 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
ENGR 211
Principles of Engineering I
(Conservation Principles in Engineering Mechanics)
Conservation of Angular Momentum
for Planar Rigid Body
 Angular Momentum of a Translating and Rotating Rigid Body
 Rate of Change of Rigid Body Angular Momentum
 Rigid Body Angular Momentum Conservation
1
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Kinetics of Planar Rigid Body Motion
Angular Momentum of a Translating and Rotating Rigid Body
For rigid bodies, we can consider motion of the rigid body as one
of three cases: pure translation, pure rotation and translation with
rotation.
y
x
z
rG
Pure translation
Pure rotation
Translation and rotation
It is convenient to think of the position of any point on the rigid
body by its position relative to the center of mass. Consider the
following vectors in order to define any point of the body in
terms of the position vector for the center of mass (G):
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3
Kinetics of Planar Rigid Body Motion

y
ua
G
rn
x
G
r
z
system
ra
G
r
G
r
O
G
r = rG + Gr
axis through the
center of mass
Read the notation in the following way:
r = position vector for any particle in the system
rG = position vector for center of mass (G)
Gr = position vector of particle relative to center of mass (G)
ra = position vector relative to G in direction of ua
G
rn = position vector relative to G in direction normal to ua
G
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Kinetics of Planar Rigid Body Motion
Review center of mass:
Consider a mass ( msys ). The center of mass (or center of
gravity) is located at rG and is defined by:
dm
_
r
_
rG
_
dm g
x
z
differential element
of mass
r  xi  y j  zk
sys
msys
y
y
rdm  rG msys
_
msys g
x
z
center of gravity
rG  xG i  yG j  zG k
 rG 
sys
rdm
msys
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Kinetics of Planar Rigid Body Motion
The differential mass can be written in terms of density and
volume as dm   dV   dxdydz so that the location of the
center of gravity becomes
rG 
sys
rdm
msys


r  dV
msys
For a system which is composed of discrete masses (or mass
lumps):
msys   dm  mi   Vi
sys
y
_
ri
_
mi g
x
z
sys rdm   ri mi   ri Vi
 ri mi  ri Vi
rG 

msys
msys
 2000, W. E. Haisler
y
rG  xG i  yG j  zG k
_
rG
yG
x
zG
z
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Kinetics of Planar Rigid Body Motion
xG
Thus we can write the
individual components
of the center of mass:
 ri mi  ri Vi


msys
msys
and r  xi  y j  zk
xG

yG


zG  
xdm  xi mi

m
m
ydm  yi mi

m
m
zdm  zi mi

m
m
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Kinetics of Planar Rigid Body Motion
7
General Approach
We now proceed to rewrite the angular momentum of the system
in terms of these vectors. Note that the position of any particle is
given by r  rG  G r . We start with the usual definition for the
angular momentum and substitute the position vector:
l   (r  v )dm   [(rG  G r )  v ]dm
  (rG  v )dm   ( G r  v )dm
1st term: Note that the position of the center of mass is given by
rG and is a constant for a given body (but it may vary with time)
and it may be placed outside the integral. Consequently the 1st
term can be rewritten as
vdm

 (rG  v )dm  rG  ( vdm)  mrG  ( m )  mrG  vG (a)
vdm

 velocity of center of mass
where vG 
m
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Kinetics of Planar Rigid Body Motion
2nd term: Now we rewrite the second term:

( G r  v )dm .
Recall the position vector for a particle in the body: r  rG  G r .
Differentiating the position vector of a point with respect to time
gives the velocity of that particle as: v  vG  G v . Thus, the 2nd
term can be written as
 ( G r  v )dm   G r  (vG  G v )dm
  ( G r  vG )dm   ( G r  G v )dm
 (  G rdm)  vG   ( G r  G v )dm
The velocity of the center of mass vG is independent of position
within the body so it can be taken outside the integral. Recalling
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Kinetics of Planar Rigid Body Motion
that the center of mass definition is rG 
sys
rdm
msys
, we note that
nd
rdm

