vault backup: 2025-07-21 09:06:49
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@ -3213,26 +3213,29 @@ $\dot{r}^{2}$ 项的存在意味着粘性力所做的功始终为负，即它们
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最后，也可能遇到扭转阻尼器；在图 3.7 中，扭转弹簧将被一个阻尼器取代，该阻尼器对刚性杆施加一个力矩，其大小是角度 $\theta$ 随时间变化率的函数。扭转阻尼器中粘性力所做的微分功为 $\mathrm{d}W=-\ell f(\dot{\theta})\mathrm{d}\theta=-M(\dot{\theta})\mathrm{d}\theta$；对于线性扭转阻尼器，$M({\dot{\theta}})=c{\dot{\theta}}$ ，其中阻尼器常数的单位现在为 $\mathbf{N}{\cdot}\mathbf{m}{\cdot}\mathbf{s}$ 。
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# The energy closure equation
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#### The energy closure equation能量守恒方程
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Consider the work and energy principle given by eq. (3.24), written as $\begin{array}{r l}{\int_{t_{i}}^{t_{f}}\underline{{F}}_{n c}^{T}\mathrm{d}\underline{{r}}=}&{{}}\end{array}$ $E_{f}-E_{i}$ . In this expression, the initial and final time instants can be selected arbitrarily; in particular, let the final time be an arbitrary time, $t_{f}=t$ , during the evolution of the system. The principle of work and energy now becomes
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考虑由公式 (3.24) 给出的功与能量原理，写成：$\begin{array}{r l}{\int_{t_{i}}^{t_{f}}\underline{{F}}_{n c}^{T}\mathrm{d}\underline{{r}}=}&{{}}\end{array}$ $E_{f}-E_{i}$ 。 在此表达式中，初始时间和最终时间可以任意选择；特别地，令最终时间为任意时间，$t_{f}=t$ ，在系统演化过程中。 功与能量原理现在变为
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$$
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E(t)-\int_{t_{i}}^{t}\underline{{F}}_{n c}^{T}\mathrm{d}\underline{{r}}=E_{i}.
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$$
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In the absence of non-conservative forces, this equation reduces to $E(t)\,=\,E_{i}$ , the statement of conservation of the total mechanical energy of the system. Even in the presence of non-conservative forces, however, equation (3.34) implies the conservation a scalar quantity, the difference between the total mechanical energy and the cumulative work done by the non-conservative forces, must remain constant. This relationship is known as the energy closure equation.
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# Example 3.1. bungee jumping
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在没有非保守力的情况下，此方程简化为 $E(t)\,=\,E_{i}$，这表达了系统总机械能守恒的原理。即使在存在非保守力的情形下，方程 (3.34) 仍然暗示着一个标量量的守恒：即总机械能与非保守力所做的累积功之差，必须保持不变。这种关系被称为能量封闭方程。
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#### Example 3.1. bungee jumping
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A man of mass $m$ is jumping off a bridge while attached to a bungee cord of unstretched length $d_{0}$ . An inertial frame, $\mathscr{F}\,=\,\left[\mathbf{O},\mathcal{T}=\left(\overline{{\iota}}_{1},\overline{{\iota}}_{2}\right)\right]$ , is attached to the bridge. The man is jumping from point $\mathbf{o}$ with an initial velocity, $v_{0}$ , oriented along horizontal axis $\bar{\imath}_{2}$ , and the acceleration of gravity is acting along vertical axis $\overline{{\iota}}_{1}$ .
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During the first part of his fall, the man is in free ﾚight under the effect of gravity, and at some instant in time, the bungee becomes taut. During the second portion of his fall, the man is subjected to the combined effects of gravity and the elastic force of the bungee. The potential of the bungee is of the following form: $V_{b}\;=\;$ $1/2\,\,k_{0}d_{0}^{2}\,\,\ln^{2}(1\,\bar{+}\,\bar{\varDelta})$ , where $\bar{\Delta}\,=\,(d\mathrm{~-~}d_{0})/d_{0}\,=\,\varDelta/d_{0}$ is the non-dimensional stretch of the bungee, and $d$ the distance from point $\mathbf{o}$ to the man. The magnitude of the force the bungee applied to the man is $F_{b}=\mathrm{d}V/\mathrm{d}\varDelta=k_{0}d_{0}\ln(1+\bar{\varDelta})/(1+\bar{\varDelta})$ . Determine the trajectory of the fall.
