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Moment of inertia of a set of spinning disks

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    The shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.0000 m and (total) mass M = 100.0 mg. The disks are uniform, and the disk arrangement can rotate about a perpendicular axis through its central disk at point O. (a) What is the rotational inertia of the arrangement about that axis? (b) If we approximated the arrangement as being a uniform rod of mass M and length L, what percentage error would we make in calculating the rotational inertia?



    M = 100.0 mg = 1×10-4 kg
    L = 1.0000 m
    # of disks = 15

    2. Relevant equations

    Inertia of a disk about a central diameter
    I = (MR2)/2

    Inertia of set of particles
    I = ∑miri2

    3. The attempt at a solution
    Itotal = (MR2)/4 + ∑miri2
    Itotal = ML2 / 4⋅302 + (2M/15)(1 + 4 + 9 + 16 + 25 + 36 + 49)(L/15)
    Itotal = 8.298 kg - m2

    Did I have the right solution?
     
    Last edited by a moderator: Apr 15, 2017
  2. jcsd
  3. Feb 16, 2017 #2

    BvU

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    Depends. They want you to compare it with ##I## for a rod with the same M and L. What did that give you ?
    Oh, sorry, that's part b. Still: a nice check...

    [edit] and for part a) I would check my dimensions if I were you. And the problem statement: are the disks really that light ?
     
  4. Feb 16, 2017 #3

    Simon Bridge

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    Hint: parallel axis theorem.
     
  5. Feb 16, 2017 #4

    BvU

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    Hi Simon !
    I think the parallel axis theorem is there, but the work is a bit sloppy. M and m , numerator squared, denominator not, etc.
     
  6. Feb 16, 2017 #5
    I tried to separate the rotational inertia of the center disk and the remaining fourteen disks. Is that how it is done? I'm still having trouble finding the rotational inertia of the arrangement.
     
  7. Feb 16, 2017 #6

    BvU

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    Ah, yes. So SImon was right. Do you know about the parallel axis theorem ? You'll need it here.
     
  8. Feb 16, 2017 #7
    Yeah, I do know about it.
    So I should just find the moment of a single disk then add it to its md2 for each disk? Is that correct?
    Making the solution for a as:
    0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142)
     
  9. Feb 16, 2017 #8

    BvU

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    Go slower. There are terms missing your last line.
    yes, if by ##d## you mean the distance from the axis to the center of mass.

    "moment of a single disk" for each disk is what I'm missing ...
     
  10. Feb 16, 2017 #9
    Yeah, what I'm referring to d is the distance of one disk to the axis of rotation O.
    What I mean about moment of a single disk is its moment when spun at its center.
    So for example, the outermost disk has a distance of 14r from its center, then can I consider its moment to be 1/2 mr^2 + m(14r)^2.
    Then next to that disk is a disk 12r from the axis of rotation, then can I use 1/2 mr^2 + m(12r)^2.
    and so on... adding those all together in the end.
     
  11. Feb 16, 2017 #10

    BvU

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    Agreed. But I don't see the ##I## for each disk around its own center, except for the disk in the middle.
     
  12. Feb 16, 2017 #11
    upload_2017-2-17_0-6-39.png
    This is what I'm referring to... The first term is for the center disk while the remaining is for the next ones, times two because there are rods from the left and right of the axis.
    This is letting m be the mass of each disk and r be the radius of each disk too.
     
  13. Feb 16, 2017 #12

    BvU

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    upload_2017-2-17_0-6-39-png.113311.png

    is the same as

    0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142) ???
     
  14. Feb 16, 2017 #13
    Sorry, I was confused at first... nevermind the former equation: 0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142)
    Is the other equation, the one in the image, the correct one I should use? Is that how parallel axis theorem applied in this problem?
     
  15. Feb 16, 2017 #14

    BvU

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    Seems to me the picture comes from the solution manual ?:) and we can't argue with that, can we :rolleyes: ?

    But already before we had
    answered with "yes" ! So the thing to do -- now that spoiler is in existence -- is check that your plan and the picture are in agreement. (And hopefully also see why I kept asking for the ##I## of the off-center disks around their own center :smile:)
     
  16. Feb 16, 2017 #15
    Found the inertia using that equation to be 8.352*10^-6 kg-m^2
    When it is considered as a thin rod of length 1 m, the inertia becomes 8.333*10^-6 kg-m^2
    The percent error becomes: 0.2217% (pretty low)

    BTW, I didn't get the image from a manual =( I used a math editor 'cause I don't know how to use LATEX here.
     
    Last edited: Feb 16, 2017
  17. Feb 16, 2017 #16

    BvU

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    So much the better !

    You sure these disks are so light ? And then you still find a total ##I## of
    8.352*10^6 kg-m^2 for such a small rod ?
     
  18. Feb 16, 2017 #17
    Oh, sorry, the exponent are raised to a negative power :smile:
     
  19. Feb 16, 2017 #18

    BvU

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    ##\LaTeX## is easy ! Much faster than [ SUB] [ /SUB] etc.:

    $## ##$ {1\over 2} mr^2 + 2 \; {1\over 2} mr^2 \left ( 1 + 2^2 \right ) + ... $## ##$ yields$$
    {1\over 2} mr^2 + 2 \; {1\over 2} mr^2 \left ( 1 + 2^2 \right ) + ... $$

    Learn by right-clicking over a formula and Pick Show math as ..##\TeX## commands
     
  20. Feb 16, 2017 #19
    a rod of mass M and length L has moment of inertia,I=ML2/12
    but it is only for a thin rod
    here the rod is a line passing through centers of disks,it's M.I constitute only small part of total M.I of the arrangement
    here,the moment of inertia the arrangement is comparable to the M.I of a rod of length L and mass M and width=2r=L/15
     
  21. Feb 16, 2017 #20
    So I should still consider the radius when calculating the moment of the rod? Why so?
     
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