Mechanics of Continua and Structures
It appears that in literature, rigid body motion is mostly studied using a Eulerian description. However, for determining the acceleration field inside the rigid solid through a finite number of acceleration meaurements, it is necessary to use a Lagrangian description. We describe the Lagrangian description in $\S$Kinematics
Consider a rigid body, free to move in \(\mathbb{R}^3\). A reference configuration \(B_0\) of the body is the closure of an open set in \(\mathbb{R}^3\) at time \(t=0\). Let \(\mathbb{E}^3\) be the three dimensional Euclidean vector space corresponding to the point space \(\mathbb{R}^3\), such that each point \(\mathcal{P}\in \mathcal{B}_0\) has associated with it a vector \(\boldsymbol{X}\in \mathbb{E}^3\), such that \(\mathcal{O}+\boldsymbol{X}=\mathcal{P}\). The body \(B_0\in \mathbb{R}^3\) is isomorphic to the set \(\mathcal{B}_0 \subset \mathbb{E}^3\). Let \(\{\hat{\boldsymbol{E}}_1, \hat{\boldsymbol{E}}_2, \hat{\boldsymbol{E}}_3\}:=\{\hat{\boldsymbol{E}}_I\}_{I=1,2,3}\) be the standard cartesian basis vectors of \(\mathbb{E}^3\). The component of \(\boldsymbol{X}\) w.r.t. \(\hat{\boldsymbol{E}}_I\) is \(X_{I}\).
A configuration of the body \(\mathcal{B}_0\) is a time-dependent family of mappings, \(\varphi_{t}: \mathcal{B}_0\to \mathbb{E}^3\). The set of all configurations of \(\mathcal{B}_0\) is denoted as \(\mathcal{C}\), or by \(\mathcal{C}(\mathcal{B}_0)\). A motion of the body \(\mathcal{B}_0\) is a curve in \(\mathcal{C}\). That is, it is a mapping \(t \mapsto \varphi_{t}\).
The motion of a rigid solid is defined through the mapping \(\boldsymbol{\varphi}_t(\boldsymbol{X}):= \boldsymbol{\varphi}(\boldsymbol{X},t)\) characterized as
\[\begin{equation} \boldsymbol{x}= \boldsymbol{Q}(t) \boldsymbol{X}+\boldsymbol{c}(t), \tag{1.1} \label{eq:KinematicsMotion} \end{equation}\]where \(\boldsymbol{x} = \boldsymbol{\varphi}(\boldsymbol{X},t)\) . $\boldsymbol{X}\in \mathbb{E}^3$ and $\boldsymbol{x} \in \mathbb{E}^3$, where $\mathbb{E}^3$ is the three dimensional Euclidean space, are the Lagrangian and Eulerian position vectors of a material particle $\mathcal{P}\in \mathcal{B}_0$. Here, $\mathcal{B}_0$ denotes the solid in its reference configuration. Mathematically speaking, $\mathcal{B}_0$ is a manifold. The tensor \(\boldsymbol{Q}(\cdot, t):\mathbb{E}^3 \to \mathbb{E}^3\) is a proper Orthogonal tensor. See $\S$Orthogonal tensors for a discussion of the various properties and characterization of proper Orthogonal tensors. Most notably, proper Orthogonal tensors satisfy the following two properties:
\(\begin{align} \boldsymbol{Q}^{\textsf{T}}\boldsymbol{Q} &= \boldsymbol{I},\tag{2.1}\\ \textsf{det} \left(\boldsymbol{Q}\right) &= +1\tag{2.2}, \end{align}\) where \(\textsf{det}(\cdot): \textsf{Lin} \to \mathbb{R}\) is the determinant operator, and \((\cdot)^{\textsf{T}}:\mathbb{E}^3\to \mathbb{E}^3\) is the transpose operator.
The material velocity and acceleration fields are as follows:
\[\begin{align} \boldsymbol{V}(\boldsymbol{X},t)&= \dot{\boldsymbol{Q}}(t)\boldsymbol{X}+\dot{c}(t) \label{eq:LagMatVel} \\ \boldsymbol{A}(\boldsymbol{X},t)&= \ddot{\boldsymbol{Q}}(t)\boldsymbol{X}+\ddot{c}(t) \label{eq:LagMatAccel} \end{align}\]The deformation gradient for the motion defined in \eqref{eq:KinematicsMotion} is
\begin{equation} \boldsymbol{F}(\boldsymbol{X},t)=\boldsymbol{Q}(t). \label{eq:DeformationGradient} \end{equation}
It follows from \eqref{eq:DeformationGradient} that \(\begin{align} \boldsymbol{C} &= \boldsymbol{I}, \label{eq:RCGD_Tensor} \tag{1.2} \\ \boldsymbol{E} &= \boldsymbol{O}, \label{eq:EL_Tensor} \tag{1.3} \\ \end{align}\) where $\boldsymbol{C}$ is the right Cauchy-Green deformation tensor and $\boldsymbol{E}$ is the Green-Lagrange strain tensor. They belong to the space containing all linear mappings from $T_{\boldsymbol{X}}$ to $T_{\boldsymbol{X}}$. Here, $T_{\boldsymbol{X}}$ is the tangent space of the manifold \(\mathcal{B}_{0}\) at the location $\boldsymbol{X}$, and $\boldsymbol{I}$ and $\boldsymbol{O}$ are, respectively, the identity and the null tensors belonging to the space $\textsf{Lin}_{\boldsymbol{X}}$.
