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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 ath 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, (202) 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+VP’+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’ + —2MV2 . (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 ath 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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This note was uploaded on 06/10/2008 for the course PHY 322 taught by Professor Lam during the Spring '08 term at Cal Poly Pomona.
 Spring '08
 LAM
 mechanics

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