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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?

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No, think e.g. on a point mass constrained to move along a horizontal rod. The rod is rotating with constant angular velocity in the horizontal plane. The constraints have explicit time dependence, but the kinetic energy of the point mass does not depend explicitly on time.

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  • $\begingroup$ 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 $H=T=\frac{1}{2}m(\dot{r}^2-r^2\omega^2)$. $\endgroup$ – mithusengupta123 Jul 28 at 2:35

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