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10-1. Angular Quantities (P235)tttddlim0 tttddlim0 Angular displacement(角位移角位移):Angular velocity :Angular acceleration :1. Angular variables(radian measure, 弧度弧度)xyO)( tP 0Pr第1页/共50页rttsdddd rv oo2. Relation of linear and angular variablesThe distance along a circular arc:Differentiating above equation with respect to time (r again holds constant)rttvdddd The linear speed (with r held constant):rs measure)(radian 第2页/共50页The tangential component at :rt Th he aadialal compo component t a an:They are valid only if angles are expressed in radian. 22 rvrvan an is alsos also callcalled a aR, aadial accal accel leatatio on, in ou boo ou book.第3页/共50页 - 0 = 0 +t - 0 = 0t + t2t 2 = 02 +2( -0) - 0 = (0 + )t0 - 0 = t - t2v = v0 + at x -x0 x - x0 = v0t + at2 v v2 = v02 + 2a(x - x0) tx - x0 = (v0 + v)t a x - x0 = vt - at2 v0Equations of motion for constant linear acceleration are analogous to the ones of constant angular acceleration: linear missing angular equation variable equation2121212121213. Kinematics equations for uniformly accelerated rotational motion (P238)第4页/共50页Example:The flywheel of a steam engine runs with a constant angular velocity of 150rev/min. When steam is shut off, the friction of the bearings (轴承轴承) and of the air stops the wheel in 2.2h. Answer:2min/rev14. 1)a( rev109 . 921)b(32 tto 2mm/s99. 0) c ( rat 22222m/s31m/s31)d( rtrraaaara (a) what is the constant angular acceleration, in rev/min2, of the wheel during the slowdown? (b) How many rotations does the wheel make before stopping? (c)At the instant the flywheel is tuning at 75rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)? 第5页/共50页10-2 Several Concepts Related to Rotation(a)(a)A igid bo body is os one that ca that can otat otate w with th all all its pats locts pats locked to togeththe a and w without thout aany cha change in its shapts shape ( (刚体是一种特殊的刚体是一种特殊的质质点系统点系统).(b)The motion of a rigid body reduces to a translation of its center of mass adds a rotation relative to center of mass.Ocv第6页/共50页(c) Translation is motion along a straight lineFeature：BABAAB /ABABAB(d) Rotation is the motion as the Fig. showing. protation axis Feature：Any point moves around one same axis.第7页/共50页zABA ABThe features of rotation about a fixed axis:The location and direction of rotation axis are fixed relative to a inertial reference frame.Every point of body moves in a circle whose center lies on rotation axis and radius different.Each rotation plane is perpendicular to rotation axis.Every point moves through the same angle during a particular time interval ( (各点各点的的 ， ， 相同相同) ).第8页/共50页sinRFwhwhe e is calls called mommoment am(t am(力臂力臂) ) -ththe distastanc ce rFR 11. Tooqu ue o on a an a axis s ( (对轴的力对轴的力矩矩) ) Its SI units: m NThe torque about a given axis is defined as:orfom the axis of otation that is pependicula to the line of action of the foce. RF10-3 Torque and Rotational Inertia of Rigid bodies(P241)第9页/共50页external force ;iFifinternal forceApplying N-II law on mi:amfFiii zOirifiF mi iiiiiitiiitrmrfrF2A rotational rigid body can be considered to be constituted by many parts. The ith part has a mass mi , :sum & ir 22. N N-II law fo otatII law fo otatio on (定轴转定轴转动定律动定律) )(P244)Resolving it into directions and :n iiitiititrmamfF :0 iiniinrfrF:n0 iiirfand第10页/共50页So, we have iiiiiitrmrF2 ththe mommoment of t of inettia a oo otatotatio onal al inettia a ( (转动惯量转动惯量) o on th the aaxis s. 2iirmI Net torque about given axis, and Net torque on a rigid body is equal to the product of rotational inertia and angular acceleration it causes about same axis.LetiiitinetrFThen,N-II law fo otationInet第11页/共50页For a continuous rigid body, the summation in the rotational inertia can be substituted by an integral ,3. The rotational inertia of body (P248)The moment of inertia I is a measure of the rotational inertia of a body, which plays the same role for rotational motion that mass does for translational motion.where,mrId2dldSdVdmlsr第12页/共50页2.It is determined by the total mass and the distribution of mass to the axis, so rotational inertia relative to different axis is different even for the same body.The rules for calculating rotational inertia:3.Descriptions：1.Rotational inertia is the measure for the inertia of rotation (转动惯量是转动惯性大小的量度转动惯量是转动惯性大小的量度). (kgm2) In general, smaller I means easier rotation.