The spatial velocity field $\boldsymbol{v}(\boldsymbol{x},t)$ is obtained as $ \boldsymbol{v}(\boldsymbol{x},t)= \boldsymbol{V}(\boldsymbol{\varphi}^{-1}(\boldsymbol{x},t),t), $ For the special motion given by \eqref{eq:KinematicsMotion}, we have that $ \boldsymbol{\varphi}^{-1}(\boldsymbol{x},t) =\boldsymbol{Q}^{-1}(t)\left(\boldsymbol{x}-\boldsymbol{c}(t)\right), $ thus we get that \(\begin{align} \boldsymbol{v}(\boldsymbol{x},t) &=\boldsymbol{V}( \boldsymbol{Q}^{-1}(t)( \boldsymbol{x}-\boldsymbol{c}(t) ), t)\\ &=\dot{\boldsymbol{Q}}(t) \boldsymbol{Q}^{-1}(t) (\boldsymbol{x}-\boldsymbol{c}(t)) +\dot{\boldsymbol{c}}(t)\\ &=\dot{\boldsymbol{Q}}(t) \boldsymbol{Q}^{\textsf{T}}(t) (\boldsymbol{x}-\boldsymbol{c}(t)) +\dot{\boldsymbol{c}}(t)\\ \end{align}\) Defining $\boldsymbol{W}(t):=\dot{\boldsymbol{Q}}(t) \boldsymbol{Q}^{\textsf{T}}(t)$ the above equation can be written as \(\begin{align} \boldsymbol{v}(\boldsymbol{x},t) &= \boldsymbol{W}(t) (\boldsymbol{x}-\boldsymbol{c}(t)) +\dot{\boldsymbol{c}}(t). \label{eq:spatial_v_spinor} \\ \end{align}\) It can be shown that the tensor $\boldsymbol{W}$ is a skew-symmetric tensor, i.e., $\boldsymbol{W}^{\textsf{T}}(t)=-\boldsymbol{W}(t)$. It is a well known result from tensor algebra that for any skew-symmetric tensor $\boldsymbol{W}: \mathcal{V}\to \mathcal{V} $ there exits a unique vector $\omega\in \mathcal{V} $ such that \(\begin{align} \boldsymbol{W}\boldsymbol{u}=\boldsymbol{\omega}\times \boldsymbol{u}, \end{align}\) for all $\boldsymbol{u}\in \mathcal{V}$. The tensor $\boldsymbol{W}$ is called the Spinor corresponding to $\boldsymbol{\omega}$, and $\boldsymbol{\omega}$ is called the axial vector corresponding to $\boldsymbol{W}$. The components of $\boldsymbol{W}$ and $\boldsymbol{\omega}$ can be computed from each other as
\(\begin{align} W_{ij}&=-\epsilon_{ijk}\omega_{k},\\ \omega_i&=-\frac{1}{2}\epsilon_{ijk}W_{jk}. \end{align}\)
Let $\boldsymbol{\omega}(t)$ be the axial vector corresponding to $\boldsymbol{W}$. Then the spatial velocity field can be written in terms of $\boldsymbol{\omega}(t)$ as
\[\begin{align} \boldsymbol{v}(\boldsymbol{x},t) &=\boldsymbol{\omega}(t) \times (\boldsymbol{x}-\boldsymbol{c}(t)) +\dot{\boldsymbol{c}}(t) \end{align}\]The spatial acceleration field is the material time derivative of the spatial velocity field. Thus, we have the spatial acceleration field as
\[\begin{align} \boldsymbol{a} &=\frac{D\boldsymbol{v}}{Dt}\\ &=\frac{\partial\boldsymbol{v}}{\partial t}+ \left(\nabla_{\boldsymbol{x}}\boldsymbol{v}\right)\boldsymbol{v} \end{align}\]The first term on the right hand side of the last equation is \(\begin{align} \frac{\partial\boldsymbol{v}}{\partial t} &=\dot{\boldsymbol{\omega}}(t) \times (\boldsymbol{x}-\boldsymbol{c}(t)) -\boldsymbol{\omega}(t) \times\dot{\boldsymbol{c}}(t) +\ddot{\boldsymbol{c}}(t)), \end{align}\)
where as the second term is
\[\begin{align} \left(\nabla_{\boldsymbol{x}}\boldsymbol{v}\right)\boldsymbol{v} &= \left( \boldsymbol{\omega}(t) \right) \times \left( \boldsymbol{\omega}(t) \times (\boldsymbol{x}-\boldsymbol{c}(t)) +\dot{\boldsymbol{c}}(t) \right)\\ &=\boldsymbol{\omega}(t) \times \boldsymbol{\omega}(t)\times (\boldsymbol{x}-\boldsymbol{c}(t)) +\boldsymbol{\omega}(t) \times\dot{\boldsymbol{c}}(t). \end{align}\]After adding the first and second term and simplifying, we get
