Particle on a Cylindrical Surface — Hamiltonian Mechanics: 17-Phys-A1 Classical Mechanics, December 2013, Q2

Question 2: Particle on a Cylindrical Surface — Hamiltonian Mechanics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A particle of mass $m$ confined to the surface of a cylinder of fixed radius $R$ (coordinates $\theta$, $z$), attracted toward the origin $O$ (not the cylinder axis) by a force proportional to its distance from $O$: $\bar F = -k\bar r$, with $\bar r$ the full 3-D position vector.

Figure 2 -- particle on a cylindrical surface of radius R, attracted to the origin.

Find. The Hamiltonian $H(\theta, p_\theta, z, p_z)$ and the resulting equations of motion.

Approach. Build the Lagrangian from the constrained kinematics ($r = R$ fixed), Legendre-transform it to the Hamiltonian using the conjugate momenta, then read the equations of motion straight off Hamilton’s canonical equations.

Kinematics and kinetic energy. On the cylinder, position $= (R\cos\theta, R\sin\theta, z)$, so velocity $= (-R\dot\theta\sin\theta,\, R\dot\theta\cos\theta,\, \dot z)$ and $$T = \tfrac{1}{2}m\left(R^2\dot\theta^2 + \dot z^2\right)$$
Potential energy. $\bar F = -k\bar r = -\nabla U \Rightarrow U = \tfrac{1}{2}k\,\bar r\cdot\bar r = \tfrac{1}{2}k\left(R^2 + z^2\right)$, using $|\bar r|^2 = x^2+y^2+z^2 = R^2 + z^2$ on the cylinder.
Lagrangian and conjugate momenta. $$L = T - U = \tfrac{1}{2}m\left(R^2\dot\theta^2 + \dot z^2\right) - \tfrac{1}{2}k\left(R^2+z^2\right)$$ $$p_\theta = \frac{\partial L}{\partial \dot\theta} = mR^2\dot\theta, \qquad p_z = \frac{\partial L}{\partial \dot z} = m\dot z$$
Legendre transform to the Hamiltonian. $H = p_\theta\dot\theta + p_z\dot z - L$, re-expressed in the momenta ($\dot\theta = p_\theta/mR^2$, $\dot z = p_z/m$) and dropping the additive constant $\tfrac12 kR^2$: $$\boxed{H(\theta,p_\theta,z,p_z) = \frac{p_\theta^2}{2mR^2} + \frac{p_z^2}{2m} + \tfrac{1}{2}kz^2}$$ $H$ does not depend on $\theta$ at all — $\theta$ is a cyclic coordinate.
Hamilton’s equations. $\dot q_i = \partial H/\partial p_i$, $\dot p_i = -\partial H/\partial q_i$: $$\dot\theta = \frac{\partial H}{\partial p_\theta} = \frac{p_\theta}{mR^2}, \qquad \dot p_\theta = -\frac{\partial H}{\partial \theta} = 0$$ $$\dot z = \frac{\partial H}{\partial p_z} = \frac{p_z}{m}, \qquad \dot p_z = -\frac{\partial H}{\partial z} = -kz$$ $$\boxed{p_\theta = \text{const} \;(\text{angular momentum about the cylinder axis}), \qquad \ddot z + \frac{k}{m}z = 0 \;(\text{SHM})}$$ Since $p_\theta$ is constant and $R$ is fixed, $\dot\theta$ is itself constant: the particle circles the cylinder at a uniform angular rate while oscillating harmonically along $z$ with angular frequency $\omega_z = \sqrt{k/m}$ — a helical, breathing-free motion.

Question 2 — results
Quantity	Value
Hamiltonian	$H = \dfrac{p_\theta^2}{2mR^2} + \dfrac{p_z^2}{2m} + \tfrac{1}{2}kz^2$
$\theta$ equation	$\dot\theta = p_\theta/(mR^2)$, $\dot p_\theta = 0$ (conserved)
$z$ equation	$\ddot z + (k/m)z = 0$, i.e. SHM at $\omega_z = \sqrt{k/m}$

Topic. Hamiltonian mechanics of a particle constrained to a surface, and the cyclic-coordinate conservation theorem.

Key relations. $H = \sum_i p_i\dot q_i - L$; canonical equations $\dot q_i = \partial H/\partial p_i$, $\dot p_i = -\partial H/\partial q_i$; a coordinate absent from $H$ (or $L$) is cyclic and its conjugate momentum is conserved.

Why this works. The constraint $r=R$ is built into $T$ and $U$ from the outset by writing the kinematics directly in cylindrical coordinates, so no Lagrange multiplier is needed here (contrast Question 1, where the constraint force itself was wanted and had to be kept explicit). Because the force is central ($\bar F=-k\bar r$, pointing at the single fixed point $O$) but the particle is confined off-axis to a cylinder, the problem separates cleanly: $\theta$ never appears in $H$ because rotating the whole cylindrical setup about its own axis changes nothing physical, so $p_\theta$ (angular momentum about that axis) is conserved by symmetry, while $z$ sees an ordinary linear restoring force and executes SHM.

Common pitfall. Writing $U = \tfrac12 kr^2$ with $r$ the cylindrical radius (fixed at $R$) instead of the full distance from the origin $|\bar r| = \sqrt{R^2+z^2}$ — that error deletes the $z$-dependence of $U$ entirely and predicts (wrongly) that $z$ does not oscillate at all.

Source. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §8.1–8.2 (Hamilton’s equations) and §2.6 (cyclic coordinates and conservation theorems).

Question 2 of 6: Particle on a Cylindrical Surface — Hamiltonian Mechanics

Question 2: Particle on a Cylindrical Surface — Hamiltonian Mechanics (20 marks)