0
.
Consequently,
the
2
term becomes
G

( G r  v )dm   ( G r  G v )dm .
We now write a particle's velocity relative to the center of mass
in terms of radial and tangential components: G v  G vR  G vT
and the tangential component in terms of angular velocity
nd
.
Now
the
2
term becomes
v



r
G T
G
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Kinetics of Planar Rigid Body Motion
 ( G r  v )dm   [ G r  ( G vR  G vT )]dm
  [ G r  ( G vR    G r )]dm
  [ G r  G vR ]dm   [ G r  (  G r )]dm
  [ G r  (  G r )]dm
(b)
Now we may combine the results for the 1st and 2nd terms of the
angular momentum (equations a & b) to obtain:
l   (r  v )dm  m(rG  vG )   [ G r  (  G r )]dm
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Kinetics of Planar Rigid Body Motion
Now we turn to the special case of plane motion about a
constant-direction axis having direction ua . ua is perpendicular
to the plane containing rG and vG . An example of such motion
is a wheel traveling on a roadway (in an x-y plane):
_
ua is perpendicular to x-y
plane and through the
center of the wheel
y


rG
x
As before we wish to determine the angular momentum about the
axis of rotation (which is through the center of mass). We take
the dot product of l and ua to obtain la
la  l  ua  mrG  vG   [ G r  (  G r )]dm  ua

As before the triple cross product can be written as

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Kinetics of Planar Rigid Body Motion
[ G r  (  G r )]  ua  [( G r  G r )  ( G r   ) G r )]  ua
 [( G r  G r ) ua  ( G r   ua ) G r )]  ua
 G r 2  G ra2  G rn2
and the integral term becomes

[ G r  (  G r )]dm]  ua  
2 dm  
r
G n

2dm   I
r
G n
G a
I 
G a 
2dm is the angular moment of inertia about
r
G n
the axis of rotation passing through the center of mass. Thus la
becomes
la  l  ua  [mrG  vG ]  ua  G I a
where
 m( rG  vG ) a  G I a
where the first term is the "a-component" of m(rG  vG ).
 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
In summary, for the special case of plane motion about a
constant-direction axis having direction ua , we obtain:
la  l  ua  [mrG  vG ]  ua  G I a
13
 m( rG  vG ) a  G I a
Consequently, the "a-component" of angular momentum of a
rigid body undergoing combined translation and rotation about a
constant direction axis can be determined from:
1) the angular momentum due to translation of the center of
mass (in terms of its position and velocity: rG and vG )
2) plus the angular momentum due to rotation of the body about
the center of mass (in terms of its angular velocity  and angular
moment of inertia G I a ).
Important: rG and vG are always in the same plane, which is
perpendicular to the axis of rotation of the body.
 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
14
Rate of Change of Rigid Body Angular Momentum
We again consider planar motion wherein the rigid body has
motion and rotation in a plane. For planar motion with rG and
vG in the same plane, this implies that the axis of rotation of the
body ( ua ) is perpendicular to this plane and passes through the
center of mass. We start with the angular momentum equation
developed previously la  m(rG  vG )a  G I a and differentiate
with respect to time:
 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
d l a d [m(rG  vG ) a  G I a ]

dt
dt
dvG 

d ( I )
 m vG  vG  rG 

G a

dt
dt

a
dvG 

d ( I )
 m  rG 

G a

dt
dt

a
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 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
Conservation of Angular Momentum for Planar Rigid Body
We now have all the components of the conservation of angular
momentum equation. For planar motion, we have
dvG 
d la

d ( I ) 
 m  rG 

(r  f )ext , a

a
G

dt 
dt
dt

a
We can simplify the above equation by introducing the position
vector r  rG  G r on the right side of the equation to give
dvG 
d la

