Merge remote-tracking branch 'gitea/master'

力学书籍/力学/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library)/auto/Dynamics of Multibody Systems (Shabana A.A.) (Z-Library).md

@@ -6597,11 +6597,13 @@ $$
 $$  
 
 Determine the global position of the tip point and the center of mass of the beam $C$ .  
+确定梁 $C$ 的尖端点和质心的全局位置。
+
 
 Solution At this instant of time, the transformation matrix A of Eq. 5 is given by  
 
 $$
-\mathbf{A}={\begin{bmatrix}{\cos\theta^{i}}&{-\sin\theta^{i}}\\ {\sin\theta^{i}}&{\cos\theta^{i}}\end{bmatrix}}={\left[\begin{array}{l l}{0.8660}&{-0.500{\big]}}\\ {0.500}&{0.8660}\end{array}\right]}
+\mathbf{A}={\begin{bmatrix}{\cos\theta}&{-\sin\theta}\\ {\sin\theta}&{\cos\theta}\end{bmatrix}}={\left[\begin{array}{l l}{0.8660}&{-0.500{\big]}}\\ {0.500}&{0.8660}\end{array}\right]}
 $$  
 
 The global position of point $A$ can then be written as  
@@ -6616,10 +6618,10 @@ $$
 \bar{\mathbf{u}}_{o}=\left[\begin{array}{l}{l}\\ {0}\end{array}\right]=\left[\begin{array}{l}{0.5}\\ {0}\end{array}\right]
 $$  
 
-The vector $\bar{\ensuremath{\mathbf{u}}}_{f}$ is the elastic deformation of point $A$ and can be evaluated, since $\xi=1$ at point $A$ , as  
+The vector $\bar{{\mathbf{u}}}_{f}$ is the elastic deformation of point $A$ and can be evaluated, since $\xi=1$ at point $A$ , as  
 
 $$
-\bar{\mathbf{u}}_{f}=\left[\!\!\begin{array}{c c}{\xi}&{0}\\ {0}&{3(\xi)^{2}-2(\xi)^{3}\!\!}\end{array}\!\!\right]\left[\!\!\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\!\!\right]=\left[\!\!\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\!\!\right]\left[\!\!\begin{array}{c}{0.001}\\ {0.01}\end{array}\!\!\right]=\left[\!\!\begin{array}{c}{0.001}\\ {0.01}\end{array}\!\!\right]
+\bar{\mathbf{u}}_{f}=\left[\begin{array}{c c}{\xi}&{0}\\ {0}&{3(\xi)^{2}-2(\xi)^{3}}\end{array}\right]\left[\begin{array}{c}{q_{f1}}\\ {q_{f2}}\end{array}\right]=\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]\left[\begin{array}{c}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{c}{0.001}\\ {0.01}\end{array}\right]
 $$  
 
 and accordingly  
@@ -6628,42 +6630,62 @@ $$
 \bar{\mathbf{u}}_{A}=\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]
 $$  
 
