 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
ENGR 211
Principles of Engineering I
(Conservation Principles in Engineering Mechanics)
Conservation of Angular Momentum - part 1
 Review of Moments, Moment Vectors, M  r  F , Couple
 Introduction to Angular Momentum
 Conservation of Angular Momentum Equations
 Steady State and Rigid Body Statics
1
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
2
Notes on determining moment (or torque). The moment of a
force with magnitude F about some point is defined to be the
perpendicular distance from the reference point to the line of
action of the force. Consider the
B
following bar AB with force F:
d
60 o
The moment of the force F about
0
t
the point 0 is given by
8f
M0 = F d
where
A
y
d = 8 sin (60)
Thus
x
M0 = [F] [8 sin (60)] = 6.93 F F
Another way: resolve F into components perpendicular and
parallel to line AB. The perpendicular component is F sin (60).
Thus
M0 = [F sin (60)] [8] = 6.93 F.
Same result!!
 2000, W. E. Haisler
3
Conservation of Angular Momentum - part 1
Moment Vectors. A moment is a vector (just as a force is a
vector) and it has three components corresponding to the moment
about each axis, e.g., the x component of the moment vector is
the moment about the x axis. The force F produces a moment
about the origin equal to M  r  F . We can use double-headed
arrows to denote moment vectors as shown below.
_
F
y
y
My _
M
_
r
x
z
=
Mx
Mz
z
x
 2000, W. E. Haisler
4
Conservation of Angular Momentum - part 1
Or, we can use moment symbols as shown below to indicate the
direction the moment is about each axis (following the right hand
rule for positive moments):
_
F
y
y
My _
M
_
r
x
=
Mz
x
Mx
z
z
 2000, W. E. Haisler
5
Conservation of Angular Momentum - part 1
Moment of a Force about some point. Consider a force vector
F which is located by position vector r as shown below.
y
Fy _
F  Fx i  Fy j  Fz k
F
_
r
Fx
Fz
r  rx i  ry j  rz k
x
ry
rz
rx
z
Now we determine the moment of each force component Fx, Fy
and Fz about the origin using the perpendicular distances rx, ry
and rz. Assume the right-hand-rule for positive moments.
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Moment components due to Fx : M y  Fx rz , M z   Fx ry
Moment components due to Fy : M x   Fy rz , M z  Fy rx
Moment components due to Fz : M x  Fz ry , M y   Fz rx
Now add up the moments about each axis to find components:
M x  Fz ry  Fy rz , M y  ( Fz ry  Fx rz ) , M z  Fy rx  Fx ry
So the moment vector is:
M Mx i M y j Mz k
And its magnitude is given by: M  M x2  M y2  M z2
The above can be written in short hand notation using what is
defined as the vector cross product:
6
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
i
M  r  F  rx
Fx
ry
i
Fy
j
ry
Fy
rz
rx
j
Fz
Fx
7
k
rz
Fz
rx
rz
k
Fx
Fz
ry
Fy
 i (ry Fz  rz Fy )  j (rx Fz  rz Fx )  k (rx Fy  ry Fx )
We have just proved the definition that M  r  F ! ! !
For a planar force located in the x-y plane ( Fz  0 , rz  0 ), we see
that the force produces moment only about the z-axis (i.e., a
moment vector in the z direction); with similar results for planar
forces in the x-y or y-z plane.
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Do the problem considered previously with vector mechanics.
The force vector is given by F  Fxi  F y j  Fzk Fj and the
radius vector with the origin at "0" is given by
r  8cos(210)i 8cos(120) j  cos(90)k . The moment about 0 is
then given by
i
M  r  F  rx
0
Fx
j
ry
Fy
k
i
j
k
rz  8cos(210) 8cos(120) 0
0
F
0
Fz
 0 i  0 j  (F[8cos(210)]) k F[8(.866)] k  (6.93F ) k
The moment of the force F in the x-y plane is a vector in the z
direction (moment about the z-axis).
8
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
9
Vector Cross Product Definition. The direction of the angular
momentum vector and torque vector (or moment vector) follows
the standard right hand rule. The vector cross product
i
T r f  r
x
f
x
j
r
y
f
y
k
r
z
f
z
produces a vector T which is perpendicular to the plane
containing r and f .
For example, if the force is located in the x-y plane only, this
will produce a torque about the z-axis only, i.e., the torque vector
has only a z component (i.e., k component).
 2000, W. E. Haisler
10
Conservation of Angular Momentum - part 1
A pure couple (moment) can be represented by equal and
opposite forces separated by a distance d (note: zero net force).
F
d/2
M=Fd
d/2
A
O
F
h
O
=
A
h
In the left figure, the moment of F about point O is
M=F(d/2)+F(d/2)=Fd (counterclockwise). Consequently, we
could replace the left figure by a moment of FD placed at point O
(the right figure). Note: net vertical force is zero in both cases!!
Now calculate the moment about point A (using the left
figure): M=F(h+d/2)-F(h-d/2)=Fd. Consequently, for a pure
couple M, the moment is the same about point O, or point A; i.e.,
it is independent of the moment reference point! A pure couple
has no net forces accociated with it.
 2000, W. E. Haisler
11
Conservation of Angular Momentum - part 1
A force may be moved to another point by replacing it with
the same force plus an equivalent couple. Consider the system
below with a 40 kg mass at point B. The force due to this mass is
392 N (40 * 9.8). The 392 N force can be placed at B.
Or, alternately at D, by placing a force of 392 N at D
plus an equivalent moment (couple) of 588 N m (392 * 1.5) at D
as shown on the right figure.
C
C
3m
3m
588 N m
A
4m
D
1.5 m
B
A
4m
D
1.5 m
392 N
B
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
12
Introduction to Angular Momentum
The angular momentum ( l ) of a system is defined by the
cross product of a particle's position vector ( r ) and its linear
momentum vector ( p  mv ): l  r (mv) . Because angular
momentum is a conserved property, there is no generation or
consumption. Thus, the conservation statement becomes









