\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

From the principle of linear momentum balance we get that \(\begin{align} \textsf{Div}\left(\boldsymbol{P}\right)+\boldsymbol{B}&=\rho_0 \frac{D^2 \boldsymbol{U}}{D t^2}, \\ \end{align}\) where $\boldsymbol{P}$ is the first Piola-Kirchhoff stress tensor, $\bs{B}$ is the (pseudo) body force, and

\(\begin{align} \label{eq:DispDef} \bs{U}(\bs{X},t)&=\boldsymbol{\varphi}(\boldsymbol{X},t)-\bs{X}, \end{align}\) is the displacement vector.

\[\begin{equation} \int_{\Omega_0}\textsf{Div}\left(\boldsymbol{P}\right)\, d\Omega_0+ \int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 =\int_{\Omega_0} \rho_0\frac{D^2 \boldsymbol{U}}{D t^2}\, d\Omega_0, \end{equation}\]

where $\Omega_0$ is the body in its reference configuration. Applying the divergence theorem on the first term we get that

[ \begin{align} \int_{\partial \Omega_0}\boldsymbol{P} \hat{\boldsymbol{N}}\, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0&=\int_{\Omega_0} \rho_0\frac{D^2 \boldsymbol{U}}{D t^2}\, d\Omega_0,
\end{align} ] where $\hat{\boldsymbol{N}}$ is the unit vector that is normal to the surface $\partial \Omega_0$, and $d\Gamma$ is an infinitesimal surface element of $\partial \Omega_0$. Noting that $\bs{P}\hat{\bs{N}}=\bs{T}\left(\hat{\boldsymbol{N}}\right)$, where $\bs{T}\left(\hat{\boldsymbol{N}}\right)$ is the body traction vector, we get that

[ \begin{align} \label{eq: LMB_mod} \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0&=\int_{\Omega_0} \rho_0\frac{D^2 \boldsymbol{U}}{D t^2}\, d\Omega_0. \end{align} ]

The derivative \(\begin{align} \frac{D(\cdot)}{Dt} &= \left. \frac{\partial(\cdot)}{\partial t}\right|_{\bs{X}} \label{eq:MatDerDef} \end{align}\) is called the material time derivative. It follows from Eq. $\eqref{eq:DispDef}$ that

\(\begin{align} \frac{D\bs{U}}{Dt}&=\frac{D\bs{x}}{Dt},\\ \label{eq: Acceleration} \frac{D^2\bs{U}}{Dt^2}&=\frac{D^2\bs{x}}{Dt^2} \end{align}\) Substituting Eq. \ref{eq: Acceleration} in Eq. \ref{eq: LMB_mod} we arrive at, \(\begin{align} \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0&=\int_{\Omega_0} \rho_0\frac{D^2 \boldsymbol{x}}{D t^2}\, d\Omega_0. \end{align}\)

The material density field \(\rho_0\) is defined as $dm=\rho_0 d\Omega_0$, where $dm$ is the mass of the reference volume element $d\Omega_0$. In the most general case the (material) mass density field $\bs{\rho}_0$ is a function of both $\bs{X}$ and $t$. However, in typical materials (i.e., non-radioactive, non-growing, etc.) the material density field is only a function of $\bs{X}$. In this note we only consider the case in which $\rho_0$ is not a function of time. Therefore, it follows from $\eqref{eq:MatDerDef}$ that \(\begin{align} \frac{D\rho_0}{Dt}&=0. \label{eq:MatDerRho} \end{align}\)

When the material density field does not even depend on $\bs{X}$ then the body is said to be a homogenous solid. In that case, $\rho_0$ is simply a scalar constant.

The material time derivative of $\rho_0 \bs{x}$ is equal to \(\begin{align} \frac{D\left(\rho_0 \bs{x}\right)}{Dt} &=\rho_0\frac{D\bs{x}}{Dt}+\frac{D\rho_0}{Dt}\bs{x},\\ \end{align}\) in which the second term vanishes on account of \(\eqref{eq:MatDerRho}\). Therefore, \(\begin{align} \frac{D\left(\rho_0 \bs{x}\right)}{Dt} &=\rho_0\frac{D\bs{x}}{Dt}\\ \end{align}\)

Similarly, it can be shown that \(\begin{align} \frac{D^2\left(\rho_0 \bs{x}\right)}{Dt^2} &=\rho_0\frac{D^2\bs{x}}{Dt^2}\\ \end{align}\)

\(\begin{align} \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 &= \int_{\Omega_0} \frac{D^2 \rho_0 \boldsymbol{x}}{D t^2}\, d\Omega_0, \\ \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 &= \frac{D^2 }{D t^2}\int_{\Omega_0} \rho_0 \boldsymbol{x}\, d\Omega_0, \end{align}\) where the second equation follows from the first one on account of the fact that $\Omega_0$ is not a function of time.

