UPSC 2026 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Principle of virtual work · Dynamics & Statics · asked 7× in 14 yrs · Read the full method →

Question

A smooth paraboloid of revolution is fixed with its axis vertical and vertex upwards; on it is placed a heavy elastic string of unstretched length $2\pi b$ . When the string is in equilibrium, apply the principle of virtual work to show that it rests in the form of a circle of radius $\dfrac{4\pi a b \lambda}{4\pi a \lambda - W b}$ , where $W$ is the weight of the string, $\lambda$ is the modulus of elasticity and $4a$ is the latus rectum of the generating parabola.

Technique

Principle of virtual work. The surface is smooth, so the reaction is normal to every virtual displacement along the surface and does no work; the only working forces are the weight of the string and its elastic tension. Give the ring one virtual increase $\delta r$ in radius and equate the work done by gravity to the work done against the tension.

Solution

Step 1 — Set up the geometry explicitly.

Take the vertex $O$ of the paraboloid as origin, with the axis vertical and the vertex upwards, so the surface falls away below $O$ . Measure $z$ downwards from the vertex and let $x$ be the horizontal distance from the axis. The generating parabola has latus rectum $4a$ and opens downwards, so in these coordinates it is

x^2 = 4a\,z .

By symmetry the string lies in a horizontal circle of some radius $r$ , whose plane is at depth

z(r) = \frac{r^2}{4a}\quad\text{below the vertex.} \tag{1}

Because the circle is horizontal and the string uniform, the centre of gravity of the whole string is on the axis at that same depth $z(r)$ — this is what lets a single coordinate $r$ describe the configuration. The system has exactly one degree of freedom, $r$ .

Figure to draw (small, labelled): the meridian section — an inverted (cap-shaped) parabola with vertex $O$ at the top and the vertical axis dashed downwards; mark a horizontal chord at depth $z=r^2/4a$ with the two half-widths labelled $r$ ; at one end of the chord draw the weight $W$ acting vertically down and the tension $T$ acting along the circle (into the page / out of the page, marked as $T$ each way); mark the normal reaction $R$ perpendicular to the parabola and annotate “smooth $\Rightarrow$ $R$ does no work”. Also mark the virtual displacement $\delta r$ horizontally outward and $\delta z$ vertically down. This one figure fixes the sign of the gravity work and is worth its space.

Step 2 — The forces that do work.

Weight $W$ , acting vertically downwards at the centre of gravity, i.e. at depth $z(r)$ on the axis.
Elastic tension $T$ in the string. The string is a closed loop of natural length $2\pi b$ (a circle of natural radius $b$ ) stretched into a circle of radius $r$ , hence of length $2\pi r$ . Its extension is $2\pi r - 2\pi b$ , and Hooke’s law with modulus $\lambda$ gives

T = \lambda\,\frac{\text{extension}}{\text{natural length}} = \lambda\cdot\frac{2\pi r - 2\pi b}{2\pi b} = \frac{\lambda\,(r-b)}{b}. \tag{2}

For the string to be taut (which equilibrium requires) we need $r>b$ .

Normal reaction $R$ of the surface. The surface is smooth, so $R$ is along the normal, while every virtual displacement consistent with the constraint lies in the tangent plane. Hence $R$ does no virtual work and drops out. (This is exactly why virtual work is the efficient method here — it never has to compute $R$ .)

Step 3 — Give the string a virtual displacement.

Let the circle slide down the surface so that its radius increases by $\delta r$ . Two things happen:

(i) Gravity does positive work. From (1), the depth increases by

\delta z = \frac{dz}{dr}\,\delta r = \frac{r}{2a}\,\delta r ,

so the centre of gravity descends by $\delta z$ and gravity does work

\delta W_{\text{grav}} = W\,\delta z = \frac{W r}{2a}\,\delta r .

(ii) Work is done against the tension. The length of the string increases by

\delta(2\pi r) = 2\pi\,\delta r ,

and the work done against a tension $T$ in stretching by $\delta(\text{length})$ is

\delta W_{\text{tens}} = T\cdot 2\pi\,\delta r = \frac{\lambda(r-b)}{b}\cdot 2\pi\,\delta r = \frac{2\pi\lambda (r-b)}{b}\,\delta r .

Step 4 — Apply the principle of virtual work.

For equilibrium the total virtual work of all the forces vanishes:

\delta W_{\text{grav}} - \delta W_{\text{tens}} = 0,

\frac{W r}{2a}\,\delta r \;-\; \frac{2\pi\lambda (r-b)}{b}\,\delta r \;=\; 0 .

Since $\delta r$ is arbitrary (and non-zero),

\frac{W r}{2a} = \frac{2\pi\lambda (r-b)}{b}. \tag{3}

Step 5 — Solve for $r$ .

Cross-multiplying (3) by $2ab$ :

W\,r\,b = 4\pi a \lambda\,(r-b) = 4\pi a\lambda\,r - 4\pi a \lambda b .

Collect the $r$ -terms:

r\,(4\pi a\lambda - Wb) = 4\pi a b \lambda,

\boxed{\,r = \frac{4\pi a b \lambda}{4\pi a\lambda - W b}\,}

which is the required result.

Step 6 — Two remarks the examiner rewards.

Existence and tautness. From the answer,

\frac{r}{b} = \frac{4\pi a\lambda}{4\pi a\lambda - Wb} > 1 \iff 4\pi a\lambda - Wb > 0 .

So a positive, stretched ( $r>b$ ) equilibrium circle exists precisely when $4\pi a\lambda > Wb$ ; if the string is too heavy or too slack ( $Wb \geq 4\pi a\lambda$ ), no equilibrium circle exists and the string slides off the paraboloid. The two conditions $r>0$ and $r>b$ coincide — a good sanity check on the algebra.

Stability. Taking the total potential energy (with $z$ measured downwards, so gravitational PE is $-Wz$ ),

V(r) = -\,W\frac{r^2}{4a} + \frac{\lambda\,(2\pi r - 2\pi b)^2}{2\,(2\pi b)} = -\frac{Wr^2}{4a} + \frac{\pi\lambda (r-b)^2}{b},

$V'(r)=0$ reproduces (3), and

V''(r) = -\frac{W}{2a} + \frac{2\pi\lambda}{b} \overset{\text{using (3)}}{=} -\frac{W}{2a} + \frac{Wr}{2a(r-b)} = \frac{W}{2a}\cdot\frac{b}{r-b} > 0 ,

so the equilibrium is stable.

Answer

\boxed{\;T=\frac{\lambda(r-b)}{b},\qquad \frac{Wr}{2a}=\frac{2\pi\lambda(r-b)}{b}\quad\Longrightarrow\quad r = \frac{4\pi a b \lambda}{4\pi a\lambda - W b}\;\;(\text{requires }4\pi a\lambda > Wb).\;}