Accumulation of AM
within system
during time period



























AM entering   AM leaving 





 system during  - system during 





time period   time period 




Accumulation of AM 
 AM in system 
 AM in system 










within system

at
end
of
at
beginning
of










 time period




during dime period 


 time period





 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Angular Momentum Possessed by Mass
 Angular Momentum may enter or leave the system with mass.
 Angular Momentum may also enter or leave the system at a
certain rate.
 When mass enters a system traveling a velocity ( v ) it adds
angular momentum to the system.
 The rate at which mass (m) enters the system is min
dm
( min  in ) and therefore the rate at which linear momentum
dt
enters the system is given by (mv)in and the rate at with
angular momentum enters the system is given by r  (mv )in .
Similarly, rate at which the angular momentum leaves the
system is r  (mv )out .
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 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
14
 The proper way to look at angular momentum rate is as follows





angular momentum    angular momentum   mass 




mass
time


  time 



If l  angular momentum , then the above can be written in


 l 
d
l
equation form as
  m m
dt  
In terms of linear momentum, we can write


l

d l    m  r(mv) m  (r v)m  r mv  r vm
m
dt  m 
 If there are multiple masses entering or leaving the system (say
n of them), then we simply add them up:
n 
 
r

vm


 
  
 i  in
i1
and
n 
 
r

vm


 
  