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一个质量为 $m$ 的人正在从一座桥上跳下，同时连接着一根未拉伸长度为 $d_{0}$ 的蹦极绳。一个惯性参考系 $\mathscr{F}\,=\,\left[\mathbf{O},\mathcal{T}=\left(\overline{{\iota}}_{1},\overline{{\iota}}_{2}\right)\right]$ 被固定在桥上。人从点 $\mathbf{o}$ 跳下，初始速度 $v_{0}$ 沿水平轴 $\bar{\imath}_{2}$ 方向，重力加速度沿垂直轴 $\overline{{\iota}}_{1}$ 方向。
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# Free fall
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在下落的第一个阶段，人处于重力作用下的自由落体状态，在某个时刻，蹦极绳绷紧。在下落的第二个阶段，人受到重力和蹦极绳的弹性力共同作用。蹦极绳的势能具有如下形式：$V_{b}\;=\;$ $1/2\,\,k_{0}d_{0}^{2}\,\,\ln^{2}(1\,\bar{+}\,\bar{\varDelta})$ ，其中 $\bar{\Delta}\,=\,(d\mathrm{~-~}d_{0})/d_{0}\,=\,\varDelta/d_{0}$ 是蹦极绳的无量纲拉伸量， $d$ 是从点 $\mathbf{o}$ 到人的距离。蹦极绳施加在人身上的力的大小为 $F_{b}=\mathrm{d}V/\mathrm{d}\varDelta=k_{0}d_{0}\ln(1+\bar{\varDelta})/(1+\bar{\varDelta})$ 。确定下落的轨迹。
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#### Free fall
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Let the man’s trajectory be denoted $\underline{{{r}}}(t)\;=\;x_{1}(t)\,\bar{\iota}_{1}\,+\,x_{2}(t)\,\bar{\iota}_{2}$ . During free fall, Newton’s second law writes $m\ddot{\underline{{r}}}\,=\,m g\bar{\imath}_{1}$ , where $g$ is the acceleration of gravity. Integration yields
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设该男性的轨迹记为 $\underline{{{r}}}(t)\;=\;x_{1}(t)\,\bar{\iota}_{1}\,+\,x_{2}(t)\,\bar{\iota}_{2}$ 。在自由落体过程中，牛顿第二定律可写为 $m\ddot{\underline{{r}}}\,=\,m g\bar{\imath}_{1}$ ，其中 $g$ 为重力加速度。积分可得
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$$
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\bar{\underline{{v}}}=\bar{g}\tau\bar{\iota}_{1}+\bar{\iota}_{2},\quad\bar{\underline{{r}}}={\frac{1}{2}}\bar{g}\tau^{2}\bar{\iota}_{1}+\tau\bar{\iota}_{2}.
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$$
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@ -3240,25 +3243,28 @@ $$
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The following non-dimensional quantities were introduced: $\underline{{\bar{r}}}\,=\,\underline{{r}}/d_{0},\,\underline{{\bar{v}}}\,=\,\underline{{\dot{r}}}/v_{0}.$ , $\tau=v_{0}t/d_{0}$ , and $\bar{g}=g d_{0}/v_{0}^{2}$ .
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The bungee cord becomes taut when $\|\underline{{r}}\|\;=\;d_{0}$ , or $\|\bar{\underline{{r}}}(\tau_{t})\|^{2}\;=\;1$ , where $\tau_{t}$ denotes the instant at which the bungee becomes taut. Introducing this condition in eq. (3.35) and solving for $\tau_{t}$ yields
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以下引入了无量纲量：$\underline{{\bar{r}}}\,=\,\underline{{r}}/d_{0},\,\underline{{\bar{v}}}\,=\,\underline{{\dot{r}}}/v_{0}$ , $\tau=v_{0}t/d_{0}$ , 和 $\bar{g}=g d_{0}/v_{0}^{2}$ 。
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当 $\|\underline{{r}}\|\;=\;d_{0}$ 时，蹦极绳绷紧，或者 $\|\bar{\underline{{r}}}(\tau_{t})\|^{2}\;=\;1$ ，其中 $\tau_{t}$ 表示蹦极绳绷紧的时刻。将此条件代入公式 (3.35) 并求解 $\tau_{t}$ 得到
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$$
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\tau_{t}=\frac{\sqrt{2}}{\bar{g}}\sqrt{\sqrt{1+\bar{g}^{2}}-1}.
