Question

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

Short answer

Answer:

Inertia tensor is $t=\left(\begin{array}{ccc}4& 0& 0\\ 0& 4& -2\\ 0& -2& 4\end{array}\right)$.

Step-by-step solution

Given Information

The given information is mentioned below:

$$\begin{gathered} m_1=m_2 \\ =1 \\ {r}_1=\left(1,1,1\right) \\ {r}_2=\left(-1,1,1\right) \end{gathered}$$

Definition of Inertia tensor.

Inertia tensor is the angular momentum of a rigid body determined by the product of its tensor and its rotation vector.

Find the eigenvalues.

Solve for eigenvalues.

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

Find principal axes and corresponding moment of inertia.

Find the moment of inertia.

$$\begin{gathered} V_1=\left(1,0,0\right) \\ {V}_2=\left(0,-1,1\right) \\ {V}_3=\left(0,1,1\right) \end{gathered}$$

Simplify Further

$$\begin{gathered} l_1=4 \\ {l}_2=6 \\ {l}_3=2 \end{gathered}$$

Hence, the inertia tensor is, $t=\left(\begin{array}{ccc}4& 0& 0\\ 0& 4& -2\\ 0& -2& 4\end{array}\right)$.