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.obsidian/plugins/copilot/data.json
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@ -285,7 +285,7 @@
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"name": "Translate to Chinese",
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"prompt": "<instruction>Translate the text below into Chinese:\n 1. Preserve the meaning and tone\n 2. Maintain appropriate cultural context\n 3. Keep formatting and structure\n \n </instruction>\n\n<text>{copilot-selection}</text>\n<restrictions>\n1. Blade翻译为叶片，flapwise翻译为挥舞，edgewise翻译为摆振，pitch angle翻译成变桨角度，twist angle翻译为扭角，rotor翻译为风轮，turbine、wind turbine翻译为机组、风电机组，span翻译为展向，deflection翻译为变形，mode翻译为模态，normal mode翻译为简正模态，jacket 翻译为导管架，superelement翻译为超单元，shaft翻译为主轴，azimuth、azimuth angle翻译为方位角，neutral axes 翻译为中性轴\n2. Return only the translated text.\n</restrictions>",
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"showInContextMenu": true,
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"modelKey": "gemini-2.5-flash|google"
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"modelKey": "gemma3:12b|ollama"
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},
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{
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"name": "Summarize",
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@ -3408,21 +3408,21 @@ In the absence of external moments, this principle implies $\begin{array}{r}{\un
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在没有外部力矩的情况下，该原理意味着 $\begin{array}{r}{\underline{{h}}_{O}(t_{f})\,=\,\underline{{h}}_{O}(t_{i}).}\end{array}$ ，即角动量始终保持不变。换句话说，当外施力矩消失时，质点的角动量矢量保持不变。
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# Example 3.2. Particle in a pinned tube
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#### Example 3.2. Particle in a pinned tube
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Figure 3.14 depicts a particle of mass $m$ connected to inertial point A by means of a spring of stiffness $k$ and dashpot of constant $c$ . At the initial time, the particle is located at $\theta\:=\:0$ , $\phi\,=\,\pi/2$ , and $r~=~r_{0}$ , which corresponds to the un-stretched configuration of the spring; $r$ , $\phi$ , and $\theta$ form a spherical coordinate system, see section 2.7.2. The initial velocity vector of the particle is $\underline{{v}}_{0}$ . Derive the equations of motion of the system.
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图 3.14 描绘了一个质量为 $m$ 的粒子，通过刚度系数为 $k$ 的弹簧和阻尼系数为 $c$ 的阻尼器连接到惯性点 A。在初始时刻，该粒子位于 $\theta\:=\:0$ , $\phi\,=\,\pi/2$ , 和 $r~=~r_{0}$ 处，这对应于弹簧的未拉伸状态；$r$ , $\phi$ , 和 $\theta$ 构成一个球坐标系，见第 2.7.2 节。粒子的初始速度矢量为 $\underline{{v}}_{0}$ 。推导该系统的运动方程。
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Fig. 3.14. Particle subjected to a central force due to a spring and dashpot.
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First, Newton’s second law is used to obtain the desired equations of motion as
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首先，利用牛顿第二定律可以获得所需的运动方程，如下所示：
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$$
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\begin{array}{r l}&{m\left[(\ddot{r}-r\dot{\phi}^{2}-r\dot{\theta}^{2}\sin^{2}\phi)\bar{e}_{1}+(r\ddot{\phi}+2\dot{r}\dot{\phi}-r\dot{\theta}^{2}\sin\phi\cos\phi)\bar{e}_{2}\right.}\\ &{\quad+\left.(r\ddot{\theta}\sin\phi+2\dot{r}\dot{\theta}\sin\phi+2r\dot{\phi}\dot{\theta}\cos\phi)\bar{e}_{3}\right]=-\left[k(r-r_{0})+c\dot{r}\right]\bar{e}_{1},}\end{array}
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$$
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where the components of the acceleration vector in the spherical coordinate system are given by eq. (2.95c). Projecting this equation along the unit vectors $\bar{e}_{1},\bar{e}_{2}$ , and $\bar{e}_{3}$ , then yields $m\bigl(\ddot{r}-r{\dot{\phi}}^{2}-r{\dot{\theta}}^{2}\sin^{2}\!\frac{\ulcorner}{\phi}\bigr)=-\dot{k}(r-r_{0})-c\dot{r}$ , $r\ddot{\phi}\!+\!2\dot{r}\dot{\phi}\!-\!r\dot{\theta}^{2}\sin\phi\cos\phi=$ 0, and $r\theta\sin\phi+2\dot{r}\dot{\theta}\sin\phi+2r\dot{\phi}\dot{\theta}\cos\phi=0$ , respectively. These three nonlinear differential equations can be solved for the coordinates of the particle, $r,\,\phi$ , and $\theta$ . Although this approach will indeed yield the solution of the problem, much information about the nature of the particle’s motion can be obtained from the application the principle of angular impulse and momentum.
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其中，在球坐标系中，加速度向量的各分量由公式 (2.95c) 给出。沿单位向量 $\bar{e}_{1},\bar{e}_{2}$ 和 $\bar{e}_{3}$ 投影该公式，分别得到 $m\bigl(\ddot{r}-r{\dot{\phi}}^{2}-r{\dot{\theta}}^{2}\sin^{2}\!\frac{\ulcorner}{\phi}\bigr)=-\dot{k}(r-r_{0})-c\dot{r}$ , $r\ddot{\phi}\!+\!2\dot{r}\dot{\phi}\!-\!r\dot{\theta}^{2}\sin\phi\cos\phi=$ 0, 和 $r\theta\sin\phi+2\dot{r}\dot{\theta}\sin\phi+2r\dot{\phi}\dot{\theta}\cos\phi=0$ 。这三个非线性微分方程可以求解出粒子的坐标 $r,\,\phi$ 和 $\theta$ 。虽然这种方法确实可以得到问题的解，但关于粒子运动性质的信息可以通过应用角动量定理来获得。
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Because the line of action of the forces applied to the particle passes through point A, the moment of these forces with respect to point $\mathbf{A}$ vanishes. The principle of angular impulse and momentum, eq. (3.44), then implies that the angular momentum must remain constant, $\underline{{H}}_{A}=\underline{{H}}_{A0}=r_{0}\widetilde{\iota}_{1}\ m\underline{{v}}_{0}$ .
