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In-Depth Information
Therefore the change in kinetic energy due to the contact impulse is
Δ
KE
=
2
m
1
2
−
2
m
1
2
.
u
new
1
2
+
2
m
2
u
new
2
u
old
1
2
+
2
m
2
u
old
2
Here and in the following, the superscript “2” indicates squaring. If you find
the scalar
J
that minimizes Δ
KE
(dissipating the most energy possible),
you will find it causes a perfectly inelastic collision with
u
new
1
n
=
u
new
2
n
.
Generalizing this to incompressible fluids, we begin with the kinetic
energy of a fluid, an integral version of the familiar
·
·
1
2
mv
2
:
KE
=
Ω
1
2
.
2
ρ
u
Here Ω is the fluid-filled domain. Note that we have included the density
ρ
inside the integral to be able to handle variable density fluids—this will
come in handy later on in the topic.
The change in energy can be measured as the difference in kinetic energy
and, if solid wall boundaries are present, the work exchanged between the
solid and the fluid. The first term, based on the intermediate velocity
u
and the pressure-updated final velocity
u
n
+1
,is
u
n
+1
2
2
.
1
1
2
ρ
−
2
ρ
u
Ω
Ω
The work exchanged between the fluid and the solid is equal to the inte-
gral of applied force dotted with the displacement (over the solid bound-
ary
S
). The force term, per unit area, is nothing more than pressure
times the normal direction,
pn
, using here the convention that the normal
points out from the fluid into the solid. The displacement can be esti-
mated as Δ
tu
solid
(note that this term then vanishes if the solid is not
moving!). Adding this in gives the total change in kinetic energy of the
system:
Δ
KE
=
Ω
2
+
S
1
u
n
+1
2
1
2
ρ
−
2
ρ
u
pn
·
Δ
tu
solid
.
Ω
Finally, we seek out the pressure, constrained to be zero on the free surface
part of the boundary
F
, that minimizes this (i.e., dissipates the maximum
possible energy). As is shown in Appendix B, such a pressure exactly solves
