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 \( I=m d^{2}\left[\begin{array}{ccc}18 & 0 & 7 \\ 0 & 19 & 0 \\ 7 & 0 & 13\end{array}\right] \) 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: \( 1.23 m d^{2} \)

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 \( = \frac{31+\sqrt{221}}{2}\, m d^2 \)