This Appendix provides the full details for deriving the full fluid moment equation (Eq.
11) beginning from Eq. (
4) in Sect.
2 of the main text:
$$\begin{aligned}&\iiint _{CV}\frac{\partial }{\partial t} (\mathbf{r }\times \mathbf{u}) \rho \mathrm{d}V + \iint _{\partial V} (\mathbf{r }\times \mathbf{u})(\mathbf{u }\cdot \mathbf{n}) \rho \mathrm{d}S\\&\quad = -\iint _{\partial V} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _{\partial V} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \\&\qquad + \, \iiint _{CV}{} \mathbf{r}\times \mathbf{g}\rho \mathrm{d}V . \end{aligned}$$
For completion, details included in Sect.
2 are repeated below. We begin by assuming a constant density flow to bring the fluid density
\(\rho\) out of the integrals:
$$\begin{aligned}&\rho \iiint _{CV}\frac{\partial }{\partial t} (\mathbf{r} \times \mathbf{u}) \mathrm{d}V + \rho \iint _{\partial V} (\mathbf{r} \times \mathbf{u})(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S \nonumber \\&\quad = -\iint _{\partial V} \mathbf{r}\times p \mathbf{n} \mathrm{d}S \nonumber \\&\qquad + \, \iint _{\partial V} \mathbf{r} \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S + \rho \iiint _{CV}{} \mathbf{r}\times \mathbf{g}\mathrm{d}V. \end{aligned}$$
(29)
We next simplify this expression by introducing the position of the control volume’s center of mass
C relative to
O,
\(\mathbf{r}^{C/O}\) (Fig.
1a). We note that while a spatial control volume does not have a center of mass, we can treat CV as a material volume by defining the control volume velocity to be equal to the material velocity at all times (Sonin
2001). As we discuss in Sect.
3,
\(\mathbf{r}^{C/O}\) depends on the position of the body (or more specifically, where the volume displaced by the body is) within the CV and can, therefore, vary in time. The position vector from the origin
O to a mass element
\(\mathrm{d}m\) located at point
P can be expressed in terms of
\(\mathbf{r}^{C/O}\) as
\(\mathbf{r } = \mathbf{r }^{P/C} + \mathbf{r }^{C/O}\) (where
\(\mathbf{r }^{P/C}\) is the position vector from
C to
\(\mathrm{d}m\)). By the definition of a center of mass,
\(\rho \iiint _{CV}\mathbf{r}^{P/C}\mathrm{d}V=\iiint _{CV}{} \mathbf{r}^{P/C}\mathrm{d}m=0\). We can thus rewrite the final gravity term as
\(\rho \iiint _{CV}{} \mathbf{r}\times \mathbf{g}\mathrm{d}V = \rho \iiint _{CV}{(\mathbf{r}^{P/C}+ \mathbf{r}^{C/O})\times \mathbf g }\mathrm{d}V = \rho \iiint _{CV}{\mathbf{r}^{C/O}\times \mathbf g }\mathrm{d}V = \mathbf{r}^{C/O}\times \mathbf{g} \rho V\), where
V is the fluid volume in the CV. Equation (
29) then becomes:
$$\begin{aligned}&\rho \iiint _{CV}\frac{\partial }{\partial t} (\mathbf{r }\times \mathbf{u}) \mathrm{d}V + \rho \iint _{\partial V} (\mathbf{r }\times \mathbf{u})(\mathbf{u }\cdot \mathbf{n}) \mathrm{d}S \nonumber \\&\quad = -\iint _{\partial V} \mathbf{r}\times p \mathbf{n} \mathrm{d}S \nonumber \\&\qquad + \, \iint _{\partial V} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S + \mathbf{r}^{C/O}\times \mathbf{g} \rho V. \end{aligned}$$
