拉格朗日力学--经典双摆的案例
<h3>一、系统建模与参数定义</h3><p>考虑由两根无质量刚性杆组成的平面双摆：</p>
<p><img src="data/attachment/forum/202510/15/132701hq25xg4z6g642q4j.webp" alt="R-C1.webp" title="R-C (1).webp" /><img src="data/attachment/forum/202510/15/132702dyqtwgyjtzpgjj03.webp" alt="5f27f23469c44620b5382accb18380e6.webp" title="5f27f23469c44620b5382accb18380e6.webp" /></p>
<p><img src="data/attachment/forum/202510/15/132702ztshdfhpfjs9tfcf.webp" alt="7d1bb93fb06e59b99a1b157c8599b6be.webp" title="7d1bb93fb06e59b99a1b157c8599b6be.webp" /></p>
<div class="language-math">上端杆长\\(L_1\\)，下端连接质量\\(m_1\\)的小球</div>
<div class="language-math">下端杆长\\(L_2\\)，末端连接质量\\(m_2\\)的小球</div>
<div class="language-math">两杆与竖直方向夹角分别为\\(\theta_1\\)、\\(\theta_2\\)（广义坐标，2 个自由度）</div>
<div class="language-math">重力加速度为\\(g)，系统无摩擦约束</div>
<h3>二、拉格朗日量构建（L=T-V）</h3>
<h4>1. 动能计算（平动动能总和）</h4>
<p>通过坐标变换将小球位置表示为广义坐标的函数：</p>
<pre><code>x₁ = L₁sinθ₁,y₁ = -L₁cosθ₁
x₂ = L₁sinθ₁+L₂sinθ₂,y₂ = -L₁cosθ₁-L₂cosθ₂
</code></pre>
<p>对时间求导得速度分量，动能为：</p>
<div class="language-math">(T = \frac{1}{2}m_1(\dot{x}_1^2+\dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2+\dot{y}_2^2))</div>
<p>化简后：</p>
<div class="language-math">(T = \frac{1}{2}(m_1+m_2)L_1^2\dot{\theta}_1^2 + \frac{1}{2}m_2L_2^2\dot{\theta}_2^2 + m_2L_1L_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2))</div>
<h4>2. 势能计算（重力势能总和）</h4>
<p>以悬挂点为势能零点：</p>
<div class="language-math">(V = m_1gL_1(1-\cos\theta_1) + m_2g)</div>
<h4>3. 拉格朗日量确立</h4>
<div class="language-math">(L = T - V = \left[ \frac{1}{2}(m_1+m\_2)L_1^2\dot{\theta}_1^2 + \frac{1}{2}m_2L_2^2\dot{\theta}_2^2 + m_2L_1L_2\dot{\theta}\_1\dot{\theta}_2\cos(\theta_1-\theta_2) \right]</div>
<div class="language-math">- \left[ (m_1+m_2)gL_1(1-\cos\theta_1) + m_2gL_2(1-\cos\theta_2) \right])</div>
<h3>三、欧拉 - 拉格朗日方程求解</h3>
<p>对两个广义坐标分别应用方程：</p>
<div class="language-math">(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}_i}\right) - \frac{\partial L}{\partial \theta_i} = 0\\)</div>
<h4>1. 关于θ1的方程</h4>
<ul>
<li>计算偏导数：</li>
</ul>
<div class="language-math">(\frac{\partial L}{\partial \dot{\theta}\_1} = (m_1+m_2)L_1^2\dot{\theta}_1 + m_2L_1L_2\dot{\theta}_2\cos(\theta_1-\theta_2)\\)</div>
<div class="language-math">\\(\frac{\partial L}{\partial \theta_1} = -m_2L_1L_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2) - (m_1+m_2)gL_1\sin\theta_1\\)</div>
<ul>
<li>时间导数与整理后得：</li>
</ul>
<div class="language-math">\\((m_1+m_2)L_1\ddot{\theta}_1 + m_2L_2\cos(\theta_1-\theta_2)\ddot{\theta}_2 + m_2L_2\sin(\theta_1-\\theta\_2)\dot{\theta}_2^2 + (m_1+m_2)g\sin\theta_1 = 0 \quad (1))</div>
<h4>2. 关于θ2的方程</h4>
<p>同理推导得：</p>
<div class="language-math">(L_2\ddot{\theta}_2 + L_1\cos(\theta_1-\theta_2)\ddot{\theta}_1 - L_1\sin(\theta_1-\theta_2)\dot{\theta}_1^2 + g\sin\theta_2 = 0 \quad (2)\\)</div>
<h3>四、物理意义与拓展</h3>
<ol>
<li><strong>非线性特性</strong>：</li>
<li>
<div class="language-math">方程 (1)(2) 含\\(\sin(\theta_1-\theta_2)\\)等非线性项，大摆角时表现混沌行为（对初始条件极度敏感）</div>
</li>
<li><strong>简化特例</strong>：</li>
<li>
<div class="language-math">当\\(m_2=0\\)时退化为单摆方程\\(L_1\ddot{\theta}_1 + g\sin\theta_1=0\\)</div>
</li>
<li><strong>工程关联</strong>：弹性双摆模型（杆具弹性形变）可用于柔性机械臂、仿生运动研究</li>
<li><strong>数值验证</strong>：通过 Python 的odeint函数求解微分方程，可绘制动图展示轨迹演化</li>
</ol>
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