学 3.万有引力的功w=cmn 万有引力所作的功只与质点的始末位置有关,与路径无关。 4.作用于转动刚体上的力的功,力偶的功 设在绕z轴转动的刚体上M点作用有力F,计算刚体转过 角度时力F所作的功。M点轨迹已知F=F+Fn+E sw=Fds=F rdo=m (f)do (9=92-):=了m(Fl 作用于转动刚体上力的功等于力矩的功。 Fb 如果作用力偶,m,且力W=」mdo dop 偶的作用面垂直转轴 M 若m=常量,则W=m(2-91) 注意:功的符号的确定。 16
16 W =F ds=F rd=mz (F)d ( ) =2 −1 = 2 1 ( ) W mz F d 作用于转动刚体上力的功等于力矩的功。 = 2 1 W md 若m = 常量, 则 ( ) W = m 2 −1 注意：功的符号的确定。 3．万有引力的功 ) 1 1 ( 2 1 0 r r W = Gmm − 万有引力所作的功只与质点的始末位置有关，与路径无关。 如果作用力偶，m , 且力 偶的作用面垂直转轴 4．作用于转动刚体上的力的功，力偶的功 设在绕 z 轴转动的刚体上M点作用有力 ，计算刚体转过 一角度 时力 所作的功。M点轨迹已知。 F F F = F + Fn + Fb
Dynamics 5)Work done by friction 1) Work done by kinetic friction: W=Mm -Fds=-Mm. f'Nds If N=const W=fNS. Thus the work depends on the actual path of the particle (2 )Work done by kinetic friction acting on a disc rolling without slipping on a fixed surface The normal reaction force w and the frictional force f act on the instantaneous center of velocity c The elementary displacement of the instantaneous center Is R dr=vdt=0.so SW=Fdr=Fvdt=0 (3)Work done by a couple with a moment m resisting rolling w=mp=-mie If m=const. then R
17 5) Work done by friction: (1) Work done by kinetic friction: = − =− 1 2 1 2 ' M M M M W F ds f Nds If N=const. W= –f´N S. Thus the work depends on the actual path of the particle. (2)Work done by kinetic friction acting on a disc rolling without slipping on a fixed surface. The normal reaction force N and the frictional force F act on the instantaneous center of velocity c. The elementary displacement of the instantaneous center is dr=vC dt=0 W =Fdr=FvC dt=0 (3) Work done by a couple with a moment m resisting rolling If m=const. then . R s W =−m =−m , so
功学 5.摩擦力的功 ()动滑动摩擦力的功W=J-Fd=,fMd N=常量时,WfNS,与质点的路径有关。 (2)圆轮沿固定面作纯滚动时,滑动摩擦力的功 正压力N,摩擦力F作用于瞬心C处,而瞬心的元位移 C=vcd=08=Fd=F·va=0 (3)滚动摩擦阻力偶m的功 O P 若m=常量则W=-m=m8 R 18
18 dr=vC dt=0 W =Fdr=FvC dt=0 正压力 N ，摩擦力 F 作用于瞬心C处，而瞬心的元位移 (2) 圆轮沿固定面作纯滚动时，滑动摩擦力的功 (3) 滚动摩擦阻力偶m的功 5．摩擦力的功 (1) 动滑动摩擦力的功 = − =− 1 2 1 2 ' M M M M W F ds f Nds N=常量时, W= –f´N S, 与质点的路径有关。 R s 若m = 常量则 W =−m =−m
Dynamic 5. Work done by internal forces W=F4+F"lB=F·Ch4-F·cB =F d(ra-rB)=F d(Ba) O B IB If the distance between two points a and b does not change, the sum of the elementary work done by the internal forces f and F ' is zero The sum of the work done by all the internal forces of a non- deformable system is zero. Special cases of such systems are a rigid body and an inextensible string 19
19 5. Work done by internal forces: If the distance between two points A and B does not change, the sum of the elementary work done by the internal forces F and F’ is zero. The sum of the work done by all the internal forces of a nondeformable system is zero. Special cases of such systems are a rigid body and an inextensible string. A B W =Fdr +F' dr A B =Fdr −Fdr ( ) A B =Fd r −r =Fd(BA)
学 五.质点系内力的功 oW=Fdra+F'-drB=F.dr-F-drB =F d(ra-rB)=Fd( BA B IB 只要A、B两点间距离保持不变,内力的元功和就等于零。 不变质点系的内力功之和等于零。刚体的内力功之和等于零。 不可伸长的绳索内力功之和等于零 20
20 五．质点系内力的功 只要A、B两点间距离保持不变,内力的元功和就等于零。 不变质点系的内力功之和等于零。刚体的内力功之和等于零。 不可伸长的绳索内力功之和等于零。 A B W =Fdr +F' dr A B =Fdr −Fdr ( ) A B =Fd r −r =Fd(BA)