← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q8a — Step-by-Step Solution

12 marks · Section B

Common catenary · Dynamics & Statics · asked 4× in 13 yrs · Read the full method →

Question

Find the length of an endless chain which will hang over a circular pulley of radius $a$ so as to be in contact with the two-thirds of the circumference of the pulley.

Technique

Catenary equation $y=c\cosh(x/c)$ for the hanging free segments; match tangent angle at the leaving point to the pulley tangent direction (giving $\psi=60°$); match horizontal coordinate of leaving point ($x_R=a\sin 60°=a\sqrt 3/2$) to determine $c$; total chain length = pulley arc + 2 catenary segments.

Solution

Setup. A heavy flexible chain hangs over a smooth circular pulley of radius $a$. The chain is “endless” (a closed loop), passing over the top of the pulley with the two ends hanging down. It is in contact with $2/3$ of the pulley’s circumference.

Contact arc: $\tfrac{2}{3}\cdot 2\pi=\dfrac{4\pi}{3}$ radians of the pulley’s circumference. Equivalently, the chain leaves the pulley at two symmetric points, with the arc above the leaving points being $\tfrac{4\pi}{3}$ rad (more than $\pi$, so the leaving points are on the lower part of the pulley).

The free chain hanging on each side from the leaving point to the bottom of the loop forms a catenary.

Step 1 — Identify the leaving points

Total circumference contact: $\tfrac{4\pi}{3}$ rad = $240°$. By symmetry, the contact arc is centred at the top of the pulley. So contact extends from angle $\theta_0$ to $2\pi-\theta_0$ measured from the top, where $2(\pi-\theta_0)=4\pi/3$.

Wait, let me re-parameterise. Let $\phi$ be measured from the downward vertical (bottom of pulley) going counter-clockwise. The chain leaves at $\phi=\phi_0$ and $\phi=-\phi_0$ (symmetric). The arc of contact = the arc not between these two leaving points (going over the top).

Going over the top: from $\phi=\phi_0$ counterclockwise to $\phi=2\pi-\phi_0$, an arc of $2\pi-2\phi_0$.

Set $2\pi-2\phi_0=\dfrac{4\pi}{3}$: $\phi_0=\dfrac{\pi}{3}=60°$.

So the chain leaves the pulley at angle $\phi_0=60°$ from the downward vertical (i.e. on the lower half, 60° on either side of straight down).

Step 2 — Position and tangent direction at the leaving point

Take pulley centre at origin. The leaving point on the right: in polar (measuring $\phi$ from downward vertical), position $(a\sin\phi_0,-a\cos\phi_0)=(a\sin 60°,-a\cos 60°)=(a\sqrt 3/2,-a/2)$.

Tangent direction at this point (going outward from pulley, i.e. away from contact): the tangent to the circle is perpendicular to the radius. The outward tangent direction (along which the chain leaves) makes the chain hang down and outward.

Slope of the tangent at the leaving point: radius direction is $(\sin\phi_0,-\cos\phi_0)=(\sqrt 3/2,-1/2)$. Tangent perpendicular: $(\cos\phi_0,\sin\phi_0)=(1/2,\sqrt 3/2)$ or $(-1/2,-\sqrt 3/2)$. The chain hangs down from this point — the chain extends in the direction with negative $y$-component, so direction $(-1/2,-\sqrt 3/2)$ for the chain going down on the right (wait, that goes leftward and down — let’s redo).

Right leaving point: $(\sqrt 3/2\cdot a,-a/2)$. The pulley occupies the region with the chain wrapping over the top. So on the right side, the chain comes off the pulley going downward and to the right (away from the pulley). The tangent direction at the leaving point, pointing away from the contact arc, points… hmm let me re-parameterise.

If $\phi$ is measured from the downward vertical, increasing counterclockwise, then on the right (going up and around) the chain wraps in the direction of increasing $\phi$ from $\phi_0$ to $2\pi-\phi_0$. At the leaving point $\phi=\phi_0$, the chain enters the pulley going in the $+\phi$ direction (counterclockwise), so it exits (going away from the pulley) in the $-\phi$ direction = clockwise = $-d/d\phi$ of the position vector.

Position: $\vec r(\phi)=(a\sin\phi,-a\cos\phi)$. $d\vec r/d\phi=(a\cos\phi,a\sin\phi)$. At $\phi=\phi_0=\pi/3$: $d\vec r/d\phi=(a/2,a\sqrt 3/2)$. The exit direction (clockwise = $-d/d\phi$) is $(-a/2,-a\sqrt 3/2)$, normalised $(-1/2,-\sqrt 3/2)$. Pointing left and down.

Hmm but on the right side the chain should go down on the right of the pulley, not the left. Let me reconsider the orientation.

