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The angular momentum of a rigid body is given by where \(\mathbf I\) is moment of inertia tensor and \(\vec \omega \) is the angular velocity.\begin{eqnarray} \vec{L}&=&\int dv \rho(\vec{X})\vec{X}\times(\vec{\omega}\times\vec{X})\\ &=&\int dV \rho(\vec{X})\Big[(\vec{X}\cdot\vec{X})\vec{\omega}-(\vec{X}\cdot\vec{\omega} )\vec{X}\Big] \end{eqnarray} or \(\vec L=\mathbf I\, \vec \omega\).
Let \(\vec X\) denote the position vector of a point on the rigid body. Then the angular momentum is given by
\begin{eqnarray} \vec{L} &=&\sum_{\alpha}m_{\alpha}\vec{x_{\alpha}}\times\dot{\vec{x_\alpha}} \to \int dm \vec{X}\times(\vec{\omega}\times\vec{X})\\ \vec{L}&=&\int dv \rho(\vec{X})\vec{X}\times(\vec{\omega}\times\vec{X})\\ &=&\int dV \rho(\vec{X})\Big[(\vec{X}\cdot\vec{X})\vec{\omega}-(\vec{X}\cdot\vec{\omega} )\vec{X}\Big] \end{eqnarray}
Taking components we get
\begin{equation}
L_j=I_{jk} \omega_k,\qquad \vec L=\mathbf I\, \vec \omega
\end{equation}
where $I_{jk}$ are the matrix elements of the moment of inertial tensor
\begin{equation} {\mathbf{I} }=\left[\begin{array}{clc} I_{11} &I_{12} &I_{13}\\ I_{21} &I_{22} &I_{23}\\ I_{31} &I_{32} &I_{33} \end{array}\right] \end{equation}
