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Friction problem on cube of mass

  1. Nov 15, 2005 #1
    a very small cube of mass m is placed on the inside wall of a funnel. The wall of the funnel makes an angle theta with the vertical axis of rotation (dotted line). The center of the cube is a distance r from the axis of rotation. the cube is held by static friciton. The funnel is then rotated about a vertical axis at an angular speed of w (omega). it is found that the angular speed can be increased to a maximum value w(max) (omega max) at which point the cube is no longer held by the static frictional force and it flies out of the funnel. calculate w(max)
    i separated everything into x and y components, but i think i did this wrong
    i got equations
    Fstatic - Fg cos theta = mw(max)^2r
    Fn - Fg sin theta = ma
    am i suppose to set a to zero? and are those even right to begin with
    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 16, 2005 #2


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    It helps to make a drawing with the forces an angles on it.

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  4. Nov 16, 2005 #3
    okay, this is what i got so far, i dont know if its right. correct me if i had a mistake

    taking xaxis is // to side of funnel, and yais is perpendicular to side of funnel, i drew fbd and the forces i got are
    x components : Fs, - mg cos theta
    y components : Fn, - mg sin theta

    so i hvae two equations:
    Fs - mg cos theta = mw^2r (1)
    Fn - mg sin theta = ma
    Fn - mg sin theta = 0
    Fn = mg sin theta
    since mew (u) = Fs/Fn
    Fs = u Fn
    Fs = u mg sin theta
    subbign that into equation (1)
    u mg sin theta - mg cos theta = mw^2r
    masses cancel
    u g sin theta - g cos theta = w^2r
    isolating for w:
    w = sqrt [g((u sin theta + cos theta)/r)]

    is that right?
  5. Nov 17, 2005 #4


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    I dont think so. Your Fs should be negative, but even more serious is that the centripetal acceleration of the cube is horizontal, pointing inwards towards the vertical axis.
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