-The position vector $\mathbf{r}_{A}$ can be then written as $\mathbf{r}_{A}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]}={\left[\begin{array}{l}{1.4289}\\ {0.75916}\end{array}\right]}$ At point $C,\xi=0.5$ and $\bar{\mathbf{u}}_{o}=[(l/2)\,0]^{\mathrm{T}}=[0.25\,0]^{\mathrm{T}}.$ . The deformation vector   
+The position vector $\mathbf{r}_{A}$ can be then written as 
-$\bar{\ensuremath{\mathbf{u}}}_{f}$ at $C$ is given by $\bar{\mathbf{u}}_{f}=\left[\begin{array}{l l}{0.5}&{0}\\ {0}&{3(0.5)^{2}-2(0.5)^{3}}\end{array}\right]\left[\begin{array}{l}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{l}{0.0005}\\ {0.005}\end{array}\right]$   
+$$
-and the local position of point $C$ is $\bar{\mathbf{u}}_{C}=\bar{\mathbf{u}}_{o}+\bar{\mathbf{u}}_{f}=\left[\begin{array}{c}{0.25}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.0005}\\ {0.005}\end{array}\right]=\left[\begin{array}{c}{0.2505}\\ {0.005}\end{array}\right]$   
+\mathbf{r}_{A}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]}={\left[\begin{array}{l}{1.4289}\\ {0.75916}\end{array}\right]}
-The global position $\mathbf{r}_{C}$ can then be determined as $\mathbf{r}_{C}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.2505}\\ {0.005}\end{array}\right]}={\left[\begin{array}{l}{1.2144}\\ {0.62958}\end{array}\right]}$  
+$$ 
+At point $C,\xi=0.5$ and $\bar{\mathbf{u}}_{o}=[(l/2)\,0]^{\mathrm{T}}=[0.25\,0]^{\mathrm{T}}.$  The deformation vector $\bar{{\mathbf{u}}}_{f}$ at $C$ is given by 
+$$
+\bar{\mathbf{u}}_{f}=\left[\begin{array}{l l}{0.5}&{0}\\ {0}&{3(0.5)^{2}-2(0.5)^{3}}\end{array}\right]\left[\begin{array}{l}{0.001}\\ {0.01}\end{array}\right]=\left[\begin{array}{l}{0.0005}\\ {0.005}\end{array}\right]
+$$   
+and the local position of point $C$ is 
+$$
+\bar{\mathbf{u}}_{C}=\bar{\mathbf{u}}_{o}+\bar{\mathbf{u}}_{f}=\left[\begin{array}{c}{0.25}\\ {0}\end{array}\right]+\left[\begin{array}{c}{0.0005}\\ {0.005}\end{array}\right]=\left[\begin{array}{c}{0.2505}\\ {0.005}\end{array}\right]
+$$   
+The global position $\mathbf{r}_{C}$ can then be determined as 
+$$
+\mathbf{r}_{C}={\left[\begin{array}{l}{1.0}\\ {0.5}\end{array}\right]}+{\left[\begin{array}{l l}{0.8660}&{-0.500}\\ {0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{0.2505}\\ {0.005}\end{array}\right]}={\left[\begin{array}{l}{1.2144}\\ {0.62958}\end{array}\right]}
+$$  
+
+Velocity Equations Differentiating Eq. 7 with respect to time yields 
+速度方程
+
+对公式 7 关于时间求导，得到：
 
-Velocity Equations Differentiating Eq. 7 with respect to time yields  
 
 $$
 \dot{\mathbf{r}}_{P}^{i}=\dot{\mathbf{R}}^{i}+\dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}+\mathbf{A}^{i}\dot{\bar{\mathbf{u}}}^{i}
 $$  
 
-where $(\,\cdot\,)$ denotes differentiation with respect to time. Using Eq. 6, one can write $\dot{\mathbf{u}}^{i}$ in terms of the time derivatives of the elastic coordinates of body $i$ as  
+where $(\,\dot\,)$ denotes differentiation with respect to time. Using Eq. 6, one can write $\dot{\mathbf{u}}^{i}$ in terms of the time derivatives of the elastic coordinates of body $i$ as  
+其中 $(\,\dot\,)$ 表示对时间求导。利用公式 6，可以将 $\dot{\mathbf{u}}^{i}$ 表示为物体 $i$ 的弹性坐标的时间导数，即
 
 $$
 \dot{\bar{\mathbf{u}}}^{i}=\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
 $$  
 
 where $\mathbf{S}^{i}=\mathbf{S}^{i}(x_{1}^{i},x_{2}^{i},x_{3}^{i})$ is the body shape matrix, and $\dot{\mathbf{q}}_{f}^{i}$ is the vector of elastic generalized velocities of body $i.$ . Substituting Eq. 14 into Eq. 13 yields  
+其中，$\mathbf{S}^{i}=\mathbf{S}^{i}(x_{1}^{i},x_{2}^{i},x_{3}^{i})$ 为人体形状矩阵，$\dot{\mathbf{q}}_{f}^{i}$ 为物体 $i$ 的弹性广义速度矢量。将公式 14 代入公式 13 得到
 
 $$
 \dot{\mathbf{r}}_{P}^{i}=\dot{\mathbf{R}}^{i}+\dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}+\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
 $$  
 
 where the equation $\dot{\bar{\mathbf{u}}}_{o}^{i}=\mathbf{0}$ is used. To isolate velocity terms, the central term on the right-hand side of Eq. 15 can, in general, be written as  
-
+其中使用方程 $\dot{\bar{\mathbf{u}}}_{o}^{i}=\mathbf{0}$。为了分离速度项，方程 15 右侧的中心项，通常可以写成
 $$
 \dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}=\mathbf{B}^{i}\dot{\theta}^{i}
 $$  
 
