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UPSC 2022 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →

Question

A particle at distance $r$ from centre of force moves under central force $F=-k/r^2$. Obtain Lagrangian and derive equations of motion.

Technique

Standard Lagrangian for central force; polar coordinates; PE from $F=-dV/dr$; EL equations for $r$ and $\theta$.

Solution

Setup. Central force $F=-k/r^2$ (attractive, like gravity). Use plane polar coordinates $(r,\theta)$ for 2D motion (central force ⇒ angular momentum conservation ⇒ motion confined to a plane).

Step 1 — Kinetic and potential energy

KE: $T=\dfrac{1}{2}m(\dot r^2+r^2\dot\theta^2)$.

PE: $V$ such that $F=-dV/dr$. From $F=-k/r^2$: $dV/dr=k/r^2$, so $V=-k/r+\text{const}$. Take const = 0:

(Note: $V\to 0$ as $r\to\infty$, $V<0$ for finite $r$ — attractive potential.)

Step 2 — Lagrangian

Step 3 — Equations of motion (Euler-Lagrange)

$r$ equation: $\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot r}\right)-\dfrac{\partial L}{\partial r}=0$.

$\partial L/\partial\dot r=m\dot r$, $d/dt$ = $m\ddot r$.

$\partial L/\partial r=m r\dot\theta^2-k/r^2$.

So $m\ddot r-mr\dot\theta^2+k/r^2=0$, i.e.

The $mr\dot\theta^2$ is the centripetal (outward) effect; $-k/r^2$ is the gravity pull inward.

$\theta$ equation: $\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot\theta}\right)-\dfrac{\partial L}{\partial\theta}=0$.

$\partial L/\partial\dot\theta=mr^2\dot\theta$. $d/dt(mr^2\dot\theta)=$ ?

$\partial L/\partial\theta=0$.

So $\dfrac{d}{dt}(mr^2\dot\theta)=0$, i.e. $mr^2\dot\theta=$ constant $=L_z$ (angular momentum).

Step 4 — Summary

Equations of motion:

Answer