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It follows that $r\widetilde{e}_{1}\;m\underline{{v}}=\underline{{H}}_{A0}$ . This vector product equation, see section 1.1.11, affords a solution if and only if the particle’s position vector, $r\bar{e}_{1}$ , and velocity vector, $\underline{v}_{\cdot}$ , are both contained in the plane normal to the initial angular momentum vector, $\underline{{H}}_{A0}$ . Because the particle’s position and velocity vectors are contained in the same plane, the particle’s motion is contained entirely in the plane normal to the angular momentum vector.
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@ -3433,7 +3433,7 @@ The solution of the problem is now considerably simplified. Without loss general
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The equations of motion then reduce to $r^{\prime\prime}\,-\,r\theta^{\prime2}\,=\,-(r\,-\,r_{0})\,-\,2\zeta r^{\prime}$ , and $r^{2}\theta^{\prime}=r_{0}^{2}\theta_{0}^{\prime}$ , where notation $(\cdot)^{\prime}$ indicates a derivative with respect to $\tau$ . Finally, the non-dimensionalposition of the particle is introduced, $\bar{r}=r/r_{0}$ , and the equations of motion simply become $\bar{r}^{\prime\prime}=\theta_{0}^{\prime2}/\bar{r}^{3}-(\bar{r}-1)-2\zeta\bar{r}^{\prime}$ and $\theta^{\prime}=\theta_{0}^{\prime}/\bar{r}^{2}$ , respectively; the initial conditions are $\bar{r}(t\,=\,0)\,=\,1$ , $\bar{r}^{\prime}(t\,=\,0)\,=\,(\bar{\iota}_{1}^{T}\underline{{{v}}}_{0})/(\varOmega r_{0})$ , and $\theta_{0}^{\prime}\,=$ $(\bar{\iota}_{2}^{T}\underline{{v}}_{0})/(\varOmega r_{0})$ .
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# Example 3.3. Particle sliding on a helix
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#### Example 3.3. Particle sliding on a helix
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Consider the motion of a particle sliding without friction along the helix depicted in fig. 2.3. Gravity acts down, in the opposite direction of axis $\bar{\iota}_{3}$ . Since the particle is
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@ -3463,7 +3463,7 @@ $$
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respectively. The normal component of the constraint force stems from the normal component of acceleration. Because the component of acceleration in the binormal direction vanishes, the corresponding component of the constraint force is solely due to the gravity component in that direction.
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# Example 3.4. Particle sliding on a spherical surface
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#### Example 3.4. Particle sliding on a spherical surface
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Consider the motion of a particle sliding on the spherical surface depicted in fig. 2.10. Gravity acts down, in the opposite direction of axis $\bar{\iota}_{3}$ . Since the particle is constrained to move on the spherical surface, a constraint force is applied to the particle. This force acts in the direction normal to the surface, i.e., it has a single component along the surface normal.
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@ -3495,21 +3495,21 @@ For small motions of the particle near the lowest point on the sphere, i.e for $
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The same results could have been obtained using spherical coordinates, see section 2.7.2, instead of surface coordinates; the fact that the particle is moving on the surface of a sphere then implies ${\dot{r}}=0$ .
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# 3.2.5 Problems
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### 3.2.5 Problems
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# Problem 3.1. Simple spring mass system
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#### Problem 3.1. Simple spring mass system
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Consider a simple spring mass system: a particle of mass $m$ is connected to a spring of stiffness $k$ and a gravity field with an acceleration $g$ is acting on the system. At time $t_{0}$ , the system is at rest and the spring is un-stretched. Consider the following two scenarios. Scenario $^{\,I}$ : the mass is released from rest and oscillates freely thereafter. Scenario 2: the mass is slowly brought to its static equilibrium position. (1) Find the maximum displacement of the particle for scenario 1. (2) Find the maximum displacement of the particle for scenario 2. (3) If there exist any difference in the maximum displacements for scenarios $^{\,l}$ and 2, give work and energy arguments to justify the discrepancy.
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# Problem 3.2. Work done by conservative forces
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考虑一个简单的弹簧-质量系统：质量为 $m$ 的粒子通过劲度系数为 $k$ 的弹簧连接，并且受到重力场作用，重力加速度为 $g$。在时间 $t_{0}$ 时，系统处于静止状态，弹簧处于未拉伸状态。考虑以下两种情况。情况 $^{\,I}$：质量从静止释放，之后自由振荡。情况 2：质量缓慢地移动到其静态平衡位置。 (1) 找出情况 1 中粒子的最大位移。 (2) 找出情况 2 中粒子的最大位移。 (3) 如果情况 $^{\,I}$ 和 2 的最大位移存在差异，请给出功和能量的论证来解释这种差异。
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### Problem 3.2. Work done by conservative forces
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Prove that the work done by a conservative force applied to a particle between times $t_{i}$ and $t_{f}$ is independent of the path of the particle during that time.
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# Problem 3.3. Is a constant force a conservative force?
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证明一个保守力作用于一个粒子，在时间 $t_{i}$ 和 $t_{f}$ 之间所做的功，与该粒子在这段时间内的路径无关。
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### Problem 3.3. Is a constant force a conservative force?
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Is a constant force a conservative force? If yes, find the potential of this force.
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# Problem 3.4. Particle subjected to friction forces
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### Problem 3.4. Particle subjected to friction forces
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Consider a particle of mass $m=1\;\mathrm{kg}$ subjected to a friction force $\underline{{F}}_{f}\,=\,-k v\underline{{v}},$ where $k$ is the friction coefficient and $v=\|\underline{{v}}\|$ . The particle is also subjected to gravity forces $(g=9.81$ $\mathrm{m}/\mathrm{s}^{2}$ ), see fig. 3.15. At time $t=0$ , the particle is launched with an initial speed $v_{0}=100\,\mathrm{m/s}$ with an angle $\theta=30$ deg with respect to the horizontal. $(I)$ Write the equations of motion for the particle. (2) Solve these equations for $k=0,0.001$ , and $0.002\:\mathrm{kg/m}$ . (3) Plot the trajectory of the particle for the three cases on the same graph. (4) Determine the distance $d$ and the maximum height $h$ from the computed trajectory. (5) Plot $d$ and $h$ as a function of the friction coefficient $k\in[0,0.003]\;\mathrm{kg/m}$ .
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@ -3519,15 +3519,15 @@ Fig. 3.15. Particle subjected to friction.
