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.

enter image description here

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

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


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 $g_{ij}$ is diagonal with $g_{\rho\rho}=g_{zz}=1$ and $g_{\theta\theta}=\rho^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.