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Homework Help: Mass moment of inertia (angular mass)

  1. Jan 5, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img232.imageshack.us/img232/4976/physicswe5.jpg [Broken]

    total mass of the rod = 4m
    m = 0.4kg; b = 0.4m; a = 1.0m
    rod is thin

    i need to find the mass moment of inertia for the rotating rod relative to z-axis..

    2. Relevant equations

    3. The attempt at a solution

    for the rod which lies on y axis, i could calculate the mass moment of inertia by using the above equation which will lead me to the following answer

    (1/12)*2*0.40*0.8² = 0.0427

    but what should i do with the other parts of the rod which are parallel with the z-axis?..

    or could i apply the following equation?:

    (1/24)*M*L²*sin (2φ)

    where φ = 45

    thx in advance..
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Where does that equation come from??

    Hint: For the rods that are parallel to the axis, all the mass has the same distance from the axis.
  4. Jan 5, 2008 #3
    it's from product of inertia to calculate centrifugal moment
    Jyz = ∫yz dm
    but i think it cant be applied for this type of question..

    so the mass moment of inertia should be like this?:

    (1/12)ML² + 2Mr²

    where r = b

    and the answer is

    0.0427 + 0.128 = 0.1707
  5. Jan 5, 2008 #4

    Doc Al

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    Staff: Mentor

    Looks good.
  6. Jan 5, 2008 #5
    thx alot
  7. Jan 6, 2008 #6
    i have another question..

    i need to find the reaction of the bearing due to the dynamic unbalance..

    is it that the dynamic unbalance is due to the rod parallel to the z-axis only?..
    and could i neglect the rod parallel to y-axis?
  8. Jan 6, 2008 #7

    Doc Al

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    Staff: Mentor

    Yes. It's the rods parallel to the z-axis that are unevenly distributed.
  9. Jan 8, 2008 #8
    here's the last question..

    i need to find the bearing reaction on A and B

    value given : n = 800/min; m = 0.4kg; a = 1.0m; b = 0.4m

    here's my approach to solve the question:

    Moment about x-axis:

    Mx = 2 Jyz*ω²
    = 2ω² ∫ b*(b/2) dm
    = 2ω² ∫ b²/2 dm
    = ω² * (b²/2) * m

    ω = 2(pi)n/60 = 83.776 (1/s)

    Mx = 83.776² * 0.4²/2 * 0.4
    = 449.18 Nm

    FA = FB =
    = 224.59 N

    i wonder if i have done the right approach.. i took b/2 as its center of mass.. so i came up with following equation

    ∫ b*(b/2) dm

    and since there's two parts which is parallel to z-axis.. i time the mass moment of inertia with 2..

    is this the way to answer the question?.. i'm kind of confused with another method to calculate moment of inertia where the rod has an angle φ to the y-axis..
  10. Jan 8, 2008 #9
    i came up with another approach

    J = Jz1 + Jz2
    = 2mr²
    = 2*0.4*0.4²
    = 0.128

    Mx = Jω²
    = 0.128 * (2(pi)n/60)²
    = 898.35 N

    FA = FB = Mx/2a
    = 449.175 N

    so..which one is the right approach? or is there any another approach?
  11. Jan 8, 2008 #10

    Doc Al

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    Staff: Mentor

    I'd say that your first solution (in post #8) looks good to me. I'm a physicist, not an engineer, so I'm a bit rusty on all the tricks for calculating these things that I suspect one learns in statics classes. (That's why I didn't even recognize that formula in your first post! :redface:) And my books are packed up, so I can't look things up.

    The way I look at it is simple. Each parallel rod requires a centripetal force due to the rotation equal to [itex]m\omega^2b[/itex] that effectively acts at its center. The associated torque for each rod is thus [itex](m/2)\omega^2b^2[/itex]. Thus the reactive force at each end point (A & B) must be [itex](m/2a)\omega^2b^2[/itex], which matches your answer.

    If I get a chance, I'll try to remember the more formal way of solving this using moment of inertia tensors. But you'd probably get quicker and better help for these kinds of questions if you posted them in the engineering help section. (I suspect there are plenty of engineers here who know this stuff cold.)
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