第13页/共50页The I of a group-body to an same axis equals the sum of the one of every rigid body.(1) For same axis, rotational inertia can be superimposed. iIIMm2L2LOmMOIII For the case shown in right figure: Thin rod plus a ball of particle,22)2(121LmML 第14页/共50页2MhIIC CCL21L21z zhTh his s is s knowown as th as the paallpaallel l-aaxis ths theooem m.Let Ic be the rotational inertia of body with respect to the center of mass, The rotational inertia I about the given axis is: and h be the perpendicular distance between given axis and the axis through center of mass (these two axes must be parallel). (2) Parallel-axis theoremSomome otat otatio onal al inettias as in Fig.10-21 of of P246246:第15页/共50页)d2,(121d202222 LLLCIIormLxxLmI20231dmLxxLmILO Solution:(1)xmddl l Cxx mxIdd2 xxd2l lxLmxd2From the parallel-axis theorem, we have I to O axis:222231)21(121mLLmmLmhIIco Try to find out I of a thin rod of mass m, length L to (1) C and (2) O axes.OxmdxLSolution:(2)Example:第16页/共50页In the Fig. block A has mass m1, B has m2. (m1 m2 ).Two pulleys (滑轮滑轮) (mounted together in horizontal frictionless bearings (轴承轴承) have rotational inertia I1 and I2. (neglect all friction forces and mass of cord(细绳细绳), no slide). Find: (1) The acceleration of A and B; (2) The tension of cord.ABc1m2mRr2Fgm2B2mSolution:1Fgm1A1ma 1F2FcoExample:第17页/共50页1111amFgm 2222amgmF )(2121IIrFRF Ra 1gRrmRmIIrmRma222121211 Neglect the mass of cord,;11FF 22FF ra 2grrmRmIIrmRma222121212 gmrmRmIIRrmrmIIF1222121222211 gmrmRmIIRrmRmIIF2222121121212 2Fgm2B2m1Fgm1A1ma 1F2Fco第18页/共50页A body rotating about an fixed axis is said to have rotational kinetic energy. By analogy(类比类比) with translational kinetic energy, we have 221iivmK 222)(2121 iiiirmrmUsUsing equatuatio on of of I I, 2iirmI221 IK 1. The kinetic energy of rotation10-4 Rotational Kinetic Energy (P254)ththen,第19页/共50页Fom thom the kinet tic thc theooem of a sm of a singl le patpaticlcle,KmvmvWif 222121KIIW 2i2f2121 OOne ca can gu uessssfo a otatfo a otatio on22. Wook-kinet tic c ene gy th theooem fo fm fo fixed-aaxis s otatotatio on d d rFrFWdd rFdcos d rdF rOOProof: dcosrF 第20页/共50页KIIW 2i2f2121 fid Wwhere the work done on a rigid body during a finite angular displacement equals the change of its rotational kinetic energy. the power for rotational motion ttWPddddBy analogy,So, we get the work-kinetic energy theorem for rigid body rotated about a fixed axis:第21页/共50页ChChimiU=033.Wook-ene gy p pinc ciplple fo f fo fixed-aaxis s otatotatio onThe principle of work-energy for particle-systems is still hold for the rigid bodies.meciiffext)()(EUKUKWWinc iiffext0whenUKUKWWinc The weight-energy of rigid body:刚体的重力势能与它的质量集中在质心时的势能相同 iiGghmUmhmhiiC ghmii)(CmghUG 第22页/共50页Solutolutio on:As in the Fig. : For the spring, k=2.0 103 N/m.For the wheel, I=0.5 kg m2 , R=30cm.mRkWhat is the speed of the body (m=60kg), when it falls 40cm. Initially, it is at rest and the spring is not stretched.2/ RImkxmga Example: Ra maFmgIkxRFRKnown: ( ), ( )?aa xvv x第23页/共50页xRImkxmgvvhvd/d020 m/s5 . 