What is the principal moment of inertia tensor for a lamina?

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SUMMARY

The principal moment of inertia tensor for a lamina is derived as follows: the tensor is represented by the matrix (1/3, -1/2, 0; -1/2, 4/3, 0; 0, 0, 5/3). The principal moment of inertia tensor is calculated by substituting I into the tensor, resulting in the determinant equation (1/3-I)(4/3-I)(5/3-I) - (-1/2)(-1/2)(5/3-I) = 0. The solutions for I are 5/3, (5+√18)/6, and (5-√18)/6. The discussion highlights confusion regarding the irrational nature of the results, which deviates from typical lamina moment of inertia expectations.

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LeoChan
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Homework Statement
A thin uniform rectangular plate (lamina) is of mass m and dimensions 2a by a. Choose a coordinate system Oxyz such that the plate lies in the xy plane with origin at a corner, the long dimension being along the x-axis.

(a) The moment of inertia tensor about the origin
(b) The principal moments of inertia about the origin
Relevant Equations
Inertia tensor
The moment of inertia tensor I found out is
(1/3) (-1/2) 0
(-1/2) (4/3) 0
0 0 (5/3)

The principal moment of inertia tensor I found out is
(1/3-I) . (-1/2) . 0
(-1/2) . (4/3-I) . 0
0 . 0 . (5/3-I)

det of principal of moment inertia = 0
So (1/3-I) (4/3-I)(5/3-I)-(-1/2)(-1/2)(5/3-I)=0
I=5/3, (5+√18)/6, (5-√18)/6

The answer look so weird to me and differ from common lamina moment of inertia
 
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Looks OK to me, except that the mass ##m## and factors of ##a## should be included.
 
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TSny said:
Looks OK to me, except that the mass ##m## and factors of ##a## should be included.
Thanks. But just didn’t expect the answer to be irrational for a lamina, it made me so confused.
 

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