[QUE/ME-14006] ME-PROBLEM

Two observers \(O\) and \(O{'}\), in frames \(K\) and \(K{'}\) with common origin, find that the components of an arbitrary vector \(\mathbf A\) in their respective frames are related by a time dependent rotation matrix \(R(t)\) \begin{equation} \widetilde{\sf A}{'}= R(t) \widetilde{\sf A}. \end{equation} It is given that components of the vector do not change w.r.t. the frame \(K\). Obtain an expression for angular velocity of frame \(K{'}\) w.r.t. frame \(K\) in terms of the rotation matrix. Verify that your answer gives correct answer correct answer when \(R(t)\) corresponds to a rotation about \(X_3\) axis.

We will use this relation \begin{equation} \dd[{\mathbf A}]{t}\Big|_K = \dd[{\mathbf A}{'}]{t}\Big|_{K{'}} + \pmb{\omega}\times{\mathbf A}{'}. \end{equation} to obtain an expression for the angular velocity. Here \(\vec{\omega}=(\omega_1, \omega_2,\omega_3)\) are the components of the angular velocity of frame \(K{'}\) w.r.t. the frame \(K\).\\ It is given that the vector components of \(\mathbf A\) w.r.t. frame \(K{'}\) do not change with time. Hence \begin{equation} \dd{t} \vec{A}{'} = - \vec{\omega}\times \vec{A}{'}. \end{equation} Writing the above equation in terms of components, we get \begin{eqnarray}\nonumber \dd{t}\begin{pmatrix}{A}_1{'} \\ {A}_2{'} \\ A_3{'}\end{pmatrix} &=&- \begin{pmatrix} \omega_2{A}_3{'} -\omega_3{A}_2{'} \\ \omega_3{A}_1{'} -\omega_1{A}_3{'} \\ \omega_1{A}_2{'} -\omega_2{A}_1{'} \end{pmatrix}\\ &=& \begin{pmatrix} 0 & \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{pmatrix}\label{EQ04} \begin{pmatrix} {A}_1{'} \\ {A}_2{'} \\ {A}_3{'}\end{pmatrix} \end{eqnarray} Using \(\widetilde{A}{'}\) and \(\widetilde{A}{'}\) to denote the column vectors of components \(\vec{A}\) and \(\vec{A}{'}\), we write the above relation as \begin{equation}\label{EQ06} \dd[\widetilde{A}{'}]{t} = \Omega \cdot \widetilde{A}{'}. \end{equation} where we have introduced the matrix \(\Omega\) defined by \begin{equation} \Omega =\begin{pmatrix} 0 & \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{pmatrix} \end{equation} The given rotation matrix \(R(t)\) relates the components of vector \(\mathbf A\) in the two frames. \begin{equation} \widetilde{A}{'} = R(t) \widetilde{A} \end{equation} Substituting in \eqRef{EQ06}, we get \begin{equation} \dd[R(t)]{t}\widetilde{A} = \Omega R(t) \widetilde{A} \end{equation} Since vector \(\vec{A}\) is arbitrary, we get the matrix relation \begin{equation} \dd[R(t)]{t} = \Omega R(t). \end{equation} Multiplying by\(R^T\) from the right, this equation implies that \begin{equation} \Omega = \dd[R(t)]{t} R^T(t).\label{EQ10} \end{equation} The last relation is the required relation. Differentiating \(R R^T= \hat{I}\) w.r.t. we get \begin{equation}\label{EQ11} \dd[R]{t} R^T(t) + R\dd[R^T]{t} =0 \Longrightarrow \dd[R]{t} R^T = - R\dd[R^T]{t}. \end{equation} Thus \(\Omega\) in \eqRef{EQ10} can be written as \begin{equation} \boxed{ \Omega = \dd[R(t)]{t} R^T(t)=-R\dd[R^T]{t}.