\[\begin{align} \boldsymbol{a} &= \dot{\boldsymbol{\omega}}(t) \times (\boldsymbol{x}-\boldsymbol{c}(t)) +\ddot{\boldsymbol{c}}(t))+ \boldsymbol{\omega}(t) \times \boldsymbol{\omega}(t)\times (\boldsymbol{x}-\boldsymbol{c}(t)) \end{align}\]Rearranging the terms in the last equation can be written in the classical form
\[\begin{align} \boldsymbol{a} &= \dot{\boldsymbol{\omega}}(t) \times (\boldsymbol{x}-\boldsymbol{c}(t)) +\boldsymbol{\omega}(t) \times \boldsymbol{\omega}(t)\times (\boldsymbol{x}-\boldsymbol{c}(t)) +\ddot{\boldsymbol{c}}(t)) \end{align}\]Let \(\begin{align} \boldsymbol{Q}(t)&=\hat{\boldsymbol{k}}(t)\otimes\hat{\boldsymbol{k}}(t)+ \left( \boldsymbol{I}-\hat{\boldsymbol{k}}(t)\otimes\hat{\boldsymbol{k}}(t) \right)\cos(\theta(t))+\left(\star \hat{\boldsymbol{k}}(t)\right)\sin(\theta(t)) \end{align}\)
In component notation, the above equation can be written as \(\begin{align} Q_{ij}(t)&= \hat{k}_{i}(t)\hat{k}_{j}(t) + \left( \delta_{ij}-\hat{k}_{i}(t)\hat{k}_{j}(t) \right) \cos(\theta(t)) -\epsilon_{ijk}\hat{k}_{k}(t)\sin(\theta(t)) \label{eq:RodriguesComponent} \end{align}\)
Using the representation \(\begin{align} \left[ \hat{\boldsymbol{k}} \right] %_{\hat{\boldsymbol{E}}_{i},~i=1,~2,~3} &=\left[ \begin{array}{c} \cos (\psi (t)) \sin (\phi (t)) \\ \cos (\psi (t)) \cos (\phi (t))\\ \sin (\psi (t)) \end{array} \right] \end{align}\) for the unit vector $\hat{\boldsymbol{k}}$ in \eqref{eq:RodriguesComponent} and then differentiating the components $Q_{ij}(t)$ w.r.t time, we get \(\dot{Q}_{ij}(t)\). Computing the components of $\boldsymbol{W}$ as \(\begin{align} W_{ij}&=\dot{Q}_{ik}(t)Q_{jk}(t) \end{align}\) we get that
\[\begin{align} W_{11}&=0,\\ W_{22}&=0,\\ W_{33}&=0,\\ W_{12}&=-W_{21}\\ &=\cos (\psi (t)) \left(2 \sin ^2\left(\frac{\theta (t)}{2}\right) \cos (\psi (t)) \phi '(t)-\sin (\theta (t)) \psi '(t)\right)-\theta '(t) \sin (\psi (t)) \\ W_{13} &=-W_{31}\\ &= \theta '(t) \cos (\psi (t)) \cos (\phi (t))+\cos (\psi (t)) \phi '(t) \left(2 \sin ^2\left(\frac{\theta (t)}{2}\right) \sin (\psi (t)) \cos (\phi (t))-\sin (\theta (t)) \sin (\phi (t))\right)-\psi '(t) \left(\sin (\theta (t)) \sin (\psi (t)) \cos (\phi (t))+2 \sin ^2\left(\frac{\theta (t)}{2}\right) \sin (\phi (t))\right) \\ W_{23} &=-W_{32}\\ &=\theta '(t) (-\cos (\psi (t))) \sin (\phi (t))-\cos (\psi (t)) \phi '(t) \left(2 \sin ^2\left(\frac{\theta (t)}{2}\right) \sin (\psi (t)) \sin (\phi (t))+\sin (\theta (t)) \cos (\phi (t))\right)+\psi '(t) \left(\sin (\theta (t)) \sin (\psi (t)) \sin (\phi (t))-2 \sin ^2\left(\frac{\theta (t)}{2}\right) \cos (\phi (t))\right) \end{align}\]It follows from \eqref{eq:WComponents} \(\begin{align} W_{ij}&=-W_{ji}, \label{eq:skewComponents} \end{align}\) for $i,j=1,2,3$. From which it follows that
\(\begin{align} W_{ij} &= -W^{\rm{T}}_{ij}, \quad~i,j=1,~2,~3,\\ \left[\boldsymbol{W}\right] &= -\left[\boldsymbol{W}^{\textsf{T}}\right], \\ %\boldsymbol{W} % &= % -\boldsymbol{W}^{\textsf{T}} \end{align}\) which implies that $\boldsymbol{W}$ is a skew-symmetric tensor.