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Chapter 7 Systems of particles 7-1 The motion of a complex object 7-2 Two-particle system 7-3 Many-particle system 7-4 Center of mass of solid objects 7-5 Conservation of momentum in a system of particles When can the object studied be regarded as a mass point? Doing only translational motion Translational motion Rotational motion Translational + Rotational motion 7-1 The motion of a complex object When we projectile a rigid body, the motion of the body looks very complicated. We can consider the motion of the rigid body to be a combination of a parabolic trajectory of a “center of mass” (rotational motion is not considered) plus a rotation about “center of mass” (translational motion is not considered). How to find the center of mass (CM) of a particle system? 7-2 Two-particle system Do an experiment to find CM in a two-particle system. As an example, we consider a baton consists of two particles m 1 , m 2 at its ends A and B, connected by a thin rigid rod of fixed length and negligible mass. A B m1 m 2 2m1 m2 We give the rod a push along the frictionless horizontal surface and examine its motion. Snapshots of the locations of points A and B at successive intervals of time. Clearly both m 1 and m 2 are accelerated, however, one point in the rod (point C) moves with constant velocity. If point c is regarded as a reference, A and B points rotate with a constant rotational speed. View the motion from the reference of point C. So point c is actually the center of mass. Fig 7-5 By building a Cartesian coordinate, position of point c is found at : rcm m 1 r1 m 2 r2 m1 m 2 y cm r1 rcm C r2 ' rcm C’ m2 O m1 (7-1) or written as: x cm y x m 1 x1 m 2 x 2 m1 m 2 m1 y1 m 2 y 2 m1 m 2 A (7-2) C B m2 m1 From Eq. (7-1), the velocity and acceleration of the CM are: v cm d rcm dt a cm d v cm m 1 v1 m 2 v 2 m1 m 2 dt m1 a1 m 2 a 2 m1 m 2 (7-4) (7-6) Can we also find a cm is zero from Eq. (7-6)? m1 a1 m 2 a 2 F1 r F2 r ac 0 F 2 r F1 r How about the motion if the system has net external forces? Suppose there is an external force on each particle in above expt., then m1 a1 m 2 a 2 F 1 F 2 F1 r F 2 r 0 , and if write m 1 a1 m 2 a 2 F ext F ext ( m 1 m 2 ) a cm F1 ext F1 r F 2 ext F 2 r F 1 ext F 2 ext F m a m2a2 a cm 1 1 m1 m 2 ext Newton’s second law for systems of particles This looks very like a particle of mass m 1 m 2 located at the center of mass. 7-3 Many-particle system Consider a system consisting of N particles of masses m 1 m 2 …… m N . The total mass is M mn (7-10) Each particle can be represented by its location rn , velocity v n and its acceleration a n . The CM of the system can be defined by logical extension of Eq(7-1): rcm m 1 r1 m 2 r2 m N rN m1 m 2 m N 1 M (7-11) m n rn In terms of components, Eq(7-11) can be written as (7-12) 1 1 1 y m y x cm m x n n , z cm m n z n . n n , cm M M M Taking the derivative of Eq(7-11) 1 v cm M m (7-13) vn n Differentiating once again: a cm Or M a cm m 1 M n an m F (7-14) an n 1 F 2 F N (7-15) By Newton’s third law, in Eq(7-15) the vector sum of all the internal forces is cancelled, and Eq(7-15) reduces to Fext M a cm (7-16) Eq(7-16) is just the Newton’s second law for the system of N particles treated as a single particle of mass M located at the center of mass( rcm ), experiencing a cm. rcm a cm 1 M 1 M m n rn m n an We can summarize this important result as follow: “The overall translational motion of a system of particles can be analyze using Newton’s law as if the mass were concentrated at the center