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$$
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# Trajectory when the bungee cord is taut
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#### Trajectory when the bungee cord is taut蹦极绳绷紧时的轨迹
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Once the bungee is taut, Newton’s second law implies $m\ddot{\underline{{r}}}=m g\bar{\iota}_{1}+F_{b}\bar{u}$ , were $F_{b}$ is the magnitude of the elastic force the bungee applies on the man and $\bar{u}$ the unit vector pointing from the man to point O. The distance from the man to point $\mathbf{o}$ is $d=d_{0}\bar{+}\,\bar{\varDelta}=\sqrt{x_{1}^{2}+x_{2}^{2}}$ , and $\bar{u}=(x_{1}\,\bar{\iota}_{1}+x_{2}\,\bar{\iota}_{2})/d$ . In non-dimensional form, the equation of motion becomes
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一旦蹦极绳绷紧，牛顿第二定律推导出 $m\ddot{\underline{{r}}}=m g\bar{\iota}_{1}+F_{b}\bar{u}$ ，其中 $F_{b}$ 是蹦极绳对人的弹性力的大小，$\bar{u}$ 是指向点 O 的单位向量。人到点 $\mathbf{o}$ 的距离为 $d=d_{0}\bar{+}\,\bar{\varDelta}=\sqrt{x_{1}^{2}+x_{2}^{2}}$ ，且 $\bar{u}=(x_{1}\,\bar{\iota}_{1}+x_{2}\,\bar{\iota}_{2})/d$ 。以无量纲形式，运动方程变为
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$$
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\bar{\underline{{r}}}^{\prime\prime}=\bar{g}\,\bar{\iota}_{1}-\bar{k}_{0}\frac{\ln(1+\bar{\varDelta})}{(1+\bar{\varDelta})^{2}}(\bar{x}_{1}\,\bar{\iota}_{1}+\bar{x}_{2}\,\bar{\iota}_{2}),
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$$
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where $(\cdot)^{\prime}$ indicates a derivative with respect to the non-dimensional time $\tau$ , $\bar{\varDelta}\,=$ $\sqrt{\bar{x}_{1}^{2}+\bar{x}_{2}^{2}}-1,\bar{k}_{0}=k_{0}d_{0}^{2}/(m v_{0}^{2}),\bar{x}_{1}=\bar{x}_{1}/d_{0}$ , and $\bar{x}_{2}=x_{2}/d_{0}$ .
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其中 $(\cdot)^{\prime}$ 表示对无量纲时间 $\tau$ 的导数，$\bar{\varDelta}\,=$ $\sqrt{\bar{x}_{1}^{2}+\bar{x}_{2}^{2}}-1$, $\bar{k}_{0}=k_{0}d_{0}^{2}/(m v_{0}^{2})$, $\bar{x}_{1}=\bar{x}_{1}/d_{0}$ , 以及 $\bar{x}_{2}=x_{2}/d_{0}$ 。
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Because the equations of motion are nonlinear, their solution can only be obtained by means of numerical methods, which often require recasting the governing equations in first-order form. In the present case, the first-order form of the equations is
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由于运动方程是非线性的，其解只能通过数值方法获得，这通常需要将控制方程改写成一阶形式。在本例中，方程的一阶形式是：
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$$
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\left\{\begin{array}{c}{\displaystyle\bar{x}_{1}}\\ {\displaystyle\bar{x}_{2}}\\ {\displaystyle\bar{v}_{1}}\\ {\displaystyle\bar{v}_{2}}\end{array}\right\}^{\prime}=\left\{\begin{array}{c}{\displaystyle\bar{v}_{1}}\\ {\displaystyle\bar{v}_{2}}\\ {\displaystyle\bar{g}-\bar{k}_{0}\frac{\ln(1+\bar{\cal A})}{(1+\bar{\cal A})^{2}}\bar{x}_{1}}\\ {\displaystyle-\,\bar{k}_{0}\frac{\ln(1+\bar{\cal A})}{(1+\bar{\cal A})^{2}}\bar{x}_{2}}\end{array}\right\},
