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Force needed to stop rotating drum

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A rotating drum is stopped by a brake actuated by a hydraulic cylinder as shown. The moment of inertia for the drum is Io=20 kg-m2. The friction coefficient between the brake shoe and the drum is 0.40. The dimensions are as given in millimeters. Find the force (in N) applied by the hydraulic cylinder to stop the drum in 75 revs if the initial drum speed, omega1, is as given below.

    Hint: Io is the moment of inertia for the drum. The initial rotational kinetic energy of the drum is 0.5Ioomega2. The rotation will stop when the work of the friction force (Work=Force x distance) is equal to this kinetic energy.

    omega1[RPM] = 252;

    http://www.mech.uq.edu.au/courses/engg1010/q/brake.jpg



    2. Relevant equations

    KE = 10*omega^2 (given)
    Distance = Circumference * Revolutions
    Ff = uN

    and finally KE = Ff*d

    3. The attempt at a solution

    I can't seem to get this question right.

    KE = 10*omega^2 (given)
    distance = circumference * revolutions = 37.5pi
    Ff = uN = 0.4*N

    and if Ff*d = KE, then:
    0.4 * N * 37.5pi = 10*omega^2
    N = 0.21220659*omega^2

    Then your mechanical advantage should end up being 3 if my maths is correct, so the answer should equal N/3.

    However, I keep being told I'm way off, is there something I'm missing?

    Cheers,

    Eddie
     
  2. jcsd
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