# Lagrangian preserve it's form for rigid bodies

Lagrangian for a particle moving under influence of conservative force given by $$\mathcal{L}=T-V$$ that is Kinetic energy minus Potential energy. Now for a system of particles I expect the same form of lagrangian. But Consider a rigid body with say $$n$$ particles. In this case I need to take care of constraints too. We know that for rigid body kinetic energy is ture out to be

$$T=\frac{\mathbf{\omega}\cdot I\cdot \mathbf{\omega}}{2}$$

For bodies without a fixed point, the most useful reference point is almost always the center of mass. We have already seen that the total kinetic energy and angular momentum then split neatly into one term relating to the translational motion of the center of mass and another involving rotation about the center of mass.

$$T=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2$$

And then in textbooks they take again the same form of langrangian to find equation of motion. But How do we know the Lagrangian will take the same form (Kinetic minus potential) for such a large system of particle and with large number of constraints.