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.

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

The two terms in the sum clearly separate the energy due to center of mass motion and the energy due to rotation.

Now we need a vector identity.

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

The kinetic energy due to rotation is not in general \bgroup\color{black}$ T={1\over 2} I\omega^2$\egroup but rather a more complicated inner product between the moment of inertia tensor \bgroup\color{black}$ I_{ij}$\egroup and two angular velocity vectors \bgroup\color{black}$ T={1\over 2}\vec{\omega}\cdot\mathbb{I}\cdot\vec{\omega}$\egroup.

Jim Branson 2012-10-21