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Circular Motion and Friction

  1. Feb 8, 2006 #1
    The coefficient of friction between a certain brass block and a large revolving turntable is µ = 0.23. How far from the axis of rotation can the block be placed before it slides off the turntable if it is rotating at 33 1/3 rev/min?

    ---

    the only force on the block is the friction in the same direction as the centripetal acceleration, right?
    so...

    f = m * a(centripetal)
    0.23 * N = m * v^2 / r = mRω^2, where ω = 2*pi*F(frequency), F = 1/(33 1/3) = 0.03 and N = mg.

    so, simplifying,
    0.23*m*g = 100/3*m*R*ω^2,
    0.23*g = 100/3*R*w^2
    0.0069*g = R(w^2)
    R = 0.06762 / w^2, where w^2 = 0.03553
    R = 1.9032 meters, which, apparently, is wrong. but why?

    my work seems right, and the answer sounds right...
    what have i done wrong?
     
  2. jcsd
  3. Feb 8, 2006 #2
    General tip - always do the algebra with letters only and substitue numerical values only at the end.
    Your calculation should look like this:
    miu * m* g = m * R * w^2
    miu * g = R * w^2
    R=(miu * g)/w^2 = (0.23 * 9.8)/(2 * pi * (33.33/60))^2=0.185m=18.5cm
     
  4. Feb 8, 2006 #3
    how did you get 33.33/60?
    where did 60 come from?
    is that converting it to rev/hour? if so, why? it's not part of the question btw. i'm just wondering. =)
     
    Last edited: Feb 8, 2006
  5. Feb 8, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    It looks like you are using F = 1/T, where T = period. No need for that, since you are given the frequency (you just have to convert it to standard units).

    33 1/3 is the frequency in rev/min; you need the frequency in rev/sec. (That's where the (33.33)/60 comes from.)
     
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