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    \textbf{\Huge\it 理论力学 \normalfont A} \\
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    \large 作业07  参考答案 \\
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    \textbf{学期：} 2025秋 \\
    \textbf{助教：} 于洪飞，张积翔 \\
    \textbf{日期：} \today \\
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\section*{第二章 拉格朗日力学}

\paragraph{10.} 在重力的作用下，质点在旋转抛物面 $z=x^2+y^2$ 上运动，其中$z$-轴方向是垂直向上. 写出拉氏函数和拉氏方程. 

\begin{solution}
使用柱坐标系 $(r, \theta, z)$. 质点坐标为 $x=r\cos\theta, y=r\sin\theta, z=z$. 
\begin{itemize}
    \item \textbf{动能 $T$}:
    $$ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2) $$
    \item \textbf{势能 $V$}:
    $$ V = mgz $$
    \item \textbf{约束条件}:
    约束方程为 $z = x^2+y^2 = r^2$. 这是一个完整约束. 
    我们选取 $r$ 和 $\theta$ 作为广义坐标. 
    对约束求导：$\dot{z} = 2r\dot{r}$. 
    \item \textbf{拉氏函数 $L$}
    将 $\dot{z}$ 和 $z$ 代入 $L = T - V$：
    \begin{align*}
        L &= \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + (2r\dot{r})^2) - mgr^2 \\
        &= \frac{1}{2}m((1+4r^2)\dot{r}^2 + r^2\dot{\theta}^2) - mgr^2
    \end{align*}
\end{itemize}
\textbf{拉氏方程}:
\begin{enumerate}
    \item \textbf{对 $\theta$}:
    $\theta$ 是循环坐标，因为 $\frac{\partial L}{\partial \theta} = 0$. 
    $$ \frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta} $$
    $$ \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{\theta}}\right) = \frac{\dd }{\dd t}(mr^2\dot{\theta}) = 0 $$
    $$ mr^2\dot{\theta} = \text{const} \quad  \text{(角动量守恒)} $$

    \item \textbf{对 $r$}:
    \begin{align*}
        \frac{\partial L}{\partial r} &= \frac{1}{2}m(8r\dot{r}^2 + 2r\dot{\theta}^2) - 2mgr = 4mr\dot{r}^2 + mr\dot{\theta}^2 - 2mgr \\
        \frac{\partial L}{\partial \dot{r}} &= m(1+4r^2)\dot{r} \\
        \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{r}}\right) &= m(8r\dot{r})\dot{r} + m(1+4r^2)\ddot{r} = 8mr\dot{r}^2 + m(1+4r^2)\ddot{r}
    \end{align*}
    应用 $\frac{\dd }{\dd t}(\frac{\partial L}{\partial \dot{r}}) - \frac{\partial L}{\partial r} = 0$:
    $$ [8mr\dot{r}^2 + m(1+4r^2)\ddot{r}] - [4mr\dot{r}^2 + mr\dot{\theta}^2 - 2mgr] = 0 $$
    $$ m(1+4r^2)\ddot{r} + 4mr\dot{r}^2 - mr\dot{\theta}^2 + 2mgr = 0 $$
    $$ (1+4r^2)\ddot{r} + 4r\dot{r}^2 - r\dot{\theta}^2 + 2gr = 0 $$
\end{enumerate}
\end{solution}

\hrule
\paragraph{11.} 质量$m$的小球，在光滑的水平桌面上作两维运动. 桌面中心有一光滑小孔O. 将一个质量为$M$的铁块放在小孔的下方，并用绳子将小球与之拴在一起，绳长为$l$. 