(30)
We first reformulate the left-hand side of Eq. (
30) into terms that can be more easily evaluated, beginning with the unsteady volume integral. The cross product of a vector with itself is zero, so we start by writing the unsteady term as:
$$\begin{aligned} \rho \iiint _{CV}\frac{\partial }{\partial t} (\mathbf{r }\times \mathbf{u}) \mathrm{d}V&= \rho \iiint _{CV}\frac{\partial \mathbf{r }}{\partial t} \times \mathbf{u} + \mathbf{r}\times \frac{\partial \mathbf u }{\partial t} \mathrm{d}V \\&= \rho \iiint _{CV}{} \mathbf{u} \times \mathbf{u} + \mathbf{r}\times \frac{\partial \mathbf u }{\partial t} \mathrm{d}V \\&=\rho \iiint _{CV}{} \mathbf{r}\times \frac{\partial \mathbf u }{\partial t} \mathrm{d}V. \end{aligned}$$
We then expand the integral by leveraging the position and velocity of the control volume’s center of mass,
C. Just as the position vector of a fluid element at
\(\textit{P}\) could be expressed as
\(\mathbf{r } = \mathbf{r }^{P/C} + \mathbf{r }^{C/O}\), the velocity of the fluid element can be written as
\(\mathbf u = \mathbf u ^{P/C} + \mathbf u ^{C/O}\), where
\(\mathbf u ^{P/C}\) is the fluid element’s velocity relative to
C, and
\(\mathbf u ^{C/O}\) is the velocity of
C relative to
O:
$$\begin{aligned}&\rho \iiint _{CV}{} \mathbf{r}\times \frac{\partial \mathbf u }{\partial t} \mathrm{d}V \\&\quad = \rho \iiint _{CV} \left( \mathbf{r}^{P/C} + \mathbf{r}^{C/O}\right) \times \frac{\partial }{\partial t}\left( \mathbf u ^{P/C}+\mathbf u ^{C/O}\right) \mathrm{d}V \\&\quad = \rho \iiint _{CV}{} \mathbf{r}^{P/C} \times \frac{\partial \mathbf u ^{P/C}}{\partial t} + \mathbf{r}^{P/C} \\&\qquad \times \, \frac{\partial \mathbf u ^{C/O}}{\partial t} + \mathbf{r}^{C/O} \times \left( \frac{\partial \mathbf u ^{P/C}}{\partial t}+\frac{\partial \mathbf u ^{C/O}}{\partial t} \right) \mathrm{d}V\\&\quad = \rho \iiint _{CV}{} \mathbf{r}^{P/C} \times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \rho \iiint _{CV} \mathbf{r}^{P/C}\\&\qquad \times \, \frac{\partial \mathbf u ^{C/O}}{\partial t} \mathrm{d}V + \rho \iiint _{CV} \mathbf{r}^{C/O} \times \frac{\partial \mathbf u }{\partial t} \mathrm{d}V, \end{aligned}$$
and bringing
\(\frac{\partial \mathbf u ^{C/O}}{\partial t}\) and
\(\mathbf{r }^{C/O}\) out of the integrals because they are the same for all fluid elements in the CV (they are specific only to
C),
$$\begin{aligned}&= \rho \iiint _{CV}{} \mathbf{r}^{P/C} \times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \rho \iiint _{CV} \mathbf{r}^{P/C}\mathrm{d}V\\&\quad \times \, \frac{\partial \mathbf u ^{C/O}}{\partial t} + \mathbf{r }^{C/O} \times \rho \iiint _{CV} \frac{\partial \mathbf u }{\partial t} \mathrm{d}V. \end{aligned}$$
We can now again take advantage of the definition of center of mass (
\(\rho \iiint _{CV}\mathbf{r }^{P/C}\mathrm{d}V=0\)) to eliminate the second term. We can also rewrite the last partial time derivative in terms of a material time derivative and surface integral (see Eq.
1 from Sect.