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\left\{\begin{array}{c}{\displaystyle\bar{x}_{1}}\\ {\displaystyle\bar{x}_{2}}\\ {\displaystyle\bar{v}_{1}}\\ {\displaystyle\bar{v}_{2}}\end{array}\right\}^{\prime}=\left\{\begin{array}{c}{\displaystyle\bar{v}_{1}}\\ {\displaystyle\bar{v}_{2}}\\ {\displaystyle\bar{g}-\bar{k}_{0}\frac{\ln(1+\bar{\varDelta})}{(1+\bar{\varDelta})^{2}}\bar{x}_{1}}\\ {\displaystyle-\,\bar{k}_{0}\frac{\ln(1+\bar{\varDelta})}{(1+\bar{\varDelta})^{2}}\bar{x}_{2}}\end{array}\right\},
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$$
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where $\bar{v}_{1}=v_{1}/v_{0}$ and $\bar{v}_{2}=v_{2}/v_{0}$ are the non-dimensional components of vertical and horizontal velocity, respectively. The first two equations, $\bar{x}_{1}^{\prime}=\bar{v}_{1}$ and $\bar{x}_{2}^{\prime}=\bar{v}_{2}$ , simply define the velocity components, $\bar{v}_{1}$ and $\bar{v}_{2}$ , and the last two equations are the actual equations of motion. In this form, many standard time integration methods such as Runge-Kutta integrators, among many others, can be used. Extensive discussion of these integrators can be found in many textbook on numerical analysis, see refs. [4, 5], for instance.
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@ -3268,7 +3274,13 @@ The following non-dimensional parameters are used for the simulation: $\bar{g}=1
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At time $\tau=1.5018$ , the bungee becomes slack again, and equation of motion, eq. (3.37), is no longer valid because it include the force stemming from the bungee cord. To continue the simulation past that time, the equation of motion for free fall under gravity, $m\ddot{\underline{{r}}}=m g\bar{\imath}_{1}$ , would be used again, with initial conditions corresponding to the man’s position and and velocity at the end of the previous phase, i.e., at time $\tau=1.5018$ .
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Figure 3.11 depicts the bungee non-dimensional force, $\bar{F}_{b}={F_{b}}/{(k_{0}d_{0})}$ , versus its non-dimensional stretch, $\bar{\varDelta}$ . The apparent stiffness, $k$ , of the bungee cord is the tangent to the force-stretch curve,
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其中 $\bar{v}_{1}=v_{1}/v_{0}$ 和 $\bar{v}_{2}=v_{2}/v_{0}$ 分别为垂直和水平速度的无量纲分量。前两方程，$\bar{x}_{1}^{\prime}=\bar{v}_{1}$ 和 $\bar{x}_{2}^{\prime}=\bar{v}_{2}$，简单地定义了速度分量 $\bar{v}_{1}$ 和 $\bar{v}_{2}$，而后两方程是实际的运动方程。 这种形式允许使用许多标准的时步积分方法，例如龙格-库塔积分器等。 关于这些积分器的详细讨论可以在许多数值分析教科书中找到，例如参考文献 [4, 5]。
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模拟中使用的无量纲参数为：$\bar{g}=12$ 和 $\bar{k}_{0}=50$。自由落体阶段的结束时间为 $\tau_{t}=0.4254$。图 3.9 显示了在自由落体和蹦极绳拉紧时的轨迹。对于所有时间 $\tau<\tau_{t}$，蹦极绳处于松弛状态，其伸长量为零；图 3.10 显示了在 $\tau\geq\tau_{t}$ 时蹦极绳的无量纲伸长量 $\bar{\varDelta}$。
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在时间 $\tau=1.5018$ 时，蹦极绳再次松弛，运动方程，即方程 (3.37)，不再有效，因为它包含了蹦极绳产生的力。为了在那个时间之后继续模拟，需要再次使用自由落体下的重力运动方程，$m\ddot{\underline{{r}}}=m g\bar{\imath}_{1}$，并使用与前一阶段结束时（即时间 $\tau=1.5018$ 时）对应的人的位移和速度作为初始条件。
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图 3.11 描绘了蹦极绳的无量纲力 $\bar{F}_{b}={F_{b}}/{(k_{0}d_{0})}$ 与其无量纲伸长量 $\bar{\varDelta}$ 的关系。蹦极绳的视杆 stiffness，$k$，是力-伸长曲线的切线。
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Fig. 3.9. Man’s trajectory. The symbols $\bigcirc$ indicate the free fall portion of the trajectory.