设铁块在一垂直平面内运动，且绳子始终处于绷直状态. 写出系统的拉氏函数和运动方程. 

\begin{solution}
这是一个具有3个自由度的系统. 我们选择广义坐标：
\begin{itemize}
    \item $r$: 小球 $m$ 离小孔的水平距离. 
    \item $\theta_1$: 小球 $m$ 在水平桌面上的极角. 
    \item $\theta_2$: 铁块 $M$ 偏离竖直线的摆角. 
\end{itemize}
绳长为 $l$. 小球 $m$ 占用的绳长为 $r$，则铁块 $M$ 的摆长为 $(l-r)$. 
\begin{itemize}
    \item \textbf{动能 $T$}:
    $T = T_m + T_M$
    $$ T_m = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}_1^2) $$
    对于 $M$，其速度由 $r$ 的变化（径向）和 $\theta_2$ 的变化（切向）合成：
    $$ v_M^2 = \dot{r}^2 + (l-r)^2\dot{\theta}_2^2 $$
    $$ T_M = \frac{1}{2}M v_M^2 = \frac{1}{2}M(\dot{r}^2 + (l-r)^2\dot{\theta}_2^2) $$
    $$ T = \frac{1}{2}(m+M)\dot{r}^2 + \frac{1}{2}mr^2\dot{\theta}_1^2 + \frac{1}{2}M(l-r)^2\dot{\theta}_2^2 $$
    \item \textbf{势能 $V$}:
    $V = V_m + V_M$. 设桌面为 $V_m=0$. 设小孔为 $V_M$ 的零点，竖直向下为正. 
    $$ V = -Mg \c\dd ot (l-r)\cos\theta_2 $$
    \item \textbf{拉氏函数 $L = T - V$}:
    $$ L = \frac{1}{2}(m+M)\dot{r}^2 + \frac{1}{2}mr^2\dot{\theta}_1^2 + \frac{1}{2}M(l-r)^2\dot{\theta}_2^2 + Mg(l-r)\cos\theta_2 $$
\end{itemize}
\textbf{拉氏方程}:
\begin{enumerate}
    \item \textbf{对 $\theta_1$}:
    $\theta_1$ 是循环坐标. 
    $$ \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{\theta}_1}\right) = \frac{\dd }{\dd t}(mr^2\dot{\theta}_1) = 0 $$
    $$ m(2r\dot{r}\dot{\theta}_1 + r^2\ddot{\theta}_1) = 0 \implies 2\dot{r}\dot{\theta}_1 + r\ddot{\theta}_1 = 0 $$
    
    \item \textbf{对 $r$}:
    \begin{align*}
        \frac{\partial L}{\partial r} &= mr\dot{\theta}_1^2 + M(l-r)(-1)\dot{\theta}_2^2 - Mg\cos\theta_2 \\
        &= mr\dot{\theta}_1^2 - M(l-r)\dot{\theta}_2^2 - Mg\cos\theta_2 \\
        \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{r}}\right) &= \frac{\dd }{\dd t}((m+M)\dot{r}) = (m+M)\ddot{r}
    \end{align*}
    $$ (m+M)\ddot{r} - mr\dot{\theta}_1^2 + M(l-r)\dot{\theta}_2^2 + Mg\cos\theta_2 = 0 $$
    
    \item \textbf{对 $\theta_2$}:
    \begin{align*}
        \frac{\partial L}{\partial \theta_2} &= -Mg(l-r)\sin\theta_2 \\
        \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{\theta}_2}\right) &= \frac{\dd }{\dd t}(M(l-r)^2\dot{\theta}_2) \\
        &= M\left[ 2(l-r)(-\dot{r})\dot{\theta}_2 + (l-r)^2\ddot{\theta}_2 \right]
    \end{align*}
    $$ M(l-r) \left[ -2\dot{r}\dot{\theta}_2 + (l-r)\ddot{\theta}_2 \right] + Mg(l-r)\sin\theta_2 = 0 $$
    $$ (l-r)\ddot{\theta}_2 - 2\dot{r}\dot{\theta}_2 + g\sin\theta_2 = 0 $$
\end{enumerate}
\end{solution}

\hrule
\paragraph{12.} （惠更斯等时摆）如图，已知摆线
$$ \begin{cases} x = a(\theta - \sin\theta) \\ y = a(1 - \cos\theta) \end{cases} $$
按此外形做成挡板. 一根长$4a$的无弹性轻绳，一端系在坐标原点O，另一端拴着质量为$m$的小球. 在$A$点，绳子与曲面相切. 