-where ${\dot{\Theta}}^{i}$ is the vector whose elements $\dot{\theta}_{k}^{i}$ are the time derivatives of the rotational coordinates of the body reference and $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ is defined as  
+where ${\dot{\theta}}^{i}$ is the vector whose elements $\dot{\theta}_{k}^{i}$ are the time derivatives of the rotational coordinates of the body reference and $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ is defined as  
+其中 ${\dot{\theta}}^{i}$ 是一个向量，其元素 $\dot{\theta}_{k}^{i}$ 是刚体参考系旋转坐标的时间导数，而 $\mathbf{B}^{i}=\mathbf{B}^{i}(\theta^{i},\mathbf{q}_{f}^{i})$ 定义如下：
 
 $$
 \mathbf{B}^{i}=\left[{\frac{\partial}{\partial\theta_{1}^{i}}}(\mathbf{A}^{i}{\bar{\mathbf{u}}}^{i})\cdot\cdot\cdot{\frac{\partial}{\partial\theta_{n_{r}}^{i}}}(\mathbf{A}^{i}{\bar{\mathbf{u}}}^{i})\right]
 $$  
 
 where $n_{r}$ is the total number of rotational coordinates of the reference of body $i$ . Equation 17 follows from using the chain rule of differentiation, which yields  
+其中，$n_{r}$ 为体 $i$ 的参考旋转坐标总数。 方程 17 由使用微分链式法则推导得出，其结果为
 
 $$
 \dot{\bf A}^{i}\bar{\bf u}^{i}=\sum_{k=1}^{n_{r}}\frac{\partial}{\partial\theta_{k}^{i}}({\bf A}^{i}\bar{\bf u}^{i})\dot{\theta}_{k}^{i}
@@ -6676,6 +6698,8 @@ $$
 $$  
 
 In partitioned form, the absolute velocity vector of Eq. 19 can be written as  
+'分块形式下，公式19中的绝对速度矢量可写为：
+
 
 $$
 \dot{\mathbf{r}}_{P}^{i}=[\mathbf{I}\quad\mathbf{B}^{i}\quad\mathbf{A}^{i}\mathbf{S}^{i}]\left[\begin{array}{l}{\dot{\mathbf{R}}^{i}}\\ {\dot{\theta}^{i}}\\ {\dot{\mathbf{q}}_{f}^{i}}\end{array}\right]
@@ -6688,18 +6712,22 @@ $$
 $$  
 
 where $\dot{\mathbf{q}}^{i}=[\dot{\mathbf{q}}_{r}^{i^{\mathrm{T}}}\,\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}]^{\mathrm{T}}=[\dot{\mathbf{R}}^{i^{\mathrm{T}}}\,\dot{\mathbf{\theta}}^{i^{\mathrm{T}}}\,\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}]^{\mathrm{T}}$ is the total vector of generalized velocities of body $i$ , and ${\bf L}^{i}$ is the matrix  
+其中 $\dot{\mathbf{q}}^{i}=[\dot{\mathbf{q}}_{r}^{i^{\mathrm{T}}}\,\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}]^{\mathrm{T}}=[\dot{\mathbf{R}}^{i^{\mathrm{T}}}\,\dot{\mathbf{\theta}}^{i^{\mathrm{T}}}\,\dot{\mathbf{q}}_{f}^{i^{\mathrm{T}}}]^{\mathrm{T}}$ 是刚体 $i$ 的广义速度矢量，而 ${\bf L}^{i}$ 是矩阵。
 
 $$
 \mathbf{L}^{i}=[\mathbf{I}\quad\mathbf{B}^{i}\quad\mathbf{A}^{i}\mathbf{S}^{i}]
 $$  
 
