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Max/min values of v before m slips

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A very small cube of mass m is placed on the inside of a funnel rotating about a vertical axis at a constant rate of v revs/sec. The wall of the funnel makes an angle theta with the horizontal. If the coefficient of the static friction between the cube and the funnel is Mu, and the centre of the cube is a distance r from the axis of rotation, what are the a) largest and b) smallest values of v for which the cube will not move with respect to the funnel?



    2. Relevant equations

    friction static = MuN
    F = Ma

    3. The attempt at a solution
    So, I found the min value of V.. I rotated my axis by theta and broke down my force vectors F = m v^2/r and W = mg. Then MuN = mgsintheta + m(v^2/r)costheta and then you can solve for v.. that's the minimum value of v.. I think I understand the question conceptually (the min value would be the friction keeping the block from falling, and the max value would be the value when which the block's frictional force overcomes everything else and starts sliding up). But how do I find the max value? thanks!
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    okay, I think I got it.. since it will be v^2/r, +/- the appropriate V will be the solutions..
    is this right? thanks
     
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