← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q6d — Step-by-Step Solution

13 marks · Section B

Rectilinear motion under variable force · Dynamics & Statics · asked 3× in 14 yrs · Read the full method →

Question

A mass starts from rest at a distance $a$ from the centre of force which attracts inversely as the distance. Find the time of arriving at the centre.

Technique

Energy first integral (multiply by $\dot r$); separation of variables; Gaussian substitution $r=ae^{-u^2}$ converts the singular endpoint at $r=0$ to a Gaussian integral on $[0,\infty)$.

Solution

Setup. Particle on a line, attracted toward origin with force magnitude $F=k/r$ (per unit mass), $k>0$. Equation of motion (taking $r>0$, attraction toward origin):

(For motion along a line, take $r$ as the signed coordinate; with $r>0$ initially and attraction toward origin, the force is in the $-r$ direction.)

Initial conditions: $r(0)=a$, $\dot r(0)=0$.

Step 1 — Energy first integral

Multiply by $\dot r$:

Integrate from $t=0$ to $t$, using $\dot r(0)=0$, $r(0)=a$:

So

Since $r$ is decreasing (particle moves toward origin), $\dot r<0$ and

Step 2 — Separate and integrate

Time to reach the centre:

Step 3 — Substitute $r=ae^{-u^2}$ (so $u\to 0$ at $r=a$, $u\to\infty$ at $r=0$)

$dr=ae^{-u^2}(-2u)\,du=-2ua e^{-u^2}\,du$.

$\ln(a/r)=\ln(e^{u^2})=u^2$.

$\sqrt{2k\ln(a/r)}=\sqrt{2k}\cdot u$.

So

The Gaussian integral: $\int_0^\infty e^{-u^2}\,du=\dfrac{\sqrt\pi}{2}$.

Answer