← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q8b — Step-by-Step Solution

15 marks · Section B

Rectilinear motion under variable force · Dynamics & Statics · asked 3× in 13 yrs · Read the full method →

Question

A particle is acted on by a force parallel to the axis of $y$ whose acceleration (always towards the axis of $x$) is $\mu y^{-2}$ and when $y=a$, it is projected parallel to the axis of $x$ with velocity $\sqrt{\dfrac{2\mu}{a}}$. Find the parametric equation of the path of the particle. Here $\mu$ is a constant.

Technique

Energy integral for the $y$-equation; eliminate $t$ to get $x$ as function of $y$; trigonometric substitution leading to cycloid form.

Solution

Setup. Force only in $y$-direction, magnitude $\mu/y^{2}$, toward $x$-axis (negative $y$ for $y>0$):

ICs at $t=0$: $x=0,\,y=a$; $\dot x=\sqrt{2\mu/a},\,\dot y=0$.

Step 1 — Horizontal motion

$\ddot x=0\Rightarrow\dot x=\sqrt{2\mu/a}$ constant, so $x=t\sqrt{2\mu/a}$.

Step 2 — Energy integral for $y$

Multiply $\ddot y=-\mu/y^{2}$ by $\dot y$ and integrate w.r.t. $t$:

IC ($t=0$, $\dot y=0,\,y=a$): $C=-\mu/a$. So

Step 3 — Eliminate $t$ to get path

The horizontal coordinate parametrised by $y$:

(Negative sign of $\dot y$ flipped the integral limits.)

Step 4 — Trigonometric substitution

Let $y'=a\sin^{2}\phi$, so $a-y'=a\cos^{2}\phi$, $dy'=a\sin 2\phi\,d\phi=2a\sin\phi\cos\phi\,d\phi$. Then $\sqrt{y'/(a-y')}=\tan\phi$.

Integrate from $\phi=\phi_y$ (where $a\sin^{2}\phi_y=y$) to $\phi=\pi/2$ (where $y'=a$):

Step 5 — Substitute $\eta=\pi-2\phi_y$ (cycloid parameter)

With $\eta=\pi-2\phi_y$, $\phi_y=(\pi-\eta)/2$: $\sin 2\phi_y=\sin(\pi-\eta)=\sin\eta$. $y=a\sin^{2}\phi_y=a(1-\cos 2\phi_y)/2=a(1-\cos(\pi-\eta))/2=a(1+\cos\eta)/2$.

And $\pi/2-\phi_y=\pi/2-(\pi-\eta)/2=\eta/2$. So

Step 6 — Parametric form

Answer