A system of three particles with masses $$ m_{1}=5 m $$ at position $$ (0,0,-d), m_{2}=7 m $$ at position $$ (d, 0, d) $$, and $$ m_{3}=6 m $$ at position $$ (0, d, 0) $$ has an inertia tensor of $$ I=m d^{2}\left[\begin{array}{ccc}18 & 0 & 7 \ 0 & 19 & 0 \ 7 & 0 & 13\end{array} ight] $$ about the origin. What are the principal moments of inertia of the system? Enter the moments of inertia from smallest to largest. First moment $$ = $$ Ex: $$ 1.23 m d^{2} $$ Second moment $$ = $$ Ex: $$ 1.23 m^{2} $$ Third moment $$ = $$ Ex: $$ 1.23 m d^{2} $$

Question

A system of three particles with masses $$ m_{1}=5 m $$ at position $$ (0,0,-d), m_{2}=7 m $$ at position $$ (d, 0, d) $$, and $$ m_{3}=6 m $$ at position $$ (0, d, 0) $$ has an inertia tensor of $$ I=m d^{2}\left[\begin{array}{ccc}18 & 0 & 7 \ 0 & 19 & 0 \ 7 & 0 & 13\end{array}ight] $$ about the origin. What are the principal moments of inertia of the system? Enter the moments of inertia from smallest to largest. First moment $$ = $$ Ex: $$ 1.23 m d^{2} $$ Second moment $$ = $$ Ex: $$ 1.23 m^{2} $$ Third moment $$ = $$ Ex: $$ 1.23 m d^{2} $$

UpStudy AI · Accepted Answer

We start with the given inertia tensor
$$
I = m d^2 \begin{pmatrix} 18 & 0 & 7 \ 0 & 19 & 0 \ 7 & 0 & 13 \end{pmatrix}.
$$
The principal moments of inertia are the eigenvalues of the matrix
$$
A = \begin{pmatrix} 18 & 0 & 7 \ 0 & 19 & 0 \ 7 & 0 & 13 \end{pmatrix}.
$$

**Step 1. Write the eigenvalue equation**

An eigenvalue $$\lambda$$ satisfies
$$
\det(A - \lambda I) = 0.
$$
Thus,
$$
\det\begin{pmatrix} 18-\lambda & 0 & 7 \ 0 & 19-\lambda & 0 \ 7 & 0 & 13-\lambda \end{pmatrix} = 0.
$$

**Step 2. Compute the determinant**

The determinant is computed by expanding along the second row (or any row/column that simplifies the calculation). Notice that the $$ a_{12} $$ and $$ a_{32} $$ elements are already zero, so we can write
$$
\det(A-\lambda I) = (19-\lambda)\det\begin{pmatrix} 18-\lambda & 7 \ 7 & 13-\lambda \end{pmatrix}.
$$
Now,
$$
\det\begin{pmatrix} 18-\lambda & 7 \ 7 & 13-\lambda \end{pmatrix} = (18-\lambda)(13-\lambda) - 7\cdot7.
$$
So the characteristic equation becomes
$$
(19-\lambda)\Bigl[(18-\lambda)(13-\lambda)-49\Bigr] = 0.
$$

**Step 3. Solve for the eigenvalues**

There are two factors:

1. $$19-\lambda=0$$ gives 
$$
\lambda = 19.
$$

2. The quadratic factor:
$$
(18-\lambda)(13-\lambda)-49=0.
$$
First, expand:
$$
(18-\lambda)(13-\lambda) = 234 - 18\lambda - 13\lambda + \lambda^2 = \lambda^2 - 31\lambda + 234.
$$
Thus, the equation becomes
$$
\lambda^2 - 31\lambda + 234 - 49 = \lambda^2 - 31\lambda + 185 = 0.
$$
We solve this quadratic by using the quadratic formula:
$$
\lambda = \frac{31 \pm \sqrt{31^2-4\cdot185}}{2}.
$$
Calculate the discriminant:
$$
31^2 - 4\cdot185 = 961 - 740 = 221.
$$
Thus,
$$
\lambda = \frac{31 \pm \sqrt{221}}{2}.
$$

**Step 4. List the principal moments in increasing order**

The three eigenvalues (and hence the principal moments of inertia) are:
$$
\lambda_1 = \frac{31-\sqrt{221}}{2} \, m d^2,\quad \lambda_2 = 19\, m d^2,\quad \lambda_3 = \frac{31+\sqrt{221}}{2}\, m d^2.
$$
To check the order, we can approximate:
$$
\sqrt{221} \approx 14.866.
$$
Then,
$$
\lambda_1 \approx \frac{31-14.866}{2} \approx \frac{16.134}{2} \approx 8.067\, m d^2,
$$
$$
\lambda_2 = 19\, m d^2,
$$
$$
\lambda_3 \approx \frac{31+14.866}{2} \approx \frac{45.866}{2} \approx 22.933\, m d^2.
$$

Thus, from smallest to largest:

- First moment $$ = \frac{31-\sqrt{221}}{2}\, m d^2 $$
- Second moment $$ = 19\, m d^2 $$
- Third moment $$ = \frac{31+\sqrt{221}}{2}\, m d^2 $$
The principal moments of inertia from smallest to largest are: - First moment $$ = \frac{31-\sqrt{221}}{2}\, m d^2 $$ - Second moment $$ = 19\, m d^2 $$ - Third moment $$ = \frac{31+\sqrt{221}}{2}\, m d^2 $$

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