Picture. The pulley is a circle, the chain drapes over the top. The two ends of the contact arc are on the lower half (since contact is $240°$ — more than half). On each side, the chain leaves the pulley and hangs down beneath the pulley.

At the right-side leaving point (which is at $\phi_0$ from straight-down, on the right) — the chain is to the outside of the pulley. The continuation of the chain after leaving the pulley goes in the tangent direction, outward — i.e. on the right side, downward and toward the right.

Tangent direction at $\vec r(\phi_0)=(a\sin\phi_0,-a\cos\phi_0)$: perpendicular to radius, can be either $(\cos\phi_0,\sin\phi_0)$ or $(-\cos\phi_0,-\sin\phi_0)$. The radius points outward at $(\sin\phi_0,-\cos\phi_0)$. The tangent that takes us away from contact (i.e., as we leave the pulley, the chain straightens to a catenary) — we want the direction where, starting from the leaving point and moving along the tangent, we go downward (chain hangs down due to gravity).

Tangent $(\cos\phi_0,\sin\phi_0)=(1/2,\sqrt 3/2)$ — up and to the right. Tangent $(-\cos\phi_0,-\sin\phi_0)=(-1/2,-\sqrt 3/2)$ — down and to the left.

Hmm, both tangents go on slants. The chain at the leaving point goes from the pulley downward as a catenary. The catenary’s tangent at the top end (where it meets the pulley) is along the tangent direction to the pulley. The chain continues downward.

Going from the leaving point $(a\sqrt 3/2, -a/2)$ in direction $(-1/2,-\sqrt 3/2)$ — this goes to $(a\sqrt 3/2-1/2\cdot t,-a/2-\sqrt 3/2\cdot t)$ for $t>0$. That’s leftward and downward — into the pulley region? Wait, the pulley centre is at origin and radius $a$. Point $(a\sqrt 3/2,-a/2)$ is on the pulley (distance $a$). Moving in $(-1/2,-\sqrt 3/2)$ direction by a small step $t$: new point $(a\sqrt 3/2-t/2,-a/2-t\sqrt 3/2)$. Distance from origin squared: $(a\sqrt 3/2-t/2)^2+(a/2+t\sqrt 3/2)^2$.

Expand: $(3a^2/4-a\sqrt 3 t/2+t^2/4)+(a^2/4+a\sqrt 3 t/2+3t^2/4)=a^2+t^2$. So distance $=\sqrt{a^2+t^2}>a$ — moving outside the pulley ✓.

So the direction $(-1/2,-\sqrt 3/2)$ goes outside the pulley, downward and to the left. But intuitively, on the right side, the chain should fall toward the right or straight down, not to the left.

Wait — the leaving point at $\phi_0=60°$ from straight-down is on the right of the pulley (since I took $\phi$ counterclockwise from downward). The chain exits going below and across to the left — because the chain is hanging in a vee shape below the pulley.

Actually, picture a chain wrapping $240°$ around a pulley (more than half). The chain goes over the top and the two ends meet under the pulley to form a vee, with the chain ends pointing inward (toward each other under the pulley). The two free segments hang in a closed loop below the pulley, meeting at the bottom of the loop.

So yes — on the right leaving point, the chain goes to the left and down (toward the centre under the pulley), where it meets the other chain segment coming from the left leaving point.

Step 3 — Set up the catenary

The free chain segment (right side) is a catenary from the right leaving point to the bottom of the loop (where the two free segments meet).

Catenary equation (standard form): $y=c\cosh(x/c)$, with $c=T_0/(\lambda g)$ (where $T_0$ is horizontal tension, $\lambda$ is linear mass density). The bottom of the catenary is at $x=0$ in this frame.

Tangent angle. At the leaving point, the catenary tangent makes angle $\psi$ with the horizontal, where $\tan\psi=dy/dx=\sinh(x/c)$ at the leaving point’s $x$-coordinate.

The chain’s tangent at the leaving point equals the tangent of the pulley circle there. Tangent direction $(-1/2,-\sqrt 3/2)$ relative to chain coordinates. In standard catenary coordinates (positive $y$ up, chain rising from the centre to the ends), this tangent direction (going from low end to high end) is opposite: $(1/2,\sqrt 3/2)$. So tangent angle with horizontal: $\tan\psi=(\sqrt 3/2)/(1/2)=\sqrt 3$, so $\psi=60°$.

(Equivalently: the tangent at the leaving point makes angle $60°$ with horizontal — and since by symmetry the leaving points are at $60°$ from vertical from the centre, the tangent there makes $60°$ with horizontal. Direct geometric consistency.)

Step 4 — Length of catenary segment

Arc length of catenary from bottom ($x=0$) to a point where tangent makes angle $\psi$ with horizontal:

(since $dy/dx=\sinh(x/c)$ at the point, equal to $\tan\psi$).

