[NOTES/ME-14003]-Moment of Inertia Tensor

1. The inertia tensor

Relative to a chosen set of axes, the moment of components of inertia tensor are defined to be \begin{equation} I_{jk}=\sum_\alpha m_\alpha(\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}). \end{equation} Here the rigid body is supposed to consist of masses \(m_\alpha\) at positions \(x_\alpha\). In case of a continuous body, the sum over \(\alpha\) should be replaced by an appropriate (volume) integral. \begin{equation} I_{jk}=\int d^3x \rho(\vec{x})(\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}). \end{equation} Frequently, we shall use a \(3\times3\) matrix notation for moment of inertia tensor and write \begin{equation} \underline{\sf I}=\begin{pmatrix} I_{11} & I_{12} & I_{13}\\ I_{21} & I_{22} & I_{23}\\ I_{31} & I_{32} & I_{33} \end{pmatrix} \end{equation} When a body rotates about a fixed axis \(\hat{n}\), with angular velocity \(\omega\), it moment of inertia about the axis is given by \begin{equation} I_{\mathbf n} = n_i I_{ij}n_j = \sum_\alpha m_\alpha \rho_\alpha^2 \end{equation} where \(\rho_\alpha = |\vec{x}_\alpha|^2 - (\hat{x}.vec{x}_\alpha)^2\) is the perpendicular distance of mass \(\alpha\) from the axis. The quantities \(I_{11},I_{22},I_{33}\) are known as the moments of inertia about the coordinate axes. The moment of inertia is additive and the inertia of the body is equal to the sum of moments of inertia of its parts. For a rigid body consisting of continuous distribution of mass the moment of inertia tensor is given by integral over the volume of the body \begin{equation} I_jk = \int d^x \rho(\vec{x}) (|\vec{x}|^2 \delta_{jk} - x_j x_k) \end{equation} The components \(I_{11}, I_{22}, I_{33}\) are called the moments of inertia about the corresponding axes.

2.Kinetic energy and angular momentum

If the angular velocity of rigid body is \(\vec{\omega}\), the kinetic energy and angular momentum are given by \begin{equation} \text{K.E.} = \omega_jI_{jk}\omega_k; \qquad L_j= I_{jk}\omega_k. \end{equation} In matrix form we shall write \begin{equation} \text{K.E.} = \underline{\pmb{\omega}}^T \,\underline{\mathbf I}\, \underline{\pmb{\omega}}; \qquad \underline{\mathbf L}= \underline{\mathbf I}\, \underline{\pmb{\omega}}. \end{equation} where a column vector notation \begin{equation} \underline{\pmb{\omega}}= \begin{pmatrix}\omega_1\\\omega_2\\\omega_3\end{pmatrix}, \qquad \underline{\mathbf L}= \begin{pmatrix}L_1\\L_2\\L_3\end{pmatrix}, \end{equation} for angular velocity and angular momentum has been used. Here \(\underline{\pmb \omega}^T\) means transpose of the matrix \(\underline{\pmb \omega}\). 

%\subsection{Parallel axes theorem}

3.Transformation under translation of axes

 \input{me-lec-14007}

4.Transformation under rotation of axes

 The moment of inertia tensors qualifies to be called a second rank tensor as defined above. Thus if \(K, K{'}\) are two system of axes with common origin, we have \begin{eqnarray} I_{jk} &=&\sum_\alpha m_\alpha (\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}),\\ I_{jk}{'} &=&\sum_\alpha m_\alpha (\delta_{jk} |\vec{x}_\alpha{'}|^2 - x_{\alpha j}{'} x_{\alpha k}{'}). \end{eqnarray} We shall prove it a little later, after some preparation, that the relation \begin{equation} I_{jk}{'} = R_{jm}R_{kn} I_{mn} \end{equation} holds under a rotation of axes. The inertia tensor qualifies to be called a tensor of rank 2.