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CMChap20

# CMChap20 - Chapter 20 Many Particle Systems Consider a...

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Unformatted text preview: Chapter 20 Many Particle Systems Consider a system of N point—particles, of masses mg), a = 1, . . . ,N in arbitrary motion. Let K be an inertial frame and K’ be an arbitrary moving frame (Fig. 20.1); Ta and 7'; be the position vectors of the a-th particle with respect to K and K’ respectively, and R be the position of the origin of K’ with respect to K. Then 7" :r'+R, (20.1) and, on differentiation of this equation with respect to the time t, the velocities of the particles are given by va = 011+ V, (20-2) Where V E dR/dt. The total linear momentum P of the system with respect to K is given by P = 27nan : 2mm; + V): Zmav; + MV = P’ + Mv, (20.3) 01 a ('1' where M E Ea ma is the total mass of the system and P’ E Email; 01 is the total linear momentum of the system with respect to K’. Suppose R is chosen so that (204) MR = Z mama . (20.5) Then dR d P—a; (gmara> 93 (20.6) / R A 94 CHAPTER 20. MANY PARTICLE SYSTEMS It follows from (20.3) that, with respect to this choice of K’, P’ = 0 . (20.7) A reference frame K’ with respect to which the total linear momentum P' vanishes is called a center of mass frame. The vector R satisfying (20.5) speciﬁes a point called the center of mass of the system of particles, and R itself is called the center of mass position vector. Consider the total energy E of the system: 1 q E = 5 gmavﬂ + U, (20.8) where the ﬁrst term on the RHS is the total kinetic energy and U is the potential energy of the system (due to both external forces and internal interaction forces between the particles). Using (20.2) we have 1 E=§Zma(v;-vﬁy+2viv’a+vg)+U 1 a (20.9) =§MV2+V-P’+E', where El Ill 1 5 Zmav'a "a; + U (2010) is the total energy of the system with respect to K’. If we choose K’ to be a center of mass frame, then (20.7) holds and we have the following transformation rule for the total energy between an inertial frame and a center of mass frame for a system of particles: 1 . E = E’ + —2-MV2 . (2011) Next we consider the total angular momentum L : Era >< pa = Zra x maria. (20.12) C! D! From (20.2) we have a a (20.13) =§_:ma(rﬁ1 +R) x vﬁx+MR>< V a =Zmarngg+Rmeavg+MRxV. (X a 95 If K’ is a center of mass frame, then 20, mav’a = P' = O, and L:L'+R><P, (20.14) where (20.15) L’ E Zmar’a x 1):, CM is the total angular momentum in the center of mass (CM) frame K’. Finally we consider the net torque 1'. Let F516) and F2?) be the external and internal force on the a-th particle, respectively. Then ’7' = Zn. >< (F55) + pg“). (20.16) We will show that the internal forces do not contribute to the net torque, pro- vided Fag (the force on ma by mg), for arbitrary a and B, is along the direction of 7'0z — 7‘3, in other words, along the straight line joining ma and mg (see Fig. 20.2). Wehave ZraxFS)=Zrax zFag: Zn,wa “ a (375“ a?” (20.17) =Z(TQXFQg+TBXFga)= Z('r‘a—1‘B)XFQB, a<ﬁ oz<ﬁ where the last equality follows from Newton’s third law, which implies that F130) 2 —Fag. This quantity obviously vanishes if Fa); is along the direction of Ta - r5. Thus 7' : Zn, x F55) . (20.18) oz Now L = Zn, x mava , (20.19) a which implies that E 2 Ta >< magi“ = n, x 17(5) + F“) = Ta x 12(6) . 20.20) dt C! a Oé G! or a It follows from (20.18) that __ dL _ E _ Note that this equation, which applies to a system of particles, looks exactly the same as (4.5), which applies to a single particle. Eq.(20.21) immediately 7‘ (20.21) moi .5 “'3‘ «A .5 rd l; ‘~ _..\ r}. m 1“ (9 Fri}. :20 2 96 CHAPTER 20. MANY PARTICLE SYSTEMS implies the principle of conservation of angular momentum (for a system of particles): The total angular momentum for a system of particles is conserved if the net torque due to external forces vanish. The net torque can also be written in terms of the position vectors r’a as follows. T:Z(R+r;)XFS)=R><ZFS)+ZT;XFS). (20.22) a Since, by Newton’s third law, all the internal forces must sum to zero, we have dP —— = F“) . 2 . dt Em: C, ( 023) Thus dP d (e) = __ : __ '2/ RXZFQ Rx dt dt(RxP), (20 4) where the last equality follows from the fact that dR P __ = 2 _ _ .2 dt V Mr (20 5) Eq.(20.24) allows us to write (20.22) as _ d I (e) T_ElZ(RxP)+za:raxFa . (20.26) On the other hand, (20.14) implies dL d dL’ — = — . 2 .2 dt dt(RxP)+ dt (0 7) Comparing with (20.26) and recalling (20.21) we obtain I dc; = Zr; >< FE) , (20.28) CY which is similar in form to (20.21) with 7‘ given by (20.18). This is interesting because the CM (center of mass) frame is in general not an inertial frame, and yet the equation of motion (20.28) retains the same form as that with respect to an inertial frame [Eq.(20.21)]. ...
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