d ( I )
 m  rG 

G a

dt
dt
dt

a
  ( rG  f )ext ,a   ( G r  f )ext ,a
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Kinetics of Planar Rigid Body Motion
From conservation of linear momentum we can write
dvG
m(
)a   ( f )ext , a . Take the cross-product with rG so
dt
dvG 

that m  rG 
  (rG  f )ext , a . Consequently, the

dt 

a
underlined terms above cancel and conservation of angular
momentum equation becomes:
d ( I )  ( r  f )
G
ext ,a
dt G a
The moment of inertia term is independent of time, so it can be
taken outside of the derivative. We define the angular
acceleration  to be   d and the conservation of angular
dt
momentum equation becomes
 G I a    ( G r  f )ext ,a
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 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
We note that for planar motion of a rigid body, the conservation
of angular momentum does not depend upon the point in space
chosen as the origin of the inertial coordinate system.
It depends only upon
 the forces acting on the body and their points of application
measured with respect to the center of mass,
 the moment of inertia about the axis of rotation measured with
respect to the center of mass, and
 the angular acceleration of the body about a point which is
fixed relative to all points in the body.
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Kinetics of Planar Rigid Body Motion
We will find it useful to define Conservation of Angular
Momentum with Respect to Different Reference Points (for
planar motion about constant axis of rotation):
 Origin of Inertial Coordinate System
 Center of Mass
 Arbitrary Point
 Arbitrary Point on Body
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Kinetics of Planar Rigid Body Motion
20
 Torques with respect to Origin of Inertial Coordinate
System
We previously obtained the conservation of angular momentum
with respect to the origin of an inertial coordinate system as:
dvG 
d la

d ( I ) 
 m  rG 

(r  f )ext , a

a
G

dt 
dt
dt

a
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Kinetics of Planar Rigid Body Motion
 Torques with respect to Center of Mass

ua
system
any particle
y
G
r
r
G
r
O
z
G
x
r = rG +
axis through the
center of mass
r
G
We previously obtained the conservation of angular momentum
with respect to the center of mass as:
d ( I ) 
( G r  f )ext , a

a
G
dt
or
G I a   ( G r  f )ext , a
where the the angular acceleration  is   d and   d .
dt
dt
 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
For a finite time period tbeg to tend, we can write
tend
( ( G r  f )ext , a )dt
G I a (end  beg )   t
beg
22
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23
Kinetics of Planar Rigid Body Motion
 Torques with respect to an Arbitrary Point (B)
It is sometimes convenient to determine external torques with
respect to a point other than the center of mass, particularly if
that point involves an unknown force. The point may be located
at any point (say B) in the inertial coordinate system.

any particle
ua
r
B
r
r = rG + Gr
G
y
G
r
rG
r
O
z
x
r
B G
B
B
axis through the
center of mass
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Kinetics of Planar Rigid Body Motion
r  position of particle
rG  position of center of mass (G)
Gr
 position of particle with respect to G
Br
 position of particle with respect to B
B rG
 position of center of mass (G) with respect to B
Note from the geometry that we can write:
r  rG  G r
or
Gr
r  rB  B r
or
B
 r  rG
r  r  rB
rG  rB  B rG
 B rG  G r
or G r  B r  B rG
We now substitute the last vector equation into the conservation
of angular momentum d ( G I a )   ( G r  f )ext , a to obtain
dt
Br
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Kinetics of Planar Rigid Body Motion
25
d ( I )  ( r  f )
 [( B r  B rG )  f )ext ,a

a
G
ext
,
a
G
dt
  ( B r  f )ext ,a   ( B rG  f )ext ,a
dvG
)   ( f )ext
From COLM: m(
dt
Cross COLM with B rG to obtain:
dvG
m( B rG 
)   ( B rG  f )ext
dt
Use the last result to replace the torque term in COAM to obtain
dvG
d ( I ) 
( B r  f )ext , a  m( B rG 
)a