2 with
\(\mathrm{CM} = \rho CV\)), which will enable us to combine it with the convective term from Eq. (
30):
$$\begin{aligned}&\rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \, 0 \, + \, \mathbf{r }^{C/O} \times \rho \iiint _{CV} \frac{\partial \mathbf u }{\partial t} \mathrm{d}V \nonumber \\&\quad = \rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V \nonumber \\&\qquad + \, \mathbf{r }^{C/O} \times \rho \left[ \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V - \iint _{\partial V} \mathbf{u (u}\cdot \mathbf{n}) \mathrm{d}S \right] \nonumber \\&\quad = \rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V \nonumber \\&\qquad + \, \mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V - \, \mathbf{r }^{C/O} \times \rho \iint _{\partial V} \mathbf{u (u}\cdot \mathbf{n}) \mathrm{d}S\nonumber \\&\quad = \rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \, \mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V \nonumber \\&\qquad - \, \rho \iint _{\partial V} (\mathbf{r }^{C/O} \times \mathbf u ) (\mathbf u \cdot \mathbf n ) \mathrm{d}S. \end{aligned}$$
(31)
We can now add in the convective term to complete our new expression for the left side of Eq. (
30):
$$\begin{aligned}&\rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \mathbf{r }^{C/O} \nonumber \\&\qquad \times \, \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V - \rho \iint _{\partial V} (\mathbf{r }^{C/O} \times \mathbf u ) (\mathbf u \cdot \mathbf n ) \mathrm{d}S\nonumber \\&\qquad + \, \rho \iint _{\partial V} (\mathbf{r }\times \mathbf{u})(\mathbf{u }\cdot \mathbf{n}) \mathrm{d}S \nonumber \\&\quad =\rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \mathbf{r }^{C/O} \times \, \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V \nonumber \\&\qquad + \, \rho \iint _{\partial V} \left( (\mathbf{r }-\mathbf{r }^{C/O}) \times \mathbf u \right) (\mathbf u \cdot \mathbf n ) \mathrm{d}S \nonumber \\&\quad =\rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V + \mathbf{r }^{C/O} \nonumber \\&\qquad \times \, \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V + \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ) (\mathbf u \cdot \mathbf n ) \mathrm{d}S. \end{aligned}$$
(32)
Before we add in the right-hand side of Eq. (
30), we will apply a few more reformulations to express the integral terms in Eq. (
32) (i.e. the left-hand side of Eq.
30) more intuitively. Beginning with the first term, we apply the product rule and the fact that the cross product of a vector with itself is zero:
$$\begin{aligned}&\rho \iiint _{CV}{} \mathbf{r}^{P/C}\times \frac{\partial \mathbf u ^{P/C}}{\partial t} \mathrm{d}V \\&\quad = \rho \iiint _{CV}\frac{\partial }{\partial t}(\mathbf{r}^{P/C}\times \mathbf{u}^{P/C}) - \left( \frac{\partial \mathbf{r}^{P/C}}{\partial t} \times \mathbf{u}^{P/C}\right) \mathrm{d}V \\&\quad = \rho \iiint _{CV}\frac{\partial }{\partial t}(\mathbf{r}^{P/C}\times \mathbf{u}^{P/C})\mathrm{d}V. \end{aligned}$$
We can also expand the third term in Eq. (
32) into two surface integrals (again using
\(\mathbf u = \mathbf u ^{P/C} + \mathbf u ^{C/O}\)), which will enable us to regroup the terms in Eq. (
32) with another material time derivative:
$$\begin{aligned}&\rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ) (\mathbf u \cdot \mathbf n ) \mathrm{d}S \\&\quad = \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{P/C}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S \\&\qquad + \, \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S. \end{aligned}$$
We can now rewrite the terms from Eq. (
32) as
$$\begin{aligned}&\rho \iiint _{CV}\frac{\partial }{\partial t}(\mathbf{r}^{P/C} \times \mathbf{u}^{P/C})\mathrm{d}V \\&\quad + \, \mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V\\&\quad + \, \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{P/C}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S\\&\quad + \, \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S, \end{aligned}$$
or rearranging,
$$\begin{aligned}&= \rho \iiint _{CV}\frac{\partial }{\partial t}(\mathbf{r}^{P/C}\times \mathbf{u}^{P/C})\mathrm{d}V \\&\quad + \, \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{P/C}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S\\&\quad + \, \rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S \\&\quad + \, \mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V. \end{aligned}$$
We see that the first two integrals are equal to the material time derivative, so Eq. (
32) can be written as:
$$\begin{aligned}&= \underbrace{\rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV}(\mathbf{r}^{P/C}\times \mathbf{u}^{P/C})\mathrm{d}V}_{\text {I}} \nonumber \\&\quad + \, \underbrace{\rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S}_{\text {II}} \nonumber \\&\quad + \, \underbrace{\mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V}_{\text {III}}. \end{aligned}$$
(33)
The first time derivative in Eq. (
33) (I) is the moment of CV about its center of mass
C,
$$\begin{aligned} \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV}(\mathbf{r}^{P/C}\times \mathbf{u}^{P/C})\mathrm{d}V=\frac {\mathrm{d}}{\mathrm{d}t}{} \mathbf H ^{\mathrm{CV}/C} = \mathbf{M }^{\mathrm{CV}/C}. \end{aligned}$$
(34)
The second term (II) can be simplified if we leverage the definition of center of mass again. In order to do so, we first apply the Reynolds transport theorem (Eq.