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@ -3280,7 +3292,7 @@ k=\frac{\mathrm{d}F_{b}}{\mathrm{d}\varDelta}=k_{0}\frac{1-\ln(1+\bar{\varDelta}
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$$
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As the stretch of the cord increases, its stiffness decreases and vanishes when $\ln(1+$ $\bar{\varDelta})=1$ , or $\bar{\varDelta}\approx1.718$ . Clearly, the parameters selected for the present simulation result in a very large stretching of the bungee cord, which would threaten the safety of the jumper.
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随着绳索的拉伸量增加，其刚度降低，并在 $\ln(1+$ $\bar{\varDelta})=1$ ，或 $\bar{\varDelta}\approx1.718$ 时消失。显然，本模拟所选取的参数导致了蹦极绳索的过度拉伸，这将威胁到跳跃者的安全。
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Fig. 3.11. Magnitude of the non-dimensional force, $F_{b}$ , in the bungee versus stretch, $\bar{\varDelta}$ .
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@ -3288,30 +3300,32 @@ Fig. 3.11. Magnitude of the non-dimensional force, $F_{b}$ , in the bungee versu
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Fig. 3.12. System energies: kinetic energy, solid line; potential energy, dashed line; total mechanical energy, dashed-dotted line.
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The kinetic energy of the system is $K\,=\,1/2\ m\dot{\underline{{r}}}^{2}$ . The potential of the gravity forces is $V_{g}~=~-m g x_{1}$ , and the potential of the elastic bungee cord $V_{b}\ =$ $\bar{1/2}~k_{0}d_{0}^{2}~\ln^{2}(\bar{1}+\bar{\varDelta})$ . In non-dimensional form, the total mechanical energy of the system becomes
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系统的动能为 $K\,=\,1/2\ m\dot{\underline{{r}}}^{2}$ 。重力势能为 $V_{g}~=~-m g x_{1}$ ，弹性蹦极绳的势能为 $V_{b}\ =$ $\bar{1/2}~k_{0}d_{0}^{2}~\ln^{2}(\bar{1}+\bar{\varDelta})$ 。以无量纲形式表示，系统的总机械能变为
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$$
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\bar{E}=\frac{E}{m v_{0}^{2}}=\frac{K}{m v_{0}^{2}}+\frac{V}{m v_{0}^{2}}=\frac{1}{2}\bar{\underline{{r}}}^{\prime2}-\bar{g}\bar{x}_{1}+\frac{1}{2}\bar{k}_{0}\ln^{2}(1+\bar{\Delta}).
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$$
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Figure 3.12 depicts the evolution of the system’s energies versus time $\tau$ . Because the forces acting on the system are conservative forces, the total mechanical energy remains constant during the simulation. This observation provides a validation of the derivation of the equation of motion and of its numerical solution.
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# Effect of drag forces.
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图 3.12 描绘了系统能量随时间 $\tau$ 的演化。由于作用于系统的力是保守力，因此在模拟过程中系统的总机械能保持恒定。这一观察结果验证了运动方程的推导及其数值解。
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#### Effect of drag forces.
