Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Conservation laws:Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Kinematics in Cartesian coordinatesdtdVadtdVadtVdayyxx ;;dtdyVdtdxVdtrdVyx ;;dtVydtVxdtaVdtaVyxyyxx;;;is given, you can find If Vand arDO NOT use these Constant acceleration formulas when acceleration is a function of time. You have to integrate or differentiate! DO NOT mix x- and y-components!)0()0(21)(2xtvtatxxx)0()(xxxvtatv ))()((2)()(121222txtxatvtvxxxKinematics in polar coordinatesrdtdrardtrdar 2;222rirrrdtdrVdtdrVr ;yxrriiiViVVrriaiaarrUniform circular motion: 2;;2 TrVrarKinematics of circular motiondtddtdzz ;rhr;dtdiidtdris given, you can find If and by integration,similarly to linear motionVectorsjAiAAyxjBiBByx? BACjCiCCyxxxxBAC yyyBAC Cross product, dot productNewton’s 2nd LawamFxxmaF yymaF Newton’s 3rd LawNNFfrFfrSolving Problems1. Sketch Isolate the bodies, draw a free-body diagram (only external forces; use 3rd law)2. Write down 2nd Newton’s law for each bodyamFChoose a coordinate system Write 2nd Newton’s law in component form: yyxxyxyxmaFmaFjmaimajFiFF,3. Solve for acceleration etc. 222222212121rVmVVmmVKErrTorque and Angular Momentum)(][)(][22rhrrFirFiFiFrFrrhrmrimriridtdrmrprLzrrzrKinetic energy:prLFrMotion in the plane, in polar coordinates:yxrriiexttotdtLdConstLtotext,0IfDynamics of rotational motionzzexttotdtdLConstLzztotext ,0IfCentral forceyxrriiFOrF||constimrLirFiFiFrFrzzrr20][dmrrmIrhrILiiizzz22);(222121ziiiIVmKE )(rhrIdtLdtotextm1m2RIRigid bodies rotating about a fixed axisConservation laws: •Momentum•Angular momentum•energy•These quantities are additive•P and L are vectors; only some of their components may be conservedWork Energy Theorem222221initialfinalrrtotalmVmVrdFW 22222121initialfinalyytotalyxxtotalxmVmVdyFdxFW )]()([12rUrUWveconservati)()(1221rUrUdrFWrrrCentral force:€ If Wnon −conservative= 0,U(r r 2) + K2= U(r r 1) + K1Mechanical energy is conserved! € If Wnon −conservative≠ 0,U(r r 2) + K2− (U(r r 1) + K1) = Wnon −conservativeKnow potential energy for familiar forces: • gravity near Earth, F = mg • spring force Fx = - k(x-L)Know how to find potential energy for unfamiliar forces:Fx = ax2 – bx etc.Graphic representation of motion for a given potentialenergy function. Identify points of equilibrium (stable and unstable) and turning points.€ U(x) = − Fxdx∫Conservation of MomentumextFdtpdOnly if the collision is perfectly elastic, the kinetic energy is conserved € Ifr F ext= 0,dr p dt= 0,r p = Constpx= Const py= Constpx(before) = px(after)py(before) = py(after)Sometimes only Fx or Fy may be equal to zero. Then only px or py is conserved.If F is not zero, but the collision is very short (Ft is small as compared to change in momentum), you can still use momentum conservation relating moments of time immediately before and after the collision.Conservation of Angular MomentumFrmriridtdrmrprLdtLdrexttot;][;2ConstLIftotext,0For symmetrical objects rotating about their axis of symmetry:2);(iiirmIrhrIL€ K =12mii∑Vi2=12MVcm2+12Icmω2)(rhrIdtLdtotextSecond Law:m1m2RIRolling without slipping€ a = αR,v = ωRFriction force is not zero but work done by this force is zero.One can apply energy conservation.Applications: rolling of rigid bodies, pulleys, yo-yos Linear velocity and acceleration of the rim with respect to the center of rotation are related to angular velocity and
View Full Document
Unlocking...