Calculating the Kinetic Energy

From Lagrangian dynamics, we know that we can extract the physics if we know the kinetic and potential energy, $\bgroup\color{black}$ T$\egroup$ and $\bgroup\color{black}$ U$\egroup$ . For now we will not be applying any potential so we only have the kinetic energy.

	$\displaystyle T=\sum\limits_\alpha T_\alpha = {1\over 2}\sum\limits_\alpha m_\a... ...imits_\alpha m_\alpha \left(\vec{V} + \vec{\omega}\times\vec{r}_\alpha\right)^2$
	$\displaystyle T = {1\over 2}\sum\limits_\alpha m_\alpha \left(V^2 + 2\vec{V}\cdot(\vec{\omega}\times\vec{r}_\alpha)+(\vec{\omega}\times\vec{r}_\alpha)^2\right)$
	$\displaystyle T = V^2{1\over 2}\sum\limits_\alpha m_\alpha + \vec{V}\cdot\left(... ...ght) +{1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r}_\alpha)^2$
	$\displaystyle \sum\limits_\alpha m_\alpha\vec{r}_\alpha = 0$
	$\displaystyle T = {1\over 2}M V^2 + {1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r}_\alpha)^2$
	$\displaystyle T = T_{CM} + T_{rot}$

$\displaystyle \left(\vec{A}\times\vec{B}\right)^2$	$\displaystyle = A_iB_j\epsilon_{ijk} A_mB_n\epsilon_{mnk}$
	$\displaystyle = A_iB_j A_mB_n\epsilon_{ijk}\epsilon_{mnk}$
	$\displaystyle = A_iB_j A_mB_n (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})$
	$\displaystyle = A^2B^2-\left(\vec{A}\cdot\vec{B}\right)^2$
$\displaystyle T_{rot}$	$\displaystyle = {1\over 2}\sum\limits_\alpha m_\alpha\left[\omega^2r_\alpha^2-\left(\vec{\omega}\cdot\vec{r}_\alpha\right)^2\right]$
$\displaystyle T_{rot}$	$\displaystyle = {1\over 2}\sum\limits_\alpha m_\alpha\left[\omega_i\omega_j\delta_{ij}r_\alpha^2 -\left(\omega_i r_{\alpha i}\omega_j r_{\alpha j}\right)\right]$
$\displaystyle T_{rot}$	$\displaystyle = {1\over 2}\sum\limits_\alpha m_\alpha\omega_i\omega_j \left[\delta_{ij}r_\alpha^2-r_{\alpha i} r_{\alpha j}\right]$
$\displaystyle T_{rot}$	$\displaystyle = {1\over 2} I_{ij} \omega_i\omega_j$
$\displaystyle I_{ij}$	$\displaystyle \equiv \sum\limits_\alpha m_\alpha\left[\delta_{ij}r_\alpha^2-r_{\alpha i} r_{\alpha j}\right]$