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UPSC 2014 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Principle of virtual work · Dynamics & Statics · asked 6× in 13 yrs · Read the full method →

Question

Two equal uniform rods $AB$ and $AC$, each of length $l$, are freely jointed at $A$ and rest on a smooth fixed vertical circle of radius $r$. If $2\theta$ is the angle between the rods, then find the relation between $l$, $r$ and $\theta$, by using the principle of virtual work.

Technique

Symmetric setup → single coordinate $\theta$; total PE function; $dU/d\theta=0$ at equilibrium.

Solution

Setup. Two uniform rods of length $l$ and weight $W$ each, jointed at $A$, resting on a smooth fixed vertical circle of radius $r$. The system is symmetric about the vertical through the centre $O$ of the circle. By symmetry $A$ lies on this vertical, at some height $h$ above $O$. Each rod makes angle $\theta$ with the downward vertical at $A$.

Step 1 — Geometry: position of $A$

Each rod is tangent to the circle (touches at one point). The perpendicular distance from $O$ to a rod must equal $r$.

The line through $A=(0,h)$ in direction $\vec d=(\sin\theta,-\cos\theta)$ has distance from origin $|\vec{OA}\times\vec d|=|h\sin\theta|=h\sin\theta$ (since $h,\sin\theta>0$).

Setting this $=r$: $h=r/\sin\theta$.

Step 2 — Centre of gravity of each rod

Each rod is uniform, CG at midpoint. Take rod $AC$ (the right rod): $A=(0,r/\sin\theta)$ to $C=A+l\vec d$, midpoint at $A+(l/2)\vec d=((l/2)\sin\theta,\,r/\sin\theta-(l/2)\cos\theta)$.

Height of midpoint above $O$: $y_{\text{CG}}=r/\sin\theta-(l/2)\cos\theta$.

Step 3 — Total potential energy

Both rods symmetric; each contributes the same to $U$:

(Taking $O$ as zero of PE.)

Step 4 — Principle of virtual work / equilibrium

Equilibrium requires $dU/d\theta=0$ (no virtual displacement does work on a frictionless system):

Rearranging:

Answer