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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

@@ -6833,15 +6833,17 @@ In this equation, $\pmb{\alpha}^{i}$ is the angular acceleration vector. The ter
 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  
-
+示例 5.2 
+示例 1 中的梁以常数角速度 $\omega={\dot{\theta}}=5$ rad/sec 旋转。确定在梁坐标、速度和加速度给定时，时间 $t$ 瞬时尖端点 $A$ 的绝对速度和绝对加速度。
 $$
 \begin{array}{r l}{\mathbf{q}=\left[\mathbf{q}_{r}^{\mathrm{T}}\quad\mathbf{q}_{f}^{\mathrm{T}}\right]^{\mathrm{T}}=\left[R_{1}\quad R_{2}\quad\theta\quad q_{f1}\quad q_{f2}\right]^{\mathrm{T}}}&{}\\ {=\left[1.0\quad0.5\quad30^{\circ}\quad0.001\quad0.01\right]^{\mathrm{T}}}&{}\\ {\dot{\mathbf{q}}=\left[\dot{\mathbf{q}}_{r}^{\mathrm{T}}\quad\dot{\mathbf{q}}_{f}^{\mathrm{T}}\right]^{\mathrm{T}}=\left[\dot{R}_{1}\quad\dot{R}_{2}\quad\dot{\theta}\quad\dot{q}_{f1}\quad\dot{q}_{f2}\right]^{\mathrm{T}}}&{}\\ {=\left[0.1\quad1.0\quad5\quad2\quad3\right]^{\mathrm{T}}}&{}\\ {\ddot{\mathbf{q}}=\left[\ddot{\mathbf{q}}_{r}^{\mathrm{T}}\quad\ddot{\mathbf{q}}_{f}^{\mathrm{T}}\right]^{\mathrm{T}}=\left[\ddot{R}_{1}\quad\ddot{R}_{2}\quad\ddot{\theta}\quad\ddot{q}_{f1}\quad\ddot{q}_{f2}\right]^{\mathrm{T}}}&{}\\ {=\left[2\quad0\quad0\quad10\quad20\right]^{\mathrm{T}}}&{}\end{array}
 $$  
 
 Solution In this case the matrix $\mathbf{L}$ of Eq. 22 is the $2\times5$ matrix  
+解决方案 在本例中，公式22中的矩阵 $\mathbf{L}$ 是一个 $2\times5$ 矩阵。
 
 $$
-{\bf L}=[{\bf I}\cdot\mathrm{~\bf~B~}\;\mathrm{~\bf~AS}]
+{\bf L}=[{\bf ~I~}~\mathrm{~\bf~B~}\;\mathrm{~\bf~AS}]
 $$  
 
 where $\mathbf{I}$ is the $2\times2$ identity matrix  
@@ -6865,7 +6867,7 @@ $$
 From Example 1, the vector $\bar{\mathbf{u}}_{A}$ , which is the position of the tip point $A$ defined in the beam coordinate system, is given by $\bar{\mathbf{u}}_{A}=[0.501\quad0.01]^{\mathrm{T}}$ . Therefore, the vector $\mathbf{B}$ can be evaluated as  
 
 $$
-\mathbf{B}=\mathbf{A}_{\theta}\bar{\mathbf{u}}_{A}=\left[{\bf\Pi}_{0.866}^{-0.5}\quad{-0.866}\right]\left[{\bf\Pi}_{0.01}^{0.501}\right]=\left[{\bf\Pi}_{0.42886}^{-0.25916}\right]
+\mathbf{B}=\mathbf{A}_{\theta}\bar{\mathbf{u}}_{A}={\left[\begin{array}{l l}{-0.5}&{-0.866}\\ {0.866}&{-0.5}\end{array}\right]}{\left[\begin{array}{l}{0.501}\\ {0.01}\end{array}\right]}={\left[\begin{array}{l}{−0.25916}\\ {0.42886}\end{array}\right]}
 $$  
 
 Since at point $A$ , $\xi=(x/l)=1$ , the shape matrix S evaluated at point $A$ is given by  
@@ -6883,7 +6885,7 @@ $$
 The matrix $\mathbf{L}$ can then be defined as  
 
 $$
-{\bf L}=[{\bf I}{\mathrm{~\boldmath~{~\cal~B~}~}}{\mathrm{~\boldmath~{~\cal~AS}}}]=\left[{\begin{array}{c c c c c}{1}&{0}&{-0.25916}&{0.8660}&{-0.5000}\\ {0}&{1}&{0.42886}&{0.5000}&{0.8660}\end{array}}\right]
+{\bf L}=[{\bf ~I~}~\mathrm{~\bf~B~}\;\mathrm{~\bf~AS}]=\left[{\begin{array}{c c c c c}{1}&{0}&{-0.25916}&{0.8660}&{-0.5000}\\ {0}&{1}&{0.42886}&{0.5000}&{0.8660}\end{array}}\right]
 $$  
 
 and accordingly, the global velocity vector of point $A$ is given by  
@@ -6902,7 +6904,7 @@ where
 
 $$
 {\begin{array}{r l}{\mathbf{L}{\ddot{\mathbf{q}}}={\left[\begin{array}{l l l l l}{1}&{0}&{-0.25916}&{0.8660}&{-0.5000}\\ {0}&{1}&{0.42886}&{0.500}&{0.8660}\end{array}\right]}{\left[\begin{array}{l}{2}\\ {0}\\ {0}\\ {10}\\ {20}\end{array}\right]}}\\ {={\left[\begin{array}{l}{0.66}\\ {22.32}\end{array}\right]}{\mathrm{m}}/{\mathrm{sec}}^{2}}\end{array}}
-$$  
+$$  One can verify that the matrix $\dot{\mathbf{L}}$ is
 
 $$
 \dot{\mathbf{L}}=[\mathbf{0}_{2}\quad\dot{\mathbf{B}}\quad\dot{\mathbf{A}}\mathbf{S}]=[\mathbf{0}_{2}\quad(-\mathbf{A}\bar{\mathbf{u}}_{A}\dot{\theta}+\mathbf{A}_{\theta}\mathbf{S}\dot{\mathbf{q}}_{f})\quad\mathbf{A}_{\theta}\mathbf{S}\dot{\theta}]