1/222 RImkhmghvor,221khmgh 221 I 221mv ddddddddvvxv vattxxd( )dv va xx2( )/mgkxaa xmI R第24页/共50页Rolling as rotation and translation combinedWhen bicycle moves forward by a distance s, we has a relation between s and the rotation angle ,Rs The linear speed v vcom of the center of wheel is ds/dt, , angular speed of wheel about its center is d /dt,Rv com11. Rollolling10-5 Rolling; Rotational + Translational Motion(P256)第25页/共50页The rolling motion of a wheel is a combination of purely translation and purely rotational motions:Middle: pure rotation around the axis, every point on the wheel rotates about the center with angular speed .Top Fig.: purely translational motion with linear speed v vcm;Bottom: actual rolling motion of wheel.Rolling as rotation + translation:第26页/共50页At any moment, the point P is stationary with linear speed v vP=0, it can be considered as an axis, and the rolling motion of the wheel as a pure rotation about this axis.The point P can be referred to as an instantaneous axis.The kinetic energy of rolling about the point P is: 221 PIK 2. Rolling as pure rotation:第27页/共50页222com2121 MRIK From the parallel-axis theorem, where I Icom is the rotational inertia about an axis through its center of mass. So the kinetic energy of rolling isA rolling object has two types of kinetic energy:A rotational K due to its rotation about its center of mass and translational K due to translation of its center of mass.2MRIIcomP222121 comcomMvIK第28页/共50页Analysis of a sphere on an incline using forces. The mass of the sphere is M, and the radius is R.2cm/1sinMRIga Ra For the case “rolling smoothly”, we have; 0cos MgFN;sinMaFMgfrcmIRFfrIt is suitable for the linear of any body rolling along an incline of with the horizontal.xcoma,第29页/共50页This is the minimum static friction force to keep pure rolling. If a static friction force Mgsin this value, then the sphere will not simply roll but slip as it moves down the plane.It is independent of mass and radius!For a sphere, ,522cmMRI sin75ga Then,The acceleration of the CM of a rolling sphere is less than that for an object sliding without friction (a=gsin ). sin7252 MgMa RRaMRRIFfr/252cm第30页/共50页(1) f on the ball from the ramp (斜坡斜坡) does not transfer any energy to thermal energy because the body does not slide; Two conditions given by the rolling smoothly:(2) v=R and a = R are valid.The speed v at the bottom is:axv2 gHHg710)sin)(sin75(2 第31页/共50页The torque acting on a particle relative to a fixed point O is a vector quantity defined as:Fr 1. Torque about a point10-6 Angular Momentum and Torque for a System of Particles; General Motion (P278)on a system of particles will be:iiFr where is the position vector of a particle relative to O, and the total torque r第32页/共50页(iii) That point must always be specified if the value of the torque is to be meaningful. Changing the point can change the torque in both magnitude and direction.(ii) This definition is for particles moving along any path relative to a fixed point.iiFr (i) Where is the position vector of the ith particle and is the net force on the ith particle.Fir第33页/共50页The linear momentum of a particle:2. Angular momentum of a particleThe Angular momentum relative to a fixed point：vmp)(vrmprl Direction：right-hand ruleMagnitude: sin|mvrll 第34页/共50页Supplementary explanation:(1) It is a vector quantity (SI unit: J.s);(2) It is not same relative to different point.(3) The angular momentum of particle relative to some point in the motion of straight line.大小：大小：Omrdv Direction:)(vrmprl mvd sin|mvrll 第35页/共50页In linear motion, we have Newtons second law for a single particle:,ddnettpF one can guess for its angular formThe time rate of change of angular momentum of a particle is equal to the net torque applied to it.3. Relation between angular momentum and torqueThe rotational equivalent of N-II law for a particle.dtl dnet第36页/共50页4. The angular momentum of a system of particlesIf the angular momenta of individual particles change with time, then we haveFor a system of particles, total angular momentum of the system is the (vector) sum of the angular momentum