Deriving Rotational Energy

From definition of $KE_{total}$ itself, $KE_{total}$ is the sum of all energies of each point mass in the rigid body.

$$\begin{equation} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 \end{equation}$$

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:

$$\begin{align} \sum^N_{i=1} \frac{1}{2}m_iv_i^2 =& \sum^N_{i=1} \frac{1}{2}m_i(v_i \cdot v_i) \\ =& \sum^N_{i=1} \frac{1}{2}m_i((\vec{V_{CM}} + \vec{v_i}') \cdot (\vec{V_{CM}} + \vec{v_i}')) \\ =& \sum^N_{i=1} \frac{1}{2}m_i(\vec{V_{CM}}^2 + 2 \times (\vec{v_i}' \cdot \vec{V_{CM}}) + \vec{v_i}'^2)) \\ =& \sum^N_{i=1} \frac{1}{2}m_i\vec{V_{CM}}^2 + \sum^N_{i=1} m_i \times (\vec{v_i}' \cdot \vec{V_{CM}}) + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \\ =& \frac{1}{2} \vec{V_{CM}}^2 \sum^N_{i=1} m_i + \vec{V_{CM}} \sum^N_{i=1} m_i \vec{v_i}' + \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \end{align}$$

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 $\sum_i m_i \vec{v_i}' = 0$.

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:

$$\begin{equation} KE_{total} = KE_{translational}+KE_{rotational} \end{equation}$$