial acceleration of -r*omega^2. From Newton's second law F=ma, so the force exerted on the walls of the ride is equal to -m*r*(omega^2) - therefore the normal reaction force should be opposite to that. As Frictional force = coefficient of friction * normal reaction force, and the frictional force has to equal the weight of the people, mu*R = mg, where R = m*r*(omega^2), the mass cancels out and so it's possible to find mu (I get a very small value of mu but as the ride's moving at an insanely fast speed through a large circle the value one gets should be rather low).

    Cheers,
    Just some guy.
     
  4. Oct 20, 2005 #3
    Ok, that makes perfect sense. But what do you mean when you say omega??

    thanks.
     
  5. Oct 20, 2005 #4
    Angular velocity in radians per second (as it makes a revolution in .5 seconds it's moving at 2 revolutions per second which is 4pi radians per second which is the angular velocity).
     
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