 Before proceeding in our development, perhaps it is important to explain the nature of the terms appearing in the right-hand side of Eq. 19. The vector $\dot{\mathbf{R}}^{i}$ is the absolute velocity vector of the origin of the body reference, while the last term, $\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}$ , is the velocity of point $P$ due to the deformation of the body, defined with respect to an observer stationed on the body. If the body were rigid, the term $\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}$ would be equal to zero. The central term, $\mathbf{B}^{i}\,\dot{\theta}^{i}$ , is the result of differentiation of the transformation matrix with respect to time. This term depends on the reference rotation as well as the elastic deformation of the body. In the case of rigid body translation this term vanishes, and accordingly the velocity of any point on the body is equal to the velocity $\dot{\mathbf{R}}^{i}$ of the origin of the body reference. In Chapter 2, it was shown that  
+在继续我们的开发之前，或许有必要解释一下公式 19 右侧出现的各项的含义。向量 $\dot{\mathbf{R}}^{i}$ 是刚体参考系的绝对速度向量，而最后一项 $\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}$ ，是由于刚体变形而导致的点 $P$ 的速度，定义为一个固定在刚体上的观察者所见的。如果刚体是完全刚性的，那么 $\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}$ 项将等于零。中心项 $\mathbf{B}^{i}\,\dot{\theta}^{i}$ 是对变换矩阵随时间求导的结果。该项取决于参考旋转以及刚体的弹性变形。在刚体平动的情况下，该项消失，因此，刚体上任意一点的速度都等于刚体参考系原点的速度 $\dot{\mathbf{R}}^{i}$。在第 2 章中，我们已经证明了…
+
 
 $$
-\dot{\bf A}\bar{\bf u}^{i}={\bf B}^{i}\dot{\boldsymbol\theta}^{i}={\bf A}^{i}(\bar{\bf\Phi}^{i}\times\bar{\bf u}^{i})=-{\bf A}^{i}(\bar{\bf u}^{i}\times\bar{\bf\Phi}\bar{\boldsymbol\omega}^{i})
+\dot{\bf A}\bar{\bf u}^{i}={\bf B}^{i}\dot{\boldsymbol\theta}^{i}={\bf A}^{i}(\bar{\boldsymbol\omega}^{i}\times\bar{\bf u}^{i})=-{\bf A}^{i}(\bar{\bf u}^{i}\times
+\bar{\boldsymbol\omega}^{i})
 $$  
 
-where $\bar{\mathbf{w}}^{i}$ is the angular velocity vector defined in the body reference. Alternatively, if we define  
+where $\bar{\boldsymbol\omega}^{i}$ is the angular velocity vector defined in the body reference. Alternatively, if we define  
 
 $$
 \mathbf{u}^{i}=\mathbf{A}^{i}\bar{\mathbf{u}}^{i}
@@ -6711,7 +6739,8 @@ $$
 \dot{\mathbf{A}}^{i}\bar{\mathbf{u}}^{i}=\mathbf{B}^{i}\dot{\boldsymbol{\theta}}^{i}=\boldsymbol{\omega}^{i}\times\mathbf{u}^{i}=-\mathbf{u}^{i}\times\boldsymbol{\omega}^{i}
 $$  
 
-where $\omega^{i}$ is the angular velocity vector defined in the global, fixed frame of reference. It is clear from Eq. 25 that the central term on the right-hand side of Eq. 19 is a vector that is perpendicular to both $\omega^{i}$ and the vector $\mathbf{u}^{i}$ , which represents the position of $P$ relative to the origin of the body reference. Knowing that  
+where $\bar{\boldsymbol\omega}^{i}$ is the angular velocity vector defined in the global, fixed frame of reference. It is clear from Eq. 25 that the central term on the right-hand side of Eq. 19 is a vector that is perpendicular to both $\omega^{i}$ and the vector $\mathbf{u}^{i}$ , which represents the position of $P$ relative to the origin of the body reference. Knowing that  
+其中 $\bar{\boldsymbol\omega}^{i}$ 是在全局、固定参考系中定义的角速度矢量。从公式 25 可以看出，公式 19 右侧的中心项是一个矢量，它垂直于 $\omega^{i}$ 和矢量 $\mathbf{u}^{i}$，后者代表了相对于刚体参考系的坐标原点，$P$ 的位置。已知
 