a
G
dt
dt
dvG
d
( G I a )  m( B rG 
)a   ( B r  f )ext , a
or
dt
dt
dvG
 aG  aB  B aG where we note that: rG  rB  B rG
Now
dt
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Kinetics of Planar Rigid Body Motion
26
Now substitute the last result into COAM to obtain
d ( I  )  m[ r  (a  a )] 
( B r  f )ext , a

a
B G
G
B
B
G
a
dt
or
d [ I   m( r  v ) ]  m( r  dvB )  ( r  f )
B G B G a
B G
ext ,a
dt G a
dt a  B
Note that the torque due to a force located at point B is equal to
zero. This is useful in the case when an unknown force exists on
a body. We simply choose to determine torques about the point
B where this unknown force is located.
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Kinetics of Planar Rigid Body Motion
 Torques with respect to Point B Fixed on the Body

ua
d
ra
G
r
G
y
r
B
B
axis of rotation
B G
rG
O
x
d
z
Consider a point B fixed on the rigid body. Then we can write
G vB  G rB   (B relative to G) and also
B vG   B rG      B rG (G relative to B).
Substituting above into COAM about point B gives
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Kinetics of Planar Rigid Body Motion
28
d [ I   m( r    r ) ]  m( r  dvB )  ( r  f )
B G
B G a
B G
ext ,a
dt G a
dt a  B
Now use the triple cross product relation:
B rG    B rG  ( B rG  G r )  ( B rG   ) B rG
For rotation about about ua ,   ua so the above becomes
B rG    B rG  ( B rG  B rG )ua  ( B rG  ua ) B rG
To obtain the "a" component of AM, dot the above with ua to
obtain
( B r    B r )  [( B r  B r ) u  ( B r   u ) B r ]  u
G
G a
G
G
a
G
a G
a
 r 2  ra2   d 2
Thus the COAM equation becomes
d [( I  md 2 ) ]  m( r  dvB ) 
( B r  f )ext , a

G
a
B G
a
dt
dt
 2000, W. E. Haisler
or,
Kinetics of Planar Rigid Body Motion
29
d [ I  ]  m( r  dvB ) 
( B r  f )ext , a

B
a
B G
a
dt
dt
The term G I a  md 2 is defined to be B I a (parallel axis theorem)
which is the moment of inertia for the body relative to the axis
through B:
2
I

I

md
B a G a
To prove this, we first note that the moment of intertia about the
center of gravity is given by G I a   G rn2dm. Recall that rn is
normal to ua . We wish to obtain the moment of inertia about a
parallel axis through point B (i.e., about a parallel axis that is a
distance d from the center of gravity). This is given by
 2000, W. E. Haisler
I 
B a 
Kinetics of Planar Rigid Body Motion
2dm . We note that r  r  d so that I
r
B n
G n
B n
B a
becomes
I 
B a 
2dm  ( r  d )2 dm
r
B n
 Gn
  ( G rn2 ) dm   2( G rn )( d ) dm   d 2dm
The first integral in the above equation is the moment of inertia
about the center of gravity G I a   G rn2dm. For the middle
integral, the constant 2d may be factored outside the integral so
that it becomes 2d  ( G rn )dm  0 . This integral is zero since
the position vector is relative to the center of mass. Thus we
obtain
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 2000, W. E. Haisler
Kinetics of Planar Rigid Body Motion
2
I

I

md
B a G a
This is called the parallel axis theorem or the transfer theorem.
It allows one to transfer the moment of inertia about an axis
through the center of gravity to a parallel axis that is a distance
"d" from the center of gravity.
31
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Kinetics of Planar Rigid Body Motion
32
A wheel with a diameter of 0.66 m and mass of 3.65 kg rolls
down the incline as shown below (without slipping).
y
5
12
x
a) At a given instant of time the wheel (contact point) is 5 m
from the origin and the angular velocity of the wheel is 40
rad/sec. Determine the angular momentum of the wheel.
b) If the wheel starts at rest from an upper portion of the incline,
what is its angular velocity and the velocity of the center of
mass after 20 seconds? What distance will it have traveled
down the slope?