\begin{enumerate}
    \item[(1)] 计算参数区间$[0, \theta]$的摆线长度. 
    \item[(2)] 记切点$A$的曲线参数为$\theta(t)$，求小球的坐标$x(\theta), y(\theta)$. 
    \item[(3)] 以$\theta(t)$为广义坐标，写出拉氏函数. 
    \item[(4)] 我们希望把拉氏函数中与广义速度有关的项化简，为此令动能为
    $$ T \triangleq \frac{1}{2}m\dot{q}^2 $$
    其中$m$是常数. 求变换后的广义坐标$q = q(\theta)$. 
    \item[(5)] 求$V = V(q)$的表达式. 
    \item[(6)] 求摆的周期. 周期与振幅是否有关？
\end{enumerate}
\begin{solution}
\begin{enumerate}
    \item[(1)] \textbf{摆线长度}
    摆线（旋轮线）方程为 $x=a(\theta-\sin\theta), y=a(1-\cos\theta)$. 
    弧长元 $\dd s = \sqrt{\left(\frac{\dd x}{\dd \theta}\right)^2 + \left(\frac{\dd y}{\dd \theta}\right)^2} \dd \theta$. 
    $$ \frac{\dd x}{\dd \theta} = a(1-\cos\theta), \quad  \frac{\dd y}{\dd \theta} = a\sin\theta, $$
    \begin{align*} \dd s & = \sqrt{a^2(1-\cos\theta)^2 + a^2\sin^2\theta} \dd \theta\\
    & = a\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta} \dd \theta\\
    & = a\sqrt{2(1-\cos\theta)} \dd \theta \\
    & = a\sqrt{2(2\sin^2(\theta/2))} \dd \theta \\
    & = 2a\sin(\theta/2) \dd \theta. \end{align*}
    从 $0$ 到 $\theta$ 的弧长 $s(\theta)$ 为：
    $$ s(\theta) = \int_0^\theta 2a\sin(u/2) \dd u = \left[ -4a\cos(u/2) \right]_0^\theta = 4a(1-\cos(\theta/2)) $$

    \item[(2)] \textbf{小球坐标}
    设 $y$ 轴竖直向下（与图示一致）. 切点 $A$ 的坐标为 $x_A=a(\theta-\sin\theta), y_A=a(1-\cos\theta)$. 
    切线斜率 $m = \frac{\dd y_A}{\dd x_A} = \frac{y'(\theta)}{x'(\theta)} = \frac{a\sin\theta}{a(1-\cos\theta)} = \cot(\theta/2)$. 
    切线与 $x$ 轴的夹角 $\alpha$ 满足 $\tan\alpha = \cot(\theta/2)$, $y$ 轴向下时 $\alpha = \pi/2 - \theta/2$. 
    自由绳长 $L' = (\text{总长}) - s(\theta) = 4a - 4a(1-\cos(\theta/2)) = 4a\cos(\theta/2)$. 
    小球 $P$ 的坐标 $(x_P, y_P)$ 为：
    \begin{align*}
        x_P&  = x_A + L'\cos\alpha \\
        & = a(\theta-\sin\theta) + 4a\cos(\theta/2) \cos(\pi/2 - \theta/2) \\
    & = a(\theta-\sin\theta) + 4a\cos(\theta/2)\sin(\theta/2)\\
    & = a(\theta-\sin\theta) + 2a\sin\theta \\
    & = a(\theta+\sin\theta), \\
    y_P &= y_A + L'\sin\alpha\\
    & = a(1-\cos\theta) + 4a\cos(\theta/2) \sin(\pi/2 - \theta/2)\\
    & = a(1-\cos\theta) + 4a\cos(\theta/2)\cos(\theta/2)\\
    &  = a(1-\cos\theta) + 2a(1+\cos\theta) \\
    &= a(3+\cos\theta).
    \end{align*}
    
