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Homework Help: Spinning carnival ride - need to find static friction coefficient, HELP!

  1. Oct 1, 2005 #1
    Hello! I have this problem I must submit by tomorrow 11 PM and I am getting the wrong answer and cannot figure out what I'm doing wrong. Any help would be great. Here is the problem:

    "In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See Fig. 5-35.) The room radius is 4.7 m, and the rotation frequency is 0.5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?"

    My work so far:

    T = 1/f, so T (period) = 1/.5 =2. To find velocity, 2(pi)(r)/T, so 2(pi)(4.7)/2 = 14.765 m/s

    3 Forces on the people in the spinner. A static friction force that goes in toward the circle made by the ride, and a weight force that goes down (mg) and a Normal Force that goes up. No movement on the y axis so mg=Normal Force.

    Static Friction = coef. of stat fric X Normal Force. or coef X mg.

    F = ma, a = v^2/r so F = m X v^2/r

    Set m X v^2/r = coeff. X mg....so mass cancels out leaving v^2/r = coeff X g. Then I plug in numbers (14.765)^2/4.7 = coeff X 9.8 I get for the coefficient 4.73 which is the WRONG ANSWER! :surprised

    Anyone have any clue what I am doing wrong? I would appreciate help so much, thank you!!
  2. jcsd
  3. Oct 1, 2005 #2


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    You have your coefficient on the wrong side of the equation. Can you see why it's the wrong side?
  4. Oct 1, 2005 #3
    Isn't the coefficient multiplied with the normal force (or mg)? What do you mean it is on the wrong side?
  5. Oct 1, 2005 #4


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    The force of friction is proportional to the normal force and the proportionality constant is the coefficient of friction. The normal force is the centripetal force and NOT the weight of the person.
  6. Oct 1, 2005 #5
    So normal force does not equal mg. How would I find the normal force then?

    Thank you for your help BTW.
  7. Oct 1, 2005 #6


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    The normal force is the centripetal force which is the mass times the centripetal acceleration.
  8. Oct 1, 2005 #7
    so normal force = m X v^2/r = m X 14.45^2/4.7 = 44.426 X m

    coeff X 44.426 X m = mg

    44.426 X coeff = 9.8

    coeff = .221?

    EDIT: nope, I'm wrong again :frown: oops i typed wrong velocity in
    Last edited: Oct 1, 2005
  9. Oct 1, 2005 #8
    OK! I got it right, it is .211 (once I typed the right velocity in, getting sleepy, LOL).

    Thank you so much for your help, Tide. I see what I was doing wrong now.
  10. Oct 1, 2005 #9


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    Looks good - but I am not responsible for numbers! :)
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