UPSC 2015 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Hamilton's equations · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

Solve the plane pendulum problem using the Hamiltonian approach and show that $H$ is a constant of motion.

Technique

Standard Lagrangian-to-Hamiltonian for a 1-DOF system; verify Hamilton’s equations recover the second-order ODE; show $dH/dt=0$ using $\partial H/\partial t=0$ (no explicit time dependence).

Solution

Setup. Pendulum of mass $m$ , rod length $l$ , swinging in a vertical plane. Generalized coordinate: $\theta$ (angle from downward vertical).

Position: $(x,y)=(l\sin\theta,-l\cos\theta)$ .

Velocity: $\dot x=l\dot\theta\cos\theta$ , $\dot y=l\dot\theta\sin\theta$ .

Step 1 — Lagrangian

Kinetic energy: $T=\tfrac{1}{2}m(\dot x^2+\dot y^2)=\tfrac{1}{2}ml^2\dot\theta^2$ .

Potential energy (taking pivot as $V=0$ , $y$ downward = negative height): $V=-mgl\cos\theta$ . Equivalently, taking lowest point as $V=0$ : $V=mgl(1-\cos\theta)$ . Either choice works; use $V=-mgl\cos\theta$ .

L=T-V=\tfrac{1}{2}ml^2\dot\theta^2+mgl\cos\theta.

Step 2 — Generalized momentum

p_\theta=\dfrac{\partial L}{\partial\dot\theta}=ml^2\dot\theta\;\Longrightarrow\;\dot\theta=\dfrac{p_\theta}{ml^2}.

Step 3 — Hamiltonian

H=p_\theta\dot\theta-L=p_\theta\cdot\dfrac{p_\theta}{ml^2}-\tfrac{1}{2}ml^2\cdot\dfrac{p_\theta^2}{m^2 l^4}-mgl\cos\theta.

Simplify:

H=\dfrac{p_\theta^2}{ml^2}-\dfrac{p_\theta^2}{2ml^2}-mgl\cos\theta=\dfrac{p_\theta^2}{2ml^2}-mgl\cos\theta.

Answer

\boxed{\;H=\dfrac{p_\theta^2}{2ml^2}-mgl\cos\theta.\;}