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Finding gravitron rotational frequency

  1. Feb 13, 2015 #1
    1. The problem statement, all variables and given/known data
    In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. What is the minimum rotational frequency, in rpm, for which the ride is safe?

    2. Relevant equations

    I'm not even certain which equations would be relevant here. Starting with a free body diagram, I can see that the weight force would point down, and the opposing force would have to be static friction so that would point up, and the centripetal acceleration would be due to the normal force from the walls. So N = mv^2/r

    (But I don't have a given mass; should I be able to find mass from the given coefficients of friction since friction counters weight in this situation?)

    Force of static friction = u * N

    (But I don't know how to find N-- is it still opposing mg if it is in the horizontal direction? If so, I still don't know how to find m…)

    v= 2π r/T

    Also, v = wr, and then I can convert to rpm… Is this even relevant here, given that I don't have a time period to work with anyway?

    Are my thoughts so far on the right track or am I way out in left field? Any nudge in the right direction would be greatly appreciated.
  2. jcsd
  3. Feb 13, 2015 #2


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    What was wrong with the following?

    The friction coefficient relates a maximal force in the tangential direction of a surface to the normal force. N is still the normal force.
  4. Feb 13, 2015 #3
    Okay... then I guess I'm stuck as to where to go from there. I don't know how'd I'd go about finding normal force without being given mass. I admit that my understanding of friction is not what it needs to be; I can visualize what you're saying, but I can't quite figure out how to relate it back to these formulas. But thank you for the quick reply!
  5. Feb 13, 2015 #4


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    But you have the normal force already as a function of the mass. Do not care about what it is at the moment, leave it in the formulas and see what you get when you impose some relevant condition for the people not falling down. This should clearly not depend on the mass or the safety of the attraction would depend on how fat the visitors are.
  6. Feb 13, 2015 #5
    Oh. You're right, they wouldn't want to exclude visitors based on mass!

    After reviewing friction a bit more, I have:

    Friction opposes gravity here, so Friction = mg

    Friction also = m(a sub c)u

    So I can substitute one of these friction equations into the other-- divide mass out of both sides so it won't matter, as we discussed-- and solve for v. I think I've got it from there. Thank you so much for your help!
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