A particle has Lagrangian L=tfrac12[f(theta)dottheta^2+2g(theta)dottheta w+w^2h(theta)]-V. Then tfrac12[f(theta)dottheta^2-w^2h(theta)]+V represents:

Q98Section II (Mathematics)98 of 180

Identifying the Hamiltonian for a reduced (effective) one-coordinate Lagrangian

A particle has Lagrangian $L=\tfrac12\left[f(\theta)\dot\theta^2+2g(\theta)\dot\theta w+w^2h(\theta)\right]-V$ . Then $\tfrac12\left[f(\theta)\dot\theta^2-w^2h(\theta)\right]+V$ represents:

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