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The developments presented in the previous paragraphs have ignored the effect of air friction on the man’s trajectory. These forces can be taken into account in an approximate manner applying to the man a drag force, $\underline{{{F}}}_{d}\;=\;-1/2~C_{d}\rho\underline{{{A}}}~\|\underline{{{v}}}\|\underline{{{v}}},$ where $C_{d}$ is the non-dimensional drag coefficient, $\rho$ the air density, and $\boldsymbol{\mathcal{A}}$ the man’s cross-sectional area. This drag force is at all times proportional to the square of the speed, aligned with the velocity vector, and oriented in the direction opposite to this vector. During free fall, the equation of motion is $m\ddot{\underline{{r}}}=m g\bar{\iota}_{1}-1/2\ C_{d}\rho\mathcal{A}\ \lVert\underline{{v}}\rVert\underline{{v}};$ as before, the bungee cord will become taut when $\|\underline{{r}}(\tau_{t})\|=1$ . Because the governing differential equation is now a nonlinear differential equation, a numerical process must be used for its solution and time $\tau_{t}$ must be determined numerically. A closed form analytical solution such as that given by eq. (3.36) no longer exists.
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When the bungee cord is taut, the differential equation governing the problem becomes
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前面几段描述的发展忽略了空气阻力对人轨迹的影响。我们可以近似地考虑这些力，通过对人施加阻力 $F_{d}\;=\;-1/2~C_{d}\rho\underline{{{A}}}~\|\underline{{{v}}}\|\underline{{{v}}},$ 其中 $C_{d}$ 是无量纲阻力系数，$\rho$ 是空气密度，$\boldsymbol{\mathcal{A}}$ 是人的横截面积。这个阻力始终与速度的平方成正比，与速度矢量对齐，并且方向与该矢量相反。在自由落体过程中，运动方程为 $m\ddot{\underline{{r}}}=m g\bar{\iota}_{1}-1/2\ C_{d}\rho\mathcal{A}\ \lVert\underline{{v}}\rVert\underline{{v}}；$ 就像之前一样，当 $\|\underline{{r}}(\tau_{t})\|=1$ 时，蹦极绳会变紧。由于控制方程现在是一个非线性微分方程，必须使用数值过程来求解，并且时间 $\tau_{t}$ 必须数值确定。不再存在像公式 (3.36) 所示的闭合形式解析解。
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当蹦极绳变紧时，控制问题的微分方程变为
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$$
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\bar{\underline{{x}}}^{\prime\prime}=\bar{g}\,\bar{\imath}_{1}-\bar{k}_{0}\frac{\ln(1+\bar{\varDelta})}{(1+\bar{\varDelta})^{2}}(\bar{x}_{1}\,\bar{\imath}_{1}+\bar{x}_{2}\,\bar{\imath}_{2})-\frac{1}{2}\bar{\mu}C_{d}\sqrt{\bar{v}_{1}^{2}+\bar{v}_{2}^{2}}\,\,(\bar{v}_{1}\,\bar{\imath}_{1}+\bar{v}_{2}\,\bar{\imath}_{2}),
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$$
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where $\bar{\mu}=\rho A d_{0}/m$ . Here again, the equation of motion is nonlinear, and its solution can be obtained only by means of numerical methods
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# 3.2.4 The principle of impulse and momentum
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其中 $\bar{\mu}=\rho A d_{0}/m$ 。 再次，运动方程是非线性的，其解只能通过数值方法获得。
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### 3.2.4 The principle of impulse and momentum冲量与动量原理
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The principle of impulse and momentum involves two sets of new quantities. First, the linear and angular momentum vectors of a particle are introduced; the angular momentum is the moment of the linear momentum vector. Next, the linear and angular impulse vectors of the externally applied forces are introduced.
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# Principle of linear impulse and momentum
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冲量与动量原理涉及两组新的量。首先，引入粒子的线动量矢量和角动量矢量，其中角动量是线动量矢量的力矩；其次，引入作用在粒子上的外力产生的线impulse矢量和角impulse矢量。
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#### Principle of linear impulse and momentum
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Figure 3.13 shows a particle of mass $m$ in motion with respect to an inertial frame $\mathcal{F}^{I}=[\mathbf{O},\mathcal{T}=(\bar{\iota}_{1},\bar{\iota}_{2},\bar{\iota}_{3})]$ . The inertial velocity vector of the particle is denoted $\underline{v}$ . The linear momentum vector of a particle is defined as the product of its mass by its inertial velocity vector
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