Theoretical Mechanics Chapter 14: Theorem of the change in the kinetic energy
1 Theoretical Mechanics
理论力学 第十
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Chapter 14: Theorem of the change in the kinetic energy 心§141 Work done by a force D8 14-2 Kinetic energy D8 14-3 Theorem of kinetic energy 心§14-4 Power, power equation 8 14-5 Conservative forces field, potential energy, the law of conservation of mechanical energy 8 14-6 General theorems of dynamics and its applications
3 § 14-1 Work done by a force § 14-2 Kinetic energy § 14-3 Theorem of kinetic energy § 14-4 Power, power equation § 14-5 Conservative forces field, potential energy, the law of conservation of mechanical energy § 14-6 General theorems of dynamics and its applications Chapter 14:Theorem of the change in the kinetic energy
第十四章动能定理 §141力的功 §14-2动能 §14-3动能定理 §144功率·功率方程 §14-5势力场·势能·机械能守恒定理 §146动力学普遍定理及综合应用
4 §14–1 力的功 §14–2 动能 §14–3 动能定理 §14–4 功率 · 功率方程 §14–5 势力场 · 势能 · 机械能守恒定理 §14–6 动力学普遍定理及综合应用 第十四章 动能定理
D In order to obtain the theorem of the change in the kinetic energy, we make use of the energy method to investigate dynamical problems. In different to the cases of the theorems of the changes in the linear and the angular moment we make use of the vector method. The method not only has important applications in the research of the mechanicalmotion, but it is also the bridge connecting mechanical motion with other forms of motion .The theorem of the change in the kinetic energy establishes the dependence between the physical quantities describing motion---kinetic energy and describing the acting force work. It is a law describing changes between different forms of energy. 8 14-1 Work done by a force Work is a measure of the accumulated effect of the action of a force on a body during a given displacement 1. Work done by a constant force W=FS cos a M2 The work done by a force is a scalar quanti Fora a, the work is negative. The unit of work in the SI system is the joule(). 1J=INIm
5 In order to obtain the theorem of the change in the kinetic energy,we make use of the energy method to investigate dynamical problems. In different to the cases of the theorems of the changes in the linear and the angular moment we make use of the vector method. The method not only has important applications in the research of the mechanical motion, but it is also the bridge connecting mechanical motion with other forms of motion.The theorem of the change in the kinetic energy establishes the dependence between the physical quantities describing motion---kinetic energy and describing the acting force--- work. It is a law describing changes between different forms of energy. Work is a measure of the accumulated effect of the action of a force on a body during a given displacement. 1. Work done by a constant force F S W FS = = cos The work done by a force is a scalar quantity. For , the work is positive, for , the work is zero, for , the work is negative. The unit of work in the SI system is the joule (J). 2 2 = 2 1J=1N1m § 14-1 Work done by a force
学 与动量定理和动量矩定理用矢量法研究不同,动能定理用 能量法研究动力学问题。能量法不仅在机械运动的研究中有重 要的应用,而且是沟通机械运动和其它形式运动的桥梁。动能 定理建立了与运动有关的物理量一动能和作用力的物理量—功 之间的联系,这是一种能量传递的规律。 §14-1力的功 力的功是力沿路程累积效应的度量。 常力的功 W=FS a Mi M M2 =F S 力的功是代数量。a时负功。 单位:焦耳(J);1J=1Nm
6 与动量定理和动量矩定理用矢量法研究不同，动能定理用 能量法研究动力学问题。能量法不仅在机械运动的研究中有重 要的应用，而且是沟通机械运动和其它形式运动的桥梁。动能 定理建立了与运动有关的物理量—动能和作用力的物理量—功 之间的联系，这是一种能量传递的规律。 § 14-1 力的功 力的功是力沿路程累积效应的度量。 F S W FS = = cos 力的功是代数量。 时,正功； 时,功为零； 时,负功。 单位：焦耳（Ｊ）； 2 2 = 2 1J=1N1m 一．常力的功
Dynamics 2. Work done by a variable force: 2 Elementary work ]w=Fcosads =Fd=F·cr Xdx+ydy+zd Myar (F=Xi +yj+Zk, dr=dxi +dv+dck Fdr= Xdx+Yay+ld)x The total work done by a force F during a finite curvilinear displacement M, M2is W=∫ Fcosads=∫Fds M (expression in the natural form) =∫Fb (vector expression) M =「Xax+1hy+z (expression in terms of rectangula lar coordinates) 7
7 2. Work done by a variable force: W =Fcosds F ds = = F dr = Xdx +Ydy + Zdz (F = Xi +Yj + Zk ,dr = dxi + dyj + dzk F dr = Xdx+Ydy + Zdz) Elementary work The total work done by a force during a finite curvilinear displacement is . F M1 M2 = = 2 1 2 1 cos M M M M W F ds F ds (expression in the natural form) = 2 1 M M F dr (vector expression) = + + 2 1 M M Xdx Ydy Zdz (expression in terms of rectangular coordinates)
学 二,变力的功 元功:W= Cosas -fds F·c ar Xdx+dv+zd (F=Xi+yj+Zk, dr=dxi+dv +dck F dr= Xdx+Yay+zdz 力F在曲线路程MM,中作功为 M W=∫ Fcosads=∫Fs(自然形式表达式) MI M F dr (矢量式) =「Xx+hy+Zh(直角坐标表达式) 8 M
8 二．变力的功 F ds = = F dr = Xdx +Ydy + Zdz (F = Xi +Yj + Zk ,dr = dxi + dyj + dzk F dr = Xdx+Ydy + Zdz) 力 F 在曲线路程 M1 M2 中作功为 = = 2 1 2 1 cos M M M M W F ds F ds （自然形式表达式） = 2 1 M M F dr （矢量式） = + + 2 1 M M Xdx Ydy Zdz （直角坐标表达式） 元功： W =Fcosds
3. Work done by a resol Dynamic Tant force If a particle is subjected to the action of n forces F..F, the resultant force isR=2F. The work done by the resultant force Ris M W=∫Rd=(F+F2+…+F)b 「Fd+「F2+…+「F=W+W2+…+Wn M1 Mi M1 le W=∑W he work done by the resultant force during a finite displacement is the arithmetical sum of the work done by all the component forces acting on the particle
9 3. Work done by a resultant force: If a particle is subjected to the action of n forces , the resultant force is . The work done by the resultant force is i.e, The work done by the resultant force during a finite displacement is the arithmetical sum of the work done by all the component forces acting on the particle. F F Fn , , , 1 2 R W R dr F F F dr n M M M M = = + ++ ( ) 2 1 2 1 1 2 F dr F dr F dr M M n M M M M = + ++ 2 1 2 1 2 1 1 2 W W +Wn = + + 1 2 W =Wi R = Fi
学 合力的功 质点M受n个力F,2…F作用合力为R=∑F则合力R 的功 M W=」Rd=(F1+F2+…+Fn)d M1 M M =「Fc+「F+…+「F1=W+W2+…+Wn 即 W=∑W 在任一路程上,合力的功等于各分力功的代数和 10
10 三．合力的功 质点M 受n个力 作用合力为 则合力 的功 F F Fn , , , 1 2 R = Fi R W R dr F F F dr n M M M M = = + ++ ( ) 2 1 2 1 1 2 F dr F dr F dr M M n M M M M = + ++ 2 1 2 1 2 1 1 2 W W +Wn = + + 1 2 即 在任一路程上，合力的功等于各分力功的代数和。 W =Wi