Chapter3Work and Energy
Chapter 3 Work and Energy
$ 1.WorkmF=F(t)F(t)dt → transport momentumF=F(r))What does F transport?dvdiF(r)·drF(r)m:dr =(dv)·mmdtdt transport energyd2dvdvdv=2m7dtdtdtdtd2=2.dcovariant under Galileao'sTransformation
F F(r) = m dt dv F r ( ) = m dr dv vm dt dv F r dr ( ) = = ( ) transport momentum . §1. Work F = F(t) F(t)dt What does F transport? transport energy. dt dv v v dt dv dt dv v dt dv = + = 2 2 dv v dv = 2 2 covariant under Galileao’s Transformation
Fm Theorem of Kinetic energyF(r).dr = d(_mv2)一drF(1)·dr - Work d(mv2) Kinetic Energy*2112["2 F(r).drmimi=Ek2-Ekl2(n)12）2Definition of workm But not r .dF Why?dw = F(r)·drF.drrelated toenergy,momentumdwdr= F(r)·dvF(r)Power:Pdtdt
) 2 1 ( ) ( 2 F r dr d mv = Theorem of Kinetic energy: ) 2 1 ( 2 d mv F r dr ( ) F dr Work Kinetic Energy r r EK EK r r F r dr mv mv 2 1 2 ( ) 2 ( ) 2 1 2 1 2 1 2 1 ( ) = − = − dw F r dr = ( ) But not Why? r dF F dr related to energy,momentum Definition of work: Power: F r dv dt dr F r dt dw P = = ( ) = ( )
$ 2.Law of conservation of mechanical energy2-1,Conservative force无法显示Conservative force Work done isindependent of path Non conservative forcefF.dr=0Uniform force,Gravity, Elastic force:Example(1).Work done by a uniform forcefF.dr=-'aF.dr +"F.drYh(1)(II)F=Fxi+F,j+Fzk dr=idx+jdy+kdz
§2.Law of conservation of mechanical energy 2-1. Conservative force Conservative force Non conservative force Example: (1).Work done by a uniform force. F dr • = 0 Work done is independent of path. Uniform force,Gravity, Elastic force: F dr = + b a a b r r r r F dr F dr (I) (II) F F i F j F k X y Z = + + dr i dx jdy kdz = + +
path-one一fF.dr-'(Fx+F,d+Fd)X()'(Fdx+F,dy+F.d)X(II)=0path-two(2).Work done by a central forcedrPolar coordinate (for simplicity)F=f(r)rIdr =drr+rdee
dx dy dz) F dr dx dy dz) (F F F (F F F y z r r x y z r r x b a a b + + + = + + (I) (II) = 0 path-two path-one (2).Work done by a central force. F f r r = ( ) Polar coordinate (for simplicity): ^ ^ dr = drr+ rd F dr
{F·dr =f f(r)r.(drr+rdee) =f f(r)dr=rnf(r)dr-fra f(r)dr(1)(II)=0Independent of pathElastic force: F(x) = -kx xmMGravitational force: F(r) = -GOnly depends on["aF-dr = U(a)-U(b)position""!
F dr = f (r)r(drr+ rd ) = f (r)dr (I) (II) = − ra rb ra rb f (r)dr f (r)dr = 0 Only depends on “ position”! Independent of path: Elastic force: Gravitational force: ^ F(x) = −kx x ^ ( ) r r mM F r = −G F dr U(a) U(b) a b r r = −
*How about Coriolis force? Introduce a function of position?U()U(r)+CF.dr =-dU(r)1-D:dU(x) =F not unique.F.dx =-dU(x)口dxGive U(x): F=?1dU(x)kx?=-kxU(x) =2dx“O"point: x-0, Uo=0. dU = -F.dx
Introduce a function of position: How about Coriolis force? F dr dU(r) = − U(r) →U(r) +C 1-D: F dx dU(x) x = − Fx dx dU x − = ( ) Give U(x): F=? 2 2 1 U(x) = k x k x dx dU x = − ( ) “O” point: x=0, =0. U0 dU F dx = − not unique
uodU=-F.dx=kxdxkx2Gravitational forceUOr8mjm2mjm2dUdrdr2T.2mim2GrU(r) = -G m,m21*In modern Physics F() (field) is replaced by U(r)U(r) describe a field
0 0 0 2 2 1 dU F dx kxdx kx U x U U = − = = − Gravitational force: dr r m m rdr G r m m dU G r r U U r r = − = 2 1 2 0 2 1 2 ˆ r m m G 1 2 = − r m m U r G 1 2 ( ) = − In modern Physics F(r) (field) is replaced by U(r). U(r) describe a field
Advanced Remarks(1).Partial derivativef(x,y,z)afafafdzdfdx+dyOzOxoyaaK(2) √+Ozxoy(3). Gradient of Φ:aaa@+kVp(x,y,2)=iD?Ozaxoy
Advanced Remarks: (1).Partial derivative f(x,y,z) dz z f dy y f dx x f df + + = (2). z k y j x i + + = z k y j x x y z i + + = ( , , ) (3). Gradient of Φ:
aa0V.FFDivergence of F:HOOz1ayxkijaaaCurl of F:axOyOzV×F=FFF(4) fF.d=0 V×F=0Curl-less forceJJ(V×F)-ds =fF-drmS
x y Fz z F y F x F + + = Divergence of F: F = i j k x y z Fx Fy Fz Curl of F: (4). = 0 F dr F = 0 Curl-less force F ds = F dr s ( )