 $$
 -\mathbf{u}^i \times \mathbf{\omega}^i = -\tilde{\mathbf{u}}^i \mathbf{\omega}^i = {\tilde{\mathbf{u}}^i}^T \mathbf{\omega}^i
@@ -6726,7 +6755,7 @@ $$
 and $u_{1}^{i},u_{2}^{i},u_{3}^{i}$ are the components of the vector $\mathbf{u}^{i}$ , one can write the velocity vector of Eq. 19 in the form  
 
 $$
-\dot{\mathbf{r}}_{p}^{i}=\dot{\mathbf{R}}^{i}+\tilde{\mathbf{u}}^{i}^{\mathrm{T}}\mathbf{\omega}\mathbf{\omega}^{i}+\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
+\dot{\mathbf{r}}_{p}^{i}=\dot{\mathbf{R}}^{i}+\tilde{\mathbf{u}}^{i^{\mathrm{T}}}\mathbf{\omega}^{i}+\mathbf{A}^{i}\mathbf{S}^{i}\dot{\mathbf{q}}_{f}^{i}
 $$  
 
 or alternatively  
@@ -6736,8 +6765,10 @@ $$
 $$  
 
 Equation 28 (or Eq. 29) defines the velocity vector in terms of the angular velocity vector $\omega^{i}$ . We will, in general, use Eq. 19 or 20 instead of Eqs. 28 and 29 since we prefer to develop our equations in terms of the generalized coordinates of the body. Therefore, the definition of the matrix $\mathbf{B}^{i}$ is important in the development that follows.  
+公式28（或公式29）用角速度向量 $\omega^{i}$ 定义了速度向量。通常，我们更倾向于使用公式19或20，而不是公式28和29，因为我们喜欢用刚体的广义坐标来推导方程。因此，矩阵 $\mathbf{B}^{i}$ 的定义在以下推导中至关重要。
 
-It was shown in Chapters 2 and 3 that irrespective of the reference rotational coordinates used, the vector $\bar{\mathbf{w}}^{i}$ in Eq. 23 can be written in terms of the rotational coordinates and velocities of the body reference as  
+It was shown in Chapters 2 and 3 that irrespective of the reference rotational coordinates used, the vector  $\bar{\boldsymbol\omega}^{i}$  in Eq. 23 can be written in terms of the rotational coordinates and velocities of the body reference as  
+第二章和第三章已经表明，无论采用何种参考转动坐标系，方程23中的矢量  $\bar{\boldsymbol\omega}^{i}$  可以表示为物体的参考坐标及其转动速度的函数。
 
 $$
 \bar{\pmb{\omega}}^{i}=\bar{\bf G}^{i}\,\dot{\pmb{\theta}}^{i}
@@ -6780,20 +6811,26 @@ $$
 $$  
 
 The form of the velocity vector of Eq. 21 with ${\bf L}^{i}$ defined by Eq. 36 will be used in the development of the kinetic energy in the following section.  
+第21式中速度矢量形式，其 ${\bf L}^{i}$ 由第36式定义，将在下一节中用于开发动能。
 