With $\psi=60°$: $s_{\text{one side}}=c\tan 60°=c\sqrt 3$.

Total free chain (both sides): $2c\sqrt 3$.

But we need $c$ in terms of $a$.

Step 5 — Determine $c$

At the leaving point, the chain has tension $T$ such that:

• Horizontal component: $T\cos\psi=T_0$ (constant horizontal tension along catenary).
• Vertical: $T\sin\psi=\lambda g\cdot s$ (weight of chain from leaving point to bottom).

So $\tan\psi=\lambda g s/T_0=s/c$, giving $s=c\tan\psi$ as above.

Now, the catenary’s position: the leaving point is at height $y=c\cosh(x/c)$ above the catenary’s lowest point (the vertex). And $x$ at the leaving point satisfies $\sinh(x/c)=\tan\psi=\sqrt 3$, so $x/c=\sinh^{-1}\sqrt 3=\ln(\sqrt 3+2)$.

Then $y$ at leaving point $=c\cosh(x/c)=c\sqrt{1+\sinh^2(x/c)}=c\sqrt{1+3}=2c$.

So vertical distance from catenary bottom to leaving point is $2c-c=c$ (above the catenary’s vertex; the vertex is at $y=c$ in the standard form).

Now match geometry. The leaving point is at height $-a/2$ (with pulley centre at origin). The catenary’s vertex (lowest point of chain) is at some height $y_v<-a/2$. The horizontal distance from the leaving point to the vertex (taking the right leaving point) is the $x$-coordinate at the leaving point in catenary coordinates.

By symmetry of the two leaving points (both on the lower half, symmetric about the vertical axis), the catenary’s vertex lies on the vertical axis (below the pulley). So in catenary coordinates (vertex at origin), the right leaving point has $x_R=$ horizontal distance from axis = $a\sin\phi_0=a\sqrt 3/2$.

And $\sinh(x_R/c)=\tan\psi=\sqrt 3$, so $x_R/c=\sinh^{-1}\sqrt 3$, giving

Hmm, but the catenary length per side is $c\sqrt 3=\dfrac{(a\sqrt 3/2)\cdot\sqrt 3}{\ln(\sqrt 3+2)}=\dfrac{3a/2}{\ln(\sqrt 3+2)}$.

But this gives a non-clean answer. Let me reconsider whether the catenary assumption is correct.

Alternative interpretation. “Endless chain hanging on a pulley” with $2/3$ in contact — UPSC standard answer style suggests cleaner numbers. Maybe the chain is not a catenary (perhaps it’s taut, just two straight segments hanging from the leaving points)?

Re-reading the problem: “endless chain which will hang over a circular pulley”. An endless chain forming a closed loop with $2/3$ in contact means the other 1/3 is the free hanging part. If the chain is in equilibrium and the pulley is smooth (frictionless), then the tension throughout the contact region is constant — and the free chain segments must be catenaries (gravity acts uniformly).

Hmm wait — let me reconsider the geometry. If the chain wraps $2/3$ of the pulley = $240°$ as I had, then the contact arc is centred at the top and the leaving points are at $60°$ from the bottom (lower-left and lower-right). The free chain segments hang below the pulley and join at a vertex on the symmetry axis.

But there’s another reading: contact is $2/3$ but centred at the bottom — the chain wraps below, with two segments going up and meeting above. That’s nonsensical physically (chain falls under gravity).

Sticking with the original interpretation.

Reconsider — at the leaving point, what’s the angle?

I argued $\psi=60°$ from the tangent direction. Let me re-derive geometrically.

Leaving point at $\phi_0=60°$ from downward vertical (on the right). The radius at this point makes angle $60°$ with the downward vertical, i.e. $30°$ with horizontal. The tangent is perpendicular to the radius, so the tangent makes angle $30°+90°=120°$ or $30°-90°=-60°$ with horizontal, i.e. makes angle $60°$ with horizontal.

Yes $\psi=60°$ ✓.

So the catenary tangent at the top end makes $60°$ with horizontal.

The clean answer. Let me re-examine: $c=(a\sqrt 3/2)/\sinh^{-1}(\sqrt 3)$. Note $\sinh^{-1}(\sqrt 3)=\ln(\sqrt 3+\sqrt{1+3})=\ln(\sqrt 3+2)$.

Length of one free segment $=c\sqrt 3=\dfrac{(a\sqrt 3/2)\sqrt 3}{\ln(\sqrt 3+2)}=\dfrac{3a/2}{\ln(2+\sqrt 3)}$.

Total free chain $=2\cdot\dfrac{3a/2}{\ln(2+\sqrt 3)}=\dfrac{3a}{\ln(2+\sqrt 3)}$.

Contact chain $=\dfrac{4\pi a}{3}$ (arc of pulley = $4\pi/3$ rad times radius $a$).

Total chain length:

Answer