29):
$$\begin{aligned}&\rho \iint _{\partial V} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) (\mathbf u \cdot \mathbf n ) \mathrm{d}S \\&\quad = \rho \frac {\mathrm{d}}{\mathrm{d}t} \iiint _{CV} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) \mathrm{d}V \\&\qquad - \, \rho \iiint _{CV}\frac{\partial }{\partial t} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) \mathrm{d}V \\&\quad = \rho \frac {\mathrm{d}}{\mathrm{d}t} \iiint _{CV} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) \mathrm{d}V \\&\qquad - \, \rho \iiint _{CV} \left( \frac{\partial \mathbf{r }^{P/C} }{\partial t} \times \mathbf u ^{C/O}+ \mathbf{r }^{P/C} \times \frac{\partial \mathbf u ^{C/O} }{\partial t} \right) \mathrm{d}V\\&\quad = \rho \frac {\mathrm{d}}{\mathrm{d}t} \iiint _{CV} (\mathbf{r }^{P/C} \times \mathbf u ^{C/O}) \mathrm{d}V \\&\qquad - \, \rho \iint _{CV} \frac{\partial \mathbf{r }^{P/C} }{\partial t} \times \mathbf u ^{C/O} \mathrm{d}V \\&\qquad - \, \rho \iiint _{CV}\mathbf{r }^{P/C} \times \frac{\partial \mathbf u ^{C/O} }{\partial t}\mathrm{d}V. \end{aligned}$$
Then pulling the terms specific only to
C \(\left( \mathbf u ^{C/O}\,\mathrm{and} \, \frac{\partial \mathbf u ^{C/O} }{\partial t}\right)\) out of the integrals,
$$\begin{aligned}&= \frac {\mathrm{d}}{\mathrm{d}t} \left( \rho \iiint _{CV} \mathbf{r }^{P/C} \mathrm{d}V \times \mathbf u ^{C/O} \right) \\&\quad - \rho \iiint _{CV} \frac{\partial \mathbf{r }^{P/C} }{\partial t} \mathrm{d}V \times \mathbf u ^{C/O} \\&\quad - \, \rho \iiint _{CV} \mathbf{r }^{P/C} \mathrm{d}V \times \frac{\partial \mathbf u ^{C/O} }{\partial t}, \end{aligned}$$
we can now utilize the center of mass definition,
\(\rho \iiint _{CV} \mathbf{r }^{P/C}\mathrm{d}V = 0\), to eliminate the first and last terms, which leaves us with:
$$\begin{aligned} =- \rho \iiint _{CV}\frac{\partial \mathbf{r }^{P/C} }{\partial t} \mathrm{d}V \times \mathbf u ^{C/O}. \end{aligned}$$
Then since
\(\frac{\partial \mathbf{r }^{P/C} }{\partial t} = \mathbf u ^{P/C} = \mathbf u ^{P/O}-\mathbf u ^{C/O}\), and again using the fact that the cross product of a vector with itself is zero, term II becomes:
$$\begin{aligned}&-\rho \iiint _{CV} (\mathbf u ^{P/O}-\mathbf u ^{C/O}) \mathrm{d}V \times \mathbf u ^{C/O}\\&\quad = -\rho \iiint _{CV} (\mathbf u ^{P/O}\times \mathbf u ^{C/O} - \mathbf u ^{C/O} \times \mathbf u ^{C/O}) \mathrm{d}V \\&\quad = -\rho \iiint _{CV} \mathbf u ^{P/O} \times \mathbf u ^{C/O} \mathrm{d}V\\&\quad = -\rho \iiint _{CV} \mathbf u ^{P/O} \mathrm{d}V \times \mathbf u ^{C/O}\\&\quad = \mathbf u ^{C/O} \times \rho \iiint _{CV} \mathbf u \mathrm{d}V. \end{aligned}$$
Our ultimate goal is to develop a control surface formulation, so we next convert the volume integral into a surface integral. For brevity, we temporarily switch to Einstein notation to show the steps involved in this transformation:
