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UPSC 2026 Maths Optional Paper 1 Q8b — Step-by-Step Solution

15 marks · Section B

Stability of equilibrium (energy criterion) · Dynamics & Statics · asked 6× in 14 yrs · Read the full method →

Question

$AB$ is a uniform rod of length $l$ and weight $W$, which can turn freely about a fixed point in its length distant $\dfrac{l}{3}$ from $A$. $AC$ and $BC$ are light strings each of length $\dfrac{5}{6}l$, attached to a particle $C$ of weight $w$. Prove that if $W$ is less than $2w$, there will be stable equilibrium with $AB$ inclined to the horizontal at an angle $\tan^{-1}\left(\dfrac{W + w}{4w}\right)$.

Technique

Energy method for a one-degree-of-freedom system. Because $AC = BC$, the rod and the particle form a rigid triangle turning about the fixed point $O$; the single coordinate is the inclination $\theta$ of $AB$ to the horizontal. Write the total potential energy $V(\theta)$ of the two weights above $O$, use $V'(\theta)=0$ for equilibrium and $V''(\theta)>0$ for stability. The hypothesis $W<2w$ is not produced by $V''$ — it is the condition that the string $BC$ remains taut, which is what makes the rigid-triangle model legitimate; that must be shown separately.

Solution

Step 1 — Geometry of the configuration (set up before any algebra)

Let $O$ be the fixed point, $M$ the mid-point of $AB$.

• $OA = \dfrac{l}{3}$, hence $OB = l - \dfrac{l}{3} = \dfrac{2l}{3}$.
• $AM = \dfrac{l}{2}$, hence $OM = \dfrac{l}{2} - \dfrac{l}{3} = \dfrac{l}{6}$, with $M$ lying on the $B$-side of $O$.
• Since $CA = CB = \dfrac{5l}{6}$, the point $C$ lies on the perpendicular bisector of $AB$, i.e. $CM \perp AB$, and

So, as long as both strings are taut, the shape $A$–$B$–$C$ is rigid: $C$ is reached from $O$ by going $\dfrac{l}{6}$ along the rod towards $B$ and then $\dfrac{2l}{3}$ perpendicular to the rod, on the side away from which the particle hangs. The whole body therefore has one degree of freedom — the inclination $\theta$ of $AB$ to the horizontal.

(Figure to draw, 4 cm: the rod $AB$ inclined with $A$ raised and $B$ lowered; mark $O$ with the hatching for a fixed pivot, mark $OA=l/3$, $OM=l/6$, $OB=2l/3$; drop $CM\perp AB$ from $M$ with $CM=2l/3$; draw the two strings $CA=CB=5l/6$; show the weights $W\downarrow$ at $M$ and $w\downarrow$ at $C$, and the angle $\theta$ between $AB$ and the horizontal dashed line through $O$. That one figure carries Steps 1–3.)

Step 2 — Coordinates

Take $O$ as origin, $x$ horizontal, $y$ vertically upwards. Let the rod turn so that $A$ rises and $B$ falls, and let $\theta$ be the angle $AB$ makes with the horizontal. Let

These are orthonormal. Then

Reading off the $y$-coordinates:

Step 3 — Potential energy function

The rod’s weight $W$ acts at $M$ and the particle’s weight $w$ at $C$; the strings are light and inextensible and do no work. Taking $O$ as the datum,

(using $\dfrac{2l}{3} = \dfrac{l}{6}\cdot 4$).

Step 4 — Equilibrium

Call this root $\alpha \in \left(0,\tfrac{\pi}{2}\right)$; it is the unique root in that range. (The other root of $\tan\theta = \tan\alpha$ in $[0,2\pi)$ is $\alpha+\pi$, the inverted position.)

Cross-check by moments about $O$. Independently of the energy method, resolving the particle’s equilibrium (Step 6) gives string tensions $T_A,T_B$, and taking moments of $W$, $T_A$, $T_B$ about $O$ yields

i.e. $\tan\theta = \dfrac{W+w}{4w}$ again. The two routes agree.

Step 5 — Stability

At $\theta = \alpha$ put $R = \sqrt{(W+w)^{2} + 16w^{2}}$, so that $\sin\alpha = \dfrac{W+w}{R}$, $\cos\alpha = \dfrac{4w}{R}$. Then

Hence $V$ has a strict minimum at $\theta=\alpha$, and the equilibrium there is stable.

Equivalent one-line argument (worth writing, it settles both steps at once): since $(W+w)\sin\theta + 4w\cos\theta = R\cos(\theta-\alpha)$ with $\tan\alpha = \dfrac{W+w}{4w}$,

which is manifestly minimum ($V=-lR/6$) at $\theta=\alpha$ and maximum at $\theta=\alpha+\pi$.

Step 6 — Where the condition $W < 2w$ comes from: both strings must be taut

Steps 1–5 assumed the triangle $ABC$ is rigid, which holds only while both strings pull. A string can exert tension but not thrust, so we must verify $T_A \ge 0$ and $T_B \ge 0$ at $\theta=\alpha$.

Resolve the equilibrium of the particle $C$ under its weight $w$ and the two tensions. From Step 2,

Equilibrium of $C$: $\;\dfrac{6}{5l}\bigl(T_A\vec{CA} + T_B\vec{CB}\bigr) + (0,-w) = \vec{0}$. Resolving along $\hat{u}$ and $\hat{v}$ (note $(0,-w)\cdot\hat{u} = w\sin\theta$, $(0,-w)\cdot\hat{v} = w\cos\theta$):

Therefore

$T_A>0$ always (for $0<\theta<\tfrac{\pi}{2}$). But

At the equilibrium angle $\tan\alpha = \dfrac{W+w}{4w}$, so

So $W<2w$ is exactly the condition that string $BC$ is genuinely taut, i.e. that the configuration assumed in Steps 1–5 actually exists. Under it, the stationary point $\alpha$ lies strictly inside the admissible range $0<\theta<\tan^{-1}\tfrac34$, and by Step 5 it is a strict minimum of $V$ — hence a stable equilibrium.

What happens otherwise. If $W = 2w$ then $\tan\alpha = \tfrac34$, $T_B = 0$ and $CA$ becomes vertical (indeed with $\cos\alpha=\tfrac45,\sin\alpha=\tfrac35$ one gets $\vec{CA} = \left(0,\tfrac{5l}{6}\right)$ and $T_A = w$): the particle hangs from $A$ alone and the equilibrium is neutral. If $W>2w$ the formula would demand $T_B<0$, impossible; $BC$ goes slack, the rigid-triangle model fails and there is no such equilibrium. Hence the strict inequality in the statement.

Step 7 — Conclusion

For $W<2w$ the system has a stable equilibrium with $AB$ inclined to the horizontal (the longer arm $OB$ downwards) at

Answer