TExample 1: As shown in the figure, the weight of theA15cmslider is P =9.8N , spring stiffinesscoefficient k =0.5N/cm ,when the slider is in position20cmA, the tension of the spring on the slider is 2.5N , theslide block moves from position A to position B in asmooth horizontal groove under the2oN pull of therope. Find the sum of the energetic work applied to the slider.P\TAnswer: The force on the slider at any instant is shown inathe figure. Because P and N are always perpendicular to the FINslider displacement, their work is zero. So you just have tocompute the works of T and F. The works of T :In motion, the magnitude of T is constant, but the direction ischanging, so the elemental work of T isSW, = T cosadxThere, cosα=(20-x)/ /(20-x)2 +152
A B T 20cm 15cm Find the sum of the energetic work applied to the slider. Example 1: As shown in the figure, the weight of the slider is , spring stiffness coefficient , when the slider is in position A, the tension of the spring on the slider is , the slide block moves from position A to position B in a smooth horizontal groove under the pull of the rope. P = 9.8N k = 0.5N cm 2.5N 20N T P F N Answer: The force on the slider at any instant is shown in the figure. Because and are always perpendicular to the slider displacement, their work is zero. So you just have to compute the works of and . The works of : P N T In motion, the magnitude of is constant, but the direction is changing, so the elemental work of is T T W T dx T = cos 2 2 There, cos = (20 − x) (20 − x) +15 F T
So the work done by T in the whole process is20-x20W, = [Tcosodx = [°20dx/(20 -x)2 +152= -20| /(20 - x)2 +152= 200N·cm10Then calculate the work of F:2.5=5cmBy the question: S,S, = 5+20=25cm0.5So the work done by F in the whole process isW, =k(8 -83)= =0.5(52 -252)=-150N ·cm22So the effective work isW =W_ +W =200-150=50N·cm
So the work done by in the whole process is T x N cm dx x x W T dx T = − − + = − + − = = 20 (20 ) 15 200 (20 ) 15 20 cos 20 2 0 0 2 2 2 0 0 2 0 0 2 2 Then calculate the work of F ： By the question： 5cm 0.5 2.5 1 = = 2 = 5+ 20 = 25cm So the work done by in the whole process is F WF = k − = 0.5(5 − 25 ) = −150N cm 2 1 ( ) 2 1 2 2 2 2 2 1 So the effective work is W =WT +WF = 200−150 = 50Ncm
I . The kinetic energy of the particle13.2If the mass of the particleis m and the velocityis ,thekineticThe kinetic energy of the particle and its systemenergyoftheparticleis_my?T2Since the velocity is instantaneous, the kinetic energy is alsoinstantaneous, which is a constant positive scalar, and the unitof kinetic energy in the SI system is the Joule (J).II. The kinetic energy of the system of particlesThe arithmetical sum of the kinetic energy of eachparticle in the particle system is called the kinetic energy of theparticle system, that isT=Z=mV2Rigid body is a common system of particles in engineeringpractice. When the motion form of rigid body is different,itskinetic energy expression is also different.1. Kinetic energy of a translational rigid body
13.2The kinetic energy of the particle and its system Ⅰ. The kinetic energy of the particle If the mass of the particle is and the velocity is , the kinetic energy of the particle is m v 2 2 1 T = mv Since the velocity is instantaneous, the kinetic energy is also instantaneous, which is a constant positive scalar, and the unit of kinetic energy in the SI system is the Joule (J). Ⅱ. The kinetic energy of the system of particles The arithmetical sum of the kinetic energy of each particle in the particle system is called the kinetic energy of the particle system, that is 2 2 1 i i T = m v Rigid body is a common system of particles in engineering practice. When the motion form of rigid body is different, its kinetic energy expression is also different. 1. Kinetic energy of a translational rigid body
=Mv?22. The kinetic energy of a rigid body rotating in a fixed axis>m.y022/3. The kinetic energy of a rigid body moving in a plane20Due to J..= Jc +md? So T :2d·Q =Vc So TDue to2That is, the kinetic energy of a rigid body moving in a plane isequal to the sum of the kinetic energy moving with the center ofmass and the kinetic energy rotating with the relative center ofmass
2 2 2 2 1 2 1 2 1 i i C mi MvC T = m v = v = 2. The kinetic energy of a rigid body rotating in a fixed axis 2 2 2 2 2 2 2 1 2 1 2 1 2 1 T = mi vi = mi ri = mi ri = Jz 3. The kinetic energy of a rigid body moving in a plane d C C C v 2 2 1 = C T J Due to 2 J JC md C = + So 2 2 2 2 ( ) 2 1 2 1 ( ) 2 1 T = JC + m d = JC + m d C d = v 2 2 2 1 2 1 T = mvC + JC That is, the kinetic energy of a rigid body moving in a plane is equal to the sum of the kinetic energy moving with the center of mass and the kinetic energy rotating with the relative center of mass. Due to So
Example 2: The length is 1 , the homogeneous rod OACof weight P rotates at an equal-angular velocity oabout the verticabaxis passing through the spherical1hinge. If the Angle between the bar and the lead lineis α , find the kinetic energy of the bar.Answer: Take a small segment dr on the rod at thedistance of r from O, as shown in the figure. It'sPmass is dm :, the distance from Oz to thisdrg1segment is rsin α, then the kinetic energy of thesegment is11 Pdr(r sin αの)?d2 gl2Psindr2glSo the kinetic energy of the whole bar isPo?sin2sin 12gl6g
O A C P O A C P z r dr Example 2: The length is , the homogeneous rod OA of weight rotates at an equal-angular velocity about the vertical axis passing through the spherical hinge. If the Angle between the bar and the lead line is , find the kinetic energy of the bar. l P Oz Answer：Take a small segment on the rod at the distance of from O, as shown in the figure. It’s mass is , the distance from to this segment is , then the kinetic energy of the segment is r dr dr gl P dm = Oz rsin r dr gl P dr r gl P dT dmv 2 2 2 2 2 2 sin ( sin ) 2 1 2 1 = = = So the kinetic energy of the whole bar is: 2 2 2 0 2 0 2 2 sin 2 6 sin l g P r dr gl P T dT T l = = =