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Rotational Inertia of a block

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    The uniform solid block in Fig. 10-37 has mass .172 kg and edge lengths a = 0.035 m, b = .084 m, and c = .014 m. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.


    2. Relevant equations

    I = Icom + M(h^2)
    I = 1/12M(a^2+b^2)

    3. The attempt at a solution

    With the pythagorean theorem, I found the distance it has shifted which was [1/2(a^2+b^2)]^1/2 = h so,

    I = 1/12M(a^2+b^2) + M(h^2) substituting h with the previous discovery giving:

    I = 1/12M(a^2+b^2) + M[1/2(a^2+b^2)] I still come out with an incorrect answer.

    I'm wondering if the c should be included in the problem or not, but when I do input it in the Icom, it still does not work. I must be missing something important

    The answer should be fitting with 4.7 x 10^-4 kg.m^2
  2. jcsd
  3. Apr 24, 2009 #2


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    Homework Helper

    I think your 1/2 is misplaced.

    1/2*(a2 + b2)1/2

    Squaring that for your || axis displacement yields

    (1/12 + 1/4)*(a2 + b2) = 1/3*(a2 + b2)
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