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Fig. 3.16. Particle in a slot on a rotating disk.
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# Problem 3.5. Particle in a slot on a rotating disk
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### Problem 3.5. Particle in a slot on a rotating disk
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Figure 3.16 depicts a disk rotating in a vertical plane at a constant angular speed, $\dot{\phi}\,=\,\Omega$ , around inertial point O. Mass $m$ is free to slide in a radial slot on the disk and is connected to the center of the disk by means of a spring of stiffness constant $k$ and a dashpot of constant $c$ . The system is subjected to gravity and a torque, $Q$ , is applied to the disk. The spring’s unstretched length is denoted $x_{0}$ . (1) Derive the equation of motion of the system in terms of distance $_x$ from point $\mathbf{o}$ to the particle. (2) Find the horizontal and vertical components of the reaction force at point O. (3) Find the applied torque, $Q$ , required to maintain this constant angular speed.
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# Problem 3.6. Free falling parachute
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### Problem 3.6. Free falling parachute
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Figure 3.17 shows a payload of mass $m\,=\,120~\mathrm{kg}$ attached to a parachute. The payload is dropped from an altitude $h\,=\,1000~\mathrm{m}$ with a horizontal velocity of magnitude $v_{0}~=~100$ $\mathrm{m/s}$ . The payload is subjected to a drag force $\underline{{{F}}}_{d}\;=\;-1/2~C_{d}\rho\underline{{{A}}}~v\underline{{{v}}}.$ where $C_{d}~=~1.42$ is the drag coefficient, $\rho\;=\;1.23~\mathrm{{kg/m^{3}}}$ the air density, $A\,=\,\pi D^{2}/4$ the cross-sectional area of the parachute, and $D$ its diameter. The velocity vector is denoted $\underline{v}$ and the speed is $v\ =\ \|\underline{{v}}\|$ . The payload is also subjected to gravity forces $\mathrm{(\it{g}\ =\ 9.81\ m/s^{2})}$ ), see fig. 3.17. (1) Write the equations of motion for the payload. (2) Solve these equations numerically for parachutes of diameter $D=3$ , 4, and $6\;\mathrm{m}$ . (3) Plot the horizontal position of the payload for the three cases on the same graph. (4) Plot the vertical position of the payload for the three cases on the same graph. (5) Plot the horizontal velocity of the payload for the three cases on the same graph. (6) Plot the vertical velocity of the payload for the three cases on the same graph. (7) Find an analytical expression for the constant horizontal velocity that is eventually reached by the payload. (8) Find an analytical expression for the constant vertical velocity that is eventually reached by the payload. (9) Based on this constant vertical velocity, find an analytical expression for the time it takes for the payload to reach the ground. (10) Compute the time to reach the ground as a function of parachute diameter $D\in[3,6]\mathrm{~m~}$ . On the same graph, plot the numerical and analytical solutions. $(I I)$ Compute the final vertical velocity as a function of parachute diameter $D\,\in\,[3,6]\mathrm{~m~}$ . On the same graph, plot the numerical and analytical solutions
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# Problem 3.7. Pendulum under gravity forces
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### Problem 3.7. Pendulum under gravity forces
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Consider a pendulum with a bob of mass $m=1.5~\mathrm{kg}$ , length $\ell=0.75\mathrm{~m~}$ and subjected to gravity forces $\mathit{\dot{g}}=9.81\;\mathrm{m/s^{2}}$ ). The pendulum is released from rest with $\theta=0$ , see fig. 3.18.
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@ -3539,7 +3539,7 @@ Fig. 3.18. Pendulum under gravity forces.
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(1) Write the equations of motion for the pendulum. (2) Solve these equations numerically. (3)Plot the angular motion $\theta$ and angular velocity $\dot{\theta}$ as functions of time on two separate graphs. (4) Compute and plot the tension in the cord as a function of time. (5) On one graph, plot the kinetic energy, potential energy, and total mechanical energy of the system versus time. (6) Plot the total mechanical energy of the system versus time. Does it remain constant? Comment on your results
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# Problem 3.8. Inverted pendulum
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### Problem 3.8. Inverted pendulum
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Consider an inverted pendulum with a bob of mass $m=1\;\mathrm{kg}$ . A massless, rigid bar of length $\ell=1\;\mathrm{m}$ supports the bob. Gravity, $g=9.81\:\mathrm{m/s^{2}}$ , acts in the direction indicated on fig. 3.19. A torsional spring of stiffness $k\,=\,10\ \mathrm{N{\cdot}m/r a d}$ is located at point $\mathbf{o}$ and applies a moment $M=-k\theta$ on the rigid bar. The pendulum is released from $\theta=0$ , with an initial speed $v_{0}=2$ $\mathrm{m/s}$ to the right, see fig. 3.19. (1) Write the equations of motion for the system. (2) Solve these equations numerically. (3) Plot the angular motion $\theta$ as a function of time. (4) Plot the angular velocity θ˙ as a function of time. (5) Plot the load in the rigid bar as a function of time. (6) On one graph, plot the kinetic energy, potential energy, strain energy, and total mechanical energy of the system versus time. (7) On one graph, plot the total mechanical energy of the system versus time. Does it remain constant? Comment on your results. (8) Consider two states ot the system: the initial configuration, $(\theta=0)$ ), and a final configuration, $(\theta=\theta_{f})$ ), where the angle $\theta_{f}$ is maximum. Find the maximum angular deﾚection, $\theta_{f}$ . Check your answer against the numerical simulation.
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@ -3549,27 +3549,27 @@ Fig. 3.19. Inverted pendulum under gravity forces.
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Fig. 3.20. Particle connected to the ground by a spring and damper.
