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Minimum angular velocity formula

  1. Mar 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A ride in an amusement park consists of a cylindrical chamber that rotates around a vertical axis as shown in the diagram below. When the angular velocity is sufficiently high, a person leaning against the wall can take his or her feet off the floor and remain "stuck" to the wall without falling.

    Construct an expression for the minimum angular velocity that the ride could rotate at such that the person remains stuck to the wall. Use the following Use the following when entering your symbolic expression:

    m : for the mass of the person
    g : for the gravitational field strength near the surface of the earth
    r : the radius of the cylindrical chamber (from the center to the walls)
    mu : for the coefficient of friction between the person's back and the wall
    pi : for π = 3.141592654...

    2. Relevant equations
    |FN| = mv2/r the inward normal force
    |Fs| = μs|FN|maximum force of static friction
    Fs| = mg

    3. The attempt at a solution
    I thought this was the answer but it is not correct.
    v = sqrt((g*r)/mu)

    ANy help would be appreciated!
  2. jcsd
  3. Mar 21, 2007 #2


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    Gold Member

    Setting the vertical forces to cancel out I get

    m*g = mu*m*r*w^2

    w = sqrt(g/(r*mu))
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