 Acceleration Equations The acceleration of point $P$ can be determined by direct differentiation of Eq. 21. This leads to  
+加速度方程
+
+点 $P$ 的加速度可以通过对公式 21 直接求导得到。这导出了
 
 $$
 \ddot{\mathbf{r}}_{P}^{i}=\dot{\mathbf{L}}^{i}\dot{\mathbf{q}}^{i}+\mathbf{L}^{i}\ddot{\mathbf{q}}^{i}
 $$  
 
 where $\dot{\mathbf{L}}^{i}\dot{\mathbf{q}}^{i}$ is a quadratic velocity vector that contains the Coriolis component. Using the identities presented in Chapter 2, one can verify that the acceleration vector of Eq. 37 can be written as  
+其中 $\dot{\mathbf{L}}^{i}\dot{\mathbf{q}}^{i}$ 是一个包含科里奥利力的二次速度矢量。利用第二章中给出的恒等式，可以验证方程 37 中的加速度矢量可以写成
 
 $$
-\ddot{\mathbf{r}}_{P}^{i}=\ddot{\mathbf{R}}^{i}+\mathbf{\omega}^{i}\times(\mathbf{\omega}\mathbf{\dot{\omega}}\times\mathbf{u}^{i})+\alpha^{i}\times\mathbf{u}^{i}+2\mathbf{\omega}^{i}\times(\mathbf{A}^{i}\dot{\mathbf{u}}^{i})+\mathbf{A}^{i}\ddot{\mathbf{u}}^{i}
+\ddot{\mathbf{r}}_{P}^{i}=\ddot{\mathbf{R}}^{i}+\mathbf{\omega}^{i}\times(\mathbf{\dot{\omega}}\times\mathbf{u}^{i})+\alpha^{i}\times\mathbf{u}^{i}+2\mathbf{\omega}^{i}\times(\mathbf{A}^{i}\dot{\mathbf{u}}^{i})+\mathbf{A}^{i}\ddot{\mathbf{u}}^{i}
 $$  
 
 In this equation, $\pmb{\alpha}^{i}$ is the angular acceleration vector. The term $\ddot{\mathbf{R}}^{i}$ is the absolute acceleration of the origin of the body reference. The second term, $\boldsymbol{\omega}^{i}\times(\boldsymbol{\omega}^{i}\times\mathbf{u}^{i})$ , is the normal component of the acceleration of point $P^{\prime}$ that instantaneously coincides with $P$ and does not undergo deformation. This component of the acceleration is directed along the straight line connecting the two points $o$ and $P$ . The third component, $\propto^{i}\times\mathbf{u}^{i}$ , is the tangential component of the acceleration of $P^{i}$ relative to $o$ . The direction of this component is perpendicular to both the angular acceleration vector $\alpha^{i}$ and the vector $\mathbf{u}^{i}$ . The fourth term, $2\omega^{i}\times(\mathbf{A}^{i}\dot{\mathbf{u}}^{i})$ , is the Coriolis component of the acceleration, and the fifth term, $\mathbf{A}^{i}\ddot{\mathbf{u}}^{i}$ is the acceleration of point $P$ due to the deformation relative to the body reference. If the body is rigid, the fourth and fifth components vanish.  
+在这个方程中，$\pmb{\alpha}^{i}$ 是角加速度矢量。$\ddot{\mathbf{R}}^{i}$ 是刚体参考系原点的绝对加速度。第二项，$\boldsymbol{\omega}^{i}\times(\boldsymbol{\omega}^{i}\times\mathbf{u}^{i})$ ，是与点 $P^{\prime}$ 瞬时重合且不发生变形的 $P$ 点的加速度的法向分量。该加速度分量沿连接两个点 $o$ 和 $P$ 的直线方向。第三项，$\propto^{i}\times\mathbf{u}^{i}$ ，是相对于 $o$ 点的 $P^{i}$ 点的切向加速度分量。该分量的方向垂直于角加速度矢量 $\alpha^{i}$ 和矢量 $\mathbf{u}^{i}$。第四项，$2\omega^{i}\times(\mathbf{A}^{i}\dot{\mathbf{u}}^{i})$ ，是科里奥利加速度分量，第五项，$\mathbf{A}^{i}\ddot{\mathbf{u}}^{i}$ 是由于变形相对于刚体参考系的 $P$ 点的加速度。如果刚体是刚性的，则第四项和第五项消失。
 
 Example 5.2 The reference of the beam of Example 1 rotates with a constant angular velocity $\omega={\dot{\theta}}=5$ rad/sec. Determine the absolute velocity and acceleration of the tip point $A$ at the instant of time $t$ at which the beam coordinates, velocities, and accelerations are given by