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The momentum $m v$ of a particle is formally the same as the derivative its translational kinetic energy $\frac{1}{2} m v^2$ with respect to $v$.

Similarly the angular momentum $I \omega$ is the derivative of its rotational energy $\frac{1}{2}I \omega^2$ with respect to $\omega$.

Does this relation has any physical interpretation?

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psst! Maybe you can raise your acceptance rate first and maybe you will get more responses. –  ja72 Jun 12 '12 at 15:48

1 Answer 1

up vote 3 down vote accepted

Indeed, in Hamiltonian Formalism this is the very definition of momentum. For example, for a free particle with one generalised coordinate $q$ and $\dot{q}\equiv v$ the Lagrangian is

$$ L=\frac{1}{2}mv^2 $$

and the momentum is (by definition) given by

$$ p =\frac{\partial L}{\partial v}$$

Check any book in classical mechanics for Lagrangian and Hamiltonian formalism for more details in the subject.

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