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Homework Help: Moment of inertia on a slab

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin rectangular slab, with dimensions 0.580 m by 0.830 m and mass 0.150 kg, is rotated about an axis passing through the slab parallel to the short edge. If the axis is 0.230 m from the short edge, what is the moment of inertia of the slab?

    2. Relevant equations


    3. The attempt at a solution

    So, I used the formula for inertia of a slap (I=1/12 x m x L^2) and plugged this in for Icm above:
    Ip=1/12 x M (L^2) + M x h^2
    Ip=1/12 (.150)(.83^2)+.150(.230^2)

    But this wasn't correct, could you tell me where I went wrong?
  2. jcsd
  3. Mar 15, 2010 #2


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    For problems like this I always find it useful to draw a sketch of my object along with the axes I'm taking moments of inertia around.

    Having done that, I think your error is that in applying Ip=Icm+mh^2 (parallel axis theorem) you've taken H to be 0.230 - which is incorrect, since H is the distance from the centre of mass to your new axis, yet 0.230 in this case is the distance from the edge of your slab to your axis.

    So you should use H = 0.185 (which is 0.830/2 - 0.230).

    EDIT: Oops. I said L, but I meant H, sorry.
    Last edited: Mar 16, 2010
  4. Mar 15, 2010 #3


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    Also you used the moment of inertia of a rod.

    You need to use the moment of inertia of a rectangular lamina which is

    [tex]I_c = \frac{1}{12}M(a^2+b^2)[/tex]
  5. Mar 15, 2010 #4
    Our book explains that I=1/12 M L^2 is also used for a slab with the axis through the center parallel to the edge. Isn't I=1/12 M (a^2 + b^2) used if the axis is through the center?
  6. Mar 15, 2010 #5


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    [tex]Rod = \frac{1}{12}ML^2[/tex]

    [tex] Rectangular Plate = \frac{1}{12}M(a^2+b^2)[/tex]
  7. Mar 15, 2010 #6
    I'm just telling you what our book says:

    http://img.photobucket.com/albums/1003/aliceinunderwear/Picture1.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  8. Mar 16, 2010 #7
    I'm afraid using the L as above, and also trying to use the I=1/12 x M (a^2+b^2) were also wrong according to the book.
  9. Mar 16, 2010 #8


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    As you said, I=1/12 M L^2 is for a slab with the axis passing through the center. In this problem the axis doesn't pass through the center.
  10. Mar 16, 2010 #9
    Yes, so I used the parallel axis theorem: Ip=Icm + Mh^2
    Was this incorrect?
  11. Mar 16, 2010 #10


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    The axis is through the centre (at least until we shift it using the parallel axis theorem), but parallel to the short edge, so the moment of inertia will actually be:


    It's easy enough to verify this from first principles (which is normally a good idea when you're unsure if an equation is valid in a certain context) by using the definition of the moment of inertia as an integral.

    Then using the parallel axis theorem and substituting:

    [tex]I_{p}=\frac{1}{12}ML^2 + MH^2[/tex]

    Which is the moment of inertia that you want - but of course you already figured that out. Note that here, L is 0.830m and H is the distance from the edge to the new axis (should be 0.830/2 - 0.230, just by geometry).

    If it's still 'wrong', then perhaps the answer in your book is misprinted? I got:

    [tex]I_{p}= 0.013745 kg m^2[/tex]

    Is that the answer in your book?
  12. Mar 16, 2010 #11


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    I misread the parallel to the edge part.
  13. Mar 18, 2010 #12
    Jmf - yes! Thank you so much!! As it turns out I actually had it right one of the many times I calculated it but I rounded too early and our HW is online and is SO tempermental about how one enters things!
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