← 2026 Paper 2

UPSC 2026 Maths Optional Paper 2 Q7c — Step-by-Step Solution

20 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 10× in 14 yrs · Read the full method →

Question

A smooth uniform rod, say $OA$, of length $2a$ and mass $m$ is pivoted at one end to a fixed point $O$. The rod is inclined at an angle $\theta$ with the downward vertical line $OZ$ and the plane $OAZ$ makes an angle $\phi$ with a fixed vertical plane. A bead of mass $\lambda m$ slides smoothly on the rod and is connected to $O$ by a light elastic string of modulus $nmg$ and natural length $a$. Show that the kinetic energy $T$ of the system is given by

where $x$ is the stretched length of the string, and derive the equations of motion.

The printed expression above is reproduced verbatim from the paper and contains two misprints. Both were re-checked at 450 dpi against the English and the Hindi text, which agree with each other. The honest derivation below produces

and Step 4 proves that the printed form cannot be a kinetic energy. The equations of motion are derived from the correct $T$. Step 4 gives the full reading of both misprints and states what to write in the examination hall.

Technique

Three degrees of freedom $(\theta,\phi,x)$, all velocities and no non-holonomic constraints — so the whole question is: build $T$ honestly, build $V$ honestly, and turn the Lagrangian crank. The recognisable trigger is spherical polar coordinates centred at the pivot $O$ with the polar axis along the downward vertical $OZ$: $\theta$ is the polar angle and $\phi$ the azimuth, exactly as the question defines them. Then the rod’s kinetic energy is $\tfrac12 I_{\perp}|\dot{\hat e}_r|^2$ with $I_\perp = \tfrac43 ma^2$ about $O$, and the bead’s is the standard spherical-polar $\tfrac12\lambda m(\dot r^2+r^2\dot\theta^2+r^2\dot\phi^2\sin^2\theta)$ with $r=x$. The trap is the string: it is elastic, so its energy is $\dfrac{\nu(\text{ext})^2}{2\times\text{natural length}}$ and only while it is taut — the slack branch is a separate case and must be stated.

Solution

Step 1 — Configuration, coordinates, degrees of freedom

Take $O$ as origin and let $OZ$ be the downward vertical, with $Ox_1$ chosen in the fixed vertical plane from which $\phi$ is measured. Use the right-handed frame $(x_1,x_2,x_3)$ with $x_3$ along $OZ$ (downwards).

• The rod $OA$ is pivoted at $O$, so $O$ is a fixed point of the rod; $A$ is at distance $2a$.
• The rod’s direction is fixed by $(\theta,\phi)$: $\theta$ from $OZ$, $\phi$ the azimuth of the plane $OAZ$.
• The bead $B$ lies on the rod. Since the rod passes through $O$, the light string from $O$ to $B$ lies along the rod, so its stretched length is exactly $OB$. Hence

• A thin rod has no moment of inertia about its own axis, and no coordinate for spin about that axis is offered; the rod’s configuration is completely described by $(\theta,\phi)$.

So the system has three degrees of freedom, $q = (\theta,\phi,x)$. Both contacts (pivot, bead-on-rod) are smooth, so no dissipation.

(Figure to draw, 5 cm — this figure earns 2–3 marks on its own. Vertical dashed line $OZ$ downwards from $O$ at the top; the rod $OA$ drawn as a solid segment from $O$ making angle $\theta$ with $OZ$, marked $2a$; the bead $B$ as a small filled circle on the rod at distance $x$ from $O$, with the elastic string drawn as a short zig-zag along $O\!B$ and labelled “natural length $a$, modulus $nmg$”; the centre of mass $G$ of the rod marked at distance $a$; a light horizontal ellipse through $O$ indicating the horizontal plane, with the azimuth $\phi$ marked between the fixed vertical plane (dashed) and the plane $OAZ$; and at $B$ the triad $\hat e_r$ along $OA$ outward, $\hat e_\theta$ in the plane $OAZ$ increasing $\theta$, $\hat e_\phi$ perpendicular to that plane.)

Step 2 — The moving frame and the key kinematic identity

Define the usual spherical unit vectors (polar axis $= OZ$ downwards):

These are orthonormal. $\hat e_r$ points from $O$ along the rod towards $A$.

Differentiating $\hat e_r$ with respect to time (chain rule in $\theta$ and $\phi$ only):

and, by orthonormality,

This is the whole of the disputed physics. The azimuthal component carries the factor $\sin\theta$ before squaring, so it enters $|\dot{\hat e}_r|^2$ as $\sin^2\theta$. Geometrically: a point of the rod at distance $s$ from $O$ lies at perpendicular distance $s\sin\theta$ from the axis $OZ$, so its azimuthal speed is $s\dot\phi\sin\theta$, and its contribution to kinetic energy is proportional to the square of that.

