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Consider a mass $m$ which is constrained to move on the frictionless surface of a vertical cone $\rho = cz$ (in cyclindrical polar coordinates $\rho, \theta, z$ with $z>0$) in a uniform gravitational field $g$ vertically down. Set up Hamilton's equations using $z$ and $\theta$ as generalized coordinates.

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The question is, how does one derive the following kinetic energy?

$$T = \frac{1}{2}m[\dot{\rho}^2 + (\rho\dot{\theta})^2 + \dot{z}^2]$$

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1 Answer 1

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One approach is to start from $T=\frac12 m[\dot{x}^2+\dot{y}^2+\dot{z}^2]$ and use $x=\rho\cos\theta$, $y=\rho\sin\theta$ so that, e.g. $\dot{x}=\dot{\rho}\cos\theta-\rho\dot{\theta}\sin\theta$. Combine terms and simplify.

Or, for the plane polar part, using $\frac{d\hat{\boldsymbol{\rho}}}{dt} =\dot{\theta}\hat{\boldsymbol{\theta}} $ (consider the change in $\hat{\boldsymbol{\rho}}$ with small $dt$ and compare to the change in $\theta$), we have $\dot{\boldsymbol{\rho}}=\dot{\rho}\hat{\boldsymbol{\rho}}+\rho \frac{d\hat{\boldsymbol{\rho}}}{dt} =\dot{\rho}\hat{\boldsymbol{\rho}}+\rho \dot{\theta}\hat{\boldsymbol{\theta}} $ and use the orthogonality of $\hat{\boldsymbol{\rho}}$ and $\hat{\boldsymbol{\theta}}$ after squaring.

An alternative is to use $T=\frac12 m \left(\frac{ds}{dt}\right)^2 = \frac12 m g_{ij}\dot{q}^i \dot{q}^j$ where the metric $q_{ij}$ is diagonal with $q_{\rho\rho}=q_{zz}=1$ and $q_{\theta\theta}=\rho^2$.

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