Thermodynamics and Chemistry

APPENDIX G FORCES, ENERGY, AND WORK

G.8 CENTER-OF-MASSLOCALFRAME 500

ri^0

ri

Rcm

b

bc

ut

Figure G.4 Position vectors in a lab frame and a center-of-mass frame. Open circle:

origin of lab frame; open triangle: center of mass; filled circle: particlei. The thin

lines represent the Cartesian axes of the two frames.

Thecenter of massof the system is a point whose position in the lab frame is defined
by

RcmdefD

P

imiri

m

(G.8.1)

wheremis the system mass:mD

P

imi. The position vector of particleiin the lab frame
is equal to the sum of the vectorRcmfrom the origin of the lab frame to the center of mass
and the vectorri^0 from the center of mass to the particle (see Fig.G.4):

riDRcmCri^0 (G.8.2)

We can use Eqs.G.8.1andG.8.2to derive several relations that will be needed presently.
Because the Cartesian axes of the lab frame and cm frame are parallel to one another (that
is, the cm frame does not rotate), we can add vectors or form scalar products using the
vector components measured in either frame. The time derivative of Eq.G.8.2is dri=dtD
dRcm=dtCdri^0 =dt, or
viDvcmCvi^0 (G.8.3)

where the vectorvcmgives the velocity of the center of mass in the lab frame. Substitution
from Eq.G.8.2into the sum

P

imirigives

P

imiri DmRcmC

P

imir
0
i, and a rear-
rangement of Eq.G.8.1gives

P

imiriDmRcm. Comparing these two equalities, we see
the sum

P

imir

0

imust be zero. Therefore the first and second derivatives of

P

imir
0
iwith
respect to time must also be zero:

X

i

miv^0 iD 0

X

i

mi

dvi^0

dt

D 0 (G.8.4)

From Eqs.G.1.2,G.6.1,G.6.2, andG.8.3we obtain

FiaccelDmi

d.v^0 ivi/

dt

Dmi

dvcm

dt

(G.8.5)

EquationG.8.5is valid only for a nonrotating cm frame.