Step 3 — Kinetic energy of the system

(a) The rod. The rod is uniform, mass $m$, length $2a$, with the fixed point $O$ at one end. Take the element between $s$ and $s+ds$ ($0\le s\le 2a$), of mass $\dfrac{m}{2a}\,ds$. Its position vector is $s\,\hat e_r$ with $s$ constant in time, so its velocity is $s\,\dot{\hat e}_r$ and, by (1), its speed squared is $s^2(\dot\theta^2+\dot\phi^2\sin^2\theta)$. Hence

(Equivalently: for a body with a fixed point, $T=\tfrac12\,\boldsymbol\omega^{\!\top}\!I\,\boldsymbol\omega$; a thin rod has $I_\parallel=0$ and $I_\perp = \tfrac13 m(2a)^2 = \tfrac43 ma^2$ about any axis through $O$ perpendicular to it, and $|\boldsymbol\omega_\perp| = |\dot{\hat e}_r|$, giving $T_{\text{rod}} = \tfrac12\cdot\tfrac43ma^2(\dot\theta^2+\dot\phi^2\sin^2\theta)$ — the same. The integration is safer to write out, because it also displays where $\tfrac43ma^2$ comes from.)

(b) The bead. Its position vector is $\vec r_B = x\,\hat e_r$, with both $x$ and $\hat e_r$ time-dependent. Therefore

and by orthonormality

(c) Total. Adding and doubling,

$\qquad\blacksquare$

Step 4 — Where this differs from the printed expression, and why the printed one cannot stand

The paper prints

Two entries are wrong. I state them at the point they arise rather than absorbing them silently.

(i) $\dot\phi^2\sin\theta$ must be $\dot\phi^2\sin^2\theta$. Three independent reasons, any one of which is fatal:

• Direct kinematics. Hold $\theta$ fixed and let the rod sweep the cone at rate $\dot\phi$. The end $A$ then travels on a horizontal circle of radius $2a\sin\theta$ with speed $2a\dot\phi\sin\theta$; its kinetic-energy contribution is proportional to the square of that, hence to $\sin^2\theta$. Equation (1) says the same thing for every element.
• Limiting case $\theta\to 0$ (decisive). With the printed form,

so the Lagrange $\theta$-equation would carry a finite generalised force tipping the rod away from the vertical even when the rod is exactly vertical. But when $\theta=0$ the rod lies along the axis $OZ$, and "$\dot\phi$" is then nothing but a spin about the rod’s own axis — a motion that, for an ideal thin rod, involves no material point moving at all. It cannot produce a tipping couple. The correct form gives $\partial T/\partial\theta \to 0$, as it must.

• Positivity / parity. $T$ must be a positive-semidefinite quadratic form in $(\dot\theta,\dot\phi,\dot x)$ at every configuration. The same physical configuration is described by $(\theta,\phi)$ and by $(-\theta,\phi+\pi)$; $\sin\theta$ is odd under $\theta\mapsto-\theta$ while $\sin^2\theta$ is even. The printed coefficient therefore makes $2T<0$ for $\theta\in(-\pi,0)$ — impossible for a kinetic energy.

(ii) $x^2\dot\theta^2\sin^2\theta$ must be $x^2\dot\phi^2\sin^2\theta$. Again three reasons:

• Direct kinematics. The bead is a single particle at $x\hat e_r$; its velocity resolves on the orthonormal triad as $(\dot x,\ x\dot\theta,\ x\dot\phi\sin\theta)$, so $v_B^2 = \dot x^2+x^2\dot\theta^2+x^2\dot\phi^2\sin^2\theta$. This is the standard spherical-polar form and there is no route to a second $\dot\theta^2$ term.
• Degenerate special case (decisive, and one line in the hall). Put $\dot\phi=0$ and $\theta=\pi/2$. The bead then moves in a fixed vertical plane, and plane polar coordinates give $v_B^2=\dot x^2+x^2\dot\theta^2$ exactly. The printed expression gives $\dot x^2+2x^2\dot\theta^2$ — a spurious factor $2$ on a term whose value is elementary and not in dispute.
• Structural inconsistency. The rod term and the bead term describe the same azimuthal sweep of the same rigid direction $\hat e_r$; whatever $\dot\phi$-dependence one has, the other must have. In the printed version $\dot\phi$ has vanished from the bead entirely, which would make $\phi$ contribute nothing from the bead to the angular momentum about the vertical — i.e. a bead of arbitrary mass at arbitrary distance could be whirled about $OZ$ at no energetic cost. Absurd.

Both slips are single-character typesetting errors ($\sin\theta$ for $\sin^2\theta$; $\dot\theta$ for $\dot\phi$) of exactly the kind that propagate to both language versions of a bilingual paper set from one source.