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# Problem 3.9. Particle connected to the ground with spring and damper
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### Problem 3.9. Particle connected to the ground with spring and damper
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A particle of mass $m$ is connected to the ground by a spring of stiffness $k$ and a damper of constant $c$ . The initial configuration of the system is indicated on fig. 3.20, and the initial velocity vector is $\underline{{v}}_{0}$ . The following quantities are defined: $\varOmega^{2}=k/\bar{m}$ and $c=2m\Omega\zeta$ . For this problem, it is convenient to use the polar coordinate system indicated on the figure. (1) Set up the equations of motion of the system. (2) Plot $\bar{r}\,=\,r/r_{0}$ as a function of the nondimensional time $\tau=\varOmega t$ for $\tau\in[0,20\pi]$ . (3) Plot $\theta(\tau)$ . (4) Plot the trajectory of the particle in space. (5) Plot the history of the non-dimensional angular velocity $\bar{\Omega}(\tau)=\dot{\theta}/\Omega$ . (6) Plot the history of the components of the velocity vector in the inertial frame, $\bar{v}_{x}\,=\,v_{x}/(\varOmega r_{0})$ and $\bar{v}_{y}\,=\,v_{y}/(\varOmega r_{0})$ . (7) Plot the history of the non-dimensional total mechanical energy of the system $\vec{E}(\tau)\,=\,E/(m\varOmega^{2}r_{0}^{2})$ ; comment your result. (8) Compute the non-dimensional cumulative energy dissipated in the damper $\bar{W}(\tau)\,=\,W/(m\Omega^{2}r_{0}^{2})$ . (9) Plot the history of the quantity $\bar{E}(\tau)+\bar{W}(\tau)$ ; comment your result. Use the following data: $\zeta=0.05$ ; at time $\tau=0$ $=0,\underline{{v}}_{0}/(\Omega r_{0})=1.2\,\bar{\iota}_{1}+0.8\,\bar{\iota}_{2}$ , $\theta_{0}=0$ and $r(t=0)/r_{0}=1$ ; the un-stretched length of the spring is $r_{0}$ .
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# Problem 3.10. Particle sliding along a curve
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### Problem 3.10. Particle sliding along a curve
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A particle of mass $m$ freely slides along a given curve $\mathbb{C}$ in three-dimensional space. A point on the curve has a position vector $\underline{{p}}_{0}(s)$ . Find the equation of motion for the particle if it is subjected to externally applied forces $\underline{{F}}(t)$ . What are the components of the constraint force acting on the particle.
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# Problem 3.11. Particle sliding along a helix
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### Problem 3.11. Particle sliding along a helix
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A particle slides along a helix and is subjected to a gravity force acting along the $\overline{{\imath}}_{1}$ direction, see fig. 2.3. Find the equation of motion for the particle in terms of the parameter $\eta$ . If the initial condition at $t\,=\,0$ are $\eta=0$ and $\dot{\eta}\,=\,v_{0}\bar{/}\sqrt{a^{2}+k^{2}}$ , find the minimum value of $v_{0}$ such that the particle proceeds along the helix with $\dot{\eta}>0$ at all time.
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# Problem 3.12. Particle sliding along a circular ring
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### Problem 3.12. Particle sliding along a circular ring
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Figure 3.21 depicts a particle of mass $m$ sliding along a circular ring under the effect of gravity. The ring rotates on two bearing about an axis parallel to $\bar{\iota}_{3}$ ; a torque $Q(t)$ , acting about axis $\bar{\iota}_{3}$ , is applied to the ring. $(I)$ Find the equations of motion for the particle. (2) Write the expression for the potential of the gravity forces. (3) Write the expression for the kinetic energy of the particle. (4) Write the expression for the work done by the applied torque $Q(t)$ .
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# Problem 3.13. Particle sliding along a circular ring
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### Problem 3.13. Particle sliding along a circular ring
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Figure 3.21 depicts a particle of mass $m$ sliding along a circular ring under the effect of gravity. The ring rotates on two bearing about an axis parallel to $\bar{\iota}_{3}$ ; a torque $Q(t)$ , acting about axis $\bar{\iota}_{3}$ , is applied to the ring. $(I)$ Find the equations of motion for the particle based on the principle of impulse and momentum.
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# Problem 3.14. Particle in a massless tube
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### Problem 3.14. Particle in a massless tube
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Figure 3.22 shows a particle of mass $m$ sliding in a massless tube is connected to a spring of stiffness $k_{r}$ and a damper of constant $c_{r}$ . The un-stretched length of the spring is $r_{0}$ . A spherical coordinate system $r$ , $\phi$ and $\theta$ with corresponding unit vectors $\bar{e}_{1},\,\bar{e}_{2}$ and $\bar{e}_{3}$ will be convenient to use. The spring/damper assembly is attached to the ground at point A by means of a joint that allows rotation about axis $\bar{e}_{3}$ . This joint features a torsional spring of stiffness $k_{\phi}$ and a torsional damper of constant $c_{\phi}$ . The torsional spring is un-stretched when $\phi\,=\,\pi/2$ . The angle $\theta$ has a prescribed schedule $\theta(t)\,=\,\omega t$ . The following quantities are defined: the non-dimensional time $\tau=\omega t$ , the axial spring frequency $\Omega_{r}\,=\,\dot{\sqrt{k_{r}/m}}$ , and its critical damping ratio $\zeta_{r}=c_{r}/(2m\varOmega_{r})$ , the torsional spring frequency $\varOmega_{\phi}=\sqrt{k_{\phi}/m r_{0}^{2}}$ and its critical damping ratio $\zeta_{\phi}\,=\,c_{\phi}/(2m r_{0}^{2}\varOmega_{\phi})$ . (1) Set up the equations of motion of the system. (2) Plot $\bar{r}\ =\ r/r_{0}$ as a function of the non-dimensional time $\tau~\in~[0,20\pi]$ . (3) Plot $\phi(\tau)$ . (4) Plot the trajectory of the particle in three-dimensional space. (5) Plot the history of the non-dimensional force $\Bar{F}_{3}\,=\,\Bar{F}_{3}/(m r_{0}\omega^{2})$ that the massless tube applies on the particle. (6) Plot the history of the non-dimensional total mechanical energy of the system $\bar{E}(\tau)=E/(m r_{0}^{2}\omega^{2})$ ; comment your result. (7) Plot the history of the cumulative nondimensional energy dissipated in the dampers $\bar{W}^{d}(\tau)=W^{d}/(m r_{0}^{2}\omega^{2})$ . (8) Plot the history of cumulative non-dimensional work $\bar{W}^{M}=W^{M}/(m r_{0}^{2}\omega^{2})$ done by the torque required to prescribe $\theta(t)=\omega t$ . (9) Plot the history of the quantity $\bar{E}(\tau)+\bar{W}^{d}(\tau)-\bar{W}^{M}(\tau)$ ; comment your result. Use the following data: $\bar{\Omega}_{r}\,=\,\bar{\Omega_{r}}/\omega\,=\,5$ , $\zeta_{r}~=~0.05$ ; $\bar{\Omega}_{\phi}\,=\,\Omega_{\phi}/\omega\,=\,1.5$ , $\zeta_{\phi}=0.05$ ; at time $\tau=0$ , $\underline{{v_{0}}}/(\omega r_{0})=0.6\;\bar{\iota}_{1}+1\;\bar{\iota}_{2}+0.75\;\bar{\iota}_{3}$ , $\phi_{0}=\pi/2$ and $r/r_{0}=1$ .
