�
�
J.
Peraire,
S.
Widnall
16.07
Dynamics
Fall
2008
Version
1.0
Lecture
L22
 2D
Rigid
Body
Dynamics:
Work
and
Energy
In
this
lecture,
we
will
revisit
the
principle
of
work
and
energy
introduced
in
lecture
L1113
for
particle
dynamics,
and
extend
it
to
2D
rigid
body
dynamics.
Kinetic
Energy
for
a
2D
Rigid
Body
We
start
by
recalling
the
kinetic
energy
expression
for
a
system
of
particles
derived
in
lecture
L11,
1
2
1
2
T
=
2
mv
G
+
2
m
i
r
˙
i
�
,
n
�
i
=1
where
n
is
the
total
number
of
particles,
m
i
denotes
the
mass
of
particle
i
,
and
r
�
i
is
the
position
vector
of
particle
i
with
respect
to
the
center
of
mass,
G
.
Also,
m
=
i
n
=1
m
i
is
the
total
mass
of
the
system,
and
v
G
is
the
velocity
of
the
center
of
mass.
The
above
expression
states
that
the
kinetic
energy
of
a
system
of
particles
equals
the
kinetic
energy
of
a
particle
of
mass
m
moving
with
the
velocity
of
the
center
of
mass,
plus
the
kinetic
energy
due
to
the
motion
of
the
particles
relative
to
the
center
of
mass,
G
.
For
a
2D
rigid
body,
the
velocity
of
all
particles
relative
to
the
center
of
mass
is
a
pure
rotation.
Thus,
we
can
write
r
˙
�
i
=
ω
×
r
�
i
.
Therefore,
we
have
n
n
n
�
1
�
1
�
1
2
m
i
r
˙
i
�
2
=
2
m
i
(
ω
×
r
�
i
)
·
(
ω
×
r
i
�
) =
2
m
i
r
i
�
2
ω
2
,
i
=1
i
=1
i
=1
where
we
have
used
the
fact
that
ω
and
r
�
i
are
perpendicular.
The
term
i
n
=1
m
i
r
i
�
2
is
easily
recognized
as
the
moment
of
inertia,
I
G
,
about
the
center
of
mass,
G
.
Therefore,
for
a
2D
rigid
body,
the
kinetic
energy
is
simply,
T
=
1
mv
2
+
1
I
G
ω
2
.
(1)
2
G
2
1
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When
the
body
is
rotating
about
a
fixed
point
O
,
we
can
write
I
2
O
=
I
G
+
mr
G
and
1
2
1
2
1
T
=
2
mv
G
+
2
(
I
O
−
mr
G
)
ω
2
=
2
I
O
ω
2
,
since
v
G
=
ωr
G
.
The
above
expression
is
also
applicable
in
the
more
general
case
when
there
is
no
fixed
point
in
the
motion,
provided
that
O
is
replaced
by
the
instantaneous
center
of
rotation.
Thus,
in
general,
1
T
=
I
C
ω
2
.
2
We
shall
see
that,
when
the
instantaneous
center
of
rotation
is
known,
the
use
of
the
above
expression
does
simplify
the
algebra
considerably.
Example
Rolling
Cylinder
Consider
the
cylinder
rolling
without
slipping
on
a
ﬂat
plane.
The
friction
between
the
plane
and
the
cylinder
insure
the
noslip
condition,
but
the
friction
force
does
no
work.
The
noslip
condition
requires
ω
=
−
V
G
/R
0
).
We
take
the
cylinder
to
have
its
center
of
mass
at
the
center
of
the
cylinder
but
we
allow
the
mass
distribution
to
be
nonunform
by
allowing
a
radius
of
gyration
k
G
.
For
a
uniform
cylinder
k
G
=
R
0
/
√
2
The
total
kinetic
energy,
from
equation
(1)
is
given
by
1
1
1
1
G
V
2
m
(
R
2
I
C
ω
2
T
=
mV
2
+
mk
2
G
/R
2
=
+
k
2
)
ω
2
=
(2)
2
G
2
0
2
0
G
2
where
the
parallel
axis
theorem
has
been
used
to
relate
the
moment
of
inertia
about
the
contact
point
C
to
the
moment
of
inertia
about
the
center
of
mass.
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 Fall '09
 widnall
 Dynamics, Kinetic Energy, Potential Energy

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