1 Position of \(m_i\)
In a rigid body consisting of \(N\) point masses, the vector to the position of \(m_i\) is defined as \(\vec{r_i(t)}\), which is defined as follows:
\begin{equation} \vec{r_i(t)} = \vec{R_{CM}(t)} + \vec{r_i}'(t) \end{equation}whereas, \(\vec{R_{CM}(t)}\) is the position vector of the center of mass of the rigid body as a whole, and \(\vec{r_i}'(t)\) the vector from the center of mass to \(m_i\).
2 Velocity of \(m_i\)
The velocity of \(m_i\) is simply determined by the first derivative of the position equation as per above. Namely, that:
\begin{equation} \vec{v_i(t)} = \vec{V_{CM}(t)} + \vec{v_i}'(t) \end{equation}where, \(\vec{v_i(t)}\) is the velocity vector of \(m_i\), and \(\vec{V_{CM}(t)}\) is the velocity vector of the center of mass of the rigid body, and \(\vec{v_i}'(t)\) is the velocity vector from center of mass to \(m_i\).
3 Deriving \(KE_{total}\)
3.1 Setting up
From definition of \(KE_{total}\) itself, \(KE_{total}\) is the sum of all energies of each point mass in the rigid body.
\begin{equation} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 \end{equation}3.2 Derivation, part 1
Expanding this equation and substituting the value of \(v_i\), and additionally setting \(M = \sum m_i\) (namely, that \(M\) represents the total mass of the rigid body) we could derive:
\begin{align} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 =& \sum^N_{i=1} \frac{1}{2}m_i(v_i \cdot v_i) \\ =& \sum^N_{i=1} \frac{1}{2}m_i((\vec{V_{CM}} + \vec{v_i}') \cdot (\vec{V_{CM}} + \vec{v_i}')) \\ =& \sum^N_{i=1} \frac{1}{2}m_i(\vec{V_{CM}}^2 + 2 \times (\vec{v_i}' \cdot \vec{V_{CM}}) + \vec{v_i}'^2)) \\ =& \sum^N_{i=1} \frac{1}{2}m_i\vec{V_{CM}}^2 + \sum^N_{i=1} m_i \times (\vec{v_i}' \cdot \vec{V_{CM}}) + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \\ =& \frac{1}{2} \vec{V_{CM}}^2 \sum^N_{i=1} m_i + \vec{V_{CM}} \sum^N_{i=1} m_i \vec{v_i}' + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \end{align}3.3 Dealing with the Middle Term
At this point, we must note that \(\sum^N_{i=1} m_i \vec{v_i}' = 0\). Per the definition of the center of mass, the following holds:
\begin{equation} \vec{r_{CM}} = (\frac{1}{M}) \sum_i m_i \vec{r_i} \end{equation}Changing reference frame to that of the center of mass itself, this equation therefore becomes:
\begin{equation} \vec{r_{CM}}' = (\frac{1}{M}) \sum_i m_i \vec{r_i}' \end{equation}It is important to realize here that \(\vec{r_{CM}}' = 0\) because of the fact that — at the reference point of the center of mass, the center of mass is at a zero-vector distance away from itself.
In order to figure a statement with respect to the velocity of \(r_i'\), we take the derivative of the previous equation with respect to time.
\begin{align} 0 =& (\frac{1}{M}) \sum_i m_i \vec{r_i}' \\ \Rightarrow& \frac{d}{dt} (\frac{1}{M}) \sum_i m_i \vec{r_i}' \\ =& (\frac{1}{M}) \sum_i m_i \vec{v_i}' \end{align}Given that \(\frac{1}{M}\) could not be zero for an object with non-zero mass, it is concluded therefore that \(\sum_i m_i \vec{v_i}' = 0\).
3.4 Derivation, part 2
As \(\sum_i m_i \vec{v_i}' = 0\), the \(KE_{total}\) work-in-progress equation's middle term (which contains the statement \(\sum_i m_i \vec{v_i}'\)) is therefore zero. Substituting that in and removing the term, we therefore result in:
\begin{equation} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 = \frac{1}{2} \vec{V_{CM}}^2 \sum^N_{i=1} m_i + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \end{equation}Replacing the definition of \(M = \sum m_i\), we result in
\begin{align} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 &= \frac{1}{2} M \vec{V_{CM}}^2 + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \\ KE_{total} &= \frac{1}{2} M \vec{V_{CM}}^2 + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \end{align}The left term of this equation (\(\frac{1}{2} M \vec{V_{CM}}^2\)) is the clear original statement for \(KE_{translational}\). As component masses of a rigid body cannot experience translational motion about its center of origin, the second term is therefore rotational only and so \(KE_{rotational}\).
Therefore:
\begin{equation} KE_{total} = KE_{translational}+KE_{rotational} \end{equation}