 i  out
i1
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
15
Angular Momentum in Transit: Forces
 Angular momentum enters the system as the result of forces
that act upon the mass in the system. By Newton's second law,
an external force f ext or torque (moment) Text acting on a
system provides angular momentum to the system.
 The torque produced by the external force f ext about some
reference point is equal to product of the perpendicular moment
arm "d" and the force, i.e. T  d f ext . Or, in vector form, the
torque is the cross product position vector r from the reference
point and f ext , i.e. T  r  f ext . These forces may act as surface
or contact forces on the system boundary (called tractions), or
they may act as body forces on the entire mass of the system
(for example, body forces due to gravity or electrostatic forces).
 2000, W. E. Haisler
16
Conservation of Angular Momentum - part 1
Schematic for Conservation of Angular Momentum
Resultant External
Torque Acting
on System
Surroundings
Angular
Momentum
Entering
System
System
Angular
Momentum
Surroundings
Angular
Momentum
Leaving
System
System
Boundary
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
 The external forces may be due to:
 pressure acting over the surface area of the system
 drag or frictional forces acting on the surface
 supporting structure which holds the system in place (which
cause reaction forces acting on the surface)
 gravitational and electrostatic body forces acting on the
system volume
17
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Conservation of Angular Momentum Equations
We can now state the rate equation for conservation of angular
momentum as:
d lsys
 (r mv) (r mv)out  (r  f )ext
in
dt
The angular momentum of the system may change with time
because of mass that enters and leaves, and because of external
torques caused by external forces that act on the system.
As mass enters the system, the amount of angular momentum
that enters the system is the [mass flow rate] times the [angular
momentum per unit mass (i.e., r v ) of the mass].
18
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
19
We must keep in mind that conservation of angular momentum is
a vector equation! Consequently, we can write in component
form. For a Cartesian coordinate system, we have
x-direction:
d (lx)sys
 [(rmv) x ]
 [(r f ) x ]ext
in / out
dt
d (l y)sys
 [(rmv) y ]
 [(r  f ) y ]ext
y-direction:
in / out
dt
d (lz)sys
z-direction:
 [(rmv) z ]
 [(r f ) z ]ext
in / out
dt
where ( l x ) sys  x component of ( r  mv ) , etc.
[(r  f ) x ]ext  x component of ( r  f ) ext , etc.
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Angular Momentum of the System
The angular momentum of the system may be written (for n
particles in the system) as
n
l sys    (r  mv )i 
sys
i 1
20
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
21
Integral (Finite Time Period) Equation
The rate form of the conservation of angular momentum
equation can be integrated over a finite time period to obtain


t
end

t
beg 
d lsys 
t
t
end
end  (r  mv )
dt

(
r

mv
)
dt

dt




in
out
dt
t
t
beg
beg


t
end
t
beg
 (r  fext )dt
The above equation can also be written for a finite time period
as
(lsys )
 (lsys)
 (rmv) (rmv)out  (r fext )t
in
end
beg
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
Steady State
The steady state condition for a system means that the system is
not changing with time and hence the accumulation within the
system is zero for any time period.
(lsys)
(l )
0
end sys beg
dlsys
0
or, for any instant of time:
dt
For a finite time period:
Through conservation of angular momentum, steady state
implies:
0  (rmv)
 (r f )ext
in / out
22
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
23
For rigid body statics (no mass enters or leaves the system and
the body is at steady state), angular momentum reduces to
0  (r f )ext
or
0  ( B r f )ext
The latter notation above is used to indicate that the torque can
be calculated about any point (say point B). We could also write:
0  Text
where Text  r  fext
Note: For many rigid body statics problems, we will employ
both the linear and angular momentum equations:
 fext  0
and
(r f )ext  0
 2000, W. E. Haisler
Conservation of Angular Momentum - part 1
24
Components of a Velocity Vector
In cases where a body is rotating with a constant angular velocity
 about the z axis, it is useful to resolve the vector into
components that are radial
z
_
x-y plane
(in the direction of the
v
_
position vector) and
vT
_
_
tangential (perpendicular to

v
_
R
the position vector) as
r
shown below. The velocity
y
can then be written as
v  v  v . v is the radial
x
R T
R
component of the velocity and v is the tangential component.
T
The tangential velocity component can be written in terms of the
angular velocity vector so that v  r .
T