The inertia tensor for a solid sphere with radius \( b \) and mass \( m \) rotated about the center of mass is \( \left[\begin{array}{ccc}\frac{2}{5} m b^{2} & 0 & 0 \\ 0 & \frac{2}{5} m b^{2} & 0 \\ 0 & 0 & \frac{2}{5} m b^{2}\end{array}\right] \). What are the eigenvalues of the tensor if \( b=5 \) and \( m=4 \) ? Enter the eigenvalues from smallest to largest. \( I_{1}= \) Ex: 1.23 \( I_{2}= \) Ex: 1.23 \( I_{3}= \) Ex: 1.23

We are given the inertia tensor: \[ \mathbf{I} = \begin{bmatrix}\frac{2}{5} m b^{2} & 0 & 0 \\0 & \frac{2}{5} m b^{2} & 0 \\0 & 0 & \frac{2}{5} m b^{2}\end{bmatrix} \] Substitute \( b = 5 \) and \( m = 4 \): \[ \frac{2}{5} m b^{2} = \frac{2}{5} \times 4 \times 5^{2} = \frac{2}{5} \times 4 \times 25 \] Calculate the multiplication: \[ \frac{2}{5} \times 4 = \frac{8}{5} \] and \[ \frac{8}{5} \times 25 = 8 \times 5 = 40 \] Thus, the inertia tensor becomes: \[ \mathbf{I} = \begin{bmatrix}40 & 0 & 0 \\0 & 40 & 0 \\0 & 0 & 40\end{bmatrix} \] Since the matrix is diagonal with the same entries, the eigenvalues are simply the diagonal elements. Therefore, the eigenvalues are: \[ I_{1} = 40, \quad I_{2} = 40, \quad I_{3} = 40 \]