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(2) Pulley Problems (Acceleration)

  1. Feb 10, 2006 #1
    I have two problems, they are labeled in the attached picture.
    [​IMG]

    Here is the first question: A rope is fastened to the ceiling and passes under a wheel and then back up again. If the mass of the wheel is 2 kg, its moment of inertia is .04 kg m^2 and the radius of the wheel is .3m, find the upward acceleration if a force of 50N is applied to the rope. (ANS 33 m/sec^2)

    Here is what I have.

    Since the pulley is supported by the ceiling, this means that half of the weight is supported by the upward force. So I subtract half of the weight from the total force

    F = 50 N - .5(2)(9.8) = 40.2 N

    I then figure out the torque caused by the moment of inertia and the rotational acceleration.

    Torque = I alpha

    a = F / m, a = 40.2 / 2, a = 20.1

    alpha = a / r, alpha = 20.1 / .3m, alpha = 67

    Torque(wheel) = .04 * 67 = 2.68

    T = rF
    Torque (caused by force) = .3 * 40.2 = 12.06

    Torque (net) = 12.06 - 2.68 = 9.38

    9.38 = .3 F, F = 31.27,

    F = ma, a = 31.27 / 2 = 15.63
    Obviously Incorrect


    Here is the second question:

    Find the rotational acceleration of the pulley. (ANS 88.35 rad / sec^2)

    So I start by figuring out the moment of inertia of the pulley, since this is a combination pulley, the moment of inertia is the sum of the individual pulleys.

    SO,
    (Inertia of a disk, I = 1/2MR^2)
    I = 1/2(1)(.02m)^2 + 1/2(4)(.06)^2 = .0002 + .0072 = .0074

    T1 =3 * 9.8 * .02 = .588
    T2 = 5 * 9.8 * .06 = 2.94

    T(net) = 2.94 - .588 = 2.352

    2.352 = .0074 (alpha)
    alpha = 317.83

    Which isn't correct either.


    Thanks,
    Any help would be appreciated.
     
  2. jcsd
  3. Feb 11, 2006 #2

    lightgrav

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    no, the pulley is not supported by the ceiling ... the ROPE end is.
    Drop the other end of the rope and the pulley falls freely.
    Also, the total upward Force is the SUM of the Tension in both rope halves.

    step 1: draw the diagram and notice what axis is being rotated around.
    [hint: one side of the rope is not moving; the pulley "doesn't slip"]
    step 2: sum torques around that axis, due to all Forces. set = I alpha.
    [hint: this is the I around that axis on the edge of the pulley, "parallel"]
    [and the rope that is connected to the ceiling has zero lever-arm]
    step 3: sum vertical Forces, due to all Forces. set = m a .
    step 4: now relate a to alpha , and solve for whichever you choose.

    In problem 2, you must include the "rotational Inertia" of the hanging masses
    [hint: 100% of each mass is going the same speed as its rope, treat as "3kg(.02^2) + 5kg(.06^2)"].
    With small pulleys, this has to reduce to Atwood's alpha=(5*6-3*2)g/(5*.0036+3*.0004)
     
    Last edited: Feb 11, 2006
  4. Feb 11, 2006 #3
    I am still confused with the first question, lightgrav. Is it possible for you to give a broader hint? Thank you.
    By the way, is the torque acting on the wheel equals to I(a/r^2)? And is the total force acting on the wheel equals to 50N-mg?
     
    Last edited: Feb 11, 2006
  5. Feb 11, 2006 #4

    lightgrav

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    When a wheel (or pulley) "rolls without slipping", the place that it is rolling on is at that moment at rest, and is the instantaneous axis of rotation. That's the left side of the pulley, where it "rolls" up the left rope side.

    So you have to add the MR^2 of the pulley's center-of-mass (around this axis)
    to its "regular" I around its center-of-mass [look up "parallel-axis"].

    The torque around this momentary axis is T_right*2R - mg*R .
    It causes I alpha; alpha = a/R (... not a/R^2 ... which has wrong units!)

    The Tension in the rope's left-hand side is ALMOST equal to the Tension in its right-hand-side. (T_right *R - T_left *R = I_around_center * alpha)
     
  6. Feb 11, 2006 #5
    So here is what I have done


    step 1: The axis of rotation is into the screen/paper.
    step 2: 19.6 From the left, 50 from the right,
    resulting is 30.4
    30.4 * .3 = I alpha
    alpha = 9.12 / I = 9.12 / .04 = 228 rad /sec ^ 2
    step 3:
    19.6 up from left, 50 up from right = 69.6 = ma
    a = 69.6 / 2 = 34.8

    step 4: now relate a to alpha , and solve for whichever you choose.

    alpha = a / r
    alpha = 34.8 / .3 = 116 rad / sec ^ 2

    alpha(net) = 228 - 116 = 112 rad / sec^2

    a = alpha r
    a = 112 rad/sec^2 * .3 = 33.6 m/sec^2..

    Does this look ok?
     
  7. Feb 11, 2006 #6

    lightgrav

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    the right-hand rule shows the axis of rotation directed OUT of the screen.

    (the 19.6 causes NEGATIVE TORQUE ... don't add Forces and multiply by R)
    To obtain the correct I, around the far left edge of the pulley, you
    have to add the MR^2 of the pulley's center-of-mass (around axis @ left)
    to its "regular" I around its center-of-mass ["parallel-axis"].

    (you can skip step 3 if you don't need to find the Force by the ceiling,
    finding a from alpha.)
     
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