作业参考答案2013年9月Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.1.设质点在势能场U(r)中运动，在笛卡尔坐 标系中写出其拉格朗日方程。解：拉格朗日方程为：L为拉格朗日函数笛卡尔坐标中的坐标变量为 ，那么Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.所以，带入拉格朗日方程得到这就是笛卡尔坐标系中的拉格朗日方程即牛顿第二定律Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.2.已知柱坐标 与笛卡尔坐标的关系是如图设质点在轴对称势能场 中运动 ，写出其拉格朗日方程。解：由柱坐标和笛卡尔坐标的关系可知等式两边同时除以dt那么，系统的动能为Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.那么，系统的拉格朗日函数为所以带入拉格朗日方程，则有：Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.3.长度为l的细绳系一小球，悬挂点按照 方式运动，如图所示，小球被限 制在 平面内运动， 时悬线竖直向下 。（a）求悬线和竖直线偏离 所对应的虚位移（b）已知在这一时刻的角速度为 ，求经过 时间后的位移 。问：当 时， 与 有何差别？（a）在任意时刻，约束所 容许的位移为虚位移，途中 的小球，受到细绳的和自身 重力的约束，在这个时刻，解 ：xEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.小球只能围绕O点作圆周运动，当偏离角为 时，对应的虚位移为 。（b）小球经过 时间后的位移，可以看作由 两部分组成：（1）小球绕O点作圆周运动所产生的位移（2）小球随O点一起作简谐运动所产生的位 移所以，小球的位移为和 的区别如图所示：xEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.虚位移和实际位移的主要区别在于 虚位移只和约束有关,某一时刻约束所允许 的位移。 实际位移除了和约束有关以外，还和物体 当前的运动状态有关；运动动方程和约约束允许 ，在时间间时间间 隔内所发生的位移。Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.长度同为l 的轻棒四根，相互连接成一个可 以无摩擦的改变顶角的菱形ABCD，AB和AD 两棒无摩擦的支于处于同一水平线上且相距 2a的两根钉上，BD之间用一根轻质棒连接， 在连接点（B和D处），各棒之间可以无摩擦 的转动，C点上系有一重物W，C点和重物受 到约束，只能上下运动，设A点两棒之间的 夹角为 ,试用虚功原理求平衡时联结棒BD 中的张力 ，讨论 的方向 与 的大小的关系。问：在 什么情况下有 ，说明其 意义。4.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.由虚功原理，在平衡状态下可得解：为了求棒中的张力，可将棒的约 束予以“释放”，以张力 作为主 动力代替棒。此时系统的自由度为 1，系统受3个外力作用：作用于B 的张力 ，作用于D的张力 ， 作用于C点的W。坐标系：两根钉连线的中点为坐标原点，连线 所在直线为x轴（向右为正），垂直连线为y轴 （向下为正），并取 为广义坐标。Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.B、D点的x坐标：C点的y坐标：Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.最后可得：即有 ： 杠对B的作用力向内杠对B的作用力向外杠对B无作用力Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.9质量为M的斜面可以无摩擦地在水平桌面 上滑动。斜面上无摩擦地放一滑块 m，如图所 示。写出拉格朗日方程，并求斜面的加速度 和滑块相对于斜面的加速度 。解：系统的拉格朗日函数为Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.即有：解之得：带入拉氏方程：Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.滑块的能量斜面的能量系统的总能量K系（桌面坐标系 ）K系（沿X方向以速度V相 对于桌面运动的坐标系）P37 第5题Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.系统的能量在k系和 系之间的变换方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.