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@ -3579,11 +3579,11 @@ Fig. 3.22. Particle connected to the ground by a spring, damper and revolute joi
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Fig. 3.21. Particle sliding on a circular ring.
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# Problem 3.15. Particle moving on a track
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### Problem 3.15. Particle moving on a track
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Figure 3.23 shows particle of mass $m$ moving on a track defined by a curve $\mathbb{C}$ while constrained to remain within a slot inside a massless arm. The massless arm is prescribed to move at a constant angular speed, ${\dot{\theta}}\,=\,\Omega$ . (1) Plot the radial location, $r/R$ , of the particle as a function of $\theta$ . (2) Plot the moment $\bar{m}\,=\,M/(m R^{2}\varOmega^{2})$ necessary to drive the system at a constant angular speed. (3) Plot the non-dimensional normal force $\bar{F}_{n}\,=\,F_{n}/(m R\varOmega^{2})$ the curved track applies on the particle. (4) Determine the minimum stiffness of the spring, i.e., the minimum non-dimensional frequency $\bar{\varOmega}$ , for which the particle remains on the track at all times. The curve is defined in the polar coordinate system as $p_{0}(\theta)\,=\,r(\theta)\bar{e}_{1}$ , where $r(\theta)\,=\,R\,-\,b\cos N\theta$ . It will be convenient to define the normal to the curve as $\bar{n}\,=\,\widetilde{\i_{3}}\bar{t}$ , where $\bar{t}$ is the tangent to the curve. Use the following data: $\bar{b}\;=\;b/R\;=\;0.25;\;N\;=\;6$ $\omega^{2}=k/m$ ; $\bar{\Omega}=\omega/\Omega=3$ . The spring is un-stretched when $r=0$ . At time $t=0,\theta=0$ .
|
||||
|
||||
# Problem 3.16. Particle moving on a track
|
||||
### Problem 3.16. Particle moving on a track
|
||||
|
||||
Figure 3.23 shows particle of mass $m$ moving on a track defined by a curve $\mathbb{C}$ while constrained to remain within a slot inside a massless arm. A moment, $M$ , is applied to the arm at point O. (1) Plot the time history of the angle $\theta$ . (2) Plot the angular speed $\dot{\theta}/\omega$ . (3) Plot the normal force $\bar{F}_{n}\,=\,F_{n}/(m R\omega^{2})$ the curved track applies on the particle. (4) Plot the total mechanical energy of the system, $\bar{E}\,=\,E/(m R^{2}\omega^{2})$ . Discuss your results. (5) Compare the responses of the system at $\bar{M}_{0}\,=\,0.75$ and 0.80. Explain your results. The curve is defined in the polar coordinate system as $p_{0}(\theta)\,=\,r(\theta)\bar{e}_{1}$ , where $r(\theta)\,=\,R\,-\,b\cos N\theta$ . It will be convenient to define the normal to the curve as $\bar{n}=\widetilde{\i_{3}}\bar{t}$ , where $\bar{t}$ is the tangent to the curve. Use the following data: $\bar{b}\,=\,b/R\,=\,0.25;\,N\,=\,6;\,\omega^{2}\,=\,k/m$ . The spring is un-stretched when $r=0$ . At time $t=0$ , $\theta=0$ and $\dot{\theta}=0$ . The applied moment is given as $\bar{M}=M/(m R^{2}\omega^{2})=\bar{M}_{0}(1-\cos\tau)$ for $\tau\leq2\pi$ and $\bar{m}=0$ for $\tau>2\pi$ , where $\tau=\omega t$ is the non-dimensional time and $\bar{M}_{0}=0.75$ . Simulate the system for $\tau\in[0,6\pi]$ .
|
||||
|
||||
@ -3593,56 +3593,62 @@ Fig. 3.23. Particle moving on a track.
|
||||
|
||||
Fig. 3.24. Particle connected to a spring with unilateral contact to a horizontal plane.
|
||||
|
||||
# Problem 3.17. Particle with unilateral contact
|
||||
### Problem 3.17. Particle with unilateral contact
|
||||
|
||||
The particle of mass $m$ depicted in fig. 3.24 is subjected to a gravity field of acceleration $g$ and is connected to a spring of stiffness $k$ and un-stretched length $h$ . The spring is attached to inertial point $\mathbf{A}$ , located a distance $h$ above point O. A unilateral contact condition is imposed on the particle by the horizontal plane $\mathcal{P}\,=\,(\mathbf{O},{\bar{\imath}}_{2})$ ; this means that the particle can only move in the half-space above this plane. At the initial time, the particle is at point $\mathbf{o}$ and has a velocity $\underline{{v}}(t=0)=v_{0}\bar{\iota}_{1}.(I)$ Write the equation of motion for the particle while it is in contact with the plane. (2) Plot the non-dimensional position of the particle as a function of nondimensional time. (3) Plot the non-dimensional velocity of the particle as a function of nondimensional time. (4) Find the time at which the particle leaves the plane and its corresponding position and velocity. (5) Under what condition will the particle always remain on the plane for any magnitude of the initial velocity $v_{0}?$ (6) Find the time at which the particle will first hit the plane after leaving it. (7) Plot the trajectory of the particle during its free ﾚight. Use the following data: $\bar{v}_{0}\,=\,v_{0}/(\omega h)\,=\,1$ ; $\bar{g}=m g/(k h)=0.25$ ; $\omega^{2}=k/m$ . Use the following non-dimensional time $\tau=\omega t$ . All lengths are non-dimensionalized by $h$ , velocities by $\omega h$ .
