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

In this section we only work with non-dimensional quantities. Recall that $\mathcal{E}_{\rm R}$ (or $\mathcal{E}_{\rm Ref}$) is the reference Euclidean point space, and $\mathcal{E}$ is the current Euclidean point space. In the reference and current spaces say the body occupies the regions $\u{\kappa}_{\rm R}\pr{\mathcal{B}}$ and $\u{\kappa}_{\u{t}}(\mathcal{B})$, respectively. We define the non-dimensional regions $\sf{B}(0)$ and $\sf{B}(\tau)$, which are subsets of $\mathbb{R}^3$, as follows

\[\begin{align} \sf{B}(0)&=\{\usf{X}:=(X_i)\in \mathbb{R}^3~|~X_i\u{E}_i\in \u{\kappa}_{\rm R}\pr{\mathcal{B}}\subset \mathbb{E}_{\rm Ref}\}\\ \sf{B}(\tau)&=\{\usf{X}:=(X_i)\in \mathbb{R}^3~|~X_i\u{e}_i\in \u{\kappa}_{\u{t}}\pr{\mathcal{B}}\subset \mathbb{E}\} \end{align}\]

In the non-dimensioan space $\mathbb{R}^3$, $\usf{B}(0)$ is the region corresponding to the reference (or initial) configuration of the body $\u{\kappa}_{\rm R}\pr{\mathcal{B}}$. And $\usf{B}(\tau)$ in $\mathbb{R}^3$ is the region corresponding to the current configuration of the body $\u{\kappa}_{\u{t}}\pr{\mathcal{B}}$ in $\mathbb{E}$.

Kinetics: Center of Mass

At $\tau=0$ in the non-dimensional space the body occupies the region $\usf{B}(0)$. The (non-dimensional) position vector of the center of mass of the body at time=0 is

\(\begin{align} \bar{\usf{X}}&= \frac{\int_{\sf{B}(0)}\usf{X}\rho(\usf{X},0)\, d\sf{V}}{\int_{\sf{B}(0)}\rho(\usf{X},0)\, d\sf{V}},\\ M \bar{\usf{X}} &= \int_{\sf{B}(0)}\usf{X}\rho(\usf{X},0)\, d\sf{V}, \end{align}\) where the density $\rho(\cdot,0)$ is defined such that $\rho(\usf{X},0)d\sf{V}$, where $\usf{X}\in \usf{B}(0)$ and $d\sf{V}\subset \usf{B}(0)$ is the infinitesimal region containing the point $\usf{X}$, gives the mass of the solid contained in the infinitesimal region $d\sf{V}$.

At time $\tau$ in the non-dimensional space the body occupies the region $\usf{B}(\tau)$. The position vector of the center of mass of the body at time $\tau$ is then \(\begin{align} \bar{\usf{x}}(\tau)&= \frac{\int_{\sf{B}(\tau)}\usf{x}\rho(\usf{x},\tau)\, d\sf{v}}{\int_{\sf{B}(\tau)}\rho(\usf{x},\tau)\, d\sf{v}},\\ M \bar{\usf{x}}(\tau) &=\int_{\sf{B}(\tau)}\usf{x}\rho(\usf{x},\tau)\, d\sf{v}, \end{align}\) where the density $\rho(\cdot,\tau)$ is defined such that $\rho(\usf{x},\tau)d\sf{v}$, where $\usf{x}\in \usf{B}(\tau)$ and $d\sf{v}\subset \usf{B}(\tau)$ is the infinitesimal region containing the point $\usf{x}$, gives the mass of the solid contained in the infinitesimal region $d\sf{v}$.

Thus, $\rho(\cdot,\tau)$ is the mass density per unit volume (non-dimensional) of the body in the reference configuration, i.e., the configuration at $\tau=0$, whereas $\rho(\cdot,\tau)$ is the mass density per unit volume of the body in the current configuration. The field $\rho(\cdot,0)$ is called the material density field, and $\rho(\cdot,\tau)$ is called the spatial density field.

If $d\sf{V}$ is an infinitesimal volume element of the solid at time=0, with the position vector $\usf{X}$, then $\rho(\usf{X},0)\, d\sf{V}$ gives the mass, $dm$, contained in that volume element. As the solid rotates and translates, after a time $\tau$, the volume element $d\sf{V}$ gets transformed into the volume element $d\sf{v}$ with position vector $\usf{x}(\usf{X},\tau)$, where, \(\begin{align} \usf{x}(\usf{X},\tau)&=\usf{R}(\tau)\usf{X}+\usf{c}(\tau). \label{eq:DefMapping} \end{align}\) The mass contained in the volume $d\sf{v}$ is from the definition of $\rho(\cdot,\tau)$ is $\rho(\usf{x},\tau)\, d\sf{v}$. However, since mass is conserved during motion, $\rho(\usf{x},\tau)\, d\sf{v}=dm$. Therefore, \(\begin{align} \rho(\usf{x},\tau)\, d\sf{v}&=\rho(\usf{X},0)\, d\sf{V} \end{align}\) Since we are considering rigid body motion, the volume measure of $d\usf{v}$ is the same as that of $d\usf{V}$ (skip the justification behind this statement since the proof behind it is a bit advanced). Therefore, we get from the above equation that \(\begin{align} \rho(\usf{x},\tau)&=\rho(\usf{X},0)\\ &=\rho(\usf{X}^{-1}(\usf{x},\tau),0) \end{align}\)

