Particle on a Cylinder Under a Central Force — Hamiltonian and Frame Invariance: 17-Phys-A1 Classical Mechanics, May 2016, Q2

Question 2: Particle on a Cylinder Under a Central Force — Hamiltonian and Frame Invariance (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 $m$ is held on the surface of a fixed cylinder of radius $R$ (its distance from the cylinder's central axis is always $R$); it is subject to an isotropic 3-D central force $\vec F=-k\vec r$ directed at the origin (the centre of the cylinder's base), with $\vec r$ the particle's full position vector. Cylindrical coordinates $(R,\phi,z)$ describe the particle, since the radial cylindrical coordinate is pinned at $R$.

Particle on the cylinder wall: position $\vec r=(R\cos\phi,R\sin\phi,z)$. The primed frame $xyZ$ shares the same $Z$-axis, rotated by a fixed angle from $XYZ$.

Find. (a) $H$ and Hamilton's equations of motion; (b) how those equations change under a fixed rotation of the azimuthal reference about $Z$; (c) an invariant common to both descriptions.

Approach. Build the constrained Lagrangian directly in $(\phi,z)$ (the two free coordinates once $\rho=R$ is fixed), Legendre-transform to $H$, and read off Hamilton's equations; then use the manifest axial symmetry of the potential to answer (b) and (c).

Part (a) — Hamiltonian and equations of motion. With $x=R\cos\phi$, $y=R\sin\phi$, $z=z$, $$T=\tfrac12 m\big(R^2\dot\phi^2+\dot z^2\big),\qquad V=\tfrac12 k\, r^2 = \tfrac12 k\big(R^2+z^2\big)\ \ (\text{since }\vec F=-\nabla V=-k\vec r),$$ $$L = \tfrac12 m R^2\dot\phi^2+\tfrac12 m\dot z^2-\tfrac12 k(R^2+z^2).$$ Generalized momenta: $p_\phi=\partial L/\partial\dot\phi = mR^2\dot\phi$, $\ p_z=\partial L/\partial\dot z=m\dot z$. Since $T,V$ are natural (no explicit-time, no $\dot q$-linear cross terms), $H=T+V$ expressed in momenta: $$\boxed{H = \dfrac{p_\phi^2}{2mR^2}+\dfrac{p_z^2}{2m}+\dfrac12 k\big(R^2+z^2\big)}$$ Hamilton's equations, $\dot q_i=\partial H/\partial p_i$, $\dot p_i=-\partial H/\partial q_i$: $$\dot\phi=\frac{p_\phi}{mR^2},\qquad \dot p_\phi = -\frac{\partial H}{\partial\phi}=0\ \Rightarrow\ p_\phi=\text{const.},$$ $$\dot z=\frac{p_z}{m},\qquad \dot p_z=-kz\ \Rightarrow\ m\ddot z+kz=0.$$ So $\phi$ is cyclic ($H$ has no explicit $\phi$-dependence) — $p_\phi$ (angular momentum about $Z$) is conserved and $\phi$ advances at the constant rate $\dot\phi=p_\phi/mR^2$ — while $z$ executes simple harmonic motion at $\omega=\sqrt{k/m}$, entirely decoupled from $\phi$.
Part (b) — the rotated frame $xyZ$. The frame $xyZ$ shares the same $Z$-axis as $XYZ$ and differs only by a fixed (constant, non-time-varying) rotation angle $\phi_0$ about that axis: it is a static relabelling of the azimuthal origin, not a rotating (non-inertial) frame. Since the constraint ($\rho=R$) and the potential ($V=\tfrac12 k(R^2+z^2)$) depend only on $R$ and $z$ — never on the azimuthal angle itself — the Lagrangian and Hamiltonian are form-invariant under $\phi\mapsto\phi'=\phi-\phi_0$: substituting $\phi=\phi'+\phi_0$ everywhere leaves $L$ and $H$ unchanged in form (the constant $\phi_0$ simply drops out of every derivative). Hamilton's equations in the $xyZ$ frame are therefore identical in form to those in $XYZ$: $\dot\phi'=p_{\phi'}/mR^2$, $p_{\phi'}=\text{const}=p_\phi$, and the same SHM equation for $z$. Nothing about the dynamics changes; only the zero-point from which $\phi$ is measured does.
Part (c) — invariant property. The system possesses continuous rotational symmetry (SO(2)) about the $Z$-axis, since neither $V$ nor the constraint distinguishes one azimuthal direction from another. By Noether's theorem this symmetry is exactly why $p_\phi$ (the $Z$-component of angular momentum) is conserved, and because the rotation between $XYZ$ and $xyZ$ is about that very axis, $p_{\phi'}=p_\phi$ identically — the same numerical value in both frames. The total energy $H$ is likewise a scalar, unchanged by any fixed orthogonal coordinate change. Invariant: the $Z$-component of angular momentum $p_\phi=mR^2\dot\phi$ (and the total energy $H$), unaffected by the arbitrary fixed azimuthal offset between the two frames.

Final results
Quantity	Result
Hamiltonian	$H=\dfrac{p_\phi^2}{2mR^2}+\dfrac{p_z^2}{2m}+\tfrac12 k(R^2+z^2)$
$\phi$-motion	uniform: $\dot\phi = p_\phi/mR^2 = \text{const}$
$z$-motion	SHM: $\ddot z + (k/m)z = 0$
Effect of fixed rotation about $Z$	none — equations form-invariant
Invariant quantity	$p_\phi$ (angular momentum about $Z$) and $H$

Topic: Hamiltonian mechanics for a constrained particle; cyclic coordinates and rotational symmetry.

Key relations: $H=\sum p_i\dot q_i - L$; Hamilton's equations $\dot q_i=\partial H/\partial p_i,\ \dot p_i=-\partial H/\partial q_i$; Noether's theorem (continuous symmetry $\Rightarrow$ conserved momentum).

Why this works: Fixing the particle to the cylinder removes the radial coordinate entirely, leaving a natural two-coordinate system in which the axisymmetric potential makes $\phi$ manifestly cyclic; the $z$-motion is then just a 1-D oscillator riding on top of a uniformly rotating azimuth.

Common pitfall: Treating "rotating a fixed angle" as introducing a rotating (non-inertial) frame, which would add centrifugal/Coriolis terms; here the rotation angle is a constant, so it is nothing more than redefining where $\phi=0$ is measured from — the physics is untouched.

Source: Goldstein, Classical Mechanics (3rd ed.), Ch. 8 (Hamilton's equations) and Ch. 2 (cyclic coordinates and conservation theorems).

Question 2 of 6: Particle on a Cylinder Under a Central Force — Hamiltonian and Frame Invariance

Question 2: Particle on a Cylinder Under a Central Force — Hamiltonian and Frame Invariance (20 marks)