|
||||
|
||||
# Problem 3.18. Particle moving on a parabolic surface of revolution
|
||||
### Problem 3.18. Particle moving on a parabolic surface of revolution
|
||||
|
||||
Figure 2.11 shows a particle sliding on a parabolic surface of revolution and subjected to a gravity force acting along the negative $\bar{\iota}_{3}$ direction. This surface is defined by the position vector of one of its points, $\underline{{p}}_{0}=r\cos\phi\,\bar{\iota}_{1}+r\sin\phi\,\bar{\iota}_{2}+a r^{2}\bar{\iota}_{3}$ , where $r\geq0$ and $0\leq\phi\leq2\pi$ . The following notation was used $\eta_{1}=r$ and $\eta_{2}=\phi.\left(I\right)$ Find the equation of motion for the particle in terms of the surface coordinates $r$ and $\phi$ . (2) Find the constraint force acting on the particle.
|
||||
|
||||
# Problem 3.19. Particle sliding on a linear spiral
|
||||
### Problem 3.19. Particle sliding on a linear spiral
|
||||
|
||||
A particle of mass $m$ is sliding along a linear spiral, as defined in example 2.2, under the effect of gravity acting down along the $\bar{\imath}_{2}$ axis, see fig. 2.4. (1) Derive the governing equation of motion using Newton’s law. (2) Plot the angle $\theta$ as a function of time. (3) Plot the θ˙ as a function of time. (4) Plot the time history of the magnitude of the normal reaction force that the spiral applies to the particle. (5) On one graph, plot the time history of the kinetic energy of the particle, its potential energy, and its total mechanical energy. (6) Derive the governing equation of motion of the particle from the principle of work and energy. Use the following data: $m=2.5\;\mathrm{kg}$ ; $a=0.2\:\mathrm{m}$ ; $g=9.81\:\mathrm{m}/\mathrm{s}^{2}$ . At time $t=0$ , $\theta=0$ and $\dot{\theta}=50\,\mathrm{rad/s}$ Present all your results for $t\in[0,15]$ s.
|
||||
|
||||
# Problem 3.20. Bungee jumping
|
||||
### Problem 3.20. Bungee jumping
|
||||
|
||||
A man of mass $m$ is jumping off a bridge while attached to a bungee cord of un-stretched length $d_{0}$ , as described in example 3.1. An inertial frame, $\mathcal{F}=[\mathbf{O},\mathcal{T}=(\bar{\iota}_{1},\bar{\iota}_{2})]$ , is attached to the bridge. The man is jumping from point 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}$ . In the developments presented in example 3.1, the effect of air friction on the man was ignored. In this problem, these forces will 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 coeffi- cient, $\rho$ the air density, and $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 the velocity vector. $(I)$ Derive the equation of motion for the free fall portion of the man’s trajectory. Solve the equations numerically to find the time $\tau_{t}$ at which the bungee becomes taught. (2) Derive the equations of motion once the bungee is taut. Solve the equations numerically. (3) On one graph, plot the components $\bar{x}_{1}$ and ${\bar{x}}_{2}$ of the man’s position vector as functions of $\tau$ . (4) Plot the trajectory of the man. (5) On one graph, plot the components $\bar{v}_{1}$ and $\bar{v}_{2}$ of the velocity vector as functions of $\tau$ . (6) Plot the stretch of the bungee as a function of $\tau$ . (7) On one graph, plot the non-dimensional kinetic energy, $\bar{K}=\bar{K}/(m v_{0}^{2})$ , potential energy, $\bar{V}=V/(m v_{0}^{2})$ , and total potential energy $\bar{E}=\bar{K}+\bar{V}$ . (8) Determine the non-dimensional time at which the bungee becomes slack again. Use the following non-dimensional quantities: $\bar{x}_{1}=x_{1}/d_{0},\bar{x}_{2}=x_{2}/d_{0};\bar{v}_{1}=v_{1}/v_{0},\bar{v}_{2}=v_{2}/v$ ; use the non-dimensional time $\tau~=~v_{0}t/d_{0}$ . Use the following data: $\bar{g}\;=\;g d_{0}/v_{0}^{2}\;=\;10$ , $\bar{k}_{0}=k_{0}d_{0}^{2}/(m v_{0}^{2})=60$ , $C_{d}=0.47$ and $\bar{\mu}=\rho A d_{0}/m=0.03$ . Present all your results for $\tau\in[0,3.5\tau_{t}]$ .
|
||||
|
||||
# 3.3 Contact forces
|
||||
## 3.3 Contact forces
|
||||
|
||||
When dealing with particle dynamics, it is often the case that the particle is in contact with another body. Contact can be of a continuous nature; for instance, a particle is moving while in continuous contact with a curve or a surface, see example 3.3 or 3.4, respectively. Contact could also be of an intermittent nature, such as, for instance, the impact of a particle on an obstacle. These contact forces are forces acting on the particle, which must therefore be included in the statement of Newton’s second law when studying the dynamic response of the particle. Both magnitude and direction of these forces must be studied to properly state Newton’s second law.
|
||||
|
||||
The kinematics of contact of a particle with a surface and a curve will be studied first in sections 3.3.1 and 3.3.2, respectively; contact forces are categorized into normal and tangential contact forces. Next, the magnitudes of these forces will be studied in section 3.3.3. Typically, constitutive laws are postulated that relate the magnitude of the contact forces to contact parameters. For instance, Coulomb’s friction law relates the friction force to both normal force and relative velocity of the particle with respect to surface it is in contact with.
|
||||
|
||||
# 3.3.1 Kinematics of particles in contact with a surface
|
||||
在处理粒子动力学时，通常情况下，粒子会与另一个物体接触。接触可以是连续的性质；例如，一个粒子在与曲线或曲面保持连续接触时移动，分别见例 3.3 或 3.4。接触也可能是间歇性的性质，例如，粒子撞击障碍物。这些接触力是作用在粒子上的力，因此在研究粒子的动态响应时，必须将其包含在牛顿第二定律的表述中。为了正确表述牛顿第二定律，必须研究这些力的 magnitude 和方向。
|
||||
|
||||
接下来，将分别在 3.3.1 和 3.3.2 节研究粒子与曲面和曲线的接触运动学；接触力被划分为法向接触力和切向接触力。随后，将在 3.3.3 节研究这些力的 magnitude。通常，人们会假设一些本构关系，将接触力的 magnitude 与接触参数联系起来。例如，库仑摩擦定律将摩擦力与法向力和粒子相对于其接触曲面的相对速度联系起来。
|
||||
### 3.3.1 Kinematics of particles in contact with a surface
|
||||
|
||||
Figure 3.25 depicts a particle of mass $m$ in continuous contact with surface $\mathbb{S}$ at point P. The differential geometry of a surface was studied in details in section 2.4, and it is assumed here that parameters $\eta_{1}$ and $\eta_{2}$ define the lines of curvature presented section 2.4.5. Unit vectors $\bar{e}_{1}$ and $\bar{e}_{2}$ given by eq. (2.55) define the plane tangent to the surface at point P. The normal to the surface is now defined as $\bar{n}\,=\,\widetilde{e}_{1}\bar{e}_{2}$ and $\mathcal{E}=(\bar{e}_{1},\bar{e}_{2},\bar{n})$ forms an orthonormal basis.
