[NOTES/ME-14013]-Angular velocity from rotation matrix

1.Analytical Method

Assume that a vector \(\mathbf A\), attached to origin of a system of coordinates axes, undergoes an arbitrary rotation. Using a matrix notation \[\widetilde{\sf A}= \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}.\] to denote the components of the vector \(\mathbf A\), the components at any time \(t\) will be related to components at initial time \(t_0\) by \begin{equation} \widetilde{\sf A}(t) = R(t) \widetilde{\sf A}(0). \end{equation} The rate of change of the vector is given by \begin{eqnarray}\nonumber \dd{t} \ \widetilde{\sf A}(t) &=& \Big(\dd[R(t)]{t}\Big) \widetilde{\sf A}(t_0)\\\nonumber &=& \Big(\dd[R(t)]{t}\Big)\Big( R^T\widetilde{\sf A}(t)\Big)\qquad \because \quad{A(t_0)=R^{-1}(t)\widetilde{\sf A}(t)= R^T \widetilde{A}(t)}\\ &\equiv& \Omega \, \widetilde{\sf A}(t). \label{EQ02} \end{eqnarray} where we have introduced \(\Omega\) to denote the matrix \(\big(\dd[R]{t}\big) R^T\). We will now prove that \(\Omega\) is an antisymmetric matrix by verifying that \(\Omega+ \Omega^T=0\). In fact \begin{equation} \Omega+ \Omega^T = \dd[R]{t}\, R^T + R \, \dd[R^T]{t} = \dd{t}\big(RR^T\big) =\dd{t} \hat{I} =0. \end{equation} Therefore \(\Omega\) is an antisymmetric matrix and is related to angular velocity. To demonstrate this relation, we write \(\Omega\) as \begin{equation}\label{EQ04} \Omega = \begin{pmatrix}0& -\omega_3& \omega_2\\ \omega_3 & 0& -\omega_1 \\-\omega_2& \omega_1&0 \end{pmatrix} \end{equation} Substituting \eqRef{EQ04} in \eqRef{EQ02}, we get % \begin{eqnarray} \begin{pmatrix}\dot{a}_1(t) \\ \dot{a}_2(t) \\ \dot{a}_3(t)\end{pmatrix} = \begin{pmatrix} 0 & -\omega_3& \omega_2\\ \omega_3 & 0 & -\omega_1\\ - \omega_2& \omega_1&0 \end{pmatrix} \begin{pmatrix}a_1(t) \\a_2(t) \\ a_3(t)\end{pmatrix} \end{eqnarray} Multiplying the matrices in the right hand side of \eqRef{EQ02}, we get \begin{eqnarray}\nonumber \dd[a_1]{t} &=& \omega_2 a_3(t) -\omega_3a_2(t)\\\nonumber \dd[a_2]{t} &=& \omega_3 a_1(t) -\omega_1a_3(t)\\\nonumber \dd[a_3]{t} &=& \omega_1 a_2(t) - \omega_2 a_1(t) \end{eqnarray} These equations are equivalent to the vector equation \begin{equation}\label{EQ05} \dd[\vec{A}(t)]{t} = \vec{\omega}\times \vec{A}. \end{equation} This shows that \((\omega_1,\omega_2,\omega_3)\) are to be identified with the components of angular velocity vector \(\pmb{\omega}\).\\ Attached is pdf file of Mathematica program where these computations have been done symbolically to get expression for angular velocity \(\pmb{\omega}\) in terms of Euler angles. Please note that, for this derivation, it is simpler to take rotations as active transformations. In the program the rotation matrices are written with a minus sign for angles, because the sign convention of angle of rotation used by mathematica is opposite to what we are following. \begin{eqnarray} \omega_1&=&\dot{\theta} \cos \psi +\dot{\phi} \sin \theta \sin \psi \\ \omega_2&=& \dot{\phi} \sin \theta \cos \psi -\dot{\theta} \sin \psi \\ \omega_3&=&\dot{ \phi} \cos \theta +\dot{\psi} \end{eqnarray} The kinetic energy of a rigid body in terms of Euler angles is given by \begin{equation} T= \frac{1}{2}I_1 \omega_1^2 + \frac{1}{2}I_2\omega_2^2 + \frac{1}{2} I_3\omega_3^2. \end{equation}