- #1

etotheipi

*no,*but am by no means certain. The total kinetic energy of a system of particles is $$T = \sum_{i} \frac{1}{2} m_{i}\vec{v_i}\cdot\vec{v_i} = \frac{1}{2} m_{i}(\vec{v_{COM}} + \vec{v_{i}^{'}})^{2} = \frac{1}{2}M\vec{v_{COM}}^{2} + \frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2}$$ where ##\vec{v_{i}^{'}}## is the velocity of the particle relative to the centre of mass.

When we speak of non-rigid systems of particles, the second term has a well understood interpretation: the microscopic contribution to the kinetic energy. If we add this to the sum of the microscopic internal potential energies (between the particles), we obtain the internal energy.

However in the context of rigid bodies (systems of particles which retain fixed relative positions), the kinetic energy relative to the centre of mass, ##\frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2}##, is simply the rotational kinetic energy, ##\frac{1}{2} I\omega^{2}##. Since rotational energy appears to be macroscopic, it seems wrong to lump this into an internal energy term. However, the internal energy is defined as the total energy minus the potential energy in external fields and the kinetic energy of the centre of mass (that is, the kinetic energy in the rest frame of the body). This might lead one to believe rotational KE does not contribute to the internal energy.

I was wondering which interpretation is more correct?