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Rotational Inertia of a block

  • #1

Homework Statement


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.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_37.gif

Homework Equations



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

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
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
I = Icom + M(h^2)
I = 1/12M(a^2+b^2)

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