\[\begin{align} \usf{X}^{-1}(\usf{x},\tau)=\usf{R}(\tau)^{\sf T}\usf{x}-\usf{R}^{\sf T}(\tau)\usf{c}(\tau) \end{align}\]

The expression for $\rho(\cdot, \tau)$ simplifies greatly for a homogenous body. In this case we get that

\[\begin{equation} \rho(\usf{x},\tau)=\rho(\usf{X},0)=\rho_0~(\text{a constant}) \end{equation}\]

Motion of material point identified by $\bar{\usf{X}}$

As we discussed before, the position vectors $\usf{X}$ are used as names for the different material points that compose the solid. A material point gets the name $\usf{X}$ iff it happened to be located at the point $\usf{X}$¹ at time=0. Now there may or may not be an actual material point located at $\bar{\usf{X}}$. For both a solid disk and circular rind the center of mass is their geometric centers, which of course is occupied by a material point only in the case of the solid disk.

An interesting question to ask is: How does the point that was the solid’s center of mass at time=0, i.e., the material point $\bar{\usf{X}}$, move? What is its postion at time=$t$?

Well, we know that the material point $\usf{X}$ at time=$t$ is located at $\usf{\varphi}_t(\usf{X})$. So,the material point $\bar{\usf{X}}$ at time=$t$ is located at $\usf{\varphi}_t(\bar{\usf{X}})$. For the specific case of rigid body motion,

\[\begin{align} \usf{\varphi}_t(\bar{\usf{X}})&=\usf{R}_t \bar{\usf{X}}+\usf{t}_t \end{align}\]

However, very interestingly,

\[\begin{align} \usf{R}_t \bar{\usf{X}}+\usf{t}_t&=\bar{\usf{x}}_t \end{align}\]

Proof: \(\begin{align} M \bar{\usf{x}}_t &=\int_{\sf{v}}\usf{x}\rho(\usf{x})\, d\sf{v}\\ &=\int_{\Omega_0}\usf{\varphi}_t(\usf{X})\rho(\usf{\varphi}_t(\usf{X}))\, d\Omega_0,\\ &=\int_{\Omega_0}\usf{\varphi}_t(\usf{X})\rho\circ\usf{\varphi}_t(\usf{X})\, d\Omega_0,\\ &=\int_{\Omega_0}\usf{\varphi}_t(\usf{X})\rho_0(\usf{X})\, d\Omega_0,\\ &=\int_{\Omega_0}(\usf{R}_t\usf{X}+\usf{t}_t)\rho_0(\usf{X})\, d\Omega_0,\\ &=\int_{\Omega_0}\usf{R}_t\usf{X}\rho_0(\usf{X})\, d\Omega_0+\int_{\Omega_0}\usf{t}_t\rho_0(\usf{X})\, d\Omega_0,\\ &=\usf{R}_t\int_{\Omega_0}\usf{X}\rho_0(\usf{X})\, d\Omega_0+\int_{\Omega_0}\usf{t}_t\rho_0(\usf{X})\, d\Omega_0,\\ &=\usf{R}_t M \bar{\usf{X}}+\usf{t}_t\int_{\Omega_0}\rho_0(\usf{X})\, d\Omega_0,\\ &=\usf{R}_t M \bar{\usf{X}}+\usf{t}_t M,\\ \bar{\usf{x}}_t &=\usf{R}_t \bar{\usf{X}}+\usf{t}_t,\\ \end{align}\)

This means that the material particle $\bar{\usf{X}}$, which was located at the solid’s center of mass at time=0, is located at the solid’s center of mass at all other time instances as well. So, in the case of rigid body motion, the solid’s center of mass is always occupied by the same material point. This implies that if we fix an accelerometer at the solid’s center of mass at time=0, i.e., attach the accelerometer to the material point $\bar{\usf{X}}$, that acclerometer will always be located the solid’s center of mass and will always report the accleration of the solid’s center of mass.

Mathematica

Here are some examples in Mathematica

1: or to be more precise $\mathcal{O}+\usf{X}$, where the point $\mathcal{O}$ is the origin.

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