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Rigid Body Kinetics
All formula which are applicable to a system of particles will also be applicable to Rigid
Bodies. The additional constraint placed on the system of particles, which requires the distances
between particles be constant, simpliﬁes the problem substantially.
Recall the equation of motion for a system of
particles is
X
~
F
=
d
dt
~
L
=
d
dt
N
X
i
=1
m
i
~v
i
!
.
(1)
But for a rigid body, one additional constraint can be
placed on the velocities, i.e.,
i
=
G
+
~ω
×
~
r
i/G
=
G
+
×
~ρ
i
(2)
where
i
=
~
r
i/G
is the position vector of mass
m
i
with respect to the center of gravity
G
. Combining
equations (1) and (2), we can write
X
~
F
=
d
dt
N
X
i
=1
m
i
G
+
N
X
i
=1
m
i
×
i
!
.
(3)
Removing now constants from the summation over
index
i
,wehave
X
~
F
=
d
dt
G
N
X
i
=1
m
i
+
×
N
X
i
=1
m
i
i
!
.
(4)
x
y
z
~r
i
A
~
F
j
G
~
R
G
i
j
i
m
i
j
Deﬁne now the total mass,
M
=
N
X
i
=1
m
i
,
and recognize that
N
X
i
=1
m
i
i
=
~
0
by the deﬁnition of the center of gravity, the equation of motion for rigid bodies in translation can
be written simply as
X
~
F
=
d
dt
(
M~
v
g
)=
M~a
G
which is identical to that for a system of particles.
–7/1–
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View Full Document Moment and Angular Momentum of a Rigid Body
Recall for a system of particles, the alternate form of Newton’s 2nd Law, that of moment and
angular momentum, can be written about the origin
O
as
X
~
M
O
=
d
dt
~
H
O
,
(1)
where the total moment about point
O
for all external applied forces is deﬁned as
X
~
M
O
=
X
j
~
r
j
×
~
F
j
(2)
and the total angular momentum about point
O
for
N
mass particles is deﬁned as
~
H
O
=
N
X
i
=1
~
r
i
×
m
i
~v
i
.
(3)
For a rigid body, the additional constraint on the velocities is such that
i
=
A
+
~ω
×
~
r
i/A
,
(4)
in which
A
is a point on the rigid body where the velocity is known. Since Newton’s Second Law
for translations utilizes the velocity and acceleration of the center of gravity
G
, it is convenient to
simplify equation (3) by substituting point
G
as point
A
in equation (4), i.e.,
i
=
G
+
×
~
r
i/G
=
G
+
×
~ρ
i
,
(5)
in which
~
r
i/G
=
~
r
i

~
R
G
=
i
.
To obtain
~
r
i
for equation (3), rewrite the above equation as
~
r
i
=
~
R
G
+
i
(6)
and then substitute equations (5) and (6) into equation (3) to yield
~
H
O
=
N
X
i
=1
(
~
R
G
+
i
)
×
m
i
(
G
+
×
i
)
=
N
X
i
=1
~
R
G
×
m
i
G
+
N
X
i
=1
i
×
m
i
G
+
N
X
i
=1
~
R
G
×
m
i
(
×
i
)+
N
X
i
=1
i
×
m
i
(
×
i
)
=
~
R
G
×
N
X
i
=1
m
i
!
G
+
N
X
i
=1
m
i
i
!
×
G
+
~
R
G
×
×
N
X
i
=1
m
i
i
!
+
N
X
i
=1
m
i
(
i
×
×
i
)
.
–7/2–
Using again the deﬁnition of the center of gravity, i.e.,
∑
m
i
~ρ
i
=
~
0
, and let
M
be the total mass
∑
m
i
, we can write the angular momentum about the origin
O
as
~
H
O
=
~
R
G
×
M~
v
G
+
N
X
i
=1
m
i
(
i
×
~ω
×
i
)
.
(7)
The above expression is written in
abstract algebra
form, which implies it is applicable for any
coordinate system.
The Moment of Inertia Matrix of a Rigid Body
To make equation (7) easier to apply numerically, the second term on the righthandside,
N
X
i
=1
m
i
(
i
×
×
i
)
,
can be expressed as the matrix product
[
I
G
]
, thus separating completely the geometrical
information contained in
from the kinematic information contained in
. Otherwise, the entire
summation has to be recalculated every instant the angular velocity
of the body changes.
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This note was uploaded on 04/16/2008 for the course CE 325 taught by Professor Docwong during the Spring '08 term at USC.
 Spring '08
 DocWong

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