2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 1
Conservation of Angular Momentum
(chapter 4)
Angular Momentum is defined as the moment of the Linear
Momentum about some spatial reference point. Thus, it is
the vector cross product with the linear momentum.
angular momentum = r linear momentum
Like linear momentum, angular momentum is conserved.
Thus, like conservation of linear momentum, we have the
same spatial and time derivatives; except they operate on the
angular momentum. Recall conservation of linear
momentum for a differential system as shown below:
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 2
v v v v v v v g t(i) t( j) t(k )
t x x
y y z z x y z
z
body
r
z
y
x
x y
Define r xi yj zk to be the position (or radius)
vector of the mass center about which we define the moment
of all forces acting on the control volume. This point is
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 3
assumed to be a fixed reference point in time, i.e., not
moving.
To obtain the conservation of angular momentum, we
determine the moment of the linear momentum vector about
the reference point (origin of coordinate system), i.e., we
take r (linear momentum vector).
Conservation of Angular Momentum becomes
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 4
(r v )
t
r v x v r r v z v
v y v
x y z
(r t ) ( r t ) (r t )
(i) ( j) (k )
r g
x y z
We now use the calculus "product rule" on the
accumulation, mass flow and traction terms to obtain
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 5
r v r ( v )
t t
r r r
x vxv y v y v z vz v
vxv
v yv
vz v
r g
r r r
x y z
t t t
r t r t r t r (i) ( j ) (k )
x (i) y ( j ) z (k ) x y z
The underlined terms are r (conservation of linear
momentum equation) and therefore zero. COAM becomes
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 6
r v r v v r v v r v v
x x y y z z
t
r t r t r t 0
x (i) y ( j ) z (k )
If the radius vector (point about which we are calculating
r
angular momentum is fixed in space), then 0.
t
Conservation of angular momentum becomes:
r r r
vxv v y v vz v
x y z
r t r t r t 0
x (i) y ( j ) z (k )
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 7
Proof of stress tensor symmetry
r r r
If r xi yj zk , then
x
i , j,
z
k.
y
Substitute above into the angular momentum equation:
i v x v j v y v k v z v i t(i) j t( j ) k t(k ) 0
Note that i 1i 0 j 0k (a unit vector in the x
direction). Doing each of the cross products in the bracket
term will allow a simplification. For the first term in
brackets:
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 8
i j k
i vxv vx (i v ) vx 1 0 0
vx vy vz
0 0 1 0
1 0
vx i j k vx jv z kv y
vy vz vx vz vx vy
Expanding the other two cross product terms in the brackets,
the […] becomes:
(v v j v v k ) (v v k v v i ) (v v j v v i ) 0
x z x y y x y z z x z y
r r r
The term v x v v y v v z v is thus zero!!
x y z
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 9
Angular momentum reduces only to the traction terms
r t r t r t 0
x (i) y ( j ) z (k )
and can be written in terms of Cauchy stresses to become
i (i ) j ( j ) k ( k ) 0
i (i xx j xy k xz )
j (i yx j yy k yz ) k (i zx j zy k zz )
k xy j xz k yx i yz j zx i zy
k ( xy yx ) j ( xz zx ) i ( yz zy ) 0
2001, W. E. Haisler Chapter 4: Conservation of Angular Momentum 10
Since the unit vectors are not zero, each term in (…..) = 0
and we conclude the following:
xy yx , xz zx , yz zy
Hence, the Cauchy stress tensor is symmetric, and has only 6
independent components.