UPSC 2020 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

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

Question

By writing down the Hamiltonian, find the equations of motion of a particle of mass $m$ constrained to move on the surface of a cylinder defined by $x^2+y^2=R^2$ , $R$ constant. The particle is subject to a force directed towards the origin and proportional to the distance $r$ of the particle from the origin, given by $\vec F=-k\vec r$ , $k$ constant.

Technique

Cylindrical coordinates with fixed $R$ ; $L=T-V$ with $V=\tfrac12kr^2$ ; Legendre transform to $H(\theta,z,p_\theta,p_z)$ ; Hamilton’s canonical equations.

Solution

Step 1 — Generalized coordinates

The constraint $x^2+y^2=R^2$ leaves two degrees of freedom. Use cylindrical coordinates with fixed radius $R$ :

x=R\cos\theta,\qquad y=R\sin\theta,\qquad z=z.

Generalized coordinates: $\theta,\ z$ .

Step 2 — Kinetic energy and potential

\dot x=-R\dot\theta\sin\theta,\quad \dot y=R\dot\theta\cos\theta,\quad \dot z=\dot z,

T=\tfrac12 m(\dot x^2+\dot y^2+\dot z^2)=\tfrac12 m\big(R^2\dot\theta^2+\dot z^2\big).

The force $\vec F=-k\vec r$ is conservative (central, linear): potential

V=\tfrac12 k\,r^2=\tfrac12 k\,(x^2+y^2+z^2)=\tfrac12 k\,(R^2+z^2),

since $\vec F=-\nabla V$ . (The constant $\tfrac12kR^2$ is dynamically irrelevant but kept for completeness.)

The Lagrangian:

L=T-V=\tfrac12 m R^2\dot\theta^2+\tfrac12 m\dot z^2-\tfrac12 k(R^2+z^2).

Step 3 — Generalized momenta

p_\theta=\frac{\partial L}{\partial\dot\theta}=mR^2\dot\theta,\qquad p_z=\frac{\partial L}{\partial\dot z}=m\dot z.

So $\dot\theta=\dfrac{p_\theta}{mR^2}$ , $\dot z=\dfrac{p_z}{m}$ .

Step 4 — Hamiltonian

H=p_\theta\dot\theta+p_z\dot z-L=T+V\quad(\text{scleronomic constraint, }V=V(q)),

\boxed{\;H=\frac{p_\theta^2}{2mR^2}+\frac{p_z^2}{2m}+\frac12 k\big(R^2+z^2\big).\;}

Step 5 — Hamilton’s equations of motion

\dot\theta=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{mR^2},\qquad \dot z=\frac{\partial H}{\partial p_z}=\frac{p_z}{m},

\dot p_\theta=-\frac{\partial H}{\partial\theta}=0,\qquad \dot p_z=-\frac{\partial H}{\partial z}=-kz.

Interpretation of the equations of motion:

$\theta$ is cyclic $\Rightarrow p_\theta=mR^2\dot\theta=\text{const}$ : the angular momentum about the cylinder axis is conserved, so $\dot\theta=$ const, i.e. the particle circles the cylinder at uniform angular speed.
From $\dot p_z=-kz$ and $p_z=m\dot z$ :

Answer

\boxed{\;m\ddot z=-kz\quad\Longrightarrow\quad \ddot z+\omega^2 z=0,\ \ \omega=\sqrt{\tfrac{k}{m}}.\;}