[NOTES/ME-14007]-Parallel Axes Theorem

Let \(I^\text{cm}_{jk}\) denote the components of moment of inertia tensor of a rigid body relative to a system of axes \(K^\text{cm}\) with center of mass chosen as the origin. Let a new set of axes \(K{'}\) are obtained by a parallel translation of \(K^\text{cm}\) to a point \(O{'}\) chosen as new origin. We shall show that the relation between the expressions for the moment of inertia tensors relative to the two sets of axes is given by \begin{equation}\label{EQ01} \boxed{(I{'})_{jk}= I^{\text{cm}}_{jk} + M(a^2\delta_{jk}- a_ja_k)}. \end{equation} where \(M\) is the total mass \(M=\sum_\alpha m_\alpha\), \(\vec{a}=(a_1,a_2,a_3)\) is the position vector of \(O{'}\) w.r.t. to \(K^\text{cm}\); and \(I{'}\) denotes the moment of inertia tensors relative to system \(K{'}\). Use \(\vec{x}_\alpha,\vec{x}{'}_\alpha\) to denote the position of a point in the rigid body. Then we have \(\vec{x}{'}_\alpha = \vec{x}_\alpha + \vec{a}\). Therefore, the moment of inertia tensor w.r.t. the systems \(K^\text{cm}\) and \(K{'}\) are given by \begin{eqnarray}\label{EQ02} I^\text{cm}_{jk} &=& \sum_\alpha m_\alpha\big\{ |\vec{x}_\alpha|^2 \delta_{jk}- x_{\alpha j} x_{\alpha k}\big\}\\ I^{'}_{jk} &=& \sum_\alpha m_\alpha\big\{ |\vec{x}{'}_\alpha|^2 \delta_{jk}- x{'}_{\alpha j} x{'}_{\alpha k}\big\}\label{EQ03} \end{eqnarray} Now use \(\vec{x}{'}_\alpha = \vec{x}_\alpha + \vec{a}\) to get \begin{eqnarray}\label{EQ04} |x{'}_\alpha|^2 &=& |x_\alpha|^2 + |\vec{a}|^2 + 2 \vec{a}\cdot \vec{x}_\alpha\\ x{'}_{\alpha j} x{'}_{\alpha k} &=& x_{\alpha j} x_{\alpha k} + x_{\alpha j} a_k + a_j x_{\alpha k} + a_j a_k\label{EQ05} \end{eqnarray} Recall that \(\sum_\alpha m_\alpha x_{\alpha i}= M X^\text{cm}_i\), where \(X^\text{cm}_i\) is the \(i^\text{th}\) component of the position of the centre of mass. When summed over \(\alpha\), \EqRef{EQ04}-\eqRef{EQ05} take the form \begin{eqnarray} \sum_\alpha m_\alpha |x{'}_\alpha|^2 &=&\sum_\alpha m_\alpha |x_\alpha|^2 + M |\vec{a}|^2 + 2 M \vec{a}\cdot \vec{X}^\text{cm} \nonumber\\ &=&\sum_\alpha m_\alpha|x_\alpha|^2 + M |\vec{a}|^2\label{EQ08A}\\ \sum_\alpha m_\alpha x{'}_{\alpha j} x{'}_{\alpha k} &=&\sum_\alpha m_\alpha x_{\alpha j} x_{\alpha k} + X^\text{cm}_j a_k + a_j X^\text{cm}_k + M a_j a_k \nonumber\\ &=&\sum_\alpha m_\alpha x_{\alpha j} x_{\alpha k} + M a_j a_k \label{EQ08B} \end{eqnarray} Here we have used the fact that the origin of the coordinate system is chosen to be at the center of mass, i.e. \(\vec{X}^\text{cm}=0\). On using EQ08A and EQ08B in EQ02 and EQ03, we get \begin{equation}\label{EQ06} I{'} - I^\text{cm} = M\big( a^2 \delta_{jk} - a_j a_k\big). \end{equation} This proves the result This result is known as  parallel axes theorem.