    \item[(3)] \textbf{拉氏函数}
    $$ \dot{x}_P = a(1+\cos\theta)\dot{\theta}, \quad  \dot{y}_P = -a\sin\theta\dot{\theta} $$
    $$ v_P^2 = \dot{x}_P^2 + \dot{y}_P^2 = a^2\dot{\theta}^2( (1+\cos\theta)^2 + \sin^2\theta ) = a^2\dot{\theta}^2(2+2\cos\theta) = 4a^2\dot{\theta}^2\cos^2(\theta/2) $$
    $$ T = \frac{1}{2}mv_P^2 = 2ma^2\dot{\theta}^2\cos^2(\theta/2) $$
    势能 $V = -mgy_P = -mga(3+\cos\theta)$ (取 $y=0$ 为势能零点). 
    $$ L = T - V = 2ma^2\dot{\theta}^2\cos^2(\theta/2) + mga(3+\cos\theta) $$

    \item[(4)] \textbf{新广义坐标 $q$}
    我们要求 $T = \frac{1}{2}m\dot{q}^2$. 
    $$ \frac{1}{2}m\dot{q}^2 = 2ma^2\dot{\theta}^2\cos^2(\theta/2) \implies \dot{q}^2 = 4a^2\dot{\theta}^2\cos^2(\theta/2) $$
    $$ \dot{q} = 2a\dot{\theta}\cos(\theta/2) $$
    $$ \frac{\dd q}{\dd t} = \frac{\dd q}{\dd \theta}\frac{\dd \theta}{\dd t} \implies \frac{\dd q}{\dd \theta} = 2a\cos(\theta/2) $$
    $$ q(\theta) = \int 2a\cos(\theta/2) \dd \theta = 4a\sin(\theta/2). $$
    
    \item[(5)] \textbf{势能 $V(q)$}
    $$ V = -mga(3+\cos\theta) $$
    利用 $q = 4a\sin(\theta/2) \implies \sin(\theta/2) = q/(4a)$. 
    $$ \cos\theta = 1 - 2\sin^2(\theta/2) = 1 - 2\left(\frac{q}{4a}\right)^2 = 1 - \frac{q^2}{8a^2} $$
    $$ V(q) = -mga\left(3 + 1 - \frac{q^2}{8a^2}\right) = -mga\left(4 - \frac{q^2}{8a^2}\right) $$
    
    \item[(6)] \textbf{摆的周期}
    以 $q$ 为广义坐标，新的拉氏函数为：
    $$ L = T - V = \frac{1}{2}m\dot{q}^2 + mga\left(4 - \frac{q^2}{8a^2}\right) $$
    对应的拉氏方程为：
    $$ \frac{\dd }{\dd t}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0 $$
    $$ m\ddot{q} - mga\left(-\frac{2q}{8a^2}\right) = 0 $$
    $$ m\ddot{q} + \frac{mg}{4a}q = 0 \implies \ddot{q} + \left(\frac{g}{4a}\right)q = 0 $$
    这是一个简谐振动方程 $\ddot{q} + \omega^2 q = 0$，其角频率 $\omega = \sqrt{g/(4a)}$. 
    周期 $T_p = 2\pi/\omega = 2\pi\sqrt{4a/g} = 4\pi\sqrt{a/g}$. 
    周期 $T_p$ 是一个常数，仅与 $a$ 和 $g$ 有关，与振幅（即 $q$ 的最大值）无关. 
\end{enumerate}
\end{solution}

\hrule
\paragraph{13.} 折线绕对称轴以固定的角速度$\omega$旋转，在$t = 0$时折线的形状为$z = |x|$. 折线的两段分别套着两个光滑小环（质量可忽略），小环用长$a$、质量为$m$均匀光滑细杆连接. 

\begin{enumerate}
    \item[(1)] 写出系统的拉氏函数和运动方程.
    \item[(2)] 细杆在折线上是否存在稳定位置？
\end{enumerate}