}\label{EQ12} \end{equation} It should be noted that \eqRef{EQ11} implies that \(\Omega=-\Omega^T\) and therefore \(\Omega\) is an antisummetric matrix. \paragraph*{Verification of \eqRef{EQ12} for rotation about \(X_3\)- axis:\\} For rotations about \(X_3\) axis by an angle \(\alpha(t)\) we have \begin{equation} \dd[R]{t} =\dd{t}\begin{pmatrix} \cos\alpha & \sin\alpha& 0\\ -\sin \alpha & \cos\alpha &0\\ 0 & 0& 1 \end{pmatrix} = \begin{pmatrix} -\dot{\alpha}\sin\alpha & \dot{\alpha}\cos\alpha& 0\\ -\dot{\alpha}\cos \alpha & -\dot{\alpha}\sin\alpha &0\\ 0 & 0& 1 \end{pmatrix} \end{equation} Therefore \begin{eqnarray}\nonumber \Omega = \dd[R]{t} R^T &=& \begin{pmatrix} -\dot{\alpha}\sin\alpha & \dot{\alpha}\cos\alpha& 0\\ -\dot{\alpha}\cos \alpha & -\dot{\alpha}\sin\alpha &0\\ 0 & 0& 1 \end{pmatrix} \begin{pmatrix} \cos\alpha & -\sin\alpha& 0\\ \sin \alpha & \cos\alpha &0\\ 0 & 0& 1 \end{pmatrix}\\ &=&\begin{pmatrix} 0 & \dot{\alpha} & 0\\ -\dot{\alpha} & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\label{EQ14} \end{eqnarray} Comparing with \eqRef{EQ04}, we \(\vec{\omega}=(0,0,\dot{\alpha})\) as expected. This completes the verification of \eqRef{EQ10} for rotations about the \(X_3\) axis. \subsection{$\pmb{\omega}$ in terms of Euler angles} Let \(\phi,\theta,\psi\) denote Euler angles of a axes fixed in body of a rigid body rotating about a fixed point. The fixed point is chosen as the common origin of the space and body fixed axes. Then the rotation matrix \(R\) giving the transformation of vectors from space fixed axes to body fixed axes is given by \begin{equation} R(\phi, \theta, \psi) = R_3(\psi) R_1(\theta) R_(\phi). \end{equation} To obtain an expression for components of the angular velocity vector we use \eqRef{EQ12}. Therefore we must compute \begin{eqnarray}\nonumber \Omega &=& -R \Big(\dd[R^T]{t}\Big)= - R\big(\Dsc R^T\big)\\\nonumber &=& - \big(R_3(\psi) R_1(\theta) R_3(\phi) \Dsc \big[ R_3(\psi) R_1(\theta) R_3(\phi)\big]^T\\\nonumber &=& - \Big\{R_3(\psi) R_1(\theta) R_3(\phi).\Big\}\,\Big\{ \Dsc [R_3^T(\phi)] R_1^T(\theta) R_3^T(\psi) \\ && \qquad + R_3^T(\phi)[\Dsc R_1^T(\theta)] R_3^T(\psi) + R_3^T(\phi) R_1^T(\theta)[\Dsc R^T_3(\psi)]\Big\}\label{EQ16}. \end{eqnarray} where \(\Dsc\) stands for the time derivative \(\dd{t}\). We will use the notation \begin{eqnarray} DR1&=& - \big\{(R_3(\psi) R_1(\theta)\big\} \textcolor{blue}{ [R_3(\phi) [\Dsc R_3^T(\phi)]] } \big\{R_1(\theta)^T R_3(\psi)^T\big\} \\ DR2&=& - \big\{R_3(\psi) R_1(\theta) R_3(\phi)\big\} \big\{R_3^T(\phi) [\Dsc R_1^T(\theta)] R_3^T(\psi)\big\}\\ DR3&=& - \big\{R_3(\psi) R_1(\theta) R_3(\phi)\big\} \big\{R_3^T(\phi)R_1^T(\theta) [\Dsc R_3^T(\psi)]\big\}. \end{eqnarray} Since \(R_1\) and \(R_3\) are orthogonal matrices, we use \(R^T_3 R_3= R_1^T R_1= \text{identity matrix, }\hat{I}\) to simplify \(DR2\) and \(DR3\) as \begin{eqnarray} DR1&=& - \big\{(R_3(\psi) R_1(\theta)\big\} \textcolor{blue}{ [R_3(\phi) \Dsc R_3^T(\phi)] } \big\{R_1^T(\theta) R_3^T(\psi)\big\} \\ DR2&=& - R_3(\psi) \textcolor{blue}{[R_1(\theta) \Dsc R_1^T(\theta)]} R_3^T(\psi)\\ DR3&=& - \textcolor{blue}{[R_3(\psi) \Dsc R_3^T(\psi)]}. \end{eqnarray} Using the form of rotation matrices for rotations about coordinate axes we will have, (see \begin{equation} R_3(\phi)\Dsc R_3^T(\phi) = \begin{pmatrix} 0 & \dot{\phi} & 0\\ -\dot{\phi} & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}, \qquad R_1(\theta)\Dsc R_1^T(\theta)= \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & \dot {\theta}\\ 0 & -\dot{\theta} & 0 \end{pmatrix} \end{equation} and expression for \( R_3(\psi)\Dsc R_3^T(\psi)\) will be similar to \(R_3(\phi)\Dsc R_3^T(\phi)\).