UPSC 2026 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Constrained motion · Dynamics & Statics · asked 8× in 14 yrs · Read the full method →

Question

A particle falls from rest at the vertex of an inverted catenary. Prove that the particle will leave the curve when the path described is $\sqrt{3}$ times the vertical distance through which the particle has fallen.

Technique

Three standard ingredients, combined: energy conservation ( $v^2=2gh$ , the curve being smooth), the intrinsic geometry of the catenary ( $s=c\tan\psi$ , $y=c\sec\psi$ , $\rho=c\sec^2\psi$ ), and the normal equation of motion $mg\cos\psi - N = mv^2/\rho$ , with the leaving condition $N=0$ . Everything collapses to a single equation in $\psi$ .

Solution

Step 1 — Set up the curve and the sign conventions.

Take the catenary $y=c\cosh(x/c)$ (parameter $c$ , directrix $y=0$ ) and invert it — turn it upside down — so that its vertex $A$ is the highest point and the curve falls away on both sides. The particle rests on the convex (outer) side, which is the upper side of the inverted curve.

Measure, from the vertex $A$ :

$s$ = the arc length along the curve (the “path described”), $s>0$ on the side of descent;
$h$ = the vertical distance fallen below $A$ ;
$\psi$ = the angle the tangent makes with the horizontal at the current point. At the vertex $\psi=0$ (the tangent there is horizontal), and $\psi$ increases as the particle descends.

Two forces act: the weight $mg$ vertically downwards, and the normal reaction $N$ , which for a particle on the outside of a curve that is concave downwards points outwards, i.e. away from the centre of curvature (upwards-and-outwards). The curve is smooth, so there is no friction.

Figure to draw (small, labelled): a cap-shaped inverted catenary with vertex $A$ at the top; a particle $P$ part-way down the right branch; from $P$ draw the tangent and mark $\psi$ between the tangent and the horizontal; draw $mg$ vertically down from $P$ , resolve it into $mg\sin\psi$ along the tangent (down-slope) and $mg\cos\psi$ along the inward normal (towards the centre of curvature $C$ below the curve); draw $N$ outward from $P$ along the same normal line, opposite to $mg\cos\psi$ ; mark the arc $AP=s$ and the vertical drop $h$ from the level of $A$ down to $P$ . The figure is what makes the sign of the normal equation obvious, and it earns its space.

Step 2 — Intrinsic relations for the catenary.

For $y=c\cosh(x/c)$ measured from the vertex:

\frac{dy}{dx} = \sinh\frac{x}{c} = \tan\psi ,

s = \int_0^x\sqrt{1+\sinh^2\frac{x}{c}}\;dx = \int_0^x \cosh\frac{x}{c}\,dx = c\sinh\frac{x}{c} \;=\; c\tan\psi , \tag{1}

y = c\cosh\frac{x}{c} = c\sqrt{1+\sinh^2\tfrac{x}{c}} = c\sqrt{1+\tan^2\psi} \;=\; c\sec\psi . \tag{2}

The radius of curvature is

\rho = \frac{\bigl(1+y'^2\bigr)^{3/2}}{|y''|} = \frac{\cosh^3(x/c)}{\tfrac1c\cosh(x/c)} = c\cosh^2\frac{x}{c} \;=\; c\sec^2\psi \;\left(=\frac{y^2}{c}\right). \tag{3}

For the inverted curve the vertex is at $y=c$ (the lowest point of the original catenary becomes the highest point), so the vertical distance fallen is

h = y - c = c\sec\psi - c = c(\sec\psi - 1). \tag{4}

(All three relations (1)–(4) are unchanged by turning the curve upside down: arc length, curvature and the angle with the horizontal are geometric.)

Step 3 — Energy conservation gives the speed.

The surface is smooth, so the reaction does no work; the particle starts from rest at the vertex. After descending a vertical height $h$ ,

\tfrac12 m v^2 = mgh \qquad\Longrightarrow\qquad v^2 = 2gh . \tag{5}

Step 4 — The normal equation of motion and the leaving condition.

Resolve along the inward normal (towards the centre of curvature, which lies below the inverted curve). The component of the weight along it is $mg\cos\psi$ ; the reaction $N$ acts outwards. The net inward force supplies the centripetal acceleration $v^2/\rho$ :

mg\cos\psi - N = \frac{m v^2}{\rho}. \tag{6}

The particle leaves the curve when the contact is lost, i.e. when $N=0$ :

g\cos\psi = \frac{v^2}{\rho}\qquad\Longrightarrow\qquad v^2 = g\rho\cos\psi . \tag{7}

Step 5 — Combine and solve for $\psi$ .

Substituting (5) and (3) into (7):

2gh = g\,\bigl(c\sec^2\psi\bigr)\cos\psi = g\,c\sec\psi ,

2h = c\sec\psi .

Now use (4), $h=c(\sec\psi-1)$ :

2c(\sec\psi - 1) = c\sec\psi \quad\Longrightarrow\quad 2\sec\psi - 2 = \sec\psi \quad\Longrightarrow\quad \sec\psi = 2 .

Hence

\cos\psi = \tfrac12,\qquad \psi = 60^\circ = \frac{\pi}{3},\qquad \tan\psi = \sqrt3 .

Step 6 — Read off $s$ and $h$ at that instant.

From (1), $s = c\tan\psi = \sqrt3\,c$ . From (4), $h = c(\sec\psi - 1) = c(2-1) = c$ .

Therefore

\frac{s}{h} = \frac{\sqrt3\,c}{c} = \sqrt3 \qquad\Longrightarrow\qquad s = \sqrt3\,h ,

i.e. the path described is $\sqrt3$ times the vertical distance fallen, as required. $\blacksquare$

(Note the elegance worth pointing out in the answer: the leaving point is at the fixed intrinsic angle $\psi=60^\circ$ and at $h=c$ , $s=\sqrt3 c$ — independent of $g$ and of the mass, and depending on the catenary only through its parameter $c$ . The stated ratio $s/h=\sqrt3$ is therefore a genuinely scale-free statement, which is why the paper poses it that way.)

Answer

\boxed{\;N=0 \iff \sec\psi = 2,\ \text{i.e. }\psi=60^\circ;\ \text{there } s = c\tan\psi=\sqrt3\,c,\quad h = c(\sec\psi-1)=c,\quad \therefore\ s=\sqrt3\,h.\;}