The position vector of the body’s center of mass is \(\begin{align} \label{eq: COM} M \bs{x}_{\textsf{c.o.m}}&=\int_{\Omega\left(t\right)}\bs{x}\rho(\bs{x},t)\, d\Omega, \end{align}\) where $M$ is the total mass of the body. From the conservation of mass for any arbitrary part of the body \(\mathcal{P}_0\subset \Omega_0\) of volume $dV_0$ we have,

\[\begin{align} \label{eq: MassConversation} \int_{\mathcal{P}_0} \rho_0 \left(\bs{X}\right) dV_0 = \int_{\mathcal{P}\left(t\right)} \rho \left(\bs{x},t\right) dV \end{align}\]

Introducing \(\tilde{\rho} \left(\bs{X},t\right) := \rho(\boldsymbol{\varphi}(\boldsymbol{X},t),t)\) and using the relation \(dV = J\left(\bs{X},t\right) dV_0\) (where \(J\left(\bs{X},t\right) = \text{det} \bs{F}\)) in Eq. (\ref{eq: MassConversation}) lead to the following: \(\begin{align} \label{eq: MassConversationMod1} \int_{\mathcal{P}_0} \rho_0 \left(\bs{X}\right) dV_0 = \int_{\mathcal{P}_0} \tilde{\rho} \left(\bs{X},t\right)J\left(\bs{X},t\right) dV_0 \end{align}\) From Eq. (\ref{eq: MassConversationMod1}) we have, \(\begin{align} \label{eq: MassConversationMod2} \int_{\mathcal{P}_0} \left(\rho_0 \left(\bs{X}\right)- \tilde{\rho} \left(\bs{X},t\right)J\left(\bs{X},t\right)\right) dV_0 = 0 \end{align}\) As \(\mathcal{P}_0\) has been chosen aribitrarily, we can write the following upon application of localization theorem on Eq. (\ref{eq: MassConversationMod2}). \(\begin{align} \label{eq: MassConversationMod3} \rho_0 \left(\bs{X}\right)= \tilde{\rho} \left(\bs{X},t\right)J\left(\bs{X},t\right) = \rho \left(\bs{x},t\right)J\left(\bs{X},t\right) \end{align}\) Using the relation given by Eq. (\ref{eq: MassConversationMod3}) in Eq. (\ref{eq: COM}) we may write,

\[\begin{align} M \bs{x}_{\textsf{c.o.m}}&=\int_{\Omega_0} \boldsymbol{\varphi}(\boldsymbol{X},t) \tilde{\rho} \left(\bs{X},t\right) J\left(\bs{X},t\right) d\Omega_0= \int_{\Omega_0} \boldsymbol{\varphi}(\boldsymbol{X},t) \rho_0 \left(\bs{X}\right) d\Omega_0 \end{align}\] \[\begin{align} \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 &= \frac{D^2 }{D t^2}\left(M \bs{x}_{\textsf{c.o.m}}\right) \\ \int_{\partial \Omega_0} \boldsymbol{T}\left(\hat{\boldsymbol{N}}\right) \, d\Gamma+\int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 &= M\frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \end{align}\]

For a free falling solid with no surface tractions, we have \(\begin{align} \int_{\Omega_0}\boldsymbol{B}\, d\Omega_0 &= M\frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \end{align}\)

The only body force in this case is $\bs{B}= g \rho_0(\bs{X})$ \(\begin{align} \int_{\Omega_0}g \rho_0(\bs{X})\, d\Omega_0 &= M\frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \\ g\int_{\Omega_0} \rho_0(\bs{X})\, d\Omega_0 &= M\frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \\ g M &= M\frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \\ g &= \frac{D^2\bs{x}_{\textsf{c.o.m}} }{D t^2} \end{align}\)