\begin{solution}
\begin{enumerate}
    \item[(1)] \textbf{拉氏函数和运动方程}
    设 $z$ 轴为对称轴，$t=0$ 时折线在 $xz$ 平面，方程为 $z=|x|$. 
    系统绕 $z$ 轴旋转，角速度为 $\omega$. 
    设细杆两端 $A, B$ 的坐标. 我们用杆 $AB$ 与 $z$ 轴的夹角 $\theta$ 作为广义坐标. 
    $A$ 在 $z=-x'$，$B$ 在 $z=x'$ （在旋转的 $x'z'$ 平面内）. 
    设杆 $AB$ 的中点为 $C(x'_C, z'_C)$. 
    $$ A = (x'_C - \frac{a}{2}\sin\theta, z'_C + \frac{a}{2}\cos\theta) $$
    $$ B = (x'_C + \frac{a}{2}\sin\theta, z'_C - \frac{a}{2}\cos\theta) $$
    代入约束 $z_A = -x'_A$ 和 $z_B = x'_B$：
    $$ z'_C + \frac{a}{2}\cos\theta = -x'_C + \frac{a}{2}\sin\theta \implies z'_C + x'_C = \frac{a}{2}(\sin\theta - \cos\theta) $$
    $$ z'_C - \frac{a}{2}\cos\theta = x'_C + \frac{a}{2}\sin\theta \implies z'_C - x'_C = \frac{a}{2}(\sin\theta + \cos\theta) $$
    解得质心 $C$ 的坐标（在旋转系中）：
    $$ z'_C = \frac{a}{2}\sin\theta, \quad  x'_C = -\frac{a}{2}\cos\theta $$
    
    \textbf{动能 $T = T_{trans} + T_{rot}$}

    $C$ 在实验室系的速度 $v_C^2$：
    $C$ 的柱坐标为 $(r_C, \phi_C, z_C) = (|x'_C|, \omega t, z'_C),$
    \begin{align*} v_C^2 & = \dot{r}_C^2 + r_C^2\dot{\phi}_C^2 + \dot{z}_C^2\\
    & = (\frac{\dd |x'_C|}{\dd t})^2 + |x'_C|^2\omega^2 + \dot{z}'_C{}^2\\
    & = (\frac{a}{2}\sin\theta\dot{\theta})^2 + (\frac{a}{2}\cos\theta)^2\omega^2 + (\frac{a}{2}\cos\theta\dot{\theta})^2\\
    & = \frac{a^2}{4}(\dot{\theta}^2\sin^2\theta + \omega^2\cos^2\theta + \dot{\theta}^2\cos^2\theta)\\
    & = \frac{a^2}{4}(\dot{\theta}^2 + \omega^2\cos^2\theta),\end{align*}
    $$T_{trans} = \frac{1}{2}mv_C^2 = \frac{1}{8}ma^2(\dot{\theta}^2 + \omega^2\cos^2\theta).$$
    
    杆绕质心的转动动能 $T_{rot}$：杆在旋转系中以 $\dot{\theta}$ 转动（绕 $y'$ 轴），且随框架绕 $z$ 轴以 $\omega$ 转动. 
    在杆的质心坐标系（1轴沿杆，3轴为 $y'$ 轴），转动惯量 $I_1=0, I_2=I_3=\frac{1}{12}ma^2$. 
    $\vec{\omega}_{\text{rod }}= (\omega\cos\theta, \omega\sin\theta, \dot{\theta})$. 
    \begin{align*} T_{rot}& = \frac{1}{2}(I_1\omega_1^2 + I_2\omega_2^2 + I_3\omega_3^2)\\ 
    & = \frac{1}{2}\left[ 0 + \frac{1}{12}ma^2(\omega\sin\theta)^2 + \frac{1}{12}ma^2\dot{\theta}^2 \right]\\
    & = \frac{1}{24}ma^2(\dot{\theta}^2 + \omega^2\sin^2\theta).\end{align*}
    
    最终系统动能：
    \begin{align*}T &= T_{trans} + T_{rot} \\
    &=\frac{1}{8}ma^2(\dot{\theta}^2 + \omega^2\cos^2\theta) + \frac{1}{24}ma^2(\dot{\theta}^2 + \omega^2\sin^2\theta)\\
    & = (\frac{1}{8}+\frac{1}{24})ma^2\dot{\theta}^2 + (\frac{1}{8}\cos^2\theta + \frac{1}{24}\sin^2\theta)ma^2\omega^2\\
    & = \frac{1}{6}ma^2\dot{\theta}^2 + \frac{1}{24}ma^2\omega^2(3\cos^2\theta + \sin^2\theta) \\
    & = \frac{1}{6}ma^2\dot{\theta}^2 + \frac{1}{24}ma^2\omega^2(2\cos^2\theta + 1).\end{align*}
    
