Question

Point masses 1 at (1, 1, -2) and 2 at (1, 1, 1).

Short answer

Answer:

Inertia tensor is $l=\left(\begin{array}{ccc}9& -3& 0\\ -3& 9& 0\\ 0& 0& 6\end{array}\right)$.

Step-by-step solution

Given Information.

The given information is mentioned below:

$$\begin{gathered} m_1=1 \\ {m}_2=2 \\ {r}_1=\left(1,1,-2\right) \\ {r}_2=\left(1,1,1\right) \end{gathered}$$

Definition of Point mass.

The concept of a physical object with non-zero mass but explicitly and particularly minuscule volume or linear dimension is known as point mass.

Find the eigenvalues.

Solve for eigenvalues.

$$\begin{gathered} l_{ij}=\left(6{\delta }_{ij}-{\left(-2\right)}^{{\delta }_{3i}+{\delta }_{3i}}\right)+2\left(3{\delta }_{ij}-1\right) \\ l=\left(\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right)-\left(\begin{array}{ccc}1& 1& -2\\ 1& 1& -2\\ -2& -2& 4\end{array}\right)+2\left(\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ -1& -1& 2\end{array}\right) \\ =\left(\begin{array}{ccc}5& -1& 2\\ -1& 5& 2\\ 2& 2& 2\end{array}\right)+\left(\begin{array}{ccc}4& -2& -2\\ -2& 4& -2\\ -2& -2& 4\end{array}\right) \\ l=\left(\begin{array}{ccc}9& -3& 0\\ -3& 9& 0\\ 0& 0& 6\end{array}\right) \end{gathered}$$

Find principal axes and corresponding moment of inertia.

Find the moment of inertia.

$$\begin{gathered} v_1=\left(-1,1,0\right) \\ {v}_2=\left(0,0,1\right) \\ {v}_3=\left(1,1,0\right) \end{gathered}$$

Simplify further.

$$\begin{gathered} l_1=12 \\ {l}_2=6 \\ {l}_3=6 \end{gathered}$$

Hence, inertia tensor is, $l=\left(\begin{array}{ccc}9& -3& 0\\ -3& 9& 0\\ 0& 0& 6\end{array}\right)$