|
||||
|
||||
图 3.25 描绘了一个质量为 $m$ 的粒子，其持续与表面 $\mathbb{S}$ 的点 P 接触。表面微分几何在第 2.4 节中进行了详细研究，此处假设参数 $\eta_{1}$ 和 $\eta_{2}$ 定义了第 2.4.5 节中呈现的曲率线。由公式 (2.55) 给出的单位向量 $\bar{e}_{1}$ 和 $\bar{e}_{2}$ 定义了在点 P 处的切平面。表面的法线现在定义为 $\bar{n}\,=\,\widetilde{e}_{1}\bar{e}_{2}$，并且 $\mathcal{E}=(\bar{e}_{1},\bar{e}_{2},\bar{n})$ 构成一个正交基。
|
||||
|
||||
Fig. 3.25. Particle moving on a surface.
|
||||
|
||||
The contact force, $\underline{{F}}^{c}$ , between the particle and the surface is conveniently divided into two components, the normal contact force, $\underline{{F}}^{n}=F^{n}{\bar{n}}$ , which acts along the normal to the surface, and the tangential contact force, $\underline{{F}}^{t}=F_{1}^{t}\bar{e}_{1}\!+\!F_{2}^{t}\bar{e}_{2}$ , which acts in the plane tangent to the surface. Hence, the contact force is written as
|
||||
粒子与表面之间的接触力，$\underline{{F}}^{c}$ ，可以方便地划分为两个分量：法向接触力，$\underline{{F}}^{n}=F^{n}{\bar{n}}$ ，它沿着垂直于表面的方向作用，以及切向接触力，$\underline{{F}}^{t}=F_{1}^{t}\bar{e}_{1}\!+\!F_{2}^{t}\bar{e}_{2}$ ，它在与表面切平的平面内作用。因此，接触力可以写成：
|
||||
|
||||
$$
|
||||
\underline{{F}}^{c}=\underline{{F}}^{n}+\underline{{F}}^{t}=F^{n}\bar{n}+(F_{1}^{t}\bar{e}_{1}+F_{2}^{t}\bar{e}_{2}).
|
||||
$$
|
||||
|
||||
Imagine first that the particle slides over the surface without any friction: the tangential contact forces vanish in eq. (3.48). Further assume that the surface on which the particle slides is a plane. If the particle slides on this plane under the effect of externally applied forces acting in the same plane, ${\underline{{F}}}^{a}=F_{1}^{a}{\bar{e}}_{1}\!+\!F_{2}^{a}{\bar{e}}_{2}$ , the normal contact force also vanishes. Indeed, Newton’s law now reduces to $F_{1}^{a}\bar{e}_{1}+F_{2}^{a}\bar{e}_{2}+$ $0\bar{n}=m(\ddot{x}_{1}\bar{e}_{1}+\ddot{x}_{2}\bar{e}_{2}+0\bar{n})$ : both externally applied forces and accelerations vanish along the normal direction.
|
||||
首先，设想该粒子在无摩擦力的情况下滑动于表面：方程 (3.48) 中切向接触力消失。进一步假设该粒子滑动的表面为平面。如果粒子在外部施加力作用下，且这些力作用在同一平面内滑动，则 ${\underline{{F}}}^{a}=F_{1}^{a}{\bar{e}}_{1}\!+\!F_{2}^{a}{\bar{e}}_{2}$ ，法向接触力也为零。实际上，牛顿定律现在简化为 $F_{1}^{a}\bar{e}_{1}+F_{2}^{a}\bar{e}_{2}+$ $0\bar{n}=m(\ddot{x}_{1}\bar{e}_{1}+\ddot{x}_{2}\bar{e}_{2}+0\bar{n})$： 外部施加力和加速度都消失在法向方向。
|
||||
|
||||
Consider now the same particle sliding on a curved surface under the effect of external forces applied in the plane tangent to the surface; Newton’s law now becomes $F_{1}^{a}\bar{e}_{1}+F_{2}^{a}\bar{e}_{2}+F^{n}\bar{n}=m(a_{1}\bar{e}_{1}+a_{2}\bar{e}_{2}+a_{n}\bar{n})$ , where the acceleration components are given by eq. (2.68). Since the particle has to follow the curvature of the surface, the acceleration component in the normal direction, $a_{n}$ , does not vanish, and the normal contact force, $F^{n}$ , is necessary to equilibrate the corresponding inertial forces. The normal contact force can be interpreted as the constraint force that constrains the particle to remain on the surface.
|
||||
现在考虑同一个粒子在外部力作用下，沿与曲面相切的平面滑动。牛顿定律变为 $F_{1}^{a}\bar{e}_{1}+F_{2}^{a}\bar{e}_{2}+F^{n}\bar{n}=m(a_{1}\bar{e}_{1}+a_{2}\bar{e}_{2}+a_{n}\bar{n})$ ，其中加速度分量由公式 (2.68) 给出。由于粒子必须跟随曲面的曲率，法向方向的加速度分量 $a_{n}$ 不为零，并且需要一个法向接触力 $F^{n}$ 来平衡相应的惯性力。法向接触力可以解释为约束力，它将粒子约束在曲面上。
|
||||
|
||||
If the interface between the particle and the surface is rough, friction forces acting in the plane tangent to the surface will appear in addition to the normal contact force. Figure 3.25 also gives free body diagrams for the particle and surface. Force components $F^{n}$ , $F_{1}^{t}$ , and $F_{2}^{t}$ are the forces the surface applies to the particle. Note that according to Newton’s third law, the particle applies equal and opposite forces to the surface.
