1549380323-Statistical Mechanics Theory and Molecular Simulation

Rigid body motion 41

ω(lz=Iω), the angular velocity must also be a vector whose direction is along the
z-axis. Thus, we write the angular velocity vector for this problem asω= (0, 0 ,θ ̇) and
l=Iω. Physically, we see that the moment of inertia plays the role of “mass” in an-
gular motion; however, its units are mass×length^2. The form of the moment of inertia
indicates that the farther away from the axis of rotation an object is, the greater will
be its angular momentum, although its angular velocity is the same at all distances
from the axis of rotation.
It is interesting to calculate the velocity in the body-fixed frame. The components
of the velocityv= (vx,vy) = ( ̇x,y ̇) are given by eqn. (1.11.8). Note, however, that
these can also be expressed in terms of the angular velocity vector,ω. In particular,
the velocity vector is expressible as a cross product

v=r ̇=ω×r. (1.11.13)

Sinceω= (0, 0 ,θ ̇), the cross product has two nonvanishing components

vx=−ωzy=−d(sinθ)θ ̇

vy=ωzx=d(cosθ)θ, ̇ (1.11.14)

which are precisely the velocity components given by eqn. (1.11.8). Eqn. (1.11.14)
determines the velocity of the relative position vectorr. In the body-fixed frame, the
velocities of atoms 1 and 2 would be−r ̇/2 andr ̇/2, respectively. If we wish to determine
the velocity of, for example, atom 1 at positionr 1 in a space-fixed frame rather than
in the body-fixed frame, we need to add back the velocity of the body-fixed frame. To
do this, write the positionr 1 as

r 1 =R+

1

2

r. (1.11.15)

Thus, the total velocityv 1 =r ̇ 1 is

r ̇ 1 =R ̇+

1

2

r ̇. (1.11.16)

The first term is clearly the velocity of the body-fixed frame, while the second term is
the velocity ofr 1 relative to the body-fixed frame. Note, however, that if the motion
ofr 1 relative to the body-fixed frame is removed, the remaining component of the
velocity is just that due to the motion of the body-fixed frame, andwe may write
(
dr 1
dt

)

body

=

dR

dt

. (1.11.17)

Sincer ̇=ω×r, the total time derivative of the vectorr 1 becomes
(
dr 1
dt

)

space

=

(

dr 1

dt

)

body

+ω×

1

2

r

(

dr 1

dt

)

space

=

(

dr 1

dt

)

body

+ω×r 1 , (1.11.18)

where the first term in the second line is interpreted as the velocity due solely to
the motion of the body-fixed frame and the second term is the rateof change ofr 1