The physical meaning of the derivative $\frac{\partial{L}}{\partial \dot q_i}$ of a Lagrangian

Question

The lagrangian is defined as

$$L = T - V$$

where $T$ is kinetic energy and $V$ potential energy.

Then the euler-lagrange-equation is

$$ \frac{d}{dt} \frac{\partial{L}}{\partial \dot q_i} = \frac{\partial{L}}{\partial q_i}. $$

Now what is the physical meaning of $\frac{\partial{L}}{\partial \dot q_i}$?

Answer 1

Actually it means the generalized momentum of the system. Because of $$V=V ( q )$$ so $$L ( q,\dot q )=\frac { 1} {2 } m \sum {{\dot q} ^2}- {V (q)}$$ then if we make a partial differentiation with respect to particular $\dot q$,that is , $\dot {q_i}$ : $$\frac {\partial L}{\partial \dot q_i}=m {\dot q_i}$$ So the Euler-Lagrange equation says us that $$\frac {d (m \dot q_i)}{dt}=-\frac {\partial V(q)}{\partial q_i}$$ so $$m\ddot{ {q_i}}=-\frac { \partial V } { \partial q_i}$$ as desired.