    \textbf{势能 $V$}:
    $V = mg z'_C = mg\frac{a}{2}\sin\theta$
    
    \textbf{拉氏函数 $L = T - V$}:
    $$ L = \frac{1}{6}ma^2\dot{\theta}^2 + \frac{1}{24}ma^2\omega^2(2\cos^2\theta + 1) - \frac{1}{2}mga\sin\theta $$
    
    \textbf{拉氏方程}: $\frac{\dd }{\dd t}(\frac{\partial L}{\partial \dot{\theta}}) - \frac{\partial L}{\partial \theta} = 0$
    $$ \frac{\partial L}{\partial \dot{\theta}} = \frac{1}{3}ma^2\dot{\theta} \implies \frac{\dd }{\dd t}(\dots) = \frac{1}{3}ma^2\ddot{\theta} $$
    $$ \frac{\partial L}{\partial \theta} = \frac{1}{24}ma^2\omega^2(4\cos\theta(-\sin\theta)) - \frac{1}{2}mga\cos\theta = -\frac{1}{6}ma^2\omega^2\sin\theta\cos\theta - \frac{1}{2}mga\cos\theta $$
    $$ \frac{1}{3}ma^2\ddot{\theta} - \left[ -\frac{1}{6}ma^2\omega^2\sin\theta\cos\theta - \frac{1}{2}mga\cos\theta \right] = 0 $$
    $$ \frac{1}{3}ma^2\ddot{\theta} + \frac{1}{6}ma^2\omega^2\sin\theta\cos\theta + \frac{1}{2}mga\cos\theta = 0 $$
    $$ \ddot{\theta} + \frac{1}{2}\omega^2\sin\theta\cos\theta + \frac{3g}{2a}\cos\theta = 0 $$
    
    \item[(2)] \textbf{稳定位置}
    稳定位置是有效势能 $V_{\mathrm{eff}}$ 的极小值点. $L = \frac{1}{2}A(\theta)\dot{\theta}^2 - V_{\mathrm{eff}}(\theta)$. 
    $$ V_{\mathrm{eff}}(\theta) = - \frac{1}{24}ma^2\omega^2(2\cos^2\theta + 1) + \frac{1}{2}mga\sin\theta $$
    平衡位置是 $F_{\mathrm{eff}} = -\frac{\dd V_{\mathrm{eff}}}{\dd \theta} = 0$ 的点. 
    $$ \frac{\dd V_{\mathrm{eff}}}{\dd \theta} = - \frac{1}{24}ma^2\omega^2(4\cos\theta(-\sin\theta)) + \frac{1}{2}mga\cos\theta = 0 $$
    $$ \frac{1}{6}ma^2\omega^2\sin\theta\cos\theta + \frac{1}{2}mga\cos\theta = 0 $$
    $$ \cos\theta \left( \frac{1}{6}ma^2\omega^2\sin\theta + \frac{1}{2}mga \right) = 0 $$
    平衡点有两种可能：
    1) $\cos\theta = 0 \implies \theta = \pi/2$ (杆水平)
    2) $\sin\theta = -\frac{mga/2}{ma^2\omega^2/6} = -\frac{3g}{a\omega^2}$，显然 $\sin\theta<0$ 没有物理意义，舍去.
    
    稳定性由 $\frac{\dd ^2V_{\mathrm{eff}}}{\dd \theta^2}$ 的符号决定 ($\frac{\dd ^2V_{\mathrm{eff}}}{\dd \theta^2} > 0$ 为稳定)，$\theta = \pi/2$ 时
    $$ \frac{\dd ^2V_{\mathrm{eff}}}{\dd \theta^2} \Big|_{\pi/2} = \frac{1}{6}ma^2\omega^2(0 - 1) - \frac{1}{2}mga(1) = -\frac{1}{6}ma^2\omega^2 - \frac{1}{2}mga < 0 $$
    因此没有稳定位置.
\end{enumerate}
\end{solution}