$$\begin{aligned} \iiint _{CV} \mathbf u \mathrm{d}V&= \iiint _{CV} u_i \mathrm{d}V = \iiint _{CV}\frac{\partial x_i}{\partial x_j}u_j \mathrm{d}V\\&= \iiint _{CV}\frac{\partial }{\partial x_j}(x_i u_j)\mathrm{d}V - \iiint _{CV}x_i \frac{\partial u_j}{\partial x_j} \mathrm{d}V. \end{aligned}$$
Applying Gauss’s theorem to the first term and using the fact that the divergence
\(\frac{\partial u_j}{\partial x_j}=0\) for incompressible flow to eliminate the second term,
$$\begin{aligned} \iiint _{CV} \mathbf u \mathrm{d}V&= \iint _{\partial V} x_i u_j n_j \mathrm{d}S = \iint _{\partial V} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S. \end{aligned}$$
Term (II) thus becomes:
$$\begin{aligned} \mathbf u ^{C/O} \times \rho \iiint _{CV} \mathbf u \mathrm{d}V = \mathbf u ^{C/O} \times \rho \iint _{\partial V} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S \end{aligned}$$
(35)
To express the third term (III) from Eq. (
33) in terms of surface integrals, we can apply the conservation of momentum:
$$\begin{aligned}&\mathbf{r }^{C/O} \times \rho \frac {\mathrm{d}}{\mathrm{d}t}\iiint _{CV} \mathbf u \mathrm{d}V \nonumber \\&\quad = \mathbf{r }^{C/O} \times \left( -\iint _{\partial V} p \mathbf{n} \mathrm{d}S + \iint _{\partial V} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right. \nonumber \\&\qquad \left. - \rho \iint _{\partial V} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S\right) , \end{aligned}$$
(36)
where
\(\mathbf v\) is the velocity of the control surface.
We can now combine the simplified forms of (I)–(III) from Eqs. (
34)–(
36) with the right-hand side of Eq. (
30) to form the full moment equation:
$$\begin{aligned}&\mathbf{M }^{\mathrm{CV}/C} + \mathbf u ^{C/O} \times \rho \iint _{\partial V} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S \nonumber \\&\qquad + \, \mathbf{r }^{C/O} \times \left( -\iint _{\partial V} p \mathbf{n} \mathrm{d}S + \iint _{\partial V} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right. \nonumber \\&\qquad \left. - \rho \iint _{\partial V} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \right) \mathrm{d}S \nonumber \\&\quad = -\iint _{\partial V} \mathbf{r}\times p \mathbf{n} \mathrm{d}S \nonumber \\&\qquad + \, \iint _{\partial V} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S + \mathbf{r}^{C/O}\times \mathbf{g} \rho V. \end{aligned}$$
(37)
Next, to isolate fluid moments acting on the body from the rest of the control mass, we follow Lentink (
2018) and consider the continuous control surface
\(\partial V\) in terms of the outer control surface CS, the inner control surface that encloses the deforming body
\(\partial B\), and a infinitesimally thin tube that connects the body and outer surface
\(\partial b\), i.e. control surface (
\(\partial V\)) = outer surface (CS) + tube (
\(\partial b\)) + body surface (
\(\partial B\)) (Fig.
