Moment of inertia tensor calculation and diagonalization

  • #1
BiGyElLoWhAt
Gold Member
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Homework Statement


Not sure if this is advanced, so move it wherever.
A certain rigid body may be represented by three point masses:
m_1 = 1 at (1,-1,-2)
m_2 = 2 at (-1,1,0)
m_3 = 1 at (1,1,-2)

a) find the moment of inertia tensor
b) diagonalize the matrix obtaining the eigenvalues and the principal axes (as orthogonal vectors)

Homework Equations


##I_{ij} = m_{\beta}(\delta_{ij}r_{\beta}^2 - x_{i\beta}x_{j\beta})##
##\vec{A} = \vec{P^{-1}}\vec{D}\vec{P}##
##I_{ij}=I_{ji}##


The Attempt at a Solution


I'm going to drop the beta's, but each xyz is associated with the mass attached to the term.
##I_{00} = m_1(x_0^2 + x_1^2 +x_2^2 - x_0^2) + m_2(...) + m_3(...)##
## = 1(5) + 2(1) + 1(5) = 12##
##I_{01} = m_1(-x_0x_1) + m_2(...) +m_3(...)##
##= -1(-1) -2(-1) -1(1) = 2##
##I_{02} = m_1(-x_0x_2) + m_2(...) + m_3(...)##
## = -1(-2) -2(0) -1(1) = 4##
##I_{11} = m_1(x_0^2+x_2^2) + m_2(...) + m_3(...)##
##= 1(5) +2(1) +1(5) = 12##
##I_{12} = m_1(-x_1x_2) + m_2(...) +m_3(...)##
##= -1(2) +2(0) - 1(-2) = 0##
##I_{22} = m_1(x_0^2 +x_1^2) + m_2(...) +m_3(...)##
##=1(2) +2(2) +1(2) = 8##
This gives me
##\vec{I} = \left (
\begin{array}{ccc}
12 & 2 & 4 \\
2 & 12 & 0 \\
4 & 0 & 8 \\
\end{array} \right )##

and

##det(I_{\lambda}) = 0 = -\lambda^3 + 32\lambda^2 - 316\lambda + 928##
I don't even know how to solve that equation, and online calculators give
diagonal(5.35139, 11.2841, 15.3627)
Did I mess up? There's nothing about using a calculator, but I don't know how else to solve this. Did I mess up? I am pretty sure I'm not supposed to end up with a bunch of stupid numbers like this.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
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For what it's worth, I got the same matrix you did.
 
  • #3
BiGyElLoWhAt
Gold Member
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116
Well, maybe I'll just roll with it then. I thought I checked all my algebra carefully. I'm pretty sure I'm supposed to use P^-1 D P to get the diagonal since he said "obtaining the eigenvectors", but maybe I'll just do row operations or something to avoid the decimals. Thanks.
 
  • #4
512
123
Not sure but it looks as if you are using vectors with respect to the origin. Shoudn't you be using vectors relative to the center of mass of the system
 
  • #5
512
123
instead (for mass 1) of x0 = 1 x1 = -1 and x2 = -2 in line 1 of your calculation: try x0 = 1 - x0cm = 1 - 0 = 1 x1 = -1 = x1 cm = -1 - 1/2 = -3/2 and x2 = -2 - x2 cm = -2 - -1 , etc.

this means a lot more calculation because you need to subtract off the center of masses but I think it will lead to a better answer. On the other hand, I'm glad I am not doing the calculation.
 
  • #6
512
123
I ended up with eigenvalues 11; 11/2 +/- sqrt(57) / 2.
 
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  • #7
BiGyElLoWhAt
Gold Member
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That is a lot neater. I didn't think about doing that. I will keep it in mind for next time.
 

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