UPSC 2014 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

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

Question

Find the equation of motion of a compound pendulum using Hamilton’s equations.

Technique

Standard Lagrangian → Hamiltonian transition; for natural systems $H=T+V$ ; Hamilton’s equations yield the equation of motion.

Solution

Setup. Rigid body of mass $M$ free to rotate about a fixed horizontal axis (pivot) through point $O$ . Let $h$ = distance from pivot to centre of gravity, $I$ = moment of inertia about the pivot axis, $\theta$ = angle of swing from the downward vertical.

Step 1 — Kinetic and potential energy

T=\dfrac{1}{2}I\dot\theta^{2},\qquad V=-Mgh\cos\theta

(with $V=0$ when the CG is at the level of the pivot; $V$ minimum at $\theta=0$ ).

Step 2 — Lagrangian and generalised momentum

L=T-V=\dfrac{1}{2}I\dot\theta^{2}+Mgh\cos\theta.

Generalised momentum:

p=\dfrac{\partial L}{\partial\dot\theta}=I\dot\theta\;\Longrightarrow\;\dot\theta=\dfrac{p}{I}.

Step 3 — Hamiltonian

For natural system (kinetic energy quadratic in velocities, holonomic time-independent constraints), $H=T+V$ expressed in $(\theta,p)$ :

H(\theta,p)=\dfrac{p^{2}}{2I}-Mgh\cos\theta.

Step 4 — Hamilton’s equations

\dot\theta=\dfrac{\partial H}{\partial p}=\dfrac{p}{I}.\tag{1}

\dot p=-\dfrac{\partial H}{\partial\theta}=-Mgh\sin\theta.\tag{2}

Step 5 — Combine to second-order equation

From (1): $p=I\dot\theta$ , so $\dot p=I\ddot\theta$ .

Substituting into (2): $I\ddot\theta=-Mgh\sin\theta$ , i.e.,

Answer

\boxed{\;\ddot\theta+\dfrac{Mgh}{I}\sin\theta=0.\;}