1a). We can, therefore, expand the control surface terms:
$$\begin{aligned}&\mathbf{M }^{\mathrm{CV}/C} + \mathbf u ^{C/O} \\&\qquad \times \, \rho \left( \iint _\mathrm{CS} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial b} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S \right. \\&\qquad \left. +\iint _{\partial B} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S \right) \\&\qquad + \, \mathbf{r }^{C/O} \times \left( -\iint _\mathrm{CS} p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right. \\&\qquad \left. - \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \right) \mathrm{d}S \\&\qquad -\iint _{\partial b} p \mathbf{n} \mathrm{d}S + \iint _{\partial b} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \\&\qquad - \, \rho \iint _{\partial b} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) ) \mathrm{d}S \\&\qquad -\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} \left( {\bar{\bar{\tau }}} \cdot \mathbf{n}\right) \mathrm{d}S \\&\qquad - \, \rho \iint _{\partial B} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) ) \mathrm{d}S \\&\quad = -\iint _\mathrm{CS} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \\&\qquad -\, \iint _{\partial b} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _{\partial b} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S\\&\qquad -\, \iint _{\partial B} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _{\partial B} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S\\&\qquad + \, \mathbf{r}^{C/O}\times \mathbf{g} \rho V. \end{aligned}$$
Moments on opposite sides of the infinitesimal tube
\(\partial b\) are equal and opposite and, therefore, cancel out (so all
\(\partial b\) integrals go to zero), and the convective term vanishes on the body surface because
\(\mathbf u -\mathbf v =0\) due to the no-flow boundary condition:
$$\begin{aligned}&\mathbf{M }^{\mathrm{CV}/C} + \mathbf u ^{C/O} \times \rho \left( \iint _\mathrm{CS} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial B} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S \right) \nonumber \\&\qquad +\, \mathbf{r }^{C/O} \times \left( -\iint _\mathrm{CS} p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right. \nonumber \\&\qquad \left. - \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S \right. \nonumber \\&\qquad \left. -\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right) \nonumber \\&\quad = -\iint _\mathrm{CS} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \nonumber \\&\qquad \underbrace{-\iint _{\partial B} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _{\partial B} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S}_{\text {net pressure and shear torque from body}} + \mathbf{r}^{C/O}\times \mathbf{g} \rho V. \end{aligned}$$
(38)
Using this new formulation, we can now derive the total moment acting on the body. The moment caused by pressure and shear on the surface cutout around the body (
\(\partial B\)) represents the net moment
from the body on the fluid. The moment on the body from the fluid will, therefore, be equal and opposite. To find the total external moment on the body with respect to the origin,
\(\mathbf{M }^O\), we must also add in the moment caused by gravity. For a body with mass
m and center of mass position
\(\mathbf{r }^{B/O}\) with respect to the origin (Fig.
1b),
$$\begin{aligned} \mathbf{M }^O= & {} -\left( -\iint _{\partial B} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _{\partial B} \mathbf{r }\times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right) \\&+ \mathbf{r }^{B/O}\times m\mathbf g . \end{aligned}$$
We can solve for
\(\mathbf{M }^O\) using Eq. (
38):
$$\begin{aligned} \mathbf{M }^O&= -\mathbf{M }^{\mathrm{CV}/C} - \mathbf u ^{C/O} \\&\quad \times \rho \left( \iint _\mathrm{CS} \mathbf{r}(\mathbf{u} \cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial B} \mathbf{r(u} \cdot \mathbf{n}) \mathrm{d}S \right) \\&\quad -\mathbf{r }^{C/O} \times \left( -\iint _\mathrm{CS} p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right. \\&\quad \left. - \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S \right. \\&\quad \left. -\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right) \\&\quad -\iint _\mathrm{CS} \mathbf{r}\times p \mathbf{n} \mathrm{d}S \\&\quad + \, \iint _\mathrm{CS} \mathbf{r} \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S + \mathbf{r}^{C/O}\times \mathbf{g} \rho V + \mathbf{r }^{B/O}\times m\mathbf g . \end{aligned}$$
Distributing
\(\mathbf{r }^{C/O}\),