|
||||
|
||||
# 3.3.2 Kinematics of particles in contact with a curve
|
||||
如果颗粒与表面之间的界面粗糙，除了垂直于表面的法向接触力之外，还会出现作用于与表面切线的平面上的摩擦力。图 3.25 也给出了颗粒和表面的受力分析图。力分量 $F^{n}$、 $F_{1}^{t}$ 和 $F_{2}^{t}$ 是表面施加在颗粒上的力。需要注意的是，根据牛顿第三定律，颗粒也施加大小相等、方向相反的力于表面。
|
||||
### 3.3.2 Kinematics of particles in contact with a curve
|
||||
|
||||
Figure 3.26 shows a common situation where a particle moves along a curve. This would be, for instance, the case of a train moving along its rails, or of a roller coaster car moving along its track. The differential geometry of a curve was studied in details in section 2.2. At point P of the curve, it is possible to define Frenet’s triad consisting of the tangent, normal, and binormal vectors, denoted $\bar{t},\bar{n}$ , and $\bar{b}$ , respectively.
|
||||
|
||||
图 3.26 所示的是一个常见情况，即一个质点沿曲线运动。例如，一列火车沿铁轨运动，或者一个过山车车厢沿轨道运动。曲线的微积分几何学在第 2.2 节中进行了详细研究。在曲线上的点 P，可以定义一个弗雷内特三联向量，分别由切线向量、法线向量和主法向量组成，分别用 $\bar{t}$、$\bar{n}$ 和 $\bar{b}$ 表示。
|
||||
|
||||
Fig. 3.26. Particle moving on a curve.
|
||||
|
||||
The contact force, $\underline{{F}}^{c}$ , between the particle and the curve is conveniently divided into two components, the normal contact force, ${\underline{{F}}}^{n}=F^{n}{\bar{n}}+F^{b}{\bar{b}}$ , which acts in the plane normal to the curve, and the tangential contact force, $\underline{{F}}^{t}\,=\,F^{t}\bar{t}$ , which acts along the tangent to the curve. Hence, the contact force is written as
|
||||
|
||||
粒子与曲线之间的接触力，$\underline{{F}}^{c}$ ，可以方便地划分为两个分量：法向接触力，${\underline{{F}}}^{n}=F^{n}{\bar{n}}+F^{b}{\bar{b}}$ ，它作用于垂直于曲线的平面上；以及切向接触力，$\underline{{F}}^{t}\,=\,F^{t}\bar{t}$ ，它作用于曲线的切线方向上。因此，接触力可以写成
|
||||
$$
|
||||
\underline{{F}}^{c}=F^{t}\bar{t}+(F^{n}\bar{n}+F^{b}\bar{b}).
|
||||
$$
|
||||
@ -3654,7 +3660,13 @@ It is important to distinguish the difference between unilateral and bilateral c
|
||||
When dealing with unilateral contact, it is often important to determine when a particle will loose contact with the surface or curve. Consider, at first, the case of a particle on a surface and assume the particle can freely move in the direction of the normal to the surface. In that case, $F^{n}$ is positive when the particle is on the surface and the unilateral contact condition cannot support a negative normal force. Clearly, the particle is about to leave the surface when $F^{n}=0$ .
|
||||
|
||||
In the case of a particle on a curve, the normal to the curve is always pointing to the concave side of the curve: the normal ﾚips direction at an inﾚection point of the curve. Due to this discontinuity, the condition $F^{n}=0$ must be applied with care when dealing with particle moving along a curve. For more details about the complex problems associated with unilateral contact conditions can be found in the textbook by Pfeiffer [6].
|
||||
注意式 (3.48) 中粒子与表面之间的接触力表达式与式 (3.49) 中粒子与曲线之间的接触力表达式的不同。对于表面，法向接触力只有一个沿表面的法向分量；而对于曲线，法向接触力则有两个分量，分别沿曲线的法线和副法线方向。另一方面，对于表面，切向接触力有两个分量，位于与表面切线的平面内；而对于曲线，切向接触力则只有一个沿曲线切线方向的分量。
|
||||
|
||||
区分单边接触和双边接触非常重要。例如，火车与铁轨之间是单边接触：火车不能穿透铁轨，但没有任何东西阻止火车在向上方向上脱离铁轨。当然，重力通常足以将火车固定在铁轨上。这与过山车不同：在这种情况下，车厢通过一组车轮连接到轨道上，这些车轮可以防止它们在任何方向上脱离轨道。
|
||||
|
||||
在处理单边接触时，通常重要的是确定粒子何时会失去与表面或曲线的接触。首先考虑粒子位于表面上的情况，并假设粒子可以在表面的法向方向上自由移动。在这种情况下，当粒子位于表面上时，$F^{n}$ 是正值，单边接触条件无法支撑负法向力。显然，当 $F^{n}=0$ 时，粒子即将离开表面。
|
||||
|
||||
对于粒子位于曲线上的情况，曲线的法线始终指向曲线的凹侧：法线在曲线的拐点处改变方向。由于存在这种不连续性，在处理沿曲线运动的粒子时，必须谨慎应用条件 $F^{n}=0$。有关与单边接触条件相关的复杂问题的更多细节，请参见 Pfeiffer 的教科书 [6]。
|
||||
# 3.3.3 Constitutive laws for tangential contact forces
|
||||
|
||||
The tangential contact forces are friction forces between the particle and surface or curve it moves on. Coulomb’s friction law is commonly used to evaluate the friction forces, and sometimes, friction forces are assumed to be of a viscous type.
|
||||
|
||||
27
软件组工作讨论/2025.7.25 软件组组会.md
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@ -0,0 +1,27 @@
|
||||
|
||||
标志性成果这页
|
||||
|
||||
展示度 不够
|
||||
图 重点考虑 虚虚实实
|
||||
|
||||
技术瓶颈
|
||||
凝练一下
|
||||
|
||||
|
||||
指标
|
||||
|
||||
现场特殊工况复现
|
||||
|
||||
|
||||
|
||||
开展了xxx,研发xxx，实现xxxx
|
||||
|
||||
三步
|
||||
|
||||
功能架构图，进展地图
|
||||
|
||||
模块进展，亮点
|
||||
|
||||
后续进展
|
||||
|
||||
下周一下午发
|
||||
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