# No explicit time-dependence in Lagrangian means constraints are explicitly time-independent?

Suppose a Lagrangian is not explicitly time-dependent. Does it mean that the constraint equations are also explicitly time-independent, and (as a result) the kinetic energy is necessarily a homogeneous quadratic function of generalized velocities?

• Thank you. The constraint equation is $\theta=\omega t$ and kinetic energy is $T=\frac{1}{2}(\dot{r}^2+r^2\omega^2)$ which is also the Lagrangian of the problem. Here, since T is not a homogeneous quadratic function of $\dot{r}$, the Hamiltonian is also different from T given by $H=T=\frac{1}{2}m(\dot{r}^2-r^2\omega^2)$. – mithusengupta123 Jul 28 at 2:35