$$\begin{aligned} \mathbf{M }^O&= -\mathbf{M }^{\mathrm{CV}/C} - \mathbf u ^{C/O} \\&\quad \times \, \rho \left( \iint _\mathrm{CS} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial B} \mathbf{r(u} \cdot \mathbf{n}) \mathrm{d}S \right) \\&\quad -\iint _\mathrm{CS} -\mathbf{r }^{C/O} \\&\quad \times \, p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} -\mathbf{r }^{C/O} \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S +\mathbf{r }^{C/O} \\&\quad \times \, \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S \\&\quad -\mathbf{r }^{C/O} \times \left( -\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \right) \\&\quad -\iint _\mathrm{CS} \mathbf{r}\times p \mathbf{n} \mathrm{d}S + \iint _\mathrm{CS} \mathbf{r} \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S\\&\quad + \, \mathbf{r}^{C/O}\times \mathbf{g} \rho V + \mathbf{r }^{B/O}\times m\mathbf g , \end{aligned}$$
and then grouping the surface integrals for CS:
$$\begin{aligned} \mathbf{M }^O&= -\mathbf{M }^{\mathrm{CV}/C} - \mathbf u ^{C/O} \\&\quad \times \, \rho \left( \iint _\mathrm{CS} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial B} \mathbf{r(u} \cdot \mathbf{n}) \mathrm{d}S \right) \\&\quad -\iint _\mathrm{CS}(\mathbf{r }
-\mathbf{r }^{C/O}) \times p \mathbf{n} \mathrm{d}S \\&\quad +\, \iint _\mathrm{CS} (\mathbf{r }-\mathbf{r }^{C/O}) \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S + \mathbf{r }^{C/O} \\&\quad \times \, \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S \\&\quad -\mathbf{r }^{C/O} \times \left( \underbrace{ -\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S}_{\text {Net pressure and shear force from body}} \right) \\&\quad +\, \mathbf{r}^{C/O}\times \mathbf{g} \rho V + \mathbf{r }^{B/O}\times m\mathbf g . \end{aligned}$$
To further simplify this equation, we can replace the body surface pressure and shear integrals by the net force that they have on the fluid in the CV, which is equal and opposite to the net fluid force acting on the body,
\(\mathbf{F } = -(-\iint _{\partial B} p \mathbf{n} \mathrm{d}S + \iint _{\partial B} ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S)\):
$$\begin{aligned} \mathbf{M }^O&= -\mathbf{M }^{\mathrm{CV}/C} - \mathbf u ^{C/O} \nonumber \\&\quad \times \, \rho \left( \iint _\mathrm{CS} \mathbf{r(u }\cdot \mathbf{n}) \mathrm{d}S + \iint _{\partial B} \mathbf{r(u} \cdot \mathbf{n}) \mathrm{d}S \right) \nonumber \\&\quad -\iint _\mathrm{CS}(\mathbf{r } -\mathbf{r }^{C/O}) \times p \mathbf{n} \mathrm{d}S \nonumber \\&\quad + \, \iint _\mathrm{CS} (\mathbf{r }-\mathbf{r }^{C/O}) \times ({\bar{\bar{\tau }}} \cdot \mathbf{n})\mathrm{d}S \nonumber \\&\quad + \mathbf{r }^{C/O} \times \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v})\cdot \mathbf{n}\right) \mathrm{d}S \nonumber \\&\quad +\mathbf{r }^{C/O} \times \mathbf F + \mathbf{r}^{C/O}\times \mathbf{g} \rho V + \mathbf{r }^{B/O}\times m\mathbf g , \end{aligned}$$
(39)
where the net fluid force on the body
\(\mathbf{F }\) is given by (Lentink
2018):
$$\begin{aligned} \mathbf{F }&= -\iint _\mathrm{CS}p\mathbf n \mathrm{d}S + \iint _\mathrm{CS} ({\bar{\bar{\tau }}} \cdot \mathbf{n}) \mathrm{d}S \nonumber \\&\quad - \rho \iint _\mathrm{CS} \mathbf{u}\left( (\mathbf{u-v}) \cdot \mathbf{n} \right) \nonumber \\&\quad - \underbrace{\rho \frac {\mathrm{d}}{\mathrm{d}t}\iint _{\partial B} \mathbf{r(u}\cdot \mathbf{n}) \mathrm{d}S}_{\text {unsteady body force, }{\mathbf{UBF }}} - \rho \frac {\mathrm{d}}{\mathrm{d}t}\iint _\mathrm{CS} \mathbf{r (u }\cdot \mathbf{n}) \mathrm{d}S. \end{aligned}$$
(40)
A formulation for the moment on a body about its center of mass
B, rather than about a theoretical origin, would be more physically meaningful and useful for interpreting the body’s rotational dynamics and stability. We, therefore, take one final step to apply the shift theorem for the moment of a set of forces (Mitiguy
2015), which enables us to find the moment on the body about
B based on the moment on the body about the origin
\(\mathbf{M }^O\), the position of the body’s center of mass relative to the origin
\(\mathbf{r }^{B/O}\), and the net force on the body
\(\mathbf{F }^B\), which includes the net fluid and gravitational forces on the body (Fig.
1b):
$$\begin{aligned} \mathbf{M }^B&= \mathbf{M }^O - \mathbf{r }^{B/O}\times \mathbf{F }^B =\mathbf{M }^O - \mathbf{r }^{B/O}\times (\mathbf{F } + m\mathbf g ) \nonumber \\&= \mathbf{M }^O - \mathbf{r }^{B/O}\times \mathbf{F } - \mathbf{r }^{B/O}\times m\mathbf g . \end{aligned}$$
(41)
From Eqs. (
39) and (
41), we arrive at the general expression for the moment on the body with respect to its center of mass (Fig.
1c):