of mass and the total external force were applied at that point.” These are general results that apply equally well to a solid object. Sample problem 7-3 A projectile(射弹) of mass 9.8kg is launched from the ground with initial velocity of 12.4m/s at 54 an angle of above the horizontal (Fig 7-11). At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass 6.5kg, is observed at 1.42s after the launch at a height of 5.9m and a horizontal distance of 13.6m from the launch point. Find the location of the second fragment at that same time. CM v0 0 m 2 m1 Solution: If the projectile had not exploded, the location of the projectile at t=1.42s should have been x v 0 x t ( 7 . 3 m / s ) 1 . 42 s 10 . 4 m y v0 y t 1 gt 2 2 (10 . 0 m / s ) (1 . 42 s ) 1 ( 9 . 80 m / s ) (1 . 42 s ) 2 2 2 4 .3 m It is the location of the CM. ? x cm m 1 x1 m 2 x 2 m1 m 2 ? y cm m1 y1 m 2 y 2 m1 m 2 By Eq(7-12) x2 Mx cm m 1 x 1 Fig 7-11 m2 ( 9 . 6 kg ) (10 . 4 m ) ( 6 . 5 kg ) (13 . 6 m ) 3 . 1kg 3 .7 m y2 My Cm v0 cm m1 y1 m2 0 .9 m 0 m2 m1 7-4 Center of mass of solid objects 1) If an object has symmetry, the CM must lie at the geometrical symmetrical center of the object. Suppose the mass is uniformly distributed. cm cm 2) If the object has no symmetry, sometimes it is also easy to find its cm position: cm (a) (b) Sample problem 7-4 Fig 7-13 shows a circular metal plate of radius 2R from which a disk of radius R has been removed. Find the cm (x) of the plate. Fig 7-13 y R D C x x Solution: Due to the mirror symmetry about the x axis, the cm must lie along the x axis. If the hole is filled with a disk of the same material of radius R, the cm of composite disk is at the origin of the coordinate system. xc 0 x x ( ? m D xD m x xx mD mx CM for the big circular plate mD mx )xD R 2 (2 R ) R 2 2 ( R ) 1 3 R 3) If we encounter solid irregular objects, we can divide infinite small elements. And the sums of Eqs(7-12) transform into integrals: x z O x cm y cm z cm y 1 lim M 1 m 0 M 1 m 0 M lim lim m 0 In vector form are x m n y n m n z n n m n rcm 1 M 1 M 1 M 1 M xdm ydm (7-18) zdm r dm (7-19) Sample problem 7-5 Fig 7-14 A thin strip of material is bent into the shape of a semicircle of radius R. Find its center of mass. y y cm 0 dm M d x Solution: The strip has symmetry about the y axis. So x c 0 y cm R 1 M ydm sin d 0 1 M 2R R sin 0 M d 0 . 637 R ( the small element of mass dm subtends an angle d . The location of the dm is y R sin ) 7-5 Conservation of momentum in a system of particles 1) For a system containing N particles, the total momentum P is ( M m i ) N N N m n vn (7-21) P Pn m n v n M M v cm n 1 Here dP dt n 1 M n 1 d v cm dt M M a cm Fext (7-22) If the net external force acting on a system is zero, then d P 0 and so the dt total linear momentum P of the system remains constant. 2) If we view the system from the cm frame, the velocity v n ' of a particle in this frame is (7-24) v ' (v v ) n n cm Then in this cm frame, the total momentum is P' N n 1 m nvn ' N n 1 m nvn M v cm M v cm 0 N n 1 m n v cm (7-25) Sample problem 7-8 As Fig 7-17 shows a cannon whose mass M is 1300kg fire a 72kg ball in a horizontal direction with a speed vbc of 55m/s relative to the cannon. The cannon is mounted and can recoil (后退) freely. (a) what is the velocity vcE of the recoiling cannon with respected to the Earth? (b) what is the initial velocity vbE of the ball with respected to the Earth? vcE M m Fig 7-17 v bc Solution: Momentum in horizontal direction is conserved. Pix P fx 0 P f Mv cE mv bE 0 v bc v bE v cE vcE M m Fig 7-17 v bc