What to do in the examination hall. Derive $T$ honestly, then spend one sentence on the discrepancy — “the printed form appears to contain typographical errors; the correct expressions are $\dot\phi^{2}\sin^{2}\theta$ and $x^{2}\dot\phi^{2}\sin^{2}\theta$, as the derivation above shows” — and move on to the equations of motion. Do not attempt to reverse-engineer the printed form. A “derivation” that arrives at $\dot\phi^{2}\sin\theta$ has necessarily fudged the kinematics somewhere, and every one of the remaining marks is then built on that fudge. On a “show that” question carrying a misprint, an examiner can credit a correct derivation with the discrepancy named; a fabricated one he cannot. All that follows uses the correct expression (2).

Step 5 — Potential energy

Measure height upwards, so a point at $x_3 = z$ (downward coordinate) has height $-z$.

(a) Gravity. The rod is uniform, so its centre of mass $G$ is at $s=a$, i.e. at $z_G = a\cos\theta$; the bead is at $z_B = x\cos\theta$. With the datum at $O$,

(b) The elastic string. State the convention — “modulus of elasticity $\nu$” means Hooke’s law in the form

(the P.E. being $\int_0^{e}\frac{\nu\xi}{a}\,d\xi$). Here $\nu = nmg$ and natural length $=a$, and the extension is $x-a$. A string pulls but never pushes, so this holds only while $x\ge a$:

Hence

Explicitly, on the taut branch,

Note $L$ contains no $\phi$ and no explicit $t$.

Step 6 — Lagrange’s equations

Write, once and for all,

(the moment of inertia of the whole system about any axis through $O$ perpendicular to the rod). Note $\dfrac{\partial L}{\partial\dot\theta} = A\dot\theta$ and $\dfrac{\partial L}{\partial\dot\phi} = A\dot\phi\sin^2\theta$.

(a) The $x$-equation.

So $\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot x}\right) - \dfrac{\partial L}{\partial x} = 0$ gives, after dividing by $\lambda m$,

with the last term replaced by $0$ while $x<a$ (string slack). Reading it: the bead’s radial acceleration is driven by the centrifugal term $x(\dot\theta^2+\dot\phi^2\sin^2\theta)$, the component of gravity along the rod, $g\cos\theta$, and the elastic pull $-\frac{nmg}{a}(x-a)$ towards $O$. The smooth rod’s reaction is perpendicular to the rod and correctly absent.

(b) The $\theta$-equation.

Hence

The term $2\lambda mx\dot x\dot\theta$ is the Coriolis-type term from the bead’s changing moment of inertia; $A\dot\phi^2\sin\theta\cos\theta$ is the centrifugal term that lifts the rod as it whirls; $mg(a+\lambda x)\sin\theta$ is the restoring gravitational moment about $O$.

(c) The $\phi$-equation. $\phi$ does not occur in $L$: it is cyclic. Therefore

which is exactly the conservation of the angular momentum of the system about the fixed vertical $OZ$ (no external moment about $OZ$: gravity is vertical, the pivot reaction acts at $O$, the string tension is along the rod through $O$). Expanded,

(d) Second first integral. $L$ has no explicit $t$ and $T$ is a homogeneous quadratic in the velocities, so the Jacobi integral is the total energy:

Two first integrals ($p_\phi$, $E$) for three degrees of freedom. $\qquad\blacksquare$

Step 7 — Sanity checks on the equations (cheap, and they buy confidence)

1. Static equilibrium, rod hanging vertically. Put $\theta\equiv 0$, $\dot\theta=\dot\phi=\dot x=0$. The $\theta$- and $\phi$-equations are satisfied identically ($\sin 0 = 0$), and the $x$-equation gives

Check independently: the extension is $\lambda a/n$, so the tension is $\dfrac{nmg}{a}\cdot\dfrac{\lambda a}{n} = \lambda mg$ — precisely the weight of the bead. ✓ 2. Small oscillation about the vertical, $\dot\phi=0$, $x$ frozen. The $\theta$-equation linearises to $A\ddot\theta + mg(a+\lambda x)\theta = 0$, a pendulum of period $2\pi\sqrt{A/\bigl(mg(a+\lambda x)\bigr)}$ — the right compound-pendulum form. ✓ 3. Slack-string consistency. If $x<a$, the $x$-equation reads $\ddot x = x(\dot\theta^2+\dot\phi^2\sin^2\theta) + g\cos\theta$, which is $>0$ whenever $|\theta|<\pi/2$; the bead accelerates outward and the string retautens. So the slack branch is transient for a rod below the horizontal — worth one sentence, because it justifies working on the taut branch. ✓ 4. Degenerate limit $\lambda\to0$ (bead removed). The $\theta$- and $\phi$-equations collapse to $\tfrac43a^2\ddot\theta - \tfrac43a^2\dot\phi^2\sin\theta\cos\theta + ga\sin\theta = 0$ and $\sin^2\theta\,\dot\phi =$ const — the classical spherical compound pendulum. ✓

Answer

Kinetic energy (corrected; the printed form carries two misprints, identified and falsified in Step 4):

Equations of motion (with $A(x) = \tfrac43 ma^{2}+\lambda mx^{2}$, and the elastic term present only while $x\ge a$):

together with the energy integral $E = T+V = \text{const}$.