of the individual particles :Llni 1 i nllllL 2 1 .,ninetn il dLd dtdt11iiis the net torque acting on the ith particle.tl iinetdd,第37页/共50页Total net torque acting on all particles of system)(dnet ILt The time rate of change of the total angular momentum of a particles system (or a rigid body) equals resultant external torque on the system.It can also be written as:It is true only when and are calculated with respect to either (1) the origin of an inertial RF or (2) the center of mass of a particles system. Langular momentum principle(质点系角动量定理) ddnetnettLFr )(dtpdF第38页/共50页If net external torque acting on particles system is zero, the total angular momentum of the system remains conserved. (合外力矩为零，质点系总角动量守恒合外力矩为零，质点系总角动量守恒!)5. The conservation law of angular momentumNo exceptions to the law of conservation of angular momentum have ever been found. , 0IfCLnet第39页/共50页Supplementary explanation:It is equivalent to three component equations relating to the conservation of angular momentum in three mutually perpendicular directions (through the reference point)1.2.It hold beyond the limitation of Ns Laws (not only for particles whose speeds approach that of light, but also for subatomic particles).3.Condition: not . It is only hold for Inertia reference frame.0net0netF第40页/共50页v1r1r2Ov2FA cord makes a ball, mass m, move in a circle (r1 v1) on the surface of frictionless table. (1) When one pulls the cord downward slowly and makes the radius equal to r2, v2=? (2) The work done by the during r1r2.F2211rmvrmv )(2112rrvv 12vv Solution:(1)Example:F is central force, and v r, L=mvr =constant第41页/共50页From work-kinetic energy theorem, we have21222121mvmvW wherervmFFn2 1)(2122121 rrmvSolution:(2) 21drrrFWe may also find work by usingLook at example 11-7 of P285 !第42页/共50页Considering a rigid body rotates about a fixed axis - axis.which moves about the axis in a circular path with a radius Ri, the magnitude of the angular momentum of this mass element is cosiiiprl Taking a typical mass element, mi, 2)cos(iiiiiRmrvm 10-7 Angular Momentum and Torque for a Rigid Body Rotating about a Fixed Axis(P280)1. Angular momentum & torque of rigid body第43页/共50页We have dropped the subscript , because here L and I (inertia) are about the same axis. iiiiiRmlL)(2 IL Using the angular momentum principle of particles system,Summing over all the particles to obtain:For a rigid body, the component along the rotation axis is: dtLdnet dtdLaxisIdtdIIdtd)(第44页/共50页constant LIf no net external torque acts on a rigid body, The total angular momentum of a rotating body remains constant if the net external torque acting on it is zero.The law of conservation of angular momentum is one of the great laws of physics! IdtdLaxisfiLL or.ffiiII2. Conservation of angular momentum(P251 & 284)第45页/共50页Supplementary explanation:1.It holds beyond the limitation of Ns Laws (not only for particles whose speeds approach that of light, but also for subatomic particles).2.Two cases for fixed axis: (1) I keep constant, then is also constant, Same as self-rotation of the Earth; (2) I is variable quantity, such as particle or particle system. As I decreases, increases, or its reverse.第46页/共50页(P252)第47页/共50页3.The direction of angular momentum cannot change for nonfixed axis(非定轴转动非定轴转动)LSpinning top (陀螺）(P286) or gyroscope(回转仪) :Under the condition that no any friction acts on the rotation axes, no any torque acts on the rotator.So, the orientation(指向指向) of the symmetry axes in space will keep constant, as long as it rotates at high speed.第48页/共50页The problems of chapter 10:14, 17, 22, 23, 24, 29, 31, 35, 37, 40, 42, 46, 49, 50, 55, 57, 60, 63, 65, 66, 71, 73, 74, 90The problems of chapter 11:10, 11, 13, 15, 16, 17, 18, 19, 20, 22, 23, 35, 36, 37, 39, 59, 61第49页/共50页感谢您的观看！第50页/共50页
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