- #1

BiGyElLoWhAt

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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.