Spin angular momentum of a system of particles : Is there any energy associated with it? Consider a system of point particles , where the mass of particle $i$ is $μ_i$ and its position vector is $\vec{r}_i$. Let $\vec{r}_\text{cm}$ is the position vector of the center of mass of the system. Considering the system from a reference frame attached to the center of mass, the system may have a spin about the center of mass and it is given by the spin angular momentum $\vec{L}_{spin}$. It is given by the expression 
$$\vec{L}_{spin} = \sum_i  \mu_i \Bigl[(\vec{r}_i - \vec{r}_\text{cm}) \times (\dot{\vec{r}}_i - \dot{\vec{r}}_\text{cm}) \Bigr]$$
The rate of change of this spin angular momentum is the total torque acting on the system about the center of mass in the center of mass reference frame.
My question is, is there any (spin kinetic (may be)) energy associated with the spin angular momentum in the center of mass reference frame ? How is it defined ?
 A: Similar to the derivation of separation of angular momentum into $L_{CM}$ and $L_{internal}$, one can derive similar expression for Energy as
$E = \frac{1}{2}M_{total}v_{CM}^{2} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$.
Proof:
$$E = \frac{1}{2}\sum \mu_{i} v_{i}^{2}$$ 
$$v_{i} = v_{CM} + v_{i}^{'}$$
$$E = \frac{1}{2}\sum \mu_{i} v_{CM}^{2} + v_{CM}\sum \mu_{i} v_{i}^{'} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$$ 
Since in CoM frame $\sum \mu_{i} (r_{i}-r_\text{cm}) =0 \to \sum \mu_{i} v_{i}^{'}=0$.
$$QED$$
L and E within Com frame can be related only if body is rigid. One can refer Klepner & Kolenkow Classical Mechanics.
A: The energy associated with the internal rotations is the rotational energy

http://en.wikipedia.org/wiki/Rotational_energy

given by the formula
$$ K = \frac 12 I \omega^2 $$
where $\omega$ is the angular frequency and $I$ is the moment of inertia with respect to the axis of rotation

http://en.wikipedia.org/wiki/Moment_of_inertia
  $$ I = \int {\mathrm d}m \, \rho^2 $$

where $\rho$ is the distance of the infinitesimal mass ${\mathrm d}m$ from the axis of rotation. Also, 
$$ I = I_{ab} n_a n_b $$
in terms of the tensorial moment of inertia $I_{ab}$; $\vec n$ is the unit vector along the axis of rotation.
