Verify that expression in Eq. (2.50) for the components of the rotational angular momentum Eq. (2.49) in terms of the inertia tensor is correct. Consider an arbitrary inertial coordinate frame its the origin at the center of rotation and basis vectors $\hat{e}_0$, $\hat{e}_1$ and $\hat{e}_2$, such that $\hat{e}_0 \times \hat{e}_1 = \hat{e}_2$. If the components of $\vec{\omega}$ in this frame are $\omega^0$, $\omega^1$ and $\omega^2$,
The inertia matrix of a body in this coordinate frame is: Applying the triple product formula, ${\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} = \mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ) }$, Applying the triple product formula again, ${\displaystyle \mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} )}$, where $I_{jk}$ are components of the inertia tensor.Exercise 2.8: Rotational angular momentum
$$
\vec{L} = \sum_\alpha m_\alpha \vec{\xi}_\alpha \times \left(\vec{\omega} \times \vec{\xi}_\alpha \right)\tag{2.49}
$$
$$
L_j = \sum_k I_{jk} \omega^k\tag{2.50}
$$
$$
\begin{align*}
\vec{L} &= \sum_\alpha m_\alpha \vec{\xi}_\alpha \times \left(\vec{\omega} \times \vec{\xi}_\alpha \right) \\
L_i = \vec{L}\cdot \hat{e}_i &= \left( \sum_\alpha m_\alpha \vec{\xi}_\alpha \times \left(\vec{\omega} \times \vec{\xi}_\alpha \right) \right) \cdot \hat{e}_i\\
\end{align*}
$$
$$
\begin{align*}
L_j &= \sum_\alpha \left( m_\alpha \overbrace{\vec{\xi}_\alpha}^{"\mathbf{a}"} \times \underbrace{\left(\vec{\omega} \times \vec{\xi}_\alpha \right)}_{"\mathbf{b}"} \right) \cdot \overbrace{\hat{e}_j}^{"\mathbf{c}"}\\
&= \sum_\alpha m_\alpha \vec{\xi}_\alpha \cdot \left[ \left(\vec{\omega} \times \vec{\xi}_\alpha \right) \times \hat{e}_j\right]\\
&= \sum_\alpha m_\alpha \vec{\xi}_\alpha \cdot \left[ \left(\sum_k \hat{e}_k \omega^k \times \vec{\xi}_\alpha \right) \times \hat{e}_j \right]\\
&= \sum_k \omega^k \sum_\alpha m_\alpha \vec{\xi}_\alpha \cdot \left[ \left(\hat{e}_k \times \vec{\xi}_\alpha \right) \times \hat{e}_j \right]\\
\end{align*}
$$
$$
\begin{align*}
L_j &= \sum_k \omega^k \sum_\alpha m_\alpha \underbrace{\vec{\xi}_\alpha}_{"\mathbf{a}"} \cdot \left[ \overbrace{\left(\hat{e}_k \times \vec{\xi}_\alpha \right)}^{"\mathbf{b}"} \times \underbrace{\hat{e}_j}_{"\mathbf{c}"} \right]\\
&= \sum_k \omega^k \sum_\alpha m_\alpha \left(\hat{e}_k \times \vec{\xi}_\alpha \right) \cdot \left[ \hat{e}_j \times \vec{\xi}_\alpha \right]\\
&= \sum_k \omega^k \underbrace{\sum_\alpha m_\alpha \left(\hat{e}_j \times \vec{\xi}_\alpha \right) \cdot \left( \hat{e}_k \times \vec{\xi}_\alpha \right)}_{I_{jk}}\\
